[{"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I have the function f of x here, and we're defining it using a polynomial expression. And what I would like to do here is take the derivative of our function, which is essentially going to make us take a derivative of this polynomial expression. And we're gonna take the derivative with respect to x. So the first thing I'm gonna do is let's take the derivative of both sides. So we could say the derivative with respect to x of f of x, of f of x, is equal to the derivative with respect to x, the derivative with respect to x, of x to the fifth, x to the fifth, plus two, plus two x to the third, minus x squared. And so the notation, just to get familiar with it, you could view this as the derivative operator. This says, look, I wanna take the derivative of whatever is inside of the parentheses with respect to x."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first thing I'm gonna do is let's take the derivative of both sides. So we could say the derivative with respect to x of f of x, of f of x, is equal to the derivative with respect to x, the derivative with respect to x, of x to the fifth, x to the fifth, plus two, plus two x to the third, minus x squared. And so the notation, just to get familiar with it, you could view this as the derivative operator. This says, look, I wanna take the derivative of whatever is inside of the parentheses with respect to x. So the derivative of f with respect to x, we could use a notation that that is just f prime of, f prime of x. And that is going to be equal to, now here we can use our derivative properties. The derivative of the sum or difference of a bunch of things is just the derivative of, is equal to the sum or the difference of the derivative of each of them."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This says, look, I wanna take the derivative of whatever is inside of the parentheses with respect to x. So the derivative of f with respect to x, we could use a notation that that is just f prime of, f prime of x. And that is going to be equal to, now here we can use our derivative properties. The derivative of the sum or difference of a bunch of things is just the derivative of, is equal to the sum or the difference of the derivative of each of them. So this is equal to the derivative, let me just, it's the derivative with respect to x of each of these three things. So the derivative with respect to x, actually let me just write it out like this, of that first term, plus the derivative with respect to x of that second term, minus the derivative with respect to x of that third term, of that third term. And I'll color code it here."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of the sum or difference of a bunch of things is just the derivative of, is equal to the sum or the difference of the derivative of each of them. So this is equal to the derivative, let me just, it's the derivative with respect to x of each of these three things. So the derivative with respect to x, actually let me just write it out like this, of that first term, plus the derivative with respect to x of that second term, minus the derivative with respect to x of that third term, of that third term. And I'll color code it here. So here I had an x to the fifth, so I'll put the x to the fifth there. Here I had a two x, here I had a two x to the third, so I'll put the two x to the third there. And here I have a x squared, I'm subtracting x squared, so I'm subtracting the derivative with respect to x of x squared."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll color code it here. So here I had an x to the fifth, so I'll put the x to the fifth there. Here I had a two x, here I had a two x to the third, so I'll put the two x to the third there. And here I have a x squared, I'm subtracting x squared, so I'm subtracting the derivative with respect to x of x squared. So notice, all that's happening here is I'm taking the derivative individually of each of these terms, and then I'm adding or subtracting them the same way that the terms were added or subtracted. And so what is this going to be equal to? Well, this is going to be equal to, for x to the fifth, we can just use the power rule."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And here I have a x squared, I'm subtracting x squared, so I'm subtracting the derivative with respect to x of x squared. So notice, all that's happening here is I'm taking the derivative individually of each of these terms, and then I'm adding or subtracting them the same way that the terms were added or subtracted. And so what is this going to be equal to? Well, this is going to be equal to, for x to the fifth, we can just use the power rule. We can bring the five out front and decrement the exponent by one, so it becomes five x, we could say to the five minus one power, which of course is just four. And then for this second one, we could do it in a few steps. Actually, let me just write it out here."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is going to be equal to, for x to the fifth, we can just use the power rule. We can bring the five out front and decrement the exponent by one, so it becomes five x, we could say to the five minus one power, which of course is just four. And then for this second one, we could do it in a few steps. Actually, let me just write it out here. So I could write, I could write the derivative with respect to x of two x to the third power is the same thing, it's equal to, the same, we could bring the constant out, the derivative with, two times the derivative with respect to x of x to the third power. This is one of our, this is one of our derivative properties. The derivative of a constant times some expression is the same thing as the constant times the derivative of that expression."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me just write it out here. So I could write, I could write the derivative with respect to x of two x to the third power is the same thing, it's equal to, the same, we could bring the constant out, the derivative with, two times the derivative with respect to x of x to the third power. This is one of our, this is one of our derivative properties. The derivative of a constant times some expression is the same thing as the constant times the derivative of that expression. And what will the derivative with respect to x of, x to the third be? Well, we would bring the three out front and decrement the exponent, and so this would be equal to this two times the three times x to the three minus one power, which is of course the second power. So this would give us six x squared."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of a constant times some expression is the same thing as the constant times the derivative of that expression. And what will the derivative with respect to x of, x to the third be? Well, we would bring the three out front and decrement the exponent, and so this would be equal to this two times the three times x to the three minus one power, which is of course the second power. So this would give us six x squared. So another way that you could have done it, I could just write, I could just write a six x squared here. So I could just, so this is going to be six x squared. And instead of going through all of this, you'll learn as you do more of these that you could have done this pretty much in your head, saying look, I have the three out here as an exponent."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this would give us six x squared. So another way that you could have done it, I could just write, I could just write a six x squared here. So I could just, so this is going to be six x squared. And instead of going through all of this, you'll learn as you do more of these that you could have done this pretty much in your head, saying look, I have the three out here as an exponent. Let me multiply the three times this coefficient, because that's what we ended up doing anyway. Three times the coefficient is six x, and then three minus one is two. So you didn't necessarily have to do this, but it's nice to see that this comes out of the derivative properties that we talk about in other videos."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And instead of going through all of this, you'll learn as you do more of these that you could have done this pretty much in your head, saying look, I have the three out here as an exponent. Let me multiply the three times this coefficient, because that's what we ended up doing anyway. Three times the coefficient is six x, and then three minus one is two. So you didn't necessarily have to do this, but it's nice to see that this comes out of the derivative properties that we talk about in other videos. And then finally, we have minus, and we use the power rule right over here. So bring the two out front and decrement the exponent. So it's going to be two."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you didn't necessarily have to do this, but it's nice to see that this comes out of the derivative properties that we talk about in other videos. And then finally, we have minus, and we use the power rule right over here. So bring the two out front and decrement the exponent. So it's going to be two. It's going to be two times x to the two minus one power, which is just one, which we could just write as two x. So just like that, we have been able to figure out the derivative of f. And you might say, well, what is this thing now? Well, now we have an expression that tells us the slope of the tangent line, or you could view it as the instantaneous rate of change with respect to x for any x value."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be two. It's going to be two times x to the two minus one power, which is just one, which we could just write as two x. So just like that, we have been able to figure out the derivative of f. And you might say, well, what is this thing now? Well, now we have an expression that tells us the slope of the tangent line, or you could view it as the instantaneous rate of change with respect to x for any x value. So if I were to say, if I were now to say f prime, let's say f prime of two, this would tell me what is the slope of the tangent line of our function when x is equal to two. And I do that by using this expression. So this is going to be five times two to the fourth plus six times two squared, six times two squared, minus two times two, minus two times two."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, now we have an expression that tells us the slope of the tangent line, or you could view it as the instantaneous rate of change with respect to x for any x value. So if I were to say, if I were now to say f prime, let's say f prime of two, this would tell me what is the slope of the tangent line of our function when x is equal to two. And I do that by using this expression. So this is going to be five times two to the fourth plus six times two squared, six times two squared, minus two times two, minus two times two. And this is going to be equal to, let's see, two to the fourth power is 16, 16 times five is 80, so that's 80. And then this is six times four, which is 24. And then we are going to subtract four."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be five times two to the fourth plus six times two squared, six times two squared, minus two times two, minus two times two. And this is going to be equal to, let's see, two to the fourth power is 16, 16 times five is 80, so that's 80. And then this is six times four, which is 24. And then we are going to subtract four. So this is 80 plus 24 is 104, minus four is equal to 100. So when x is equal to two, this curve is really steep. The slope is 100."}, {"video_title": "Differentiating polynomials example   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we are going to subtract four. So this is 80 plus 24 is 104, minus four is equal to 100. So when x is equal to two, this curve is really steep. The slope is 100. If you were to graph the tangent line when x is equal to two, for every positive movement in the x direction by one, you're going to move up in the y direction by 100. So it's really steep there, and it makes sense. This is a pretty high degree."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "A particle moves along a straight line. Its speed is inversely proportional to the square of the distance s it has traveled. Which equation describes this relationship? So I'm not gonna even look at these choices, and I'm just gonna try to parse this sentence up here and see if we can come up with an equation. So they tell us its speed is inversely proportional to what? To the square of the distance s it has traveled. So s is equal to distance."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm not gonna even look at these choices, and I'm just gonna try to parse this sentence up here and see if we can come up with an equation. So they tell us its speed is inversely proportional to what? To the square of the distance s it has traveled. So s is equal to distance. S is equal to distance. And how would we denote speed then, if s is distance? Well, speed is the rate of change of distance with respect to time."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So s is equal to distance. S is equal to distance. And how would we denote speed then, if s is distance? Well, speed is the rate of change of distance with respect to time. So our speed would be the rate of distance with respect to time. The rate of change of distance with respect to time. So this is going to be our speed."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, speed is the rate of change of distance with respect to time. So our speed would be the rate of distance with respect to time. The rate of change of distance with respect to time. So this is going to be our speed. So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional. So it's inversely proportional. I'll write a proportionality constant over what?"}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be our speed. So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional. So it's inversely proportional. I'll write a proportionality constant over what? It's inversely proportional to what? To the square of the distance. To the square of the distance it has traveled."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll write a proportionality constant over what? It's inversely proportional to what? To the square of the distance. To the square of the distance it has traveled. So there you go. This is an equation that I think is describing a differential equation, really, that's describing what we have up here. Now let's see which of these choices match that."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "To the square of the distance it has traveled. So there you go. This is an equation that I think is describing a differential equation, really, that's describing what we have up here. Now let's see which of these choices match that. Well, actually, this one is exactly what we wrote. The speed, the rate of change of distance with respect to time is inversely proportional to the square of the distance. Now just to make sure we understand these other ones, let's just interpret them."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's see which of these choices match that. Well, actually, this one is exactly what we wrote. The speed, the rate of change of distance with respect to time is inversely proportional to the square of the distance. Now just to make sure we understand these other ones, let's just interpret them. This is saying that the distance, which is a function of time, is inversely proportional to the time squared. That's not what they told us. This is saying that the distance is inversely proportional to the distance squared."}, {"video_title": "Writing a differential equation   Differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now just to make sure we understand these other ones, let's just interpret them. This is saying that the distance, which is a function of time, is inversely proportional to the time squared. That's not what they told us. This is saying that the distance is inversely proportional to the distance squared. That one is especially strange. And this is saying that the distance with respect to time, the change in distance with respect to time, the derivative of the distance with respect to time, ds dt, or the speed, is inversely proportional to time squared. Well, that's not what they said."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is get some more practice identifying when to use u-substitution and picking an appropriate u. So let's say we have the indefinite integral of natural log of x to the 10th power, all of that over x, dx. Does u-substitution apply? And if so, how would we make that substitution? Well, the key for u-substitution is to see do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if so, how would we make that substitution? Well, the key for u-substitution is to see do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx. Now, it's clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx. Now, it's clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution. We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative. In this case, it's its exact derivative."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could make the substitution. We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative. In this case, it's its exact derivative. And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. And so here you have it. This right over here is du, and then this right over here is our u."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "In this case, it's its exact derivative. And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. And so here you have it. This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, and to actually evaluate this indefinite integral. Let's do another one."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, and to actually evaluate this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something interesting here. Let's say the integral of tangent x, dx. Does u-substitution apply here?"}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do another one. Let's say that we have the integral of, let's do something interesting here. Let's say the integral of tangent x, dx. Does u-substitution apply here? And at first you say, well, I just have a tangent of x. Where is its derivative? But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Does u-substitution apply here? And at first you say, well, I just have a tangent of x. Where is its derivative? But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine. So we could write this as the integral of sine of x over cosine of x, dx. And now you might say, well, where does u-substitution apply here? Well, there's a couple of ways to think about it."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine. So we could write this as the integral of sine of x over cosine of x, dx. And now you might say, well, where does u-substitution apply here? Well, there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting, you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting, you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice. So we could say the negative of the negative sine of x. And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice. So we could say the negative of the negative sine of x. And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties. This is equivalent. I can put a negative on the outside and a negative on the inside so that this is the derivative of cosine of x. And so now this is interesting."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties. This is equivalent. I can put a negative on the outside and a negative on the inside so that this is the derivative of cosine of x. And so now this is interesting. In fact, let me rewrite this. This is going to be equal to negative, the negative integral of one over cosine of x times negative sine of x dx. Now, does it jump out at you what our u might be?"}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now this is interesting. In fact, let me rewrite this. This is going to be equal to negative, the negative integral of one over cosine of x times negative sine of x dx. Now, does it jump out at you what our u might be? Well, I have a cosine of x in a denominator and I have its derivative. So what if I made u equal to cosine of x? u is equal to cosine of x and then du dx would be equal to negative sine of x."}, {"video_title": "_-substitution defining _ (more examples)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, does it jump out at you what our u might be? Well, I have a cosine of x in a denominator and I have its derivative. So what if I made u equal to cosine of x? u is equal to cosine of x and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx. And just like that, I have my du here. And this, of course, is my u."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let R be the region in the first quadrant enclosed by the graphs of f of x is equal to 8x to the third, and g of x is equal to sine of pi x, as shown in the figure above. And they drew the figure right over here. Part A, write the equation for the line tangent to the graph of f at x is equal to 1 half. So let me just redraw it here, just so that I like drawing on the black background, I guess is the main reason why I'm redrawing it. So the function f of x is equal to 8x to the third, looks like this. It looks like that. This is our f of x axis, and this is our x axis."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me just redraw it here, just so that I like drawing on the black background, I guess is the main reason why I'm redrawing it. So the function f of x is equal to 8x to the third, looks like this. It looks like that. This is our f of x axis, and this is our x axis. And we want the equation for the line tangent at x is equal to 1 half. So this is x is equal to 1 half. If you go up here, if you evaluate f of 1 half, you get 8 times 1 half to the third, which is 8 times 1 eighth, which is 1."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our f of x axis, and this is our x axis. And we want the equation for the line tangent at x is equal to 1 half. So this is x is equal to 1 half. If you go up here, if you evaluate f of 1 half, you get 8 times 1 half to the third, which is 8 times 1 eighth, which is 1. And they actually gave us that already on this point. This is the point 1 half comma 1. And we need to find the equation for the tangent line."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you go up here, if you evaluate f of 1 half, you get 8 times 1 half to the third, which is 8 times 1 eighth, which is 1. And they actually gave us that already on this point. This is the point 1 half comma 1. And we need to find the equation for the tangent line. So the tangent line will look something like that. And to figure out this equation, we just really figure out its slope, and then we know a point that it's on, and we could either use point slope, or you could just use your kind of standard slope intercept form to give an equation for that line. So the first part, let's figure out its slope."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we need to find the equation for the tangent line. So the tangent line will look something like that. And to figure out this equation, we just really figure out its slope, and then we know a point that it's on, and we could either use point slope, or you could just use your kind of standard slope intercept form to give an equation for that line. So the first part, let's figure out its slope. And the slope of the tangent line is going to be the same slope as the slope of our function at that point. Or another way to think about it, it is going to be f prime of 1 half, or the derivative evaluated at 1 half. The derivative gives us the slope of that line at any point."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first part, let's figure out its slope. And the slope of the tangent line is going to be the same slope as the slope of our function at that point. Or another way to think about it, it is going to be f prime of 1 half, or the derivative evaluated at 1 half. The derivative gives us the slope of that line at any point. So what is f prime of x? f prime of x is just the derivative of this. So 3 times 8 is 24 times x squared."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative gives us the slope of that line at any point. So what is f prime of x? f prime of x is just the derivative of this. So 3 times 8 is 24 times x squared. 24x squared, f prime of 1 half is equal to 24 times 1 half squared, which is equal to 24 times 1 fourth, which is equal to 6. So the slope of this line is equal to 6. I'll use m for slope."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 3 times 8 is 24 times x squared. 24x squared, f prime of 1 half is equal to 24 times 1 half squared, which is equal to 24 times 1 fourth, which is equal to 6. So the slope of this line is equal to 6. I'll use m for slope. That's the convention that we use when we first learn it in algebra. So the slope is going to be 6. So the general equation for this line is y is equal to mx plus b."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll use m for slope. That's the convention that we use when we first learn it in algebra. So the slope is going to be 6. So the general equation for this line is y is equal to mx plus b. This is the slope, this is the y-intercept. We already know that the slope is 6. And then we can use the fact that the line goes through the point 1 half 1 to figure out what b is."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the general equation for this line is y is equal to mx plus b. This is the slope, this is the y-intercept. We already know that the slope is 6. And then we can use the fact that the line goes through the point 1 half 1 to figure out what b is. So when y is 1, 1 is equal to our slope times x. x is 1 half. Or another way to say it, when x is 1 half, y is 1, plus some y-intercept. So if I take x as 1 half, I multiply it times the slope, plus the y-intercept, I should get 1."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can use the fact that the line goes through the point 1 half 1 to figure out what b is. So when y is 1, 1 is equal to our slope times x. x is 1 half. Or another way to say it, when x is 1 half, y is 1, plus some y-intercept. So if I take x as 1 half, I multiply it times the slope, plus the y-intercept, I should get 1. And so I get 1 is equal to 3 plus b. I can subtract 3 from both sides, and I get negative 2 is equal to b. So the equation of the line is going to be y is equal to 6x minus 2. That is the equation of the tangent line."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I take x as 1 half, I multiply it times the slope, plus the y-intercept, I should get 1. And so I get 1 is equal to 3 plus b. I can subtract 3 from both sides, and I get negative 2 is equal to b. So the equation of the line is going to be y is equal to 6x minus 2. That is the equation of the tangent line. Now part b, find the area of r. So r is this region right over here. It's bounded above by g of x, which they've defined as sine of pi x. It's bounded below by f of x, or 8x to the third."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is the equation of the tangent line. Now part b, find the area of r. So r is this region right over here. It's bounded above by g of x, which they've defined as sine of pi x. It's bounded below by f of x, or 8x to the third. So the area is going to be, actually let me just do it, I'll scroll down a little bit. I still want to be able to see this graph right over here. Part b, the area of r is going to be equal to the integral from 0, that's this point of intersection right over here, to 1 half."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's bounded below by f of x, or 8x to the third. So the area is going to be, actually let me just do it, I'll scroll down a little bit. I still want to be able to see this graph right over here. Part b, the area of r is going to be equal to the integral from 0, that's this point of intersection right over here, to 1 half. So let me make it clear, this is 0 to 1 half. And then the function on the top, so we could just take the area of that, but then we're going to have to subtract from that the area underneath the function below. Or one way to think about it is, the integral from 0 to 1 half of the top function is g of x, which is sine of pi x."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Part b, the area of r is going to be equal to the integral from 0, that's this point of intersection right over here, to 1 half. So let me make it clear, this is 0 to 1 half. And then the function on the top, so we could just take the area of that, but then we're going to have to subtract from that the area underneath the function below. Or one way to think about it is, the integral from 0 to 1 half of the top function is g of x, which is sine of pi x. So sine of pi x. But if we just evaluated this integral, let me just put a dx over here. If we just evaluated this, we would get the area of this entire region."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or one way to think about it is, the integral from 0 to 1 half of the top function is g of x, which is sine of pi x. So sine of pi x. But if we just evaluated this integral, let me just put a dx over here. If we just evaluated this, we would get the area of this entire region. But what we need to do is subtract out the area underneath the second, underneath f of x. We need to subtract out the area under that. So we just subtract from that f of x."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we just evaluated this, we would get the area of this entire region. But what we need to do is subtract out the area underneath the second, underneath f of x. We need to subtract out the area under that. So we just subtract from that f of x. And f of x, we already saw, is 8x to the third power. And now we can just evaluate this. So let me draw a little line here."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just subtract from that f of x. And f of x, we already saw, is 8x to the third power. And now we can just evaluate this. So let me draw a little line here. It's getting a little bit messy. I'll just do it down here. So we need to take the antiderivative of sine of pi x."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw a little line here. It's getting a little bit messy. I'll just do it down here. So we need to take the antiderivative of sine of pi x. The derivative of cosine of x is negative sine x. The derivative of cosine of pi x is negative pi cosine of pi x. So the antiderivative of sine of pi x is going to be negative 1 over pi cosine of pi x."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we need to take the antiderivative of sine of pi x. The derivative of cosine of x is negative sine x. The derivative of cosine of pi x is negative pi cosine of pi x. So the antiderivative of sine of pi x is going to be negative 1 over pi cosine of pi x. And you can verify it for yourself. And you might say, wait, how did you know it was a negative? Well, I put the negative there so that when I take the derivative of the cosine of pi x, I would get a negative sign."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the antiderivative of sine of pi x is going to be negative 1 over pi cosine of pi x. And you can verify it for yourself. And you might say, wait, how did you know it was a negative? Well, I put the negative there so that when I take the derivative of the cosine of pi x, I would get a negative sign. But that negative will cancel out the negative to give me a positive here. And you say, why did you put a 1 over pi here? Well, when you take the derivative of this thing using the chain rule, you take the derivative of the pi x, you'll get pi that you would multiply everything by."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I put the negative there so that when I take the derivative of the cosine of pi x, I would get a negative sign. But that negative will cancel out the negative to give me a positive here. And you say, why did you put a 1 over pi here? Well, when you take the derivative of this thing using the chain rule, you take the derivative of the pi x, you'll get pi that you would multiply everything by. And then you would get negative sine of pi x. And that pi doesn't show up here, so I need something for it to cancel out with. And that's what this 1 over pi is going to cancel out with."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, when you take the derivative of this thing using the chain rule, you take the derivative of the pi x, you'll get pi that you would multiply everything by. And then you would get negative sine of pi x. And that pi doesn't show up here, so I need something for it to cancel out with. And that's what this 1 over pi is going to cancel out with. And you could do u substitution and all the rest if you found something like that useful. But it's in general a good habit, or I guess it's good to be able to do this almost by sight. And you can verify that this derivative is equal to sine of pi x."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's what this 1 over pi is going to cancel out with. And you could do u substitution and all the rest if you found something like that useful. But it's in general a good habit, or I guess it's good to be able to do this almost by sight. And you can verify that this derivative is equal to sine of pi x. So the antiderivative of sine of pi x is this. The antiderivative of negative 8x to the third power is negative 8. I'm going to divide it by 4, so negative 2x to the fourth power."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can verify that this derivative is equal to sine of pi x. So the antiderivative of sine of pi x is this. The antiderivative of negative 8x to the third power is negative 8. I'm going to divide it by 4, so negative 2x to the fourth power. And all I did is I incremented the 3 to a 4, and then I divided the 8 by the 4. And you could take the derivative of this to verify that it is the same thing as negative 8x to the third power. And we're going to have to evaluate that from 0 to 1 half."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm going to divide it by 4, so negative 2x to the fourth power. And all I did is I incremented the 3 to a 4, and then I divided the 8 by the 4. And you could take the derivative of this to verify that it is the same thing as negative 8x to the third power. And we're going to have to evaluate that from 0 to 1 half. When you evaluate it 1 half, so I'm going to get negative 1 over pi cosine of pi over 2 minus 2 times 1 half to the fourth power is 1 sixteenth. So that's it evaluated at 1 half. And then from that I'm going to subtract negative 1 over pi cosine of 0."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to have to evaluate that from 0 to 1 half. When you evaluate it 1 half, so I'm going to get negative 1 over pi cosine of pi over 2 minus 2 times 1 half to the fourth power is 1 sixteenth. So that's it evaluated at 1 half. And then from that I'm going to subtract negative 1 over pi cosine of 0. Let me just write it out. So minus negative 1 over pi cosine of pi times 0 minus 2 times 0 to the fourth. So that's just going to be minus 0."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then from that I'm going to subtract negative 1 over pi cosine of 0. Let me just write it out. So minus negative 1 over pi cosine of pi times 0 minus 2 times 0 to the fourth. So that's just going to be minus 0. So let's evaluate this. So to simplify it, we have a cosine of pi over 2. Cosine of pi over 2 is just going to be 0."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's just going to be minus 0. So let's evaluate this. So to simplify it, we have a cosine of pi over 2. Cosine of pi over 2 is just going to be 0. So this whole thing just becomes 0. And then you have a negative 2 divided by 16. That's negative 1 eighth."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of pi over 2 is just going to be 0. So this whole thing just becomes 0. And then you have a negative 2 divided by 16. That's negative 1 eighth. And then from that I'm going to subtract this business here. Cosine of 0, this is 1. So this is just a negative 1 over pi."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's negative 1 eighth. And then from that I'm going to subtract this business here. Cosine of 0, this is 1. So this is just a negative 1 over pi. And then I have a 0 there, so I can ignore that. So this is equal to negative 1 over 8 plus 1 over pi. And we are done."}, {"video_title": "2011 Calculus AB free response #3 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is just a negative 1 over pi. And then I have a 0 there, so I can ignore that. So this is equal to negative 1 over 8 plus 1 over pi. And we are done. This part of it you're not allowed to use a calculator. So this is about as far as I would expect them to get. About as far as I would expect them to expect you to get."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I have two different expressions here that I want to take the derivative of. And what I want you to do is pause the video and think about how you would first approach taking the derivative of this expression and how that might be the same or different as your approach in taking the derivative of this expression. The goal here isn't to compute the derivatives all the way, but really to just think about how we identify what strategies to use. Okay, so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is let's look at the outside rather than the inside details. So if we look at the outside here, we have the sign of something."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "Okay, so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is let's look at the outside rather than the inside details. So if we look at the outside here, we have the sign of something. So there's a sign of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we look at the outside here, we have the sign of something. So there's a sign of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting it into a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting it into a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would want to use, in this case, the chain rule, CR for chain rule. And how would we apply it?"}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would want to use, in this case, the chain rule, CR for chain rule. And how would we apply it? Well, we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "And how would we apply it? Well, we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something. I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can write this as the derivative with respect to that something. I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now, what would that be tangibly in this case? Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now, what would that be tangibly in this case? Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course, I wouldn't be done yet. I have more derivative taking to do."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course, I wouldn't be done yet. I have more derivative taking to do. Here, now I would look at the big structure of what's going on. And I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "I have more derivative taking to do. Here, now I would look at the big structure of what's going on. And I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well, this looks a lot more like this step of the first problem than the beginning of the original problem."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well, this looks a lot more like this step of the first problem than the beginning of the original problem. Here, I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here, I have the product of two expressions, just like we saw over here. We have this expression being multiplied by this expression."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this looks a lot more like this step of the first problem than the beginning of the original problem. Here, I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here, I have the product of two expressions, just like we saw over here. We have this expression being multiplied by this expression. So my brain just says, okay, I have two expressions. Then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have this expression being multiplied by this expression. So my brain just says, okay, I have two expressions. Then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case, it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case, it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression. Just going to do that with the orange circle times the second expression. I'm going to do that with the blue circle plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of the second expression. Once again, here, this is just the product rule."}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that tells me that this is going to be the derivative with respect to x of the first expression. Just going to do that with the orange circle times the second expression. I'm going to do that with the blue circle plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of the second expression. Once again, here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle. And you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about, well, do I use the chain rule first and then use the product rule here?"}, {"video_title": "Differentiating using multiple rules strategy   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle. And you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about, well, do I use the chain rule first and then use the product rule here? Or in this case, do I use the product rule first? And even once you do this, you're not going to be done. Then to compute this derivative, you're going to have to use the chain rule."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when I say get a sense, we're gonna do that by seeing what values for this expression we get as x gets closer and closer to three. Now one thing that you might wanna try out is well what happens to this expression when x is equal to three? Well then it's going to be three to the third power minus three times three squared over five times three minus 15. So at x equals three, this expression's gonna be and you see the numerator, you have 27 minus 27, zero over 15 minus 15 over zero. So this expression is actually not defined at x equals three we get this indeterminate form, we get zero over zero. But let's see, even though the function, even though the expression is not defined, let's see if we can get a sense of what the limit might be. And to do that, I'm gonna set up a table."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So at x equals three, this expression's gonna be and you see the numerator, you have 27 minus 27, zero over 15 minus 15 over zero. So this expression is actually not defined at x equals three we get this indeterminate form, we get zero over zero. But let's see, even though the function, even though the expression is not defined, let's see if we can get a sense of what the limit might be. And to do that, I'm gonna set up a table. So let me set up a table here. And actually I'm gonna set up two tables. So this is x and this is x to the third minus three x squared over five x minus 15."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to do that, I'm gonna set up a table. So let me set up a table here. And actually I'm gonna set up two tables. So this is x and this is x to the third minus three x squared over five x minus 15. And actually I'm gonna do that again and I'll tell you why in a second. So this is gonna be x and this is x to the third minus three x squared over five x minus 15. The reason why I set up two tables, I didn't have to do two tables, I could have done it all in one table but hopefully this will make it a little bit more intuitive what I'm trying to do is on this left table, I'm gonna, let's try out x values that get closer and closer to three from the left, from values that are less than three."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is x and this is x to the third minus three x squared over five x minus 15. And actually I'm gonna do that again and I'll tell you why in a second. So this is gonna be x and this is x to the third minus three x squared over five x minus 15. The reason why I set up two tables, I didn't have to do two tables, I could have done it all in one table but hopefully this will make it a little bit more intuitive what I'm trying to do is on this left table, I'm gonna, let's try out x values that get closer and closer to three from the left, from values that are less than three. So for example, we could go to 2.9 and figure out what the expression equals when x is 2.9. But then we could try to get even a little bit closer than that. We could go to 2.99."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The reason why I set up two tables, I didn't have to do two tables, I could have done it all in one table but hopefully this will make it a little bit more intuitive what I'm trying to do is on this left table, I'm gonna, let's try out x values that get closer and closer to three from the left, from values that are less than three. So for example, we could go to 2.9 and figure out what the expression equals when x is 2.9. But then we could try to get even a little bit closer than that. We could go to 2.99. And then we could go even closer than that. We could go to 2.999. And so one way to think about it here is as we try to figure out what this expression equals as we get closer and closer to three, we're trying to approximate the limit from the left."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could go to 2.99. And then we could go even closer than that. We could go to 2.999. And so one way to think about it here is as we try to figure out what this expression equals as we get closer and closer to three, we're trying to approximate the limit from the left. So limit from the left. And why do I say the left? Well, if you think about this on a coordinate plane, these are the x values that are to the left of three but we're getting closer and closer and closer."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so one way to think about it here is as we try to figure out what this expression equals as we get closer and closer to three, we're trying to approximate the limit from the left. So limit from the left. And why do I say the left? Well, if you think about this on a coordinate plane, these are the x values that are to the left of three but we're getting closer and closer and closer. We're moving to the right but these are the x values that are on the left side of three, they're less than three. But we also, in order for the limit to exist, we have to be approaching the same thing from both sides, from both the left and the right. So we could also try to approximate the limit from the right."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if you think about this on a coordinate plane, these are the x values that are to the left of three but we're getting closer and closer and closer. We're moving to the right but these are the x values that are on the left side of three, they're less than three. But we also, in order for the limit to exist, we have to be approaching the same thing from both sides, from both the left and the right. So we could also try to approximate the limit from the right. And so what values would those be? Well, those would be x values larger than three. So we could say 3.1 but then we might wanna get a little bit closer."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could also try to approximate the limit from the right. And so what values would those be? Well, those would be x values larger than three. So we could say 3.1 but then we might wanna get a little bit closer. We could go 3.01 but then we might wanna get even closer to three, 3.001. And every time we get closer and closer to three, we're gonna get a better approximation for, or we're gonna get a better sense of what we are actually approaching. So let's get a calculator out and do this."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say 3.1 but then we might wanna get a little bit closer. We could go 3.01 but then we might wanna get even closer to three, 3.001. And every time we get closer and closer to three, we're gonna get a better approximation for, or we're gonna get a better sense of what we are actually approaching. So let's get a calculator out and do this. And you could keep going, 2.99999, 3.00001. Now one key idea here to point out before I even calculate what these are going to be, sometimes when people say the limit from both sides or the limit from the left or the limit from the right, they imagine that the limit from the left is negative values and the limit from the right are positive values. But as you can see here, the limit from the left are to the left of the x value that you're trying to find the limit at."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's get a calculator out and do this. And you could keep going, 2.99999, 3.00001. Now one key idea here to point out before I even calculate what these are going to be, sometimes when people say the limit from both sides or the limit from the left or the limit from the right, they imagine that the limit from the left is negative values and the limit from the right are positive values. But as you can see here, the limit from the left are to the left of the x value that you're trying to find the limit at. So these aren't negative values, these are just approaching the three right over here from values less than three. This is approaching the three from values larger than three. So now let's fill out this table."}, {"video_title": "Approximating limits using tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But as you can see here, the limit from the left are to the left of the x value that you're trying to find the limit at. So these aren't negative values, these are just approaching the three right over here from values less than three. This is approaching the three from values larger than three. So now let's fill out this table. And I'm speeding up my work so that you don't have to sit through me typing everything into a calculator. So based on what we're seeing here, I would make the estimate that this looks like it's approaching 1.8. So is this equal to 1.8?"}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Using the, I guess we could call it the washer method or the ring method, we were able to come up with a definite integral for the volume of this solid of revolution right over here. So this is equal to the volume. And so in this video, let's actually evaluate this integral. So the first thing that we could do is maybe factor out this pi. So this is going to be equal to pi times the definite integral from 0 to 1. And then let's square this stuff that we have right here in green. So 2 squared is going to be 4."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first thing that we could do is maybe factor out this pi. So this is going to be equal to pi times the definite integral from 0 to 1. And then let's square this stuff that we have right here in green. So 2 squared is going to be 4. And then we're going to have 2 times the product of both of these terms. So 2 times negative 2y squared times 2 is going to be negative 4y squared. And then negative y squared squared is plus y to the fourth."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 2 squared is going to be 4. And then we're going to have 2 times the product of both of these terms. So 2 times negative 2y squared times 2 is going to be negative 4y squared. And then negative y squared squared is plus y to the fourth. And then from that, we are going to subtract this thing squared. We are going to subtract this business square, which is going to be 4 minus 4 square roots of y plus the square root of y squared is just going to be y. And then all of that dy."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then negative y squared squared is plus y to the fourth. And then from that, we are going to subtract this thing squared. We are going to subtract this business square, which is going to be 4 minus 4 square roots of y plus the square root of y squared is just going to be y. And then all of that dy. Let me write that in that same color. All of that dy. And so this is going to be equal to pi times the definite integral from 0 to 1."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then all of that dy. Let me write that in that same color. All of that dy. And so this is going to be equal to pi times the definite integral from 0 to 1. And let's see if we can simplify this. We have a positive 4 here, but then when you distribute this negative, you're going to have a negative 4. So that cancels with that."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to pi times the definite integral from 0 to 1. And let's see if we can simplify this. We have a positive 4 here, but then when you distribute this negative, you're going to have a negative 4. So that cancels with that. And let's see. The highest degree term here is going to be our y to the fourth. So we have a y to the fourth."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that cancels with that. And let's see. The highest degree term here is going to be our y to the fourth. So we have a y to the fourth. I'll write it in that same color. y to the fourth. And so the next highest degree term right here is this negative 4y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have a y to the fourth. I'll write it in that same color. y to the fourth. And so the next highest degree term right here is this negative 4y squared. So then you have negative. Let me do that same color. We have negative 4y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the next highest degree term right here is this negative 4y squared. So then you have negative. Let me do that same color. We have negative 4y squared. That's that one right over there. And then we have this y, but remember we have this negative out front. So it's a negative y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have negative 4y squared. That's that one right over there. And then we have this y, but remember we have this negative out front. So it's a negative y. So this one right over here is a negative y. And then we have a negative times a negative, which is going to give us a positive 4 square roots of y. So this is going to end up being a positive 4 square roots of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's a negative y. So this one right over here is a negative y. And then we have a negative times a negative, which is going to give us a positive 4 square roots of y. So this is going to end up being a positive 4 square roots of y. And actually, just to make it clear, when we take the antiderivative, I'm going to write that as 4y to the 1 half power. And we're going to multiply all of that stuff by dy. Now we're ready to take the antiderivative."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to end up being a positive 4 square roots of y. And actually, just to make it clear, when we take the antiderivative, I'm going to write that as 4y to the 1 half power. And we're going to multiply all of that stuff by dy. Now we're ready to take the antiderivative. So it's going to be equal to pi times the antiderivative of y to the fourth is y to the fifth over 5. Antiderivative of negative 4y squared is negative 4 thirds y to the third power. Antiderivative of the negative y is negative y squared over 2."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we're ready to take the antiderivative. So it's going to be equal to pi times the antiderivative of y to the fourth is y to the fifth over 5. Antiderivative of negative 4y squared is negative 4 thirds y to the third power. Antiderivative of the negative y is negative y squared over 2. And then the antiderivative of 4y to the 1 half, let's see, we're going to increment. It's going to be y to the 3 halves. Multiply by 2 thirds."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Antiderivative of the negative y is negative y squared over 2. And then the antiderivative of 4y to the 1 half, let's see, we're going to increment. It's going to be y to the 3 halves. Multiply by 2 thirds. We're going to get 8 thirds plus 8 thirds y to the 3 halves. And let's see. Yep, that all works out."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Multiply by 2 thirds. We're going to get 8 thirds plus 8 thirds y to the 3 halves. And let's see. Yep, that all works out. And we're going to evaluate this at 1 and at 0. And lucky for us, when you evaluate at 0, this whole thing turns out to be 0. So this is all going to be equal to pi times evaluating all this business at 1."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yep, that all works out. And we're going to evaluate this at 1 and at 0. And lucky for us, when you evaluate at 0, this whole thing turns out to be 0. So this is all going to be equal to pi times evaluating all this business at 1. So that's going to be 1 fifth minus 4 thirds. I'll do that in green color, minus 4 thirds minus 1 half. So whenever you evaluate at 1, it's just going to be."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is all going to be equal to pi times evaluating all this business at 1. So that's going to be 1 fifth minus 4 thirds. I'll do that in green color, minus 4 thirds minus 1 half. So whenever you evaluate at 1, it's just going to be. So plus 8 thirds. Plus 8 thirds. Plus 8 thirds."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So whenever you evaluate at 1, it's just going to be. So plus 8 thirds. Plus 8 thirds. Plus 8 thirds. And let's see. What's the least common multiple over here? Let's see."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus 8 thirds. And let's see. What's the least common multiple over here? Let's see. 5, a 3, and a 2. Looks like we're going to have a denominator of 30. So we could rewrite this as equal to pi."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see. 5, a 3, and a 2. Looks like we're going to have a denominator of 30. So we could rewrite this as equal to pi. And we could put everything over a denominator of 30. 1 fifth is 6 over 30. 4 thirds is 40 over 30."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could rewrite this as equal to pi. And we could put everything over a denominator of 30. 1 fifth is 6 over 30. 4 thirds is 40 over 30. So this is minus 40. That's the different shade of green. Well, actually, let me make that other shade of green."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "4 thirds is 40 over 30. So this is minus 40. That's the different shade of green. Well, actually, let me make that other shade of green. So this is minus 40 over 30. Negative 1 half, that's minus 15 over 30. And then finally, 8 thirds is the same thing as 80 over 30."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, actually, let me make that other shade of green. So this is minus 40 over 30. Negative 1 half, that's minus 15 over 30. And then finally, 8 thirds is the same thing as 80 over 30. So that's plus 80. So this simplifies to. So let's see."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, 8 thirds is the same thing as 80 over 30. So that's plus 80. So this simplifies to. So let's see. We have 86 minus 55. Actually, let me make sure I'm doing the math right over here. So 80 minus 40 gets us 40."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see. We have 86 minus 55. Actually, let me make sure I'm doing the math right over here. So 80 minus 40 gets us 40. Plus 6 is 46. Minus 15 is 31. So this is equal to 31 pi over 30."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 80 minus 40 gets us 40. Plus 6 is 46. Minus 15 is 31. So this is equal to 31 pi over 30. I have a suspicion that I might have done something shady in this last part right over here. So this is going to be. Let's see."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to 31 pi over 30. I have a suspicion that I might have done something shady in this last part right over here. So this is going to be. Let's see. 86 negative 51 plus 80. I think that seems right. I'm going to do it one more time."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see. 86 negative 51 plus 80. I think that seems right. I'm going to do it one more time. Let's see. 80 minus 40 is 40. 46 minus 10 is 36."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You are likely already familiar with the idea of a slope of a line. If you're not, I encourage you to review it on Khan Academy. But all it is, it's describing the rate of change of a vertical variable with respect to a horizontal variable. So for example, here I have our classic y-axis in the vertical direction and x-axis in the horizontal direction. And if I wanted to figure out the slope of this line, I could pick two points, say that point and that point. I could say, okay, from this point to this point, what is my change in x? Well, my change in x would be this distance right over here."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, here I have our classic y-axis in the vertical direction and x-axis in the horizontal direction. And if I wanted to figure out the slope of this line, I could pick two points, say that point and that point. I could say, okay, from this point to this point, what is my change in x? Well, my change in x would be this distance right over here. Change in x, the Greek letter delta, this triangle here, it's just shorthand for change. So change in x. And I could also calculate the change in y."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, my change in x would be this distance right over here. Change in x, the Greek letter delta, this triangle here, it's just shorthand for change. So change in x. And I could also calculate the change in y. So this point going up to that point, our change in y would be this right over here, our change in y. And then we would define slope, or we have defined slope, as change in y over change in x. So slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I could also calculate the change in y. So this point going up to that point, our change in y would be this right over here, our change in y. And then we would define slope, or we have defined slope, as change in y over change in x. So slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable. It's sometimes described as rise over run. And for any line, it's associated with a slope because it has a constant rate of change. If you took any two points on this line, no matter how far apart or no matter how close together, anywhere they sit on the line, if you were to do this calculation, you would get the same slope."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable. It's sometimes described as rise over run. And for any line, it's associated with a slope because it has a constant rate of change. If you took any two points on this line, no matter how far apart or no matter how close together, anywhere they sit on the line, if you were to do this calculation, you would get the same slope. That's what makes it a line. But what's fascinating about calculus is we're going to build the tools so that we can think about the rate of change not just of a line, which we've called slope in the past, we can think about the rate of change, the instantaneous rate of change of a curve, of something whose rate of change is possibly constantly changing. So for example, here's a curve where the rate of change of y with respect to x is constantly changing."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you took any two points on this line, no matter how far apart or no matter how close together, anywhere they sit on the line, if you were to do this calculation, you would get the same slope. That's what makes it a line. But what's fascinating about calculus is we're going to build the tools so that we can think about the rate of change not just of a line, which we've called slope in the past, we can think about the rate of change, the instantaneous rate of change of a curve, of something whose rate of change is possibly constantly changing. So for example, here's a curve where the rate of change of y with respect to x is constantly changing. Even if we wanted to use our traditional tools, if we said, okay, we can calculate the average rate of change, let's say between this point and this point, well, what would it be? Well, the average rate of change between this point and this point would be the slope of the line that connects them. So it'd be the slope of this line, of the secant line."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, here's a curve where the rate of change of y with respect to x is constantly changing. Even if we wanted to use our traditional tools, if we said, okay, we can calculate the average rate of change, let's say between this point and this point, well, what would it be? Well, the average rate of change between this point and this point would be the slope of the line that connects them. So it'd be the slope of this line, of the secant line. But if we pick two different points, if we pick this point and this point, the average rate of change between those points all of a sudden looks quite different. It looks like it has a higher slope. So even when we take the slopes between two points on the line, the secant lines, you can see that those slopes are changing."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it'd be the slope of this line, of the secant line. But if we pick two different points, if we pick this point and this point, the average rate of change between those points all of a sudden looks quite different. It looks like it has a higher slope. So even when we take the slopes between two points on the line, the secant lines, you can see that those slopes are changing. But what if we wanted to ask ourselves an even more interesting question? What is the instantaneous rate of change at a point? So for example, how fast is y changing with respect to x exactly at that point, exactly when x is equal to that value?"}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So even when we take the slopes between two points on the line, the secant lines, you can see that those slopes are changing. But what if we wanted to ask ourselves an even more interesting question? What is the instantaneous rate of change at a point? So for example, how fast is y changing with respect to x exactly at that point, exactly when x is equal to that value? Let's call it x one. Well, one way you could think about it is what if we could draw a tangent line to this point, a line that just touches the graph right over there? And we can calculate the slope of that line."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, how fast is y changing with respect to x exactly at that point, exactly when x is equal to that value? Let's call it x one. Well, one way you could think about it is what if we could draw a tangent line to this point, a line that just touches the graph right over there? And we can calculate the slope of that line. Well, that should be the rate of change at that point, the instantaneous rate of change. So in this case, the tangent line might look something like that. If we know the slope of this, well, then we could say that that's the instantaneous rate of change at that point."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can calculate the slope of that line. Well, that should be the rate of change at that point, the instantaneous rate of change. So in this case, the tangent line might look something like that. If we know the slope of this, well, then we could say that that's the instantaneous rate of change at that point. Why do I say instantaneous rate of change? Well, think about the video on the sprinters, the Usain Bolt example. If we wanted to figure out the speed of Usain Bolt at a given instant, well, maybe this describes his position with respect to time."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we know the slope of this, well, then we could say that that's the instantaneous rate of change at that point. Why do I say instantaneous rate of change? Well, think about the video on the sprinters, the Usain Bolt example. If we wanted to figure out the speed of Usain Bolt at a given instant, well, maybe this describes his position with respect to time. If y was position and x is time, usually you would see t as time, but let's say x is time. So then if we're talking about right at this time, we're talking about the instantaneous rate. And this idea is a central idea of differential calculus, and it's known as a derivative."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we wanted to figure out the speed of Usain Bolt at a given instant, well, maybe this describes his position with respect to time. If y was position and x is time, usually you would see t as time, but let's say x is time. So then if we're talking about right at this time, we're talking about the instantaneous rate. And this idea is a central idea of differential calculus, and it's known as a derivative. The slope of the tangent line, which you could also view as the instantaneous rate of change, I'm putting exclamation mark because it's so conceptually important here. So how can we denote a derivative? One way is known as Leibniz's notation, and Leibniz is one of the fathers of calculus along with Isaac Newton."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this idea is a central idea of differential calculus, and it's known as a derivative. The slope of the tangent line, which you could also view as the instantaneous rate of change, I'm putting exclamation mark because it's so conceptually important here. So how can we denote a derivative? One way is known as Leibniz's notation, and Leibniz is one of the fathers of calculus along with Isaac Newton. And his notation, you would denote the slope of the tangent line as equaling dy over dx. Now why do I like this notation? Because it really comes from this idea of a slope, which is change in y over change in x."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "One way is known as Leibniz's notation, and Leibniz is one of the fathers of calculus along with Isaac Newton. And his notation, you would denote the slope of the tangent line as equaling dy over dx. Now why do I like this notation? Because it really comes from this idea of a slope, which is change in y over change in x. As you'll see in future videos, one way to think about the slope of the tangent line is, well, let's calculate the slope of secant lines. Let's say between that point and that point. But then let's get even closer."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because it really comes from this idea of a slope, which is change in y over change in x. As you'll see in future videos, one way to think about the slope of the tangent line is, well, let's calculate the slope of secant lines. Let's say between that point and that point. But then let's get even closer. Let's say that point and that point, and then let's get even closer, and that point and that point, and then let's get even closer. And let's see what happens as the change in x approaches zero. And so using these d's instead of deltas, this was Leibniz's way of saying, hey, what happens if my changes in, say, x become close to zero?"}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then let's get even closer. Let's say that point and that point, and then let's get even closer, and that point and that point, and then let's get even closer. And let's see what happens as the change in x approaches zero. And so using these d's instead of deltas, this was Leibniz's way of saying, hey, what happens if my changes in, say, x become close to zero? So this idea, this is known as sometimes differential notation, Leibniz's notation, is instead of just change in y over change in x, super small changes in y for a super small change in x, especially as the change in x approaches zero. And as you will see, that is how we will calculate the derivative. Now there's other notations."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so using these d's instead of deltas, this was Leibniz's way of saying, hey, what happens if my changes in, say, x become close to zero? So this idea, this is known as sometimes differential notation, Leibniz's notation, is instead of just change in y over change in x, super small changes in y for a super small change in x, especially as the change in x approaches zero. And as you will see, that is how we will calculate the derivative. Now there's other notations. If this curve is described as y is equal to f of x, the slope of the tangent line at that point could be denoted as equaling f prime of x one. So this notation, it takes a little bit of time getting used to, the Lagrange notation. It's saying f prime is representing the derivative."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now there's other notations. If this curve is described as y is equal to f of x, the slope of the tangent line at that point could be denoted as equaling f prime of x one. So this notation, it takes a little bit of time getting used to, the Lagrange notation. It's saying f prime is representing the derivative. It's telling us the slope of the tangent line for a given point. So if you input an x into this function, into f, you're getting the corresponding y value. If you input an x into f prime, you're getting the slope of the tangent line at that point."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's saying f prime is representing the derivative. It's telling us the slope of the tangent line for a given point. So if you input an x into this function, into f, you're getting the corresponding y value. If you input an x into f prime, you're getting the slope of the tangent line at that point. Now another notation that you'll see less likely in a calculus class, but you might see in a physics class, is the notation y with a dot over it. So you could write this as y with a dot over it, which also denotes the derivative. You might also see y prime."}, {"video_title": "Derivative as a concept   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you input an x into f prime, you're getting the slope of the tangent line at that point. Now another notation that you'll see less likely in a calculus class, but you might see in a physics class, is the notation y with a dot over it. So you could write this as y with a dot over it, which also denotes the derivative. You might also see y prime. This would be more common in a math class. Now as we march forward in our calculus adventure, we will build the tools to actually calculate these things. And if you're already familiar with limits, they will be very useful, as you can imagine, because we're really going to be taking the limit of our change in y over change in x as our change in x approaches zero."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say I have some function f of x that is defined as being equal to x squared minus 6x plus 8 for all x. And what I want to do is show that for this function we can definitely find a c in an interval where the derivative at the point c is equal to the average rate of change over that interval. So let's give ourselves an interval right over here. Let's say we care about the interval between 2 and 5. And this function is clearly both continuous over this closed interval and it's also differentiable over it. It just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, such that the derivative at c is equal to the average rate of change over this interval or is equal to the slope of the secant line between the two endpoints of the interval."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say we care about the interval between 2 and 5. And this function is clearly both continuous over this closed interval and it's also differentiable over it. It just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, such that the derivative at c is equal to the average rate of change over this interval or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well, to do that, let's just calculate what this has to be."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, such that the derivative at c is equal to the average rate of change over this interval or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well, to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see, f of 5 minus f of 2. f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8, that's going to be 0."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see, f of 5 minus f of 2. f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8, that's going to be 0. So this is equal to 3 over 3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well, let's see, f prime of x is equal to 2x minus 6."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's 4 minus 12 plus 8, that's going to be 0. So this is equal to 3 over 3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well, let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be within this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be within this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7 halves, which is the same thing as 3 and a half. So it's definitely in this interval right over here. So we've just found our c. c is equal to 7 halves."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7 halves, which is the same thing as 3 and a half. So it's definitely in this interval right over here. So we've just found our c. c is equal to 7 halves. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis. And then this right over here is our x-axis."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we've just found our c. c is equal to 7 halves. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis. And then this right over here is our x-axis. Looks like all the action is happening in the first and fourth quadrants. So that is our x-axis. Let's say this is 1, 2, 3, 4, and 5."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this right over here is our x-axis. Looks like all the action is happening in the first and fourth quadrants. So that is our x-axis. Let's say this is 1, 2, 3, 4, and 5. So we already know that 2 is one of the 0s here. So we know that our function, if we wanted to graph it, it intersects the x-axis right over here. And you can factor this out as x minus 2 times x minus 4."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say this is 1, 2, 3, 4, and 5. So we already know that 2 is one of the 0s here. So we know that our function, if we wanted to graph it, it intersects the x-axis right over here. And you can factor this out as x minus 2 times x minus 4. So the other place where our function hits 0 is when x is equal to 4 right over here. Our vertex is going to be right in between at x is equal to 3. When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can factor this out as x minus 2 times x minus 4. So the other place where our function hits 0 is when x is equal to 4 right over here. Our vertex is going to be right in between at x is equal to 3. When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1. So you have the point 3, negative 1 on it. So that's enough for us to graph it. And we also know at 5 we're at 3."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1. So you have the point 3, negative 1 on it. So that's enough for us to graph it. And we also know at 5 we're at 3. So 1, 2, 3. So at 5 we are right over here. So over the interval that we care about, our graph looks something like this."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we also know at 5 we're at 3. So 1, 2, 3. So at 5 we are right over here. So over the interval that we care about, our graph looks something like this. Our graph looks something like this. So that's the interval that we care about. And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over the interval that we care about, our graph looks something like this. Our graph looks something like this. So that's the interval that we care about. And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points. And if I were to just visually look at it, I'd say, well, yeah, it looks like right around there, just based on my drawing, the slope of the tangent line looks like it's parallel. It looks like it has the same slope. It looks like the tangent line is parallel to the secant line."}, {"video_title": "Mean value theorem example polynomial   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points. And if I were to just visually look at it, I'd say, well, yeah, it looks like right around there, just based on my drawing, the slope of the tangent line looks like it's parallel. It looks like it has the same slope. It looks like the tangent line is parallel to the secant line. And that looks like it's right at 3 1\u20442 or 7 halves. So it makes sense. So this right over here is our c. And it's equal to 7 halves."}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "At a certain instant, t sub zero, the first car is a distance x sub t sub zero, or x of t sub zero, of half a kilometer from the intersection, and the second car is a distance y of t sub zero of 1.2 kilometers from the intersection. What is the rate of change of the distance, d of t, between the cars at that instant, so at t sub zero? Which equation should be used to solve the problem? And they give us a choice of four equations right over here. So you could pause the video and try to work through it on your own, but I'm about to do it as well. So let's just draw what's going on. That's always a healthy thing to do."}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they give us a choice of four equations right over here. So you could pause the video and try to work through it on your own, but I'm about to do it as well. So let's just draw what's going on. That's always a healthy thing to do. So two cars are driving towards an intersection from perpendicular directions. So let's say that this is one car right over here, and it is moving in the x direction towards that intersection, which is right over there. And then you have another car that is moving in the y direction."}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's always a healthy thing to do. So two cars are driving towards an intersection from perpendicular directions. So let's say that this is one car right over here, and it is moving in the x direction towards that intersection, which is right over there. And then you have another car that is moving in the y direction. So let's say it's moving like this. So this is the other car. I should've maybe done a top view."}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you have another car that is moving in the y direction. So let's say it's moving like this. So this is the other car. I should've maybe done a top view. Oh, here we go. This square represents the car, and it is moving in that direction. Now, they say at a certain instant, t sub zero."}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I should've maybe done a top view. Oh, here we go. This square represents the car, and it is moving in that direction. Now, they say at a certain instant, t sub zero. So let's draw that instant. So the first car is a distance x of t sub zero of 0.5 kilometers. So this distance right over here, let's just call this x of t, and let's call this distance right over here y of t. Now, how does the distance between the cars relate to x of t and y of t?"}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, they say at a certain instant, t sub zero. So let's draw that instant. So the first car is a distance x of t sub zero of 0.5 kilometers. So this distance right over here, let's just call this x of t, and let's call this distance right over here y of t. Now, how does the distance between the cars relate to x of t and y of t? Well, we could just use the distance formula, which is essentially just the Pythagorean theorem, to say, well, the distance between the cars would be the hypotenuse of this right triangle. Remember, they're traveling from perpendicular directions, so that's a right triangle there. So this distance right over here would be x of t squared plus y of t squared and the square root of that, and that's just the Pythagorean theorem right over here."}, {"video_title": "Analyzing related rates problems equations (Pythagoras)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this distance right over here, let's just call this x of t, and let's call this distance right over here y of t. Now, how does the distance between the cars relate to x of t and y of t? Well, we could just use the distance formula, which is essentially just the Pythagorean theorem, to say, well, the distance between the cars would be the hypotenuse of this right triangle. Remember, they're traveling from perpendicular directions, so that's a right triangle there. So this distance right over here would be x of t squared plus y of t squared and the square root of that, and that's just the Pythagorean theorem right over here. This would be d of t, or we could say that d of t squared is equal to x of t squared plus y, let me, too many parentheses, plus y of t squared. So that's the relationship between d of t, x of t, and y of t, and it's useful for solving this problem because now we can take the derivative of both sides of this equation with respect to t, and we'd be using various derivative rules, including the chain rule, in order to do it, and then that would give us a relationship between the rate of change of d of t, which would be d prime of t, and the rate of change of x of t, y of t, and x of t and y of t themselves. And so if we look at these choices right over here, we indeed see that d sets up that exact same relationship that we just did ourselves, that it shows that the distance squared between the cars is equal to that x distance from the intersection squared plus the y distance from the intersection squared, and then we can take the derivative of both sides to actually figure out this related rates question."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's take the derivative of this. So we can view this as the product of two functions. So the product rule tells us that this is going to be the derivative with respect to x of e to the cosine of x, e to the cosine of x times cosine, times cosine of e to the x plus, plus the first function, just e to the cosine of x, e to the cosine of x times the derivative of the second function, times the derivative with respect to x of cosine of e to the x, cosine of e to the x. And so we just need to figure out what these two derivatives are. And so you can imagine the chain rule might be applicable here. So let me make it clear. This we got from the product rule."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we just need to figure out what these two derivatives are. And so you can imagine the chain rule might be applicable here. So let me make it clear. This we got from the product rule. Product, product rule. But then to evaluate each of these derivatives, we need to use the chain rule. So let's think about this a little bit."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This we got from the product rule. Product, product rule. But then to evaluate each of these derivatives, we need to use the chain rule. So let's think about this a little bit. So the derivative, let me copy and paste this so I don't have to rewrite it. So copy and paste. So let's think about what the derivative of e to the cosine of x is."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So the derivative, let me copy and paste this so I don't have to rewrite it. So copy and paste. So let's think about what the derivative of e to the cosine of x is. E to the cosine of x. So we could view our outer function as e to the something, as e to the something. And the derivative of e to the something with respect to something is just going to be e to that something."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what the derivative of e to the cosine of x is. E to the cosine of x. So we could view our outer function as e to the something, as e to the something. And the derivative of e to the something with respect to something is just going to be e to that something. So it's going to be e to the cosine of x. So let me do that in that same blue color. So it's going to be e, actually let me do it in a new color."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the derivative of e to the something with respect to something is just going to be e to that something. So it's going to be e to the cosine of x. So let me do that in that same blue color. So it's going to be e, actually let me do it in a new color. Let me do it in magenta. So the derivative of e to the something with respect to something is just e to the something. It's just e to the cosine of x."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be e, actually let me do it in a new color. Let me do it in magenta. So the derivative of e to the something with respect to something is just e to the something. It's just e to the cosine of x. And we have to multiply that times the derivative of the something with respect to x. So what's the derivative of cosine of x with respect to x? Well, that's just negative sine of x."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's just e to the cosine of x. And we have to multiply that times the derivative of the something with respect to x. So what's the derivative of cosine of x with respect to x? Well, that's just negative sine of x. So it's times negative sine of x. And so we figured out this first derivative. Let me make it clear."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's just negative sine of x. So it's times negative sine of x. And so we figured out this first derivative. Let me make it clear. This right over here is the derivative of e to the cosine of x with respect to cosine of x. And this right over here is the derivative of cosine of x with respect to x. And we just took the product of the two."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make it clear. This right over here is the derivative of e to the cosine of x with respect to cosine of x. And this right over here is the derivative of cosine of x with respect to x. And we just took the product of the two. That's what the chain rule tells us. Fair enough. Now let's figure out this derivative out here."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we just took the product of the two. That's what the chain rule tells us. Fair enough. Now let's figure out this derivative out here. So we want to find the derivative with respect to x of cosine e to the x. So once again, let me copy and paste it. So we need to figure out this thing right over here."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's figure out this derivative out here. So we want to find the derivative with respect to x of cosine e to the x. So once again, let me copy and paste it. So we need to figure out this thing right over here. So first, just like we did, we're just going to apply the chain rule again. We need to figure out the derivative of cosine of something, in this case, e to the x, with respect to that something. So this is going to be equal to derivative of cosine of something with respect to that something is equal to the negative sine of that something."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we need to figure out this thing right over here. So first, just like we did, we're just going to apply the chain rule again. We need to figure out the derivative of cosine of something, in this case, e to the x, with respect to that something. So this is going to be equal to derivative of cosine of something with respect to that something is equal to the negative sine of that something. Negative sine of e to the x. Once again, we can view this as the derivative of cosine of e to the x with respect to e to the x. And then we multiply that times the derivative of the something with respect to x."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to derivative of cosine of something with respect to that something is equal to the negative sine of that something. Negative sine of e to the x. Once again, we can view this as the derivative of cosine of e to the x with respect to e to the x. And then we multiply that times the derivative of the something with respect to x. So let me do this in this. I'm running out of colors. Let me do this in this green color."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we multiply that times the derivative of the something with respect to x. So let me do this in this. I'm running out of colors. Let me do this in this green color. So times the derivative of e to the x with respect to x is just e to the x. So that right over there is the derivative of e to the x with respect to x. And so we're essentially done."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do this in this green color. So times the derivative of e to the x with respect to x is just e to the x. So that right over there is the derivative of e to the x with respect to x. And so we're essentially done. We just have to substitute what we found using the chain rule back into our original expression. The derivative of this business up here is going to be equal to, let me just copy and paste everything just to make everything nice and clean. So copy and paste."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we're essentially done. We just have to substitute what we found using the chain rule back into our original expression. The derivative of this business up here is going to be equal to, let me just copy and paste everything just to make everything nice and clean. So copy and paste. So that is going to be equal to this times cosine e to the x. So this is going to be, let's see, we could put the e to the x out front. We could put the negative out front."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So copy and paste. So that is going to be equal to this times cosine e to the x. So this is going to be, let's see, we could put the e to the x out front. We could put the negative out front. So we could write it as negative e to the cosine x times sine of x times cosine of e to the x. So that's this first term here. Plus e to the cosine x times all of this stuff."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could put the negative out front. So we could write it as negative e to the cosine x times sine of x times cosine of e to the x. So that's this first term here. Plus e to the cosine x times all of this stuff. And so let's see, we could put the negative out front again. So let's put that negative out front. So we have a negative."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus e to the cosine x times all of this stuff. And so let's see, we could put the negative out front again. So let's put that negative out front. So we have a negative. Now we have e to the cosine of x times e to the x. So I could write it this way. e to the x times e to the cosine x."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have a negative. Now we have e to the cosine of x times e to the x. So I could write it this way. e to the x times e to the cosine x. And you could simplify that or combine it since you're multiplying two things with the same base. But I'll just leave it like this. e to the x times e to the cosine x times the sine."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "e to the x times e to the cosine x. And you could simplify that or combine it since you're multiplying two things with the same base. But I'll just leave it like this. e to the x times e to the cosine x times the sine. We already have the negative. So then we have sine of e to the x. Sine of e to the x."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "e to the x times e to the cosine x times the sine. We already have the negative. So then we have sine of e to the x. Sine of e to the x. Let me write it over here. Times sine of e to the x. We had negative sine e to the x times e to the x."}, {"video_title": "Derivative of e_____cos(e_)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine of e to the x. Let me write it over here. Times sine of e to the x. We had negative sine e to the x times e to the x. Negative sine of e to the x times e to the x. And then that was multiplied by e to the cosine of x. So we have the exact same thing right over here."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The graph of g is given right over here, given below. How many inflection points does the graph of g have? And so let's just remind ourselves what are inflection points. So inflection points are where we change concavity. So we go from concave, concave upwards, upwards, actually let me just draw it graphically. We're going from concave upwards to concave downwards, or concave downwards to concave upwards. So, or another way you could think about it, you could say we're going from our slope increasing, increasing, increasing to our slope decreasing, to our slope decreasing, or the other way around."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So inflection points are where we change concavity. So we go from concave, concave upwards, upwards, actually let me just draw it graphically. We're going from concave upwards to concave downwards, or concave downwards to concave upwards. So, or another way you could think about it, you could say we're going from our slope increasing, increasing, increasing to our slope decreasing, to our slope decreasing, or the other way around. Any points where your slope goes from decreasing, our slope goes from decreasing to increasing, to increasing. So let's think about that. So as we start off right over here, so at the extreme left, it seems like we have a very high slope."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, or another way you could think about it, you could say we're going from our slope increasing, increasing, increasing to our slope decreasing, to our slope decreasing, or the other way around. Any points where your slope goes from decreasing, our slope goes from decreasing to increasing, to increasing. So let's think about that. So as we start off right over here, so at the extreme left, it seems like we have a very high slope. It's a very steep curve. And then it stays increasing, but it's getting less positive. So it's getting a little bit, it's getting a little bit flatter."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as we start off right over here, so at the extreme left, it seems like we have a very high slope. It's a very steep curve. And then it stays increasing, but it's getting less positive. So it's getting a little bit, it's getting a little bit flatter. So our slope is at a very high level, but it's decreasing. It's decreasing, decreasing, decreasing. Slope is decreasing, decreasing even more."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's getting a little bit, it's getting a little bit flatter. So our slope is at a very high level, but it's decreasing. It's decreasing, decreasing, decreasing. Slope is decreasing, decreasing even more. It's even more. And then it's actually going to zero. Our slope is zero."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Slope is decreasing, decreasing even more. It's even more. And then it's actually going to zero. Our slope is zero. And then it becomes negative. So our slope is still decreasing. And then it's becoming more and more and more negative."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our slope is zero. And then it becomes negative. So our slope is still decreasing. And then it's becoming more and more and more negative. And then right around, and then right around here, it looks like it starts becoming less negative, or it starts increasing. So our slope is increasing, increasing. It's really just becoming less and less negative."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it's becoming more and more and more negative. And then right around, and then right around here, it looks like it starts becoming less negative, or it starts increasing. So our slope is increasing, increasing. It's really just becoming less and less negative. And then it's going close to zero, approaching zero. It looks like our slope is zero right over here. But then it looks like right over there, our slope begins decreasing again."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's really just becoming less and less negative. And then it's going close to zero, approaching zero. It looks like our slope is zero right over here. But then it looks like right over there, our slope begins decreasing again. So it looks like our slope is decreasing again. So it looks like our slope is decreasing. It's becoming more and more and more and more negative."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then it looks like right over there, our slope begins decreasing again. So it looks like our slope is decreasing again. So it looks like our slope is decreasing. It's becoming more and more and more and more negative. And so it looks like something interesting happened right over there. We had a transition point. And then right around here, it looks like it starts, the slope starts increasing again."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's becoming more and more and more and more negative. And so it looks like something interesting happened right over there. We had a transition point. And then right around here, it looks like it starts, the slope starts increasing again. So it looks like the slope starts increasing. It's negative, but it's becoming less and less and less negative. And then it becomes zero."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then right around here, it looks like it starts, the slope starts increasing again. So it looks like the slope starts increasing. It's negative, but it's becoming less and less and less negative. And then it becomes zero. And then it becomes positive. And then more and more and more and more positive. So inflection points are where we go from slope increasing to slope decreasing."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it becomes zero. And then it becomes positive. And then more and more and more and more positive. So inflection points are where we go from slope increasing to slope decreasing. So concave upwards to concave downwards. And so slope increasing was here to slope decreasing. So this was an inflection point."}, {"video_title": "Inflection points (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So inflection points are where we go from slope increasing to slope decreasing. So concave upwards to concave downwards. And so slope increasing was here to slope decreasing. So this was an inflection point. And also from slope decreasing to slope increasing. So that's slope decreasing to slope increasing. And this is also slope decreasing to slope increasing."}, {"video_title": "Undefined limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if we can figure out the limit of x over natural log of x as x approaches one. And like always, pause this video and see if you can figure it out on your own. Well, we know from our limit properties this is going to be the same thing as the limit as x approaches one of x over, over the limit, the limit as x approaches one of the natural log of x. Now, this top limit, the one I have in magenta, this is pretty straightforward. This, if we had the graph of y equals x, that would be continuous everywhere. It's defined for all real numbers and it's continuous at all real numbers. And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one."}, {"video_title": "Undefined limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, this top limit, the one I have in magenta, this is pretty straightforward. This, if we had the graph of y equals x, that would be continuous everywhere. It's defined for all real numbers and it's continuous at all real numbers. And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. So this is just going to be one. We just put a one in for this x. So the numerator here, we just evaluate to a one."}, {"video_title": "Undefined limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. So this is just going to be one. We just put a one in for this x. So the numerator here, we just evaluate to a one. And then the denominator. Natural log of x is not defined for all x's and therefore it isn't continuous everywhere, but it is continuous at x equals one. And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one."}, {"video_title": "Undefined limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the numerator here, we just evaluate to a one. And then the denominator. Natural log of x is not defined for all x's and therefore it isn't continuous everywhere, but it is continuous at x equals one. And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. So this is just going to be the natural log, the natural log of one, which of course is zero. E to the zero power is one. So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero."}, {"video_title": "Undefined limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. So this is just going to be the natural log, the natural log of one, which of course is zero. E to the zero power is one. So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. And now we face a bit of a conundrum. One over zero is not defined. If it was zero over zero, we wouldn't necessarily be done yet."}, {"video_title": "Undefined limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. And now we face a bit of a conundrum. One over zero is not defined. If it was zero over zero, we wouldn't necessarily be done yet. That's indeterminate form. As we will learn in the future, there are tools we can apply when we're trying to find limits and we evaluate it like this and we get zero over zero. But one over zero, this is undefined, which tells us that this limit does not exist."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have the graph of y equals f of x right over here, and we want to figure out three different limits. And like always, pause this video and see if you can figure it out on your own before we do it together. Alright, now first let's think about what's the limit of f of x as x approaches six. So as x, let me do this in a color you can see, as x approaches six from both sides, well as we approach six from the left-hand side, from values less than six, it looks like our f of x is approaching one, and as we approach x equals six from the right-hand side, it looks like our f of x is once again approaching one. And in order for this limit to exist, we need to be approaching the same value from both the left and the right-hand side. And so here, at least graphically, so you never are sure with the graph, but this is a pretty good estimate, it looks like we are approaching one. Right over there, let me do it in a darker color."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as x, let me do this in a color you can see, as x approaches six from both sides, well as we approach six from the left-hand side, from values less than six, it looks like our f of x is approaching one, and as we approach x equals six from the right-hand side, it looks like our f of x is once again approaching one. And in order for this limit to exist, we need to be approaching the same value from both the left and the right-hand side. And so here, at least graphically, so you never are sure with the graph, but this is a pretty good estimate, it looks like we are approaching one. Right over there, let me do it in a darker color. Now let's do this next one. The limit of f of x as x approaches four. So as we approach four from the left-hand side, what is going on?"}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right over there, let me do it in a darker color. Now let's do this next one. The limit of f of x as x approaches four. So as we approach four from the left-hand side, what is going on? Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three. Remember, you can have a limit exist at an x value where the function itself is not defined. The function, if you said what is f of four, it's not defined."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as we approach four from the left-hand side, what is going on? Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three. Remember, you can have a limit exist at an x value where the function itself is not defined. The function, if you said what is f of four, it's not defined. But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three. And when we approach four from the right, once again, it looks like our function is approaching three. So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The function, if you said what is f of four, it's not defined. But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three. And when we approach four from the right, once again, it looks like our function is approaching three. So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. Now let's think about the limit as x approaches two. So this is interesting. The function is defined there."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. Now let's think about the limit as x approaches two. So this is interesting. The function is defined there. F of two is two. But see, when we approach from the left-hand side, it looks like our function is approaching the value of two. But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The function is defined there. F of two is two. But see, when we approach from the left-hand side, it looks like our function is approaching the value of two. But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five. And since we are approaching two different values from the left-hand side and the right-hand side as x approaches two from the left-hand side and the right-hand side, we would say that this limit does not exist. So does not exist, which is interesting."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five. And since we are approaching two different values from the left-hand side and the right-hand side as x approaches two from the left-hand side and the right-hand side, we would say that this limit does not exist. So does not exist, which is interesting. In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six. Here, the function was not defined at x equals four, but the limit does exist. Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So does not exist, which is interesting. In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six. Here, the function was not defined at x equals four, but the limit does exist. Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. Let's do another function, just to get more cases of looking at graphical limits. So here, we have the graph of y is equal to g of x. And once again, pause this video and have a go at it."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. Let's do another function, just to get more cases of looking at graphical limits. So here, we have the graph of y is equal to g of x. And once again, pause this video and have a go at it. See if you can figure out these limits graphically. So first, we have the limit as x approaches five of g of x. So as we approach five from the left-hand side, it looks like we are approaching this value."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, pause this video and have a go at it. See if you can figure out these limits graphically. So first, we have the limit as x approaches five of g of x. So as we approach five from the left-hand side, it looks like we are approaching this value. So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value. And as we approach five from the right-hand side, it also looks like we are approaching that same value. And so this value, just eyeballing it off of here, it looks like it's about.4."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as we approach five from the left-hand side, it looks like we are approaching this value. So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value. And as we approach five from the right-hand side, it also looks like we are approaching that same value. And so this value, just eyeballing it off of here, it looks like it's about.4. So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise. So I would say it's approximately 0.4. It might be 0.41."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this value, just eyeballing it off of here, it looks like it's about.4. So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise. So I would say it's approximately 0.4. It might be 0.41. It might be 0.41456789. We don't know exactly just looking at this graph. But it looks like a value roughly around there."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It might be 0.41. It might be 0.41456789. We don't know exactly just looking at this graph. But it looks like a value roughly around there. Now let's think about the limit of g of x as x approaches seven. So let's do the same exercise. What happens as we approach from the left from values less than seven?"}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it looks like a value roughly around there. Now let's think about the limit of g of x as x approaches seven. So let's do the same exercise. What happens as we approach from the left from values less than seven? 6.9, 6.99, 6.999. Well, it looks like the value of our function is approaching two. It doesn't matter that the actual function is defined."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What happens as we approach from the left from values less than seven? 6.9, 6.99, 6.999. Well, it looks like the value of our function is approaching two. It doesn't matter that the actual function is defined. The g of seven is five. But as we approach from the left, as x goes 6.9, 6.99, and so on, it looks like our value of our function is approaching two. And as we approach x equals seven from the right-hand side, it seems like the same thing is happening."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It doesn't matter that the actual function is defined. The g of seven is five. But as we approach from the left, as x goes 6.9, 6.99, and so on, it looks like our value of our function is approaching two. And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. It seems like we are approaching two. And so I would say that this is going to be equal to two. And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. It seems like we are approaching two. And so I would say that this is going to be equal to two. And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. Now let's do one more. What's the limit as x approaches one? Well, we'll do the same thing."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. Now let's do one more. What's the limit as x approaches one? Well, we'll do the same thing. From the left-hand side, it looks like we're going unbounded. As x goes.9, 0.99, 0.999, 0.99999, it looks like we're just going unbounded towards infinity. And as we approach from the right-hand side, it looks like the same thing, same thing is happening."}, {"video_title": "Limits from graphs   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we'll do the same thing. From the left-hand side, it looks like we're going unbounded. As x goes.9, 0.99, 0.999, 0.99999, it looks like we're just going unbounded towards infinity. And as we approach from the right-hand side, it looks like the same thing, same thing is happening. We're going unbounded to infinity. So formally, sometimes informally people might say, oh, it's approaching infinity or something like that, but if we wanna be formal about what a limit means in this context, because it is unbounded, we would say that it does not exist. It does not exist."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "He was a contemporary of Isaac Newton. These two gentlemen together were really the founding fathers of calculus, and they did some of their, most of their major work in the late 1600s. And this right over here is Usain Bolt, Jamaican sprinter, who's continuing to do some of his best work in 2012. And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might not, you might have not made the association with these three gentlemen. You might not think that they have a lot in common, but they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might not, you might have not made the association with these three gentlemen. You might not think that they have a lot in common, but they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? And in the case of Usain Bolt, how fast is he going right now? Not just what his average speed was for the last second, or his average speed over the next 10 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? And in the case of Usain Bolt, how fast is he going right now? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about, instantaneous rates of change. Differential calculus."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about, instantaneous rates of change. Differential calculus. Calculus. It's all, and Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Differential calculus. Calculus. It's all, and Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. In this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis, on this axis, I'll have distance."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it's all about what's happening in this instant. In this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis, on this axis, I'll have distance. So and I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on this axis, on this axis, I'll have distance. So and I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. And I could say t is equal to time, but I'll just say x is equal to time. X is equal to time."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We'll say y is equal to distance. And in this axis, we'll say time. And I could say t is equal to time, but I'll just say x is equal to time. X is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero, he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "X is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero, he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way. His average speed is just going to be his change in distance over his change in time. And using the variables over here, we're saying y is distance, so this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you from basic algebra."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write it this way. His average speed is just going to be his change in distance over his change in time. And using the variables over here, we're saying y is distance, so this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you from basic algebra. This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this might look somewhat familiar to you from basic algebra. This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run, you might have heard in your algebra class, is going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run, you might have heard in your algebra class, is going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds. And the slope is essentially just the rate of change, or you could view it as the average rate of change, between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is 100 meters over 9.58 seconds. And the slope is essentially just the rate of change, or you could view it as the average rate of change, between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction. And we can figure out what that is. Let me get a calculator out. Let me get the calculator on the screen."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would be velocity if we also specified the direction. And we can figure out what that is. Let me get a calculator out. Let me get the calculator on the screen. So we're going 100 meters in 9.58 seconds. So it's 10.4. I'll just write 10.4."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me get the calculator on the screen. So we're going 100 meters in 9.58 seconds. So it's 10.4. I'll just write 10.4. I'll round to 10.4. So it's approximately 10.4. And then the units are meters per second."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll just write 10.4. I'll round to 10.4. So it's approximately 10.4. And then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed, how it's different than the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed, how it's different than the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you want to know how many meters he's going in an hour, well, there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you want to know how many meters he's going in an hour, well, there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if you were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then if you were to say how many miles per hour, there's roughly 1,600."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if you were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then if you were to say how many miles per hour, there's roughly 1,600. And I don't know the exact number, but roughly 1,600 meters per mile. So let's divide it by 1,600. And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then if you were to say how many miles per hour, there's roughly 1,600. And I don't know the exact number, but roughly 1,600 meters per mile. So let's divide it by 1,600. And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. So this is approximately, I'll write it this way, this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. So this is approximately, I'll write it this way, this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast. Now, to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But relative to me, extremely fast. Now, to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. He's going to have to accelerate. So at first, he's going to start off going a little bit slower. So his slope is going to be a little bit lower than the average slope."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. He's going to have to accelerate. So at first, he's going to start off going a little bit slower. So his slope is going to be a little bit lower than the average slope. He's going to go a little bit slower. Then he's going to start accelerating. And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So his slope is going to be a little bit lower than the average slope. He's going to go a little bit slower. Then he's going to start accelerating. And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. And then maybe near the end, he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. And then maybe near the end, he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. But we could see at any given moment, the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we calculated here is just the average slope across this change in time. But we could see at any given moment, the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates. Over here, it seems like his rate of change of distance, which would be roughly, or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. And when you all average it out, it gets to 23 and 1 1 half miles per hour."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then over here, then he accelerates. Over here, it seems like his rate of change of distance, which would be roughly, or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. And when you all average it out, it gets to 23 and 1 1 half miles per hour. And I looked it up. Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when you all average it out, it gets to 23 and 1 1 half miles per hour. And I looked it up. Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds, is closer to 30 miles per hour. But you see, it's not a trivial thing to do. You could say, OK, let me try to approximate the slope right over here."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds, is closer to 30 miles per hour. But you see, it's not a trivial thing to do. You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x and figure out what the change of y is around it, or as we go past that. So you get that."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x and figure out what the change of y is around it, or as we go past that. So you get that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you get that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change. You could view it as the instantaneous slope at that point in the curve, or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when you do that, you're going to approach that instantaneous rate of change. You could view it as the instantaneous slope at that point in the curve, or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative. And the notation we use for the derivative is dy over dx. And that's why I reserved the letter y. And you say, well, how does this relate to the word differential?"}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the instantaneous slope is the derivative. And the notation we use for the derivative is dy over dx. And that's why I reserved the letter y. And you say, well, how does this relate to the word differential? Well, the word differential is relating this dy is a differential. dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you say, well, how does this relate to the word differential? Well, the word differential is relating this dy is a differential. dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope, or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is 0, you're going to get something that's undefined."}, {"video_title": "Newton, Leibniz, and Usain Bolt   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope, or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is 0, you're going to get something that's undefined. You can't divide by 0. So we take the limit as it approaches 0. And we'll define that more rigorously in the next few videos."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the way I'd like to think about it, let's think about what the volume, if I were to just take the bottom function, if I were to just take f of x, if I were to just take f of x, and if I were to rotate that function around y equals x, if I were to rotate this thing around, sorry, around y is equal to one, what would the volume of that be? And then I'm going to subtract from that the volume if I were to take the top function, if I were to take g of x and rotate it around. So let's first of all think about what volume I would get, and this is really the disk method, and I go into it in much more detail earlier in the calculus playlist. But let's think about the volume if f of x is rotated around that axis. And to do that, let's imagine each sliver of that volume. So this is, let me draw a little thing right over here. And you can imagine once this little sliver is rotated, it forms, it forms, or you can imagine this length is the radius of a disk."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's think about the volume if f of x is rotated around that axis. And to do that, let's imagine each sliver of that volume. So this is, let me draw a little thing right over here. And you can imagine once this little sliver is rotated, it forms, it forms, or you can imagine this length is the radius of a disk. And so, and just to imagine that, let me draw the entire disk. So if this is rotated around, if this is rotated around, it will become a disk. It will become a disk."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can imagine once this little sliver is rotated, it forms, it forms, or you can imagine this length is the radius of a disk. And so, and just to imagine that, let me draw the entire disk. So if this is rotated around, if this is rotated around, it will become a disk. It will become a disk. It will become a disk that looks something like that. And I'll just call the depth of the disk, so the disk, if you imagine a coin, this is kind of the side of the coin, the depth of the coin, the depth of the coin right over there. I know I can draw that better than that."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It will become a disk. It will become a disk that looks something like that. And I'll just call the depth of the disk, so the disk, if you imagine a coin, this is kind of the side of the coin, the depth of the coin, the depth of the coin right over there. I know I can draw that better than that. Let me, so the depth of the coin is just like that. It's a fixed depth, and I'm going to call that dx, so it's just this distance right over here. It is dx."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I know I can draw that better than that. Let me, so the depth of the coin is just like that. It's a fixed depth, and I'm going to call that dx, so it's just this distance right over here. It is dx. And what is going to be the area of that coin? Well, the area of the surface of this coin, so let me do it like this. I want to do a different color."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is dx. And what is going to be the area of that coin? Well, the area of the surface of this coin, so let me do it like this. I want to do a different color. So I want to do it in blue. The area of this coin is just pi times the radius of that coin squared. And what is the radius of the coin?"}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I want to do a different color. So I want to do it in blue. The area of this coin is just pi times the radius of that coin squared. And what is the radius of the coin? Well, the radius of the coin is this height, is this height right over here. And what is that height? Well, it is one minus f of x."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what is the radius of the coin? Well, the radius of the coin is this height, is this height right over here. And what is that height? Well, it is one minus f of x. So that is equal to the radius. So the area, the area of kind of the face of this coin is going to be pi, the area of the face of this coin is going to be pi times the radius squared, which is equal to pi times one minus f of x, one minus f of x squared. That's this blue area right over here."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it is one minus f of x. So that is equal to the radius. So the area, the area of kind of the face of this coin is going to be pi, the area of the face of this coin is going to be pi times the radius squared, which is equal to pi times one minus f of x, one minus f of x squared. That's this blue area right over here. And then if I want to find the volume of this coin, I would multiply it by the depth of the coin. So times dx. And if I wanted to find the volume of this entire solid, this entire rotated solid, I would want to find the sum of all of these volumes."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's this blue area right over here. And then if I want to find the volume of this coin, I would multiply it by the depth of the coin. So times dx. And if I wanted to find the volume of this entire solid, this entire rotated solid, I would want to find the sum of all of these volumes. So this is just the disk right over here, but I could have another disk, a similar disk that I do right over here. I could have another disk right over here. And I want to take the sum of all of those disks."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if I wanted to find the volume of this entire solid, this entire rotated solid, I would want to find the sum of all of these volumes. So this is just the disk right over here, but I could have another disk, a similar disk that I do right over here. I could have another disk right over here. And I want to take the sum of all of those disks. So I want to take, so the volume is going to be the sum over all of those disks. So x goes from zero, which is this bounding point, to x is equal to 1 1\u20442, times pi times one minus f of x squared. This is the area of the face of each of those disks."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I want to take the sum of all of those disks. So I want to take, so the volume is going to be the sum over all of those disks. So x goes from zero, which is this bounding point, to x is equal to 1 1\u20442, times pi times one minus f of x squared. This is the area of the face of each of those disks. And then I multiply it times the depth of each of those disks. Now this is the volume of each of those disks, and I'm taking the sum of all of them. So this is the volume, this is the volume if I were to just rotate f of x around y is equal to one."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the area of the face of each of those disks. And then I multiply it times the depth of each of those disks. Now this is the volume of each of those disks, and I'm taking the sum of all of them. So this is the volume, this is the volume if I were to just rotate f of x around y is equal to one. And actually I should just write dx here. And so this right over here, this expression, I just did that, so they really are equal. This is obviously just the volume of each of those disks."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the volume, this is the volume if I were to just rotate f of x around y is equal to one. And actually I should just write dx here. And so this right over here, this expression, I just did that, so they really are equal. This is obviously just the volume of each of those disks. So this is the volume if I were to take the f of x around y equals one. Let's figure out, so let me call this volume of f of x. And by the same logic, the same exact logic, we can figure out the volume if we take g of x, if we rotate g of x around, if we rotate g of x, if we construct disks like this and rotate them around y is equal to one."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is obviously just the volume of each of those disks. So this is the volume if I were to take the f of x around y equals one. Let's figure out, so let me call this volume of f of x. And by the same logic, the same exact logic, we can figure out the volume if we take g of x, if we rotate g of x around, if we rotate g of x, if we construct disks like this and rotate them around y is equal to one. And so the volume, if I take g of x around y equals one, would be zero to 1 1\u20442 times pi times one minus g of x, because one minus g of x is each of these radiuses right over here, each of these radiuses. That squared dx. And so the volume of what they're asking us, the volume of the solid generated when r is rotated, well, r is kind of the space in between f of x and g of x."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And by the same logic, the same exact logic, we can figure out the volume if we take g of x, if we rotate g of x around, if we rotate g of x, if we construct disks like this and rotate them around y is equal to one. And so the volume, if I take g of x around y equals one, would be zero to 1 1\u20442 times pi times one minus g of x, because one minus g of x is each of these radiuses right over here, each of these radiuses. That squared dx. And so the volume of what they're asking us, the volume of the solid generated when r is rotated, well, r is kind of the space in between f of x and g of x. So it's going to be, the volume is going to be the difference between these volumes. It's going to be this volume, this is kind of the outer volume, and we're gonna take out its hollow core. We're gonna hollow it out by subtracting out this volume."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the volume of what they're asking us, the volume of the solid generated when r is rotated, well, r is kind of the space in between f of x and g of x. So it's going to be, the volume is going to be the difference between these volumes. It's going to be this volume, this is kind of the outer volume, and we're gonna take out its hollow core. We're gonna hollow it out by subtracting out this volume. So the volume of that region is going to be the integral, I'll do this in a new color, integral from zero to 1 1\u20442 of pi times one minus f of x squared dx minus the integral from zero to 1 1\u20442 of pi times one minus g of x squared dx. And this is a completely valid answer, but you might want to simplify it. We have the same bounds of integration."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna hollow it out by subtracting out this volume. So the volume of that region is going to be the integral, I'll do this in a new color, integral from zero to 1 1\u20442 of pi times one minus f of x squared dx minus the integral from zero to 1 1\u20442 of pi times one minus g of x squared dx. And this is a completely valid answer, but you might want to simplify it. We have the same bounds of integration. We have the same variable of integration. And actually we have this pi over here, so we could factor that out. And so this is the same thing as pi times the integral from zero to 1 1\u20442 of one minus f of x, one minus f of x squared minus one minus g of x, one minus g of x squared, and then all of that dx."}, {"video_title": "2011 Calculus AB free response #3 (c)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have the same bounds of integration. We have the same variable of integration. And actually we have this pi over here, so we could factor that out. And so this is the same thing as pi times the integral from zero to 1 1\u20442 of one minus f of x, one minus f of x squared minus one minus g of x, one minus g of x squared, and then all of that dx. And then, and actually you probably would want to do this, you probably would want to do this if you're taking the AP exam, not just leave it in terms of f of x and g of x. You would actually want to write the expressions for what f of x, f of x, and g of x are. So really the best answer would probably be pi times the integral from zero to 1 1\u20442 times one minus, well f of x is eight x to the third power, eight x to the third power squared, minus one minus g of x, g of x is sine of pi x, that squared, that squared, times dx."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "Let f be the function given by f of x is equal to the square root of x plus 4 minus 3 over x minus 5, if x does not equal 5, and it's equal to c if x equals 5. And say if f is continuous at x equals 5, what is the value of c? So if we know that f is continuous at x equals 5, that means that the limit, the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9, the square root of that is positive 3, the principal root is positive 3, 3 minus 3 is 0, so you get a 0 in the numerator, and then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0 over 0."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9, the square root of that is positive 3, the principal root is positive 3, 3 minus 3 is 0, so you get a 0 in the numerator, and then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0 over 0. And in the future we will see that we do have a tool that allows us, or gives us an option to attempt to find limits when we get this indeterminate form. It's called L'Hopital's Rule. But we can actually tackle this with a little bit of fancy algebra."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "So you get this indeterminate form of 0 over 0. And in the future we will see that we do have a tool that allows us, or gives us an option to attempt to find limits when we get this indeterminate form. It's called L'Hopital's Rule. But we can actually tackle this with a little bit of fancy algebra. And to do that I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "But we can actually tackle this with a little bit of fancy algebra. And to do that I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical what you can do is multiply by the radical doing the, or if you have a radical minus 3 you multiply by the radical plus 3. So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical what you can do is multiply by the radical doing the, or if you have a radical minus 3 you multiply by the radical plus 3. So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3 then we would do a minus 3 here. This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators. It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3 then we would do a minus 3 here. This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators. It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. But if you multiply this out, and I encourage you to do it, you notice this has the pattern that you learned in algebra class. This is the difference of squares. Something minus something times something plus something."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. But if you multiply this out, and I encourage you to do it, you notice this has the pattern that you learned in algebra class. This is the difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. The second term is going to be the second something or you're going to subtract the second something squared."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. The second term is going to be the second something or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. In the denominator you're of course going to have x minus 5 times the square root of x plus 4 plus 3. So this has, I guess you could say simplified to, although not arguably any simpler, but at least we've gotten our radical."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "The second term is going to be the second something or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. In the denominator you're of course going to have x minus 5 times the square root of x plus 4 plus 3. So this has, I guess you could say simplified to, although not arguably any simpler, but at least we've gotten our radical. We're really just playing around with it algebraically to see if we can then substitute x equals 5 or if we can somehow simplify it to figure out what the limit is. When you simplify the numerator up here, you get x plus 4 minus 9. Well that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "So this has, I guess you could say simplified to, although not arguably any simpler, but at least we've gotten our radical. We're really just playing around with it algebraically to see if we can then substitute x equals 5 or if we can somehow simplify it to figure out what the limit is. When you simplify the numerator up here, you get x plus 4 minus 9. Well that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3. And now it pops out at you. Both the numerator and the denominator are now divisible by x minus 5. So you can have a completely identical expression if you say that this is the same thing."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "Well that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3. And now it pops out at you. Both the numerator and the denominator are now divisible by x minus 5. So you can have a completely identical expression if you say that this is the same thing. You can divide the numerator and the denominator by x minus 5 if you assume x does not equal 5. So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5. Which is fine because in the first part of this function definition, this is for the case for x does not equal 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "So you can have a completely identical expression if you say that this is the same thing. You can divide the numerator and the denominator by x minus 5 if you assume x does not equal 5. So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5. Which is fine because in the first part of this function definition, this is for the case for x does not equal 5. So we could actually replace this, and this is a simpler expression, with 1 over square root of x plus 4 plus 3. And so now when we take the limit as x approaches 5, we're going to get closer and closer to 5. We're going to get x values closer and closer to 5, but not quite at 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "Which is fine because in the first part of this function definition, this is for the case for x does not equal 5. So we could actually replace this, and this is a simpler expression, with 1 over square root of x plus 4 plus 3. And so now when we take the limit as x approaches 5, we're going to get closer and closer to 5. We're going to get x values closer and closer to 5, but not quite at 5. We can use this expression right over here. So the limit of f of x as x approaches 5 is going to be the same thing as the limit of 1 over the square root of x plus 4 plus 3 as x approaches 5. And now we can substitute a 5 in here."}, {"video_title": "Fancy algebra to find a limit and make a function continuous   Differential Calculus   Khan Academy.mp3", "Sentence": "We're going to get x values closer and closer to 5, but not quite at 5. We can use this expression right over here. So the limit of f of x as x approaches 5 is going to be the same thing as the limit of 1 over the square root of x plus 4 plus 3 as x approaches 5. And now we can substitute a 5 in here. It's going to be 1 over 5 plus 4 is 9, the principal root of that is 3, 3 plus 3 is 6. So if c is equal to 1 6, then the limit of our function as x approaches 5 is going to be equal to f of 5, and we are continuous at x equals 5. So it's 1 6."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause this video and see if you can figure this one out from this graph. All right, we're going from x equals negative six to x equals negative two, and the definite integral is going to be the area below our graph and above the x-axis. So it's going to be this area right over here. And how do we figure that out? Well, this is a semicircle, and we know how to find the area of a circle if we know its radius, and this circle has radius two, has a radius of two. No matter what direction we go in from the center, it has a radius of two. And so the area of a circle is pi r squared, so it'd be pi times our radius, which is two squared, but this is a semicircle, so I'm gonna divide by two."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "And how do we figure that out? Well, this is a semicircle, and we know how to find the area of a circle if we know its radius, and this circle has radius two, has a radius of two. No matter what direction we go in from the center, it has a radius of two. And so the area of a circle is pi r squared, so it'd be pi times our radius, which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only half the area of the full circle. So this is going to be four pi over two, which is equal to two pi. All right, let's do another one."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the area of a circle is pi r squared, so it'd be pi times our radius, which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only half the area of the full circle. So this is going to be four pi over two, which is equal to two pi. All right, let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if you can figure that out. All right, let's do it together."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if you can figure that out. All right, let's do it together. So we're going from negative two to one, and so we have to be a little bit careful here. So the definite integral, you could view it as the area below the function and above the x-axis. But here, the function is below the x-axis."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's do it together. So we're going from negative two to one, and so we have to be a little bit careful here. So the definite integral, you could view it as the area below the function and above the x-axis. But here, the function is below the x-axis. And so what we can do is we can figure out this area, just knowing what we know about geometry, and then we have to realize that this is going to be a negative value for the definite integral because our function is below the x-axis. So what's the area here? Well, there's a couple of ways to think about it."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "But here, the function is below the x-axis. And so what we can do is we can figure out this area, just knowing what we know about geometry, and then we have to realize that this is going to be a negative value for the definite integral because our function is below the x-axis. So what's the area here? Well, there's a couple of ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid, or you could just split it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1 1\u20442."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, there's a couple of ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid, or you could just split it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1 1\u20442. So this has an area of one. This rectangle right over here has an area of two times one. So it has an area of two."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you split it up like this, this triangle right over here has an area of one times two times 1 1\u20442. So this has an area of one. This rectangle right over here has an area of two times one. So it has an area of two. And then this triangle right over here is the same area as the first one. It's going to have a base of one, a height of two. So it's one times two times 1 1\u20442."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it has an area of two. And then this triangle right over here is the same area as the first one. It's going to have a base of one, a height of two. So it's one times two times 1 1\u20442. Remember, the area of a triangle is 1 1\u20442 base times height, so it's one. So if you add up those areas, one plus two plus one is four. And so you might be tempted to say, oh, is this going to be equal to four?"}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's one times two times 1 1\u20442. Remember, the area of a triangle is 1 1\u20442 base times height, so it's one. So if you add up those areas, one plus two plus one is four. And so you might be tempted to say, oh, is this going to be equal to four? But remember, our function is below the x-axis here. And so this is going to be a negative four. All right, let's do another one."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you might be tempted to say, oh, is this going to be equal to four? But remember, our function is below the x-axis here. And so this is going to be a negative four. All right, let's do another one. So now we're gonna go from one to four of f of x dx. So pause the video and see if you can figure that out. So we're gonna go from here to here."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's do another one. So now we're gonna go from one to four of f of x dx. So pause the video and see if you can figure that out. So we're gonna go from here to here. And so it's gonna be this area right over there. So how do we figure that out? Well, just the formula for the area of a triangle, base times height times 1 1\u20442."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna go from here to here. And so it's gonna be this area right over there. So how do we figure that out? Well, just the formula for the area of a triangle, base times height times 1 1\u20442. So, or you could say 1 1\u20442 times our base, which is a length of, see, we have a base of three right over here, we go from one to four. So 1 1\u20442 times three times our height, which is one, two, three, four, times four. Well, this is just going to get us six."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, just the formula for the area of a triangle, base times height times 1 1\u20442. So, or you could say 1 1\u20442 times our base, which is a length of, see, we have a base of three right over here, we go from one to four. So 1 1\u20442 times three times our height, which is one, two, three, four, times four. Well, this is just going to get us six. All right, last but not least, if we are going from four to six of f of x dx, so that's going to be this area right over here. But we have to be careful. Our function is below the x-axis."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is just going to get us six. All right, last but not least, if we are going from four to six of f of x dx, so that's going to be this area right over here. But we have to be careful. Our function is below the x-axis. So we'll figure out this area, and then it's going to be negative. So this is a half of a circle of radius one. And so the area of a circle is pi times r squared, so it's pi times one squared."}, {"video_title": "Finding definite integrals using area formulas   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our function is below the x-axis. So we'll figure out this area, and then it's going to be negative. So this is a half of a circle of radius one. And so the area of a circle is pi times r squared, so it's pi times one squared. That would be the area if we went all the way around like that. But this is only half of the circle, so divided by two. And since this area is above the function and below the x-axis, it's going to be negative."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we've got this function f of x that is piecewise continuous, it's defined over several intervals here. For x being, or for zero less than x, being less than or equal to two, f of x is natural log of x. For any x is larger than two, well then f of x is going to be x squared times the natural log of x. And what we want to do is, we want to find the limit of f of x as x approaches two. And what's interesting about the value two is that that's essentially the boundary between these two intervals. If we wanted to evaluate it at two, we would fall into this first interval, f of two. Well, two is less than or equal to two, and it's greater than zero."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we want to do is, we want to find the limit of f of x as x approaches two. And what's interesting about the value two is that that's essentially the boundary between these two intervals. If we wanted to evaluate it at two, we would fall into this first interval, f of two. Well, two is less than or equal to two, and it's greater than zero. So f of two, f of two would be pretty straightforward, that would just be natural log of two. But that's not necessarily what the limit is going to be. For to figure out what the limit is going to be, we should think about, well what's the limit as we approach from the left?"}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, two is less than or equal to two, and it's greater than zero. So f of two, f of two would be pretty straightforward, that would just be natural log of two. But that's not necessarily what the limit is going to be. For to figure out what the limit is going to be, we should think about, well what's the limit as we approach from the left? What's the limit as we approach from the right? And do those exist? And if they do exist, are they the same thing?"}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "For to figure out what the limit is going to be, we should think about, well what's the limit as we approach from the left? What's the limit as we approach from the right? And do those exist? And if they do exist, are they the same thing? And if they are the same thing, well then we have a well-defined limit. So let's do that. Let's first think about the limit, the limit of f of x as we approach two from the left, from values lower than two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if they do exist, are they the same thing? And if they are the same thing, well then we have a well-defined limit. So let's do that. Let's first think about the limit, the limit of f of x as we approach two from the left, from values lower than two. Well, this is gonna be the case where we're gonna be operating in this interval right over here. We're operating from values less than two, and we're going to be approaching two from the left. And so we'll fall under this clause."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's first think about the limit, the limit of f of x as we approach two from the left, from values lower than two. Well, this is gonna be the case where we're gonna be operating in this interval right over here. We're operating from values less than two, and we're going to be approaching two from the left. And so we'll fall under this clause. And so since this clause or case is continuous over the interval in which we're operating, and for sure between, or for all values greater than zero and less than or equal to two, this limit is going to be equal to just this clause evaluated at two, because it's continuous over the interval. So this is just going to be the natural log of two. All right, so now let's think about the limit from the right-hand side, from values greater than two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we'll fall under this clause. And so since this clause or case is continuous over the interval in which we're operating, and for sure between, or for all values greater than zero and less than or equal to two, this limit is going to be equal to just this clause evaluated at two, because it's continuous over the interval. So this is just going to be the natural log of two. All right, so now let's think about the limit from the right-hand side, from values greater than two. So the limit, the limit of f of x as x approaches two from the right-hand side. Well, even though two falls into this clause, as soon as we go anything greater than two, we fall in this clause. So we're gonna be approaching two essentially using this case."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so now let's think about the limit from the right-hand side, from values greater than two. So the limit, the limit of f of x as x approaches two from the right-hand side. Well, even though two falls into this clause, as soon as we go anything greater than two, we fall in this clause. So we're gonna be approaching two essentially using this case. And once again, this case here is continuous for all x values not only greater than two, actually greater than or equal to two. And so for this one over here, we can make the same argument that this limit is going to be this clause evaluated at two. Because once again, if we were just evaluated the function at two, it falls under this clause."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna be approaching two essentially using this case. And once again, this case here is continuous for all x values not only greater than two, actually greater than or equal to two. And so for this one over here, we can make the same argument that this limit is going to be this clause evaluated at two. Because once again, if we were just evaluated the function at two, it falls under this clause. But if we're approaching from the right, well, if we're approaching from the right, those are x values greater than two, so this clause is what's at play. So we'll evaluate this clause at two. So because it is continuous."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because once again, if we were just evaluated the function at two, it falls under this clause. But if we're approaching from the right, well, if we're approaching from the right, those are x values greater than two, so this clause is what's at play. So we'll evaluate this clause at two. So because it is continuous. So this is going to be two squared times the natural log of two. And so this is equal to four times the natural log of two. Four times the natural log of two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So because it is continuous. So this is going to be two squared times the natural log of two. And so this is equal to four times the natural log of two. Four times the natural log of two. So the right-hand limit does exist. The left-hand limit does exist. But the thing that might jump out at you is that these are two different values."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Four times the natural log of two. So the right-hand limit does exist. The left-hand limit does exist. But the thing that might jump out at you is that these are two different values. We approach a different value from the left as we do from the right. If you were to graph this, you would see a jump in the actual graph. You would see a discontinuity occurring there."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the thing that might jump out at you is that these are two different values. We approach a different value from the left as we do from the right. If you were to graph this, you would see a jump in the actual graph. You would see a discontinuity occurring there. And so for this one in particular, you have that jump discontinuity. This limit would not exist because the left-hand limit and the right-hand limit go to two different values. So limit, so this does not exist."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we know from the definition of the derivative that the derivative of the function square root of x, that is equal to, let me switch colors just for variety, that's equal to the limit as delta x approaches 0. And some people say h approaches 0 or d approaches 0. I just use delta x. So the change in x approaches 0. And then we say f of x plus delta x. So in this case, this is f of x. So it's the square root of x plus delta x minus f of x, in this case, the square root of x."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the change in x approaches 0. And then we say f of x plus delta x. So in this case, this is f of x. So it's the square root of x plus delta x minus f of x, in this case, the square root of x. All of that over the change in x, over delta x. So what I'm going to do, right now when I look at that, there's not much simplification I can do to make this come out with something meaningful. I'm going to multiply this fraction times, I'm going to multiply the numerator and the denominator by the conjugate of the numerator."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's the square root of x plus delta x minus f of x, in this case, the square root of x. All of that over the change in x, over delta x. So what I'm going to do, right now when I look at that, there's not much simplification I can do to make this come out with something meaningful. I'm going to multiply this fraction times, I'm going to multiply the numerator and the denominator by the conjugate of the numerator. So what do I mean by that? Let me rewrite it. Limit as delta x approaches 0."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I'm going to multiply this fraction times, I'm going to multiply the numerator and the denominator by the conjugate of the numerator. So what do I mean by that? Let me rewrite it. Limit as delta x approaches 0. I'm just rewriting what I have here. So I said the square root of x plus delta x minus square root of x, all of that over delta x. And I'm going to multiply that, after switching colors, times square root of x plus delta x plus the square root of x over the square root of x plus delta x plus the square root of x, right?"}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Limit as delta x approaches 0. I'm just rewriting what I have here. So I said the square root of x plus delta x minus square root of x, all of that over delta x. And I'm going to multiply that, after switching colors, times square root of x plus delta x plus the square root of x over the square root of x plus delta x plus the square root of x, right? This is just 1. So I could, of course, multiply that times, if we assume that x and delta x aren't both 0, this is a defined number. This would be 1."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And I'm going to multiply that, after switching colors, times square root of x plus delta x plus the square root of x over the square root of x plus delta x plus the square root of x, right? This is just 1. So I could, of course, multiply that times, if we assume that x and delta x aren't both 0, this is a defined number. This would be 1. And we can do that. This is 1 over 1. We're just multiplying it times this equation."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This would be 1. And we can do that. This is 1 over 1. We're just multiplying it times this equation. And we get limit as delta x approaches 0. Well, if you view this as a minus b times a plus b, right? Let me do a little aside here."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We're just multiplying it times this equation. And we get limit as delta x approaches 0. Well, if you view this as a minus b times a plus b, right? Let me do a little aside here. Let me say a plus b times a minus b is equal to a squared minus b squared, right? So this is a plus b times a minus b. So it's going to be equal to a squared."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do a little aside here. Let me say a plus b times a minus b is equal to a squared minus b squared, right? So this is a plus b times a minus b. So it's going to be equal to a squared. So what's this quantity squared or this quantity squared, either one? These are my a's. Well, it's just going to be x plus delta x, right?"}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be equal to a squared. So what's this quantity squared or this quantity squared, either one? These are my a's. Well, it's just going to be x plus delta x, right? So you get x plus delta x. And then what's b squared? So minus square root of x is b in this analogy."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, it's just going to be x plus delta x, right? So you get x plus delta x. And then what's b squared? So minus square root of x is b in this analogy. So square root of x squared is just x. And all of that over delta x times square root of x plus delta x plus the square root of x. Let's see what simplification we can do."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So minus square root of x is b in this analogy. So square root of x squared is just x. And all of that over delta x times square root of x plus delta x plus the square root of x. Let's see what simplification we can do. Well, we have an x and then a minus x. So those cancel out. So we have delta x minus x."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's see what simplification we can do. Well, we have an x and then a minus x. So those cancel out. So we have delta x minus x. And then we're left in the numerator and the denominator. All we have is a delta x here and a delta x here. So let's divide the numerator and the denominator by delta x."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we have delta x minus x. And then we're left in the numerator and the denominator. All we have is a delta x here and a delta x here. So let's divide the numerator and the denominator by delta x. So this goes to 1. And so this equals the limit. I'll write smaller because I'm running out of space."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's divide the numerator and the denominator by delta x. So this goes to 1. And so this equals the limit. I'll write smaller because I'm running out of space. Limit as delta x approaches 0 of 1 over. And of course, we can only do this assuming that delta, well, we're dividing by delta x to begin with. So we know it's not 0."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I'll write smaller because I'm running out of space. Limit as delta x approaches 0 of 1 over. And of course, we can only do this assuming that delta, well, we're dividing by delta x to begin with. So we know it's not 0. It's just approaching 0. So we get square root of x plus delta x plus the square root of x. And now we can just directly take the limit as it approaches 0."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we know it's not 0. It's just approaching 0. So we get square root of x plus delta x plus the square root of x. And now we can just directly take the limit as it approaches 0. We can just set delta x is equal to 0. That's what it's approaching. So that that equals 1 over the square root of x. Delta x is 0, so we can ignore that."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And now we can just directly take the limit as it approaches 0. We can just set delta x is equal to 0. That's what it's approaching. So that that equals 1 over the square root of x. Delta x is 0, so we can ignore that. We can take the limit all the way to 0. And then this is, of course, just a square root of x here plus the square root of x. And that equals 1 over 2 square root of x."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So that that equals 1 over the square root of x. Delta x is 0, so we can ignore that. We can take the limit all the way to 0. And then this is, of course, just a square root of x here plus the square root of x. And that equals 1 over 2 square root of x. And that equals 1 half x to the negative 1 half. So we just proved that x to the 1 half power, the derivative of it, is 1 half x to the negative 1 half. And so it is consistent with the general property that the derivative of x to the n is equal to nx to the n minus 1."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And that equals 1 over 2 square root of x. And that equals 1 half x to the negative 1 half. So we just proved that x to the 1 half power, the derivative of it, is 1 half x to the negative 1 half. And so it is consistent with the general property that the derivative of x to the n is equal to nx to the n minus 1. Even in this case where n was 1 half. Well, hopefully that's satisfying. I didn't prove it for all fractions, but this is a start."}, {"video_title": "Proof d dx(sqrt(x))   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And so it is consistent with the general property that the derivative of x to the n is equal to nx to the n minus 1. Even in this case where n was 1 half. Well, hopefully that's satisfying. I didn't prove it for all fractions, but this is a start. This is a common one you see. Square root of x. And it's hopefully not too complicated of a proof."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is a mouse. And it's diving straight down near a streetlight. Let's get some information about what's going on. So the streetlight right over here is 20 feet high. So this is a 20 foot high street lamp. And right at this moment, and I haven't drawn it completely to scale, the owl is 15 feet above the mouse. So this distance right over here is 15 feet."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the streetlight right over here is 20 feet high. So this is a 20 foot high street lamp. And right at this moment, and I haven't drawn it completely to scale, the owl is 15 feet above the mouse. So this distance right over here is 15 feet. And the mouse itself is 10 feet from the base of the lamp. Let me draw that. So the mouse is 10 feet from the base of the lamp."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this distance right over here is 15 feet. And the mouse itself is 10 feet from the base of the lamp. Let me draw that. So the mouse is 10 feet from the base of the lamp. And we also know, we have our little radar gun out, we know that this owl is driving straight down. And right now, it is going 20 feet per second. So right now, this is going down at 20 feet per second."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the mouse is 10 feet from the base of the lamp. And we also know, we have our little radar gun out, we know that this owl is driving straight down. And right now, it is going 20 feet per second. So right now, this is going down at 20 feet per second. Now, what we're curious about is we have the light over here. Light is coming from the street lamp in every direction. And it creates a shadow of the owl."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So right now, this is going down at 20 feet per second. Now, what we're curious about is we have the light over here. Light is coming from the street lamp in every direction. And it creates a shadow of the owl. So right now, the shadow is out here. And as the owl goes further and further down, the shadow is going to move to the left like that. And so given everything that we've set up right over here, the question is, at what rate is the shadow moving?"}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it creates a shadow of the owl. So right now, the shadow is out here. And as the owl goes further and further down, the shadow is going to move to the left like that. And so given everything that we've set up right over here, the question is, at what rate is the shadow moving? So let's think about what we know and what we don't know. And to do that, let's set up some variables. So let me draw the same thing a little bit more geometrically."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so given everything that we've set up right over here, the question is, at what rate is the shadow moving? So let's think about what we know and what we don't know. And to do that, let's set up some variables. So let me draw the same thing a little bit more geometrically. So let's say that this right over here is the street light that is 20 feet tall. And then this right over here is the height of the owl right at this moment. So this is 15 feet."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw the same thing a little bit more geometrically. So let's say that this right over here is the street light that is 20 feet tall. And then this right over here is the height of the owl right at this moment. So this is 15 feet. The distance between the base of the lamp and where the owl is going, where that mouse is right now, this is 10 feet. And if I were to think about where the shadow is, well, the light's emitting from right over here. The light's emitting right over here."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is 15 feet. The distance between the base of the lamp and where the owl is going, where that mouse is right now, this is 10 feet. And if I were to think about where the shadow is, well, the light's emitting from right over here. The light's emitting right over here. And so the owl blocks the light right over there. So the shadow is going to be right over there. So if you just draw a straight line from the source of light through the owl, and you just keep going, and you hit the ground, you're going to figure out where the shadow is."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The light's emitting right over here. And so the owl blocks the light right over there. So the shadow is going to be right over there. So if you just draw a straight line from the source of light through the owl, and you just keep going, and you hit the ground, you're going to figure out where the shadow is. So the shadow is going to be right over here. It's going to be right over there. And we need to figure out how quickly is that moving."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you just draw a straight line from the source of light through the owl, and you just keep going, and you hit the ground, you're going to figure out where the shadow is. So the shadow is going to be right over here. It's going to be right over there. And we need to figure out how quickly is that moving. And it's going to be moving in the leftward direction. So let's set up some variables over here. So let's say, so what's changing?"}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we need to figure out how quickly is that moving. And it's going to be moving in the leftward direction. So let's set up some variables over here. So let's say, so what's changing? Well, we know that the height of the owl is changing. So let's call that y. Right at this moment, it's equal to 15."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say, so what's changing? Well, we know that the height of the owl is changing. So let's call that y. Right at this moment, it's equal to 15. But it is actually changing. And let's call the distance between the shadow and the mouse x. Now, given this setup, can we come up with a relationship between x and y?"}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right at this moment, it's equal to 15. But it is actually changing. And let's call the distance between the shadow and the mouse x. Now, given this setup, can we come up with a relationship between x and y? And then using that relationship, what we're really trying to come up with is what is the rate at which x is changing with respect to time? We know what y is right at this moment. We know what dy dt is right at this moment."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, given this setup, can we come up with a relationship between x and y? And then using that relationship, what we're really trying to come up with is what is the rate at which x is changing with respect to time? We know what y is right at this moment. We know what dy dt is right at this moment. Can we come up with a relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx dt is at a given moment in time? Well, both of these triangles, and when I say both of these triangles, let me be clear what I'm talking about. This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what dy dt is right at this moment. Can we come up with a relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx dt is at a given moment in time? Well, both of these triangles, and when I say both of these triangles, let me be clear what I'm talking about. This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle. It's a similar triangle to this larger triangle that I am tracing in blue. It's similar to this larger one. How do I know that?"}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle. It's a similar triangle to this larger triangle that I am tracing in blue. It's similar to this larger one. How do I know that? Well, they both have a right angle right over here. They both share this angle. So if they have two angles in common, then all three angles must be in common."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "How do I know that? Well, they both have a right angle right over here. They both share this angle. So if they have two angles in common, then all three angles must be in common. So they are similar triangles, which means the ratio between corresponding sides must be the same. So we know that the ratio of x to y must be the ratio of this entire base, which is x plus 10, to the height of the larger triangle, 220. And right there, we have a relationship between x and y."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if they have two angles in common, then all three angles must be in common. So they are similar triangles, which means the ratio between corresponding sides must be the same. So we know that the ratio of x to y must be the ratio of this entire base, which is x plus 10, to the height of the larger triangle, 220. And right there, we have a relationship between x and y. And if we take the derivative of both sides with respect to t, we're probably doing pretty well. Now, before taking the derivative with respect to t, I could do it right over here just to simplify things a little bit. Let me just cross multiply."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And right there, we have a relationship between x and y. And if we take the derivative of both sides with respect to t, we're probably doing pretty well. Now, before taking the derivative with respect to t, I could do it right over here just to simplify things a little bit. Let me just cross multiply. So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator. So on the left-hand side, it simplifies to 20x. I don't want to write over it."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just cross multiply. So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator. So on the left-hand side, it simplifies to 20x. I don't want to write over it. Well, I'll just write 20x. And on the left-hand side, it is 20x. And then on the right-hand side, let's see, this cancels with that."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I don't want to write over it. Well, I'll just write 20x. And on the left-hand side, it is 20x. And then on the right-hand side, let's see, this cancels with that. We have xy plus 10y. And now let me take the derivative of both sides with respect to time. So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then on the right-hand side, let's see, this cancels with that. We have xy plus 10y. And now let me take the derivative of both sides with respect to time. So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20. That's the derivative of 20x with respect to x times dx, the derivative of x with respect to t, is equal to. Now, over here, we're going to have to break out a little bit of the product rule. So first, we want to figure out the derivative of x with respect to time."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20. That's the derivative of 20x with respect to x times dx, the derivative of x with respect to t, is equal to. Now, over here, we're going to have to break out a little bit of the product rule. So first, we want to figure out the derivative of x with respect to time. So the derivative of x with respect to time. So the derivative of the first thing times the second thing times y, plus just the first thing times the derivative of the second thing. So the derivative of y with respect to t is just dy dt."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So first, we want to figure out the derivative of x with respect to time. So the derivative of x with respect to time. So the derivative of the first thing times the second thing times y, plus just the first thing times the derivative of the second thing. So the derivative of y with respect to t is just dy dt. And then finally, right over here, the derivative of 10y with respect to t is the derivative of 10y with respect to y, which is just 10, times the derivative of y with respect to t, which is dy dt. And there you have it. You have your relationship between dx dt, dy dt, and x and y."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of y with respect to t is just dy dt. And then finally, right over here, the derivative of 10y with respect to t is the derivative of 10y with respect to y, which is just 10, times the derivative of y with respect to t, which is dy dt. And there you have it. You have your relationship between dx dt, dy dt, and x and y. So let's just make sure we have everything. This is what we're trying to solve for, dx dt. And let's see, we have another dx dt here."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have your relationship between dx dt, dy dt, and x and y. So let's just make sure we have everything. This is what we're trying to solve for, dx dt. And let's see, we have another dx dt here. We're going to try to solve for that. We know what y is. y is equal to 15."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we have another dx dt here. We're going to try to solve for that. We know what y is. y is equal to 15. We know what dy dt is. dy dt, if we make the convention since y is decreasing, we can say it's negative 20. So we know what this is."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "y is equal to 15. We know what dy dt is. dy dt, if we make the convention since y is decreasing, we can say it's negative 20. So we know what this is. And so if we just know what x is, we can solve for dx dt. So what is x right at this moment? Well, we can use this first equation."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know what this is. And so if we just know what x is, we can solve for dx dt. So what is x right at this moment? Well, we can use this first equation. We could actually use this one up here, but this one is simplified a little bit to actually solve for x. So let's do that, and then we'll substitute back into this thing where we've taken the derivative. So we get 20 times x is equal to x times y. y is 15."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can use this first equation. We could actually use this one up here, but this one is simplified a little bit to actually solve for x. So let's do that, and then we'll substitute back into this thing where we've taken the derivative. So we get 20 times x is equal to x times y. y is 15. And just remember, I could have used this equation, but this is just one step further. We've already cross-multiplied. So it's x times y. y is 15."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get 20 times x is equal to x times y. y is 15. And just remember, I could have used this equation, but this is just one step further. We've already cross-multiplied. So it's x times y. y is 15. So it's x times 15 plus 10 times y. Plus 10 times 15. Did I do that right?"}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's x times y. y is 15. So it's x times 15 plus 10 times y. Plus 10 times 15. Did I do that right? 20x is equal to x times 15 plus 10 times 15. So let's see. If you subtract, so this is 20x is equal to 15x plus 150."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Did I do that right? 20x is equal to x times 15 plus 10 times 15. So let's see. If you subtract, so this is 20x is equal to 15x plus 150. Subtract 15x from both sides, you get 5x is equal to 30. 5x is equal to 150. My brain is getting ahead."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you subtract, so this is 20x is equal to 15x plus 150. Subtract 15x from both sides, you get 5x is equal to 30. 5x is equal to 150. My brain is getting ahead. 5x is equal to 150. Divide both sides by 5, you get x is equal to 30 feet. x is equal to 30 feet right at this moment."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "My brain is getting ahead. 5x is equal to 150. Divide both sides by 5, you get x is equal to 30 feet. x is equal to 30 feet right at this moment. So this distance, just going back to our original diagram, this distance right over here is 30 feet. So let's substitute all the values we know back into this equation to actually solve for dx dt. So we have, let me do it right over here."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "x is equal to 30 feet right at this moment. So this distance, just going back to our original diagram, this distance right over here is 30 feet. So let's substitute all the values we know back into this equation to actually solve for dx dt. So we have, let me do it right over here. We have 20 times dx dt. I'll do that in orange. We'll solve for that."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have, let me do it right over here. We have 20 times dx dt. I'll do that in orange. We'll solve for that. Actually, I already used orange. So let's say dx dt, I'll use this pink. 20 times dx dt is equal to dx dt times y. y right now is 15 feet."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We'll solve for that. Actually, I already used orange. So let's say dx dt, I'll use this pink. 20 times dx dt is equal to dx dt times y. y right now is 15 feet. So times 15 times, I didn't want to do that color, times 15 plus x. We already know that x is 30. Plus 30 times dy dt."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "20 times dx dt is equal to dx dt times y. y right now is 15 feet. So times 15 times, I didn't want to do that color, times 15 plus x. We already know that x is 30. Plus 30 times dy dt. What is dy dt? dy dt we could say is negative 20 feet per second. y is decreasing."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus 30 times dy dt. What is dy dt? dy dt we could say is negative 20 feet per second. y is decreasing. The bird is diving down to get its dinner. So times 20 feet per second, so that's that right over there, plus 10 times dy dt. So plus 10 times negative 20 feet per second."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "y is decreasing. The bird is diving down to get its dinner. So times 20 feet per second, so that's that right over there, plus 10 times dy dt. So plus 10 times negative 20 feet per second. And now we just solve for dx dt. So let's see, what do we have? We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So plus 10 times negative 20 feet per second. And now we just solve for dx dt. So let's see, what do we have? We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation. And we get 5 dx dts. 5 dx dts, I just subtracted this from both sides of the equation. This is 15 dx dts, this is 20."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation. And we get 5 dx dts. 5 dx dts, I just subtracted this from both sides of the equation. This is 15 dx dts, this is 20. So we have 5 dx dts is equal to, this part right over here is negative 600. And this part right over here is negative 200. So it's equal to negative 800 feet per second."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 15 dx dts, this is 20. So we have 5 dx dts is equal to, this part right over here is negative 600. And this part right over here is negative 200. So it's equal to negative 800 feet per second. Or negative 800, and actually this will actually be in feet per second. And so dx dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is negative 160 feet per second."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's equal to negative 800 feet per second. Or negative 800, and actually this will actually be in feet per second. And so dx dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is negative 160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing."}, {"video_title": "Related rates shadow   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is negative 160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing. And we see that. That's why we have this negative sign here. The value of x is decreasing."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "In the last video, we attempted to approximate the area under a curve by constructing four rectangles of equal width and using the left boundary of each rectangle, the function evaluated at the left boundary, to determine the height. And we came up with an approximation. What I want to do in this video is generalize things a bit, using the exact same method, but doing it for an arbitrary function with arbitrary boundaries and an arbitrary number of rectangles. So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a. And this right over here is x. This is x equals b. So this is b."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is x equals a. And this right over here is x. This is x equals b. So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering?"}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3?"}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area?"}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1."}, {"video_title": "Riemann sums in summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1. So that's all we did right over there, times delta x. And then you sum all of these from the first rectangle all the way to the n-th. So hopefully that makes you a little bit more comfortable with this notation."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that f of x is equal to x squared plus x minus 6 over x minus 2. And we're curious about what the limit of f of x as x approaches 2 is equal to. Now, the first attempt that you might want to do right when you see something like this is to see what happens. What is f of 2? Now, this won't always be the limit even if it's defined, but it's a good place to start just to see if something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator we're going to get 2 squared plus 2 minus 6. So that's going to be 4 plus 2, which is 6, minus 6."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is f of 2? Now, this won't always be the limit even if it's defined, but it's a good place to start just to see if something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator we're going to get 2 squared plus 2 minus 6. So that's going to be 4 plus 2, which is 6, minus 6. You're going to get 0 in the numerator, and you're going to get 0 in the denominator. So we don't have the function is not defined. So not defined at x is equal to 2. f not defined."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's going to be 4 plus 2, which is 6, minus 6. You're going to get 0 in the numerator, and you're going to get 0 in the denominator. So we don't have the function is not defined. So not defined at x is equal to 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then actually the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly that the function is not defined here."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So not defined at x is equal to 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then actually the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly that the function is not defined here. So let's see if we can simplify this, and we'll also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we see very clearly that the function is not defined here. So let's see if we can simplify this, and we'll also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. Well, that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that, over x minus 2. So as long as x does not equal 2, these two things will cancel out."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. Well, that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that, over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all x's, except for x is equal to 2, as long as x does not equal 2. So that's another way of looking at it. Another way we could rewrite our f of x. I'm doing it in blue just to change the colors."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all x's, except for x is equal to 2, as long as x does not equal 2. So that's another way of looking at it. Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. We could rewrite f of x. This is the exact same function. f of x is equal to x plus 3 when x does not equal 2."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. We could rewrite f of x. This is the exact same function. f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is not anywhere near being a straight line. That is much better. So let's call this the y-axis."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's try to do it. So that is not anywhere near being a straight line. That is much better. So let's call this the y-axis. I'll call it y equals f of x. And then let's, over here, let me make a horizontal line. That is my x-axis."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's call this the y-axis. I'll call it y equals f of x. And then let's, over here, let me make a horizontal line. That is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3, and then the slope is 1. The slope is 1."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3, and then the slope is 1. The slope is 1. But it's defined for all x's, except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2, it is undefined."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope is 1. But it's defined for all x's, except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2, it is undefined. So let me make sure I can. So it's undefined right over there. It's undefined right over there."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x is equal to 2, it is undefined. So let me make sure I can. So it's undefined right over there. It's undefined right over there. So this is what f of x looks like. Now, given this, let's try to answer our question. What is the limit of f of x as x approaches 2?"}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's undefined right over there. So this is what f of x looks like. Now, given this, let's try to answer our question. What is the limit of f of x as x approaches 2? Well, we can look at this graphically. As x approaches 2 from lower values than 2, so this right over here is x is equal to 2. If we get to maybe, let's say this is 1.7, we see that our f of x is right over there."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the limit of f of x as x approaches 2? Well, we can look at this graphically. As x approaches 2 from lower values than 2, so this right over here is x is equal to 2. If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. If we get even closer to 2, our f of x is right over there. Once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. If we get even closer to 2, our f of x is right over there. Once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. If we ride this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. This is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be this value right over here is equal to 5."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. If we ride this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. This is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be this value right over here is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with a y-intercept of 3, this value right over here is 5. Now, we can also try to do this numerically. So let's try to do that."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is essentially going to be this value right over here is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with a y-intercept of 3, this value right over here is 5. Now, we can also try to do this numerically. So let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, this is almost obvious, 1.9999 plus 3, well, that gets you pretty darn close to 5."}, {"video_title": "Limits by factoring   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, this is almost obvious, 1.9999 plus 3, well, that gets you pretty darn close to 5. If I put even more 9s here, got even closer to 2, we'd get even closer to 5 here. If we approach 2 from the positive direction, and then we once again, we're getting closer and closer to 5 from the positive direction. If we were even closer to 2, we'd be even closer to 5."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we've done this multiple times. Let's think about what g of x approaches as x approaches five from the left. G of x is approaching negative six. As x approaches five from the right, g of x looks like it's approaching negative six. So a reasonable estimate based on looking at this graph is that as x approaches five, g of x is approaching negative six. And it's worth noting that that's not what g of five is. G of five is a different value."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "As x approaches five from the right, g of x looks like it's approaching negative six. So a reasonable estimate based on looking at this graph is that as x approaches five, g of x is approaching negative six. And it's worth noting that that's not what g of five is. G of five is a different value. But the whole point of this video is to appreciate all that a limit does. A limit only describes the behavior of a function as it approaches a point. It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of five is a different value. But the whole point of this video is to appreciate all that a limit does. A limit only describes the behavior of a function as it approaches a point. It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. For example, I could construct many different functions for which the limit as x approaches five is equal to negative six, and they would look very different from g of x. For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x. And in fact, if you're up for it, pause this video and see if you could do the same."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. For example, I could construct many different functions for which the limit as x approaches five is equal to negative six, and they would look very different from g of x. For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x. And in fact, if you're up for it, pause this video and see if you could do the same. If you have some graph paper, even just sketch it. Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six. So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in fact, if you're up for it, pause this video and see if you could do the same. If you have some graph paper, even just sketch it. Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six. So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. As we approach from the left, we're approaching negative six as we approach from the right, we are approaching negative six You could have a function like this. Let's say the limit, let's call it h of x as x approaches five is equal to negative six. You could have a function like this."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. As we approach from the left, we're approaching negative six as we approach from the right, we are approaching negative six You could have a function like this. Let's say the limit, let's call it h of x as x approaches five is equal to negative six. You could have a function like this. Maybe it's defined up to there, then it's, you have a circle there, then it keeps going. Maybe it's not defined at all for any of these values. And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could have a function like this. Maybe it's defined up to there, then it's, you have a circle there, then it keeps going. Maybe it's not defined at all for any of these values. And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. So notice, all of these functions as x approaches five, they all have the limit defined and it's equal to negative six, but these functions all look very, very, very different. Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value. So for example, x approaches five, five is interesting right over here because we have this point discontinuity."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. So notice, all of these functions as x approaches five, they all have the limit defined and it's equal to negative six, but these functions all look very, very, very different. Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value. So for example, x approaches five, five is interesting right over here because we have this point discontinuity. But you could take the limit on an infinite number of points for this function right over here. You could say the limit of g of x as x approaches, not x equals, as x approaches one. What would that be?"}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, x approaches five, five is interesting right over here because we have this point discontinuity. But you could take the limit on an infinite number of points for this function right over here. You could say the limit of g of x as x approaches, not x equals, as x approaches one. What would that be? Pause the video and try to figure it out. Let's see. As x approaches one from the left-hand side, it looks like we are approaching this value here."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What would that be? Pause the video and try to figure it out. Let's see. As x approaches one from the left-hand side, it looks like we are approaching this value here. And as x approaches one from the right-hand side, it looks like we are approaching that value there. So that would be equal to g of one. That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "As x approaches one from the left-hand side, it looks like we are approaching this value here. And as x approaches one from the right-hand side, it looks like we are approaching that value there. So that would be equal to g of one. That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. And if we were to estimate that, g of one looks like it's approximately negative 5.1 or 5.2, negative 5.1. We could find the limit of g of x as x approaches pi. So pi is right around there."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. And if we were to estimate that, g of one looks like it's approximately negative 5.1 or 5.2, negative 5.1. We could find the limit of g of x as x approaches pi. So pi is right around there. As x approaches pi from the left, we're approaching that value, which looks actually pretty close to the one we just thought about. As we approach from the right, we're approaching that value. And once again, in this case, this is gonna be equal to g of pi."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pi is right around there. As x approaches pi from the left, we're approaching that value, which looks actually pretty close to the one we just thought about. As we approach from the right, we're approaching that value. And once again, in this case, this is gonna be equal to g of pi. We don't have any interesting discontinuities there or anything like that. So there's two big takeaways here. You can construct many different functions that would have the same limit at a point."}, {"video_title": "Connecting limits and graphical behavior   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, in this case, this is gonna be equal to g of pi. We don't have any interesting discontinuities there or anything like that. So there's two big takeaways here. You can construct many different functions that would have the same limit at a point. And for a given function, you can take the limit at many different points. In fact, an infinite number of different points. And it's important to point that out, no pun intended, because oftentimes we get used to seeing limits only at points where something strange seems to be happening."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now at first when you see this, you're like, wow, this is strange. I have a function that is being defined by an integral, a definite integral, but one of its bounds are x. And you should just say, well, this is okay. A function can be defined any which way. And as we'll see, it's actually quite straightforward to evaluate this. So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two. So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "A function can be defined any which way. And as we'll see, it's actually quite straightforward to evaluate this. So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two. So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. Well, x is now negative two. This is now negative two. And so how do we figure out what this is?"}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. Well, x is now negative two. This is now negative two. And so how do we figure out what this is? Now before we even look at this graph, you might say, okay, this is the region under, the area of the region under the graph y equals f of t between negative two and zero, but you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so how do we figure out what this is? Now before we even look at this graph, you might say, okay, this is the region under, the area of the region under the graph y equals f of t between negative two and zero, but you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. And so when you swap the bounds, this is going to be equal to negative definite integral from negative two, negative two to zero of f of t dt. And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero. So between negative two and zero."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. And so when you swap the bounds, this is going to be equal to negative definite integral from negative two, negative two to zero of f of t dt. And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero. So between negative two and zero. So that is this area right over here that we care about. Now what is that going to be? Well, you could, there's a bunch of different ways that you could do this."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So between negative two and zero. So that is this area right over here that we care about. Now what is that going to be? Well, you could, there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four. It's two by two."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you could, there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four. It's two by two. And just make sure to, make sure you look at the units. Sometimes each square doesn't represent one square unit, but in this case it does. So that's four."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's two by two. And just make sure to, make sure you look at the units. Sometimes each square doesn't represent one square unit, but in this case it does. So that's four. And then up here, this is half of four, right? If it was all of this, that would be four. This triangle is half of four."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's four. And then up here, this is half of four, right? If it was all of this, that would be four. This triangle is half of four. So this is two right over there. Or you could view this as base times height times 1 1 2, which is gonna be two times two times 1 1 2. And so this area right over here is six."}, {"video_title": "Functions defined by integrals switched interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "This triangle is half of four. So this is two right over there. Or you could view this as base times height times 1 1 2, which is gonna be two times two times 1 1 2. And so this area right over here is six. So this part is six, but we can't forget that negative sign. So this is going to be equal to negative, negative six. And so g of negative two is negative six."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Justify your answer. Well, to think about whether we have a relative minimum or a relative maximum, we can see, well, what's the derivative at that point? If it's zero, then it's a good candidate that we're dealing with a, that it could be a relative minimum or a maximum. If it's not zero, then it's neither. And then if it is zero, if we want to figure out relative minimum or relative maximum, we can evaluate the sign of the second derivative. So let's just think about this. So we want to evaluate f prime, we want to figure out what f prime of, f prime of two is equal to."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If it's not zero, then it's neither. And then if it is zero, if we want to figure out relative minimum or relative maximum, we can evaluate the sign of the second derivative. So let's just think about this. So we want to evaluate f prime, we want to figure out what f prime of, f prime of two is equal to. So we know that f prime of x, f prime of x, which is the same thing as dy, dx, is equal to two times x minus y. We saw that in the last problem. And so f prime of two, right this way, f prime of two is going to be equal to two times two, two times two, minus whatever y is when x is equal to two."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to evaluate f prime, we want to figure out what f prime of, f prime of two is equal to. So we know that f prime of x, f prime of x, which is the same thing as dy, dx, is equal to two times x minus y. We saw that in the last problem. And so f prime of two, right this way, f prime of two is going to be equal to two times two, two times two, minus whatever y is when x is equal to two. Well, do we know what y is when x equals to two? Sure, they tell us right over here. Y is equal to f of x."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so f prime of two, right this way, f prime of two is going to be equal to two times two, two times two, minus whatever y is when x is equal to two. Well, do we know what y is when x equals to two? Sure, they tell us right over here. Y is equal to f of x. So when x is equal to two, when x is equal to two, y is equal to three. So two times two minus three, and so this is going to be equal to four minus three is equal to one. And so since the derivative at two is not zero, this is not going to be a minimum, a relative minimum, or a relative maximum."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Y is equal to f of x. So when x is equal to two, when x is equal to two, y is equal to three. So two times two minus three, and so this is going to be equal to four minus three is equal to one. And so since the derivative at two is not zero, this is not going to be a minimum, a relative minimum, or a relative maximum. So we could say since f prime of two, f prime of two does not equal zero, this, we have a, f has, right this way, f has neither, neither minimum, or relative minimum I guess I could say, relative min, or relative max at x equals two. Alright, let's do the next one. Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so since the derivative at two is not zero, this is not going to be a minimum, a relative minimum, or a relative maximum. So we could say since f prime of two, f prime of two does not equal zero, this, we have a, f has, right this way, f has neither, neither minimum, or relative minimum I guess I could say, relative min, or relative max at x equals two. Alright, let's do the next one. Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation. Alright, this one is interesting. So let's actually, let's just write down everything we know before we even think that y equals mx plus b could be a solution to the differential equation. So we know that, we know that dy over dx is equal to two x minus y."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Find the values of the constants m and b for which y equals mx plus b is a solution to the differential equation. Alright, this one is interesting. So let's actually, let's just write down everything we know before we even think that y equals mx plus b could be a solution to the differential equation. So we know that, we know that dy over dx is equal to two x minus y. They told us that. We also know that the second derivative, the second derivative of y with respect to x is equal to two minus dy dx. We figured that out in part b of this problem."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that, we know that dy over dx is equal to two x minus y. They told us that. We also know that the second derivative, the second derivative of y with respect to x is equal to two minus dy dx. We figured that out in part b of this problem. And then we could also express this, we saw that we could also write that as two minus two x plus y if you just substitute, if you substitute this in for that. So it's two minus two x plus y. So let me write it that way."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We figured that out in part b of this problem. And then we could also express this, we saw that we could also write that as two minus two x plus y if you just substitute, if you substitute this in for that. So it's two minus two x plus y. So let me write it that way. So this is also equal to two minus two x plus y. So that's everything we know before we even thought that maybe there's a solution of y equals mx plus b. So now let's start with y equals mx plus b."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write it that way. So this is also equal to two minus two x plus y. So that's everything we know before we even thought that maybe there's a solution of y equals mx plus b. So now let's start with y equals mx plus b. So if y is equal to mx plus b, y equals mx plus b, so this is the equation of a line, then dy dx is going to be equal to, well the derivative of this with respect to x is just m, derivative of this with respect to x, it's a constant, it's not going to change with respect to x, it's just a zero. And that makes sense. The rate of change of y with respect to x is the slope of our line."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now let's start with y equals mx plus b. So if y is equal to mx plus b, y equals mx plus b, so this is the equation of a line, then dy dx is going to be equal to, well the derivative of this with respect to x is just m, derivative of this with respect to x, it's a constant, it's not going to change with respect to x, it's just a zero. And that makes sense. The rate of change of y with respect to x is the slope of our line. So can we use, and this is really all that we know, we could keep, actually we could go even further. We could take the second derivative here. The second derivative of y with respect to x, well that's going to be zero."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The rate of change of y with respect to x is the slope of our line. So can we use, and this is really all that we know, we could keep, actually we could go even further. We could take the second derivative here. The second derivative of y with respect to x, well that's going to be zero. The second derivative of a linear function, well it's going to be zero, we see that here. So this is all of the information that we have. We get this from the previous parts of the problem and we get this just taking the first and second derivatives of y equals mx plus b."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The second derivative of y with respect to x, well that's going to be zero. The second derivative of a linear function, well it's going to be zero, we see that here. So this is all of the information that we have. We get this from the previous parts of the problem and we get this just taking the first and second derivatives of y equals mx plus b. So given this, can we figure out, can we figure out what m and b are? Alright, so we could, if we said m is equal to two x minus y, that doesn't seem right. This one is a tricky one."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get this from the previous parts of the problem and we get this just taking the first and second derivatives of y equals mx plus b. So given this, can we figure out, can we figure out what m and b are? Alright, so we could, if we said m is equal to two x minus y, that doesn't seem right. This one is a tricky one. Let's see, we know that the second derivative is going to be equal to zero. We know that this is going to be equal to zero for this particular solution. And we know dy dx is equal to m. We know this is m. And so there you have it, we have enough information to solve for m. We know that zero is equal to two minus m. So zero is equal to two minus m. And so we can add m to both sides and we get m is equal to two."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This one is a tricky one. Let's see, we know that the second derivative is going to be equal to zero. We know that this is going to be equal to zero for this particular solution. And we know dy dx is equal to m. We know this is m. And so there you have it, we have enough information to solve for m. We know that zero is equal to two minus m. So zero is equal to two minus m. And so we can add m to both sides and we get m is equal to two. So that by itself was quite useful. And then what we could say, let's see, can we solve this further? Well we know that this right over here, dy dx, this is m. This is m. And it's equal to two."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know dy dx is equal to m. We know this is m. And so there you have it, we have enough information to solve for m. We know that zero is equal to two minus m. So zero is equal to two minus m. And so we can add m to both sides and we get m is equal to two. So that by itself was quite useful. And then what we could say, let's see, can we solve this further? Well we know that this right over here, dy dx, this is m. This is m. And it's equal to two. So we could say that two is equal to two x minus y. Two is equal to two x minus y. And let's see, if we solve for y, add y to both sides, subtract two from both sides, we get y is equal to two x minus two."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well we know that this right over here, dy dx, this is m. This is m. And it's equal to two. So we could say that two is equal to two x minus y. Two is equal to two x minus y. And let's see, if we solve for y, add y to both sides, subtract two from both sides, we get y is equal to two x minus two. And there we have our whole solution. And so you have your m right over there. That is m. And then we also have our b."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, if we solve for y, add y to both sides, subtract two from both sides, we get y is equal to two x minus two. And there we have our whole solution. And so you have your m right over there. That is m. And then we also have our b. This one was a tricky one. Anytime that you have to do something like this and it doesn't just jump out at you, and if it wasn't obvious it didn't jump out at me at first when I looked at this problem, I said well let me just write down everything that they told us. So they wrote this before."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is m. And then we also have our b. This one was a tricky one. Anytime that you have to do something like this and it doesn't just jump out at you, and if it wasn't obvious it didn't jump out at me at first when I looked at this problem, I said well let me just write down everything that they told us. So they wrote this before. And then we say okay, this is going to be a solution. And so let me see if I can somehow solve, so let's see what I didn't use. I didn't use that."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they wrote this before. And then we say okay, this is going to be a solution. And so let me see if I can somehow solve, so let's see what I didn't use. I didn't use that. I did use this. I absolutely used that. I did use that."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I didn't use that. I did use this. I absolutely used that. I did use that. I did use that. And I did use that. So this was a little bit of a fun little puzzle where I just wrote down all the information they gave us and I tried to figure out, based on that, whether I could figure out m and b."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I did use that. I did use that. And I did use that. So this was a little bit of a fun little puzzle where I just wrote down all the information they gave us and I tried to figure out, based on that, whether I could figure out m and b. And this is pretty neat. This is a solution, 2x minus 2. If we go to our slope field above, it wouldn't have jumped out at me, but if you think about, so 2x minus 2, its y-intercept would be negative 2 like that."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this was a little bit of a fun little puzzle where I just wrote down all the information they gave us and I tried to figure out, based on that, whether I could figure out m and b. And this is pretty neat. This is a solution, 2x minus 2. If we go to our slope field above, it wouldn't have jumped out at me, but if you think about, so 2x minus 2, its y-intercept would be negative 2 like that. Let me do this in a different color. And so the line would look something like this. The line would look something like this."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we go to our slope field above, it wouldn't have jumped out at me, but if you think about, so 2x minus 2, its y-intercept would be negative 2 like that. Let me do this in a different color. And so the line would look something like this. The line would look something like this. And you can verify that any one of these points, at any one of these points, the slope is equal to 2. If we're at the point 2, 2, well it's going to be 2 times 2 minus 2 is 2. 1, 0, 2 times 1 minus 0 is 2."}, {"video_title": "2015 AP Calculus AB BC 4cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The line would look something like this. And you can verify that any one of these points, at any one of these points, the slope is equal to 2. If we're at the point 2, 2, well it's going to be 2 times 2 minus 2 is 2. 1, 0, 2 times 1 minus 0 is 2. Negative 2, 0, negative 2, well 0 minus negative 2, that's 2. So you see this is pretty neat. The slope is changing all around it, but this is a linear solution to that original differential equation."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you learn multiple notations for this. For example, if you know that y is equal to f of x, you might write this as y prime. You might write this as dy dx, which you'll often hear me say is the derivative of y with respect to x, and that you could use the derivative of f with respect to x, because y is equal to our function. But then later on, especially when you start getting into differential equations, you see people start to treat this notation as an actual algebraic expression. For example, you will learn, or you might have already seen, if you're trying to solve the differential equation, the derivative of y with respect to x is equal to y, so the rate of change of y with respect to x is equal to the value of y itself. This is one of the most basic differential equations you might see. You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then later on, especially when you start getting into differential equations, you see people start to treat this notation as an actual algebraic expression. For example, you will learn, or you might have already seen, if you're trying to solve the differential equation, the derivative of y with respect to x is equal to y, so the rate of change of y with respect to x is equal to the value of y itself. This is one of the most basic differential equations you might see. You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression. So you multiply both sides by dx, and then you have, so that would cancel out algebraically. And so you see people treat it like that. So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression. So you multiply both sides by dx, and then you have, so that would cancel out algebraically. And so you see people treat it like that. So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do. Y is an algebraic expression. So if you divide both sides by y, you get one over y dy is equal to dx. And then folks will integrate both sides to find a general solution to this differential equation."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do. Y is an algebraic expression. So if you divide both sides by y, you get one over y dy is equal to dx. And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy. And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand-wavy."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy. And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand-wavy. It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions. And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x. And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so to just feel reasonably okay about doing this, this is a little bit hand-wavy. It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions. And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x. And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y. So that's one way that you can feel a little bit better of, and this is actually one of the justifications for this type of notation, is you could view this, what's the resulting super small, or what's the super small change in y for a given super small change in x, which is giving us the sense of what's the limiting value of the slope as we go from the slope of a secant line to a tangent line. And if you view it that way, you might feel a little bit better about using the differentials or treating them algebraically, whereas, okay, let me just multiply both sides by that super small change in x. So the big picture is this is a technique that you will often see in introductory differential equations classes, introductory multivariable classes, and introductory calculus classes, but it's not very mathematically rigorous to just treat differentials like algebraic expressions, but even though it's not very mathematically rigorous to do it willy-nilly like that, it has proven to be very useful."}, {"video_title": "Addressing treating differentials algebraically   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y. So that's one way that you can feel a little bit better of, and this is actually one of the justifications for this type of notation, is you could view this, what's the resulting super small, or what's the super small change in y for a given super small change in x, which is giving us the sense of what's the limiting value of the slope as we go from the slope of a secant line to a tangent line. And if you view it that way, you might feel a little bit better about using the differentials or treating them algebraically, whereas, okay, let me just multiply both sides by that super small change in x. So the big picture is this is a technique that you will often see in introductory differential equations classes, introductory multivariable classes, and introductory calculus classes, but it's not very mathematically rigorous to just treat differentials like algebraic expressions, but even though it's not very mathematically rigorous to do it willy-nilly like that, it has proven to be very useful. Now, as you get more sophisticated in your mathematics, there are rigorous definitions of a differential where you can get a better sense of where it is mathematically rigorous to use it and where it isn't. But the whole point here is if you felt a little weird feeling about multiplying both sides by dx or dividing both sides by dx or dy, your feeling is mathematically justified because it's not a very rigorous thing to do, at least until you have more rigor behind it. But I will tell you that if you're an introductory student, it is a reasonable thing to do as you explore and manipulate some of these basic differential equations."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope between the two endpoints is gonna be your change in y, which is going to be your change in your function value, so f of b minus f of a over, over b minus a. And once again, we do this, we go into much more depth in this when we covered it the first time in differential calculus, but just to give you a visualization of it, because I think it's always handy, the mean value theorem that we learned in differential calculus just tells us, hey look, you know, if this is a, this is b, I've got my function doing something interesting, so this is f of a, this is f of b, so this quantity right over here, where you're taking the change in the value of our function divided by, so this right over here is f of b, f of b minus f of a is this change in the value of our function divided by the change in our x-axis, so it's our change in y over change in x, that gives us the slope, this right over here gives us the slope of this line, the slope of the line that connects, that connects these two points, that's this quantity, and this, the mean value theorem tells us that there's some c in between a and b where you're gonna have the same slope, so it might be at least one place, so it might be right over there where you have the exact same slope, there exists a c where the slope of the tangent line at that point is going to be the same, so this would be a c right over there, and we actually might have a couple of c's, that's another candidate c, there's at least one c where the slope of the tangent line is the same as the average slope across the interval, and once again, we have to assume that f is continuous and f is differentiable. Now, when you see this, it, something might, it might evoke some similarities with what we saw when we saw the, how we defined, I guess you could say, or the formula for the average value of a function. Remember, what we saw for the average value of a function, we said the average, the average value of a function is going to be equal to one over b minus a, notice, one over b minus a, you have a b minus a in the denominator here, times the definite integral from a to b of f of x dx. Now, this is interesting, because here we have a derivative, here we have an integral, but maybe we could connect these, maybe we could connect these two things. Well, one thing that might jump out at you is maybe we could rewrite, maybe we could rewrite this numerator right over here in this form somehow, and I encourage you to pause the video and see if you can, and I'll give you actually a quite huge hint, instead of it being an f of x here, what happens if there's an f prime of x there? So I encourage you to try to do that."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, what we saw for the average value of a function, we said the average, the average value of a function is going to be equal to one over b minus a, notice, one over b minus a, you have a b minus a in the denominator here, times the definite integral from a to b of f of x dx. Now, this is interesting, because here we have a derivative, here we have an integral, but maybe we could connect these, maybe we could connect these two things. Well, one thing that might jump out at you is maybe we could rewrite, maybe we could rewrite this numerator right over here in this form somehow, and I encourage you to pause the video and see if you can, and I'll give you actually a quite huge hint, instead of it being an f of x here, what happens if there's an f prime of x there? So I encourage you to try to do that. So once again, this is, let me rewrite all of this. This is going to be equal to, this over here is the exact same thing as the definite integral from a to b of f prime of x dx. Think about it."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I encourage you to try to do that. So once again, this is, let me rewrite all of this. This is going to be equal to, this over here is the exact same thing as the definite integral from a to b of f prime of x dx. Think about it. You're gonna take the antiderivative of f prime of x, which is going to be f of x, and you're going to evaluate it at b, f of b, and then from that, you're going to subtract it, evaluate it at a, minus f of a. These two things are identical, and then you can, of course, divide by b minus a. Now this is starting to get interesting."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Think about it. You're gonna take the antiderivative of f prime of x, which is going to be f of x, and you're going to evaluate it at b, f of b, and then from that, you're going to subtract it, evaluate it at a, minus f of a. These two things are identical, and then you can, of course, divide by b minus a. Now this is starting to get interesting. One way to think about it is there must be a c, if we, there must be a c that takes on the average value of, there must be a c that when you evaluate the derivative at c, it takes on the average value of the derivative. Or another way to think about it, another way to think about it, if we were to just write, if we were to just write, let me just say g of x is equal to f prime of x, then we get very close to what we have over here, because this right over here is going to be g of c. Remember, f prime of c is the same thing as g of c, is equal to one over, is equal to one over b minus a, one over b minus a. So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this is starting to get interesting. One way to think about it is there must be a c, if we, there must be a c that takes on the average value of, there must be a c that when you evaluate the derivative at c, it takes on the average value of the derivative. Or another way to think about it, another way to think about it, if we were to just write, if we were to just write, let me just say g of x is equal to f prime of x, then we get very close to what we have over here, because this right over here is going to be g of c. Remember, f prime of c is the same thing as g of c, is equal to one over, is equal to one over b minus a, one over b minus a. So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx. F prime of x is the same thing as g of x. So another way of thinking about it, and this is actually another form of the mean value theorem, it's called the mean value theorem for integrals. Mean value theorem for integrals."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx. F prime of x is the same thing as g of x. So another way of thinking about it, and this is actually another form of the mean value theorem, it's called the mean value theorem for integrals. Mean value theorem for integrals. Let me, so this is the mean, I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, and to give it a slightly more formal sense, is if you have some function g, so if g is, let me actually go down a little bit, which tells us that if g of x is continuous, continuous on this closed interval, going from a to b, then there exists, then there exists a c in this interval where g of c is equal to, what is this? This is the average value of our function. There exists a c where g of c is equal to the average value of your function over the interval."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Mean value theorem for integrals. Let me, so this is the mean, I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, and to give it a slightly more formal sense, is if you have some function g, so if g is, let me actually go down a little bit, which tells us that if g of x is continuous, continuous on this closed interval, going from a to b, then there exists, then there exists a c in this interval where g of c is equal to, what is this? This is the average value of our function. There exists a c where g of c is equal to the average value of your function over the interval. This was our definition of the average value of a function. So anyway, this is just another way of saying, you might see some of the mean value theorem of integrals, and just to show you that it's really closely tied, it's using different notation, but it's usually, it's essentially the same exact idea as the mean value theorem you learned in differential calculus, but now you have different notation, and I guess you can have a slightly different interpretation. When we're thinking about it in differential calculus, we're thinking about having a point where the slope of the tangent line of the function at that point is the same as the average rate, so that's when we had our kind of differential mode, and we were kind of thinking in terms of slopes and slopes of tangent lines, and now when we're in integral mode, we're thinking much more in terms of average value, average value of the function."}, {"video_title": "Mean value theorem for integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "There exists a c where g of c is equal to the average value of your function over the interval. This was our definition of the average value of a function. So anyway, this is just another way of saying, you might see some of the mean value theorem of integrals, and just to show you that it's really closely tied, it's using different notation, but it's usually, it's essentially the same exact idea as the mean value theorem you learned in differential calculus, but now you have different notation, and I guess you can have a slightly different interpretation. When we're thinking about it in differential calculus, we're thinking about having a point where the slope of the tangent line of the function at that point is the same as the average rate, so that's when we had our kind of differential mode, and we were kind of thinking in terms of slopes and slopes of tangent lines, and now when we're in integral mode, we're thinking much more in terms of average value, average value of the function. So there's some c where g of c, there's some c where the function evaluated at that point is equal to the average value. So another way of thinking about it, if I were to draw, if I were to draw g of, if I were to draw g of x, if I were to draw g of x, so that's x, that is my y-axis, this is the graph of y is equal to g of x, which of course was the same thing as f prime of x, but we've, I guess, we've just rewritten it now to be more consistent with our average value formula, and we're talking about the interval from a to b. We've already seen how to calculate the average value."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now ready for part B, and I've wiped all the salt off of my fingers, and I've had a glass of water, so now I'm ready for some business. Find the second derivative of W with respect to T. In terms of W, use this second derivative to determine whether your answer in part A is an underestimate or an overestimate for the amount of solid waste that the landfill contains at time T equals 1 fourth. So let's do the first things first. Let's find the second derivative in terms of W. And we already have the first derivative over here, and in part A I rewrote it just with slightly different notation, but I'll just use this just because I need to write down here and I can still refer to this thing. So let's just take the derivative of both sides of this differential equation with respect to T. So you take the derivative of this left-hand side with respect to T, you get the second derivative of W with respect to T as a function of T is equal to, and you take the derivative of this with respect to T, this is the same thing as 1 over 25 WT minus 12. And so when you take the derivative, that constant part drops off and you're just left with 1 over 25 times the derivative of WT. And let me make that clear."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's find the second derivative in terms of W. And we already have the first derivative over here, and in part A I rewrote it just with slightly different notation, but I'll just use this just because I need to write down here and I can still refer to this thing. So let's just take the derivative of both sides of this differential equation with respect to T. So you take the derivative of this left-hand side with respect to T, you get the second derivative of W with respect to T as a function of T is equal to, and you take the derivative of this with respect to T, this is the same thing as 1 over 25 WT minus 12. And so when you take the derivative, that constant part drops off and you're just left with 1 over 25 times the derivative of WT. And let me make that clear. This is the same thing as 1 over 25 W as a function of T minus 12. 1 25th of 300 is 12. And you take the derivative of this, you get 1 25th times the first derivative of W, and then the derivative of this with respect to T is just 0."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me make that clear. This is the same thing as 1 over 25 W as a function of T minus 12. 1 25th of 300 is 12. And you take the derivative of this, you get 1 25th times the first derivative of W, and then the derivative of this with respect to T is just 0. A constant obviously does not change with respect to T. And so we get this right over here. Now this is the second derivative in terms of the first derivative. But the question asks us, write the second derivative in terms of W. But lucky for us, we know how to express this as a function of W. They gave that to us in the problem."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you take the derivative of this, you get 1 25th times the first derivative of W, and then the derivative of this with respect to T is just 0. A constant obviously does not change with respect to T. And so we get this right over here. Now this is the second derivative in terms of the first derivative. But the question asks us, write the second derivative in terms of W. But lucky for us, we know how to express this as a function of W. They gave that to us in the problem. This was given. This is just me rewriting it in a different notation. So this is going to be the same thing as 1 over 25 times the derivative of W. And the derivative of W is this."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the question asks us, write the second derivative in terms of W. But lucky for us, we know how to express this as a function of W. They gave that to us in the problem. This was given. This is just me rewriting it in a different notation. So this is going to be the same thing as 1 over 25 times the derivative of W. And the derivative of W is this. The differential equation literally tells us the derivative of W is this over here. So 1 over 25 times 1 over 25 WT, or the function W as a function of T, minus 300. And so we can say the second derivative of W as a function of T is equal to 1 over 625 times W as a function of T, which is a function of T, minus 300."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be the same thing as 1 over 25 times the derivative of W. And the derivative of W is this. The differential equation literally tells us the derivative of W is this over here. So 1 over 25 times 1 over 25 WT, or the function W as a function of T, minus 300. And so we can say the second derivative of W as a function of T is equal to 1 over 625 times W as a function of T, which is a function of T, minus 300. So we've done the first part. We've found out the second derivative of W in terms of just W. Now let's try to address the second part of their question. So we did the first part."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can say the second derivative of W as a function of T is equal to 1 over 625 times W as a function of T, which is a function of T, minus 300. So we've done the first part. We've found out the second derivative of W in terms of just W. Now let's try to address the second part of their question. So we did the first part. Now use what we just figured out to determine whether your answer in part A is an underestimate or an overestimate of the amount of solid waste that the landfill contains at time 1 fourth. So in part A, we found the slope of the tangent line at time equals 0. And we use that slope to extrapolate out to 3 months, or time equals 1 fourth, 1 fourth of a year."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we did the first part. Now use what we just figured out to determine whether your answer in part A is an underestimate or an overestimate of the amount of solid waste that the landfill contains at time 1 fourth. So in part A, we found the slope of the tangent line at time equals 0. And we use that slope to extrapolate out to 3 months, or time equals 1 fourth, 1 fourth of a year. Now, if the function's W's slope over that fourth was exactly the same as the slope of the tangent line, or if that slope did not change, then our extrapolation would be exactly right. If W's slope is increasing over that time, then our estimate would be an underestimate of the correct thing. And if the slope is decreasing over that time, then our estimate would be an overestimate of the actual amount in the landfill."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we use that slope to extrapolate out to 3 months, or time equals 1 fourth, 1 fourth of a year. Now, if the function's W's slope over that fourth was exactly the same as the slope of the tangent line, or if that slope did not change, then our extrapolation would be exactly right. If W's slope is increasing over that time, then our estimate would be an underestimate of the correct thing. And if the slope is decreasing over that time, then our estimate would be an overestimate of the actual amount in the landfill. And to figure out whether the slope is increasing or decreasing, we just have to look at the value of the second derivative. If the second derivative is positive, that means our slope is increasing, which means that our extrapolation would be an understatement. Let me just be clear here."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if the slope is decreasing over that time, then our estimate would be an overestimate of the actual amount in the landfill. And to figure out whether the slope is increasing or decreasing, we just have to look at the value of the second derivative. If the second derivative is positive, that means our slope is increasing, which means that our extrapolation would be an understatement. Let me just be clear here. Let me draw this. Let me make it very clear. So let me just draw a random function."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just be clear here. Let me draw this. Let me make it very clear. So let me just draw a random function. So let's say that's W. So this is a case where W's slope is increasing faster, or W's slope is increasing from that starting point. So our starting point, this was our slope, and then W's slope keeps increasing from there. And in this case, our estimate is going to be an underestimate of where W actually is after a fourth of the year."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me just draw a random function. So let's say that's W. So this is a case where W's slope is increasing faster, or W's slope is increasing from that starting point. So our starting point, this was our slope, and then W's slope keeps increasing from there. And in this case, our estimate is going to be an underestimate of where W actually is after a fourth of the year. If W's slope is exactly the same as our function over the course of the year, over the course of the first 3 months, so it would look something like that. Maybe it diverges later on. And in this case, our approximation would be really good."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in this case, our estimate is going to be an underestimate of where W actually is after a fourth of the year. If W's slope is exactly the same as our function over the course of the year, over the course of the first 3 months, so it would look something like that. Maybe it diverges later on. And in this case, our approximation would be really good. It would probably be exact for W. And if W's slope, for whatever reason, goes negative after that point, and they already tell us this is an increasing function, so that is not likely. Well, it doesn't have to even go negative. If W's slope decreases, it could still stay positive."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in this case, our approximation would be really good. It would probably be exact for W. And if W's slope, for whatever reason, goes negative after that point, and they already tell us this is an increasing function, so that is not likely. Well, it doesn't have to even go negative. If W's slope decreases, it could still stay positive. And the way I drew it doesn't, let me draw it like this, just to make it clear. So let's say that, let's say we find out that the slope looks something like this. This is the slope of the tangent line at time equals 0."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "If W's slope decreases, it could still stay positive. And the way I drew it doesn't, let me draw it like this, just to make it clear. So let's say that, let's say we find out that the slope looks something like this. This is the slope of the tangent line at time equals 0. If W's slope increases from that point, then W might look something like that. And then, our answer to part A would be an understatement, would be an underestimate of where W actually is after a fourth of a year. Let me make it clear to you what I'm doing."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the slope of the tangent line at time equals 0. If W's slope increases from that point, then W might look something like that. And then, our answer to part A would be an understatement, would be an underestimate of where W actually is after a fourth of a year. Let me make it clear to you what I'm doing. So this is my W axis. So this is right at our initial condition of 1400. And this right over here is our time axis."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make it clear to you what I'm doing. So this is my W axis. So this is right at our initial condition of 1400. And this right over here is our time axis. And this is at 1 fourth. So in the last video, we said, hey, this is sitting right at 1411. If W's slope increases from that point, then this is an underestimate."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this right over here is our time axis. And this is at 1 fourth. So in the last video, we said, hey, this is sitting right at 1411. If W's slope increases from that point, then this is an underestimate. If W's slope stays the same, then this is actually a very good estimate. Because then we're going to hit that point directly with W. And if W's slope decreases, so you can imagine maybe W looks something like this. It has that slope of the tangent line when it starts, but then the slope decreases."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "If W's slope increases from that point, then this is an underestimate. If W's slope stays the same, then this is actually a very good estimate. Because then we're going to hit that point directly with W. And if W's slope decreases, so you can imagine maybe W looks something like this. It has that slope of the tangent line when it starts, but then the slope decreases. And it's still an increasing function, but the slope is decreasing. And in that case, we would have an overestimate. And this is a situation, so this first situation, slope is increasing, that means W prime prime is positive."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "It has that slope of the tangent line when it starts, but then the slope decreases. And it's still an increasing function, but the slope is decreasing. And in that case, we would have an overestimate. And this is a situation, so this first situation, slope is increasing, that means W prime prime is positive. The slope is increasing. The second derivative is positive. This means, or this would be a byproduct, or this would cause the second derivative to be negative."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is a situation, so this first situation, slope is increasing, that means W prime prime is positive. The slope is increasing. The second derivative is positive. This means, or this would be a byproduct, or this would cause the second derivative to be negative. Our slope is decreasing. And this would be our second derivative. Our second derivative is 0."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "This means, or this would be a byproduct, or this would cause the second derivative to be negative. Our slope is decreasing. And this would be our second derivative. Our second derivative is 0. If your slope isn't changing, if your slope is constant, if your first derivative is constant, your second derivative is going to be 0. So let's just see what our second derivative is at our initial condition, and then we'll have a pretty good sense of whether we have an overestimate or an underestimate for part A. So let's just figure out what our second derivative is at time 0."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our second derivative is 0. If your slope isn't changing, if your slope is constant, if your first derivative is constant, your second derivative is going to be 0. So let's just see what our second derivative is at our initial condition, and then we'll have a pretty good sense of whether we have an overestimate or an underestimate for part A. So let's just figure out what our second derivative is at time 0. It's going to be equal to 1 over 625 times W of 0, which we know, minus 300. Well, W of 0, the amount of waste we have at time 0, they told us in the problem, is 1400 tons. 1400 minus 1300 is 1100."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just figure out what our second derivative is at time 0. It's going to be equal to 1 over 625 times W of 0, which we know, minus 300. Well, W of 0, the amount of waste we have at time 0, they told us in the problem, is 1400 tons. 1400 minus 1300 is 1100. And then 1100 divided by 625 is a small number, it's one point something, but it is a positive number. And that's the important thing here. So this thing, all of this business, it is positive."}, {"video_title": "2011 Calculus AB free response #5b   AP Calculus AB   Khan Academy.mp3", "Sentence": "1400 minus 1300 is 1100. And then 1100 divided by 625 is a small number, it's one point something, but it is a positive number. And that's the important thing here. So this thing, all of this business, it is positive. So the second derivative is positive, which means the slope is increasing, at least right at our starting point our slope is increasing, which tells us that since our slope is increasing from that point, and frankly the fact that it's increasing at all, and that we know that W is an increasing function, tells us that our estimate in part A is an underestimate. So this is the case that we're dealing with. So our estimate in part A is an underestimate of the actual amount of solid waste in the landfill at time t equals 1 fourth."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "And what I want to do is figure out what is the limit of g of h of x as x approaches one. Pause this video and see if you can figure that out. All right, now let's do this together. Now the first thing that you might try to say is, all right, let's just figure out first the limit as x approaches one of h of x. And when you look at that, what is that going to be? Well, as we approach one from the left, it looks like h of x is approaching two. And as we approach from the right, it looks like h of x is approaching two."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "Now the first thing that you might try to say is, all right, let's just figure out first the limit as x approaches one of h of x. And when you look at that, what is that going to be? Well, as we approach one from the left, it looks like h of x is approaching two. And as we approach from the right, it looks like h of x is approaching two. So it looks like this is just going to be two. And then we say, okay, well maybe we can then just input that into g. So what is g of two? Well, g of two is zero, but the limit doesn't seem defined."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "And as we approach from the right, it looks like h of x is approaching two. So it looks like this is just going to be two. And then we say, okay, well maybe we can then just input that into g. So what is g of two? Well, g of two is zero, but the limit doesn't seem defined. It looks like when we approach two from the right, we're approaching zero. And when we approach two from the left, we're approaching negative two. So maybe this limit doesn't exist."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "Well, g of two is zero, but the limit doesn't seem defined. It looks like when we approach two from the right, we're approaching zero. And when we approach two from the left, we're approaching negative two. So maybe this limit doesn't exist. But if you're thinking that, we haven't fully thought through it. Because what we could do is think about this limit in terms of both left-handed and right-handed limits. So let's think of it this way."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So maybe this limit doesn't exist. But if you're thinking that, we haven't fully thought through it. Because what we could do is think about this limit in terms of both left-handed and right-handed limits. So let's think of it this way. First, let's think about what is the limit as x approaches one from the left-hand side of g of h of x. All right, when you think about it this way, if we're approaching one from the left, right over here, we see that we are approaching two from the left, I guess you could say, or we're approaching two from below. And so the thing that we are inputting into g of x is approaching two from below."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So let's think of it this way. First, let's think about what is the limit as x approaches one from the left-hand side of g of h of x. All right, when you think about it this way, if we're approaching one from the left, right over here, we see that we are approaching two from the left, I guess you could say, or we're approaching two from below. And so the thing that we are inputting into g of x is approaching two from below. So the thing that we are inputting into g is approaching two from below. So if you approach two from below, right over here, what is g approaching? It looks like g is approaching negative two."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "And so the thing that we are inputting into g of x is approaching two from below. So the thing that we are inputting into g is approaching two from below. So if you approach two from below, right over here, what is g approaching? It looks like g is approaching negative two. So this looks like it is going to be equal to negative two, at least this left-handed limit. Now let's do a right-handed limit. What is the limit as x approaches one from the right-hand of g of h of x?"}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "It looks like g is approaching negative two. So this looks like it is going to be equal to negative two, at least this left-handed limit. Now let's do a right-handed limit. What is the limit as x approaches one from the right-hand of g of h of x? Well, we can do the same exercise. As we approach one from the right, it looks like h is approaching two from below, from values less than two. And so if we are approaching two from below, because remember, whatever h is outputting is the input into g. So if the thing that we're inputting into g is approaching two from below, that means that g once again is going to be approaching negative two."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "What is the limit as x approaches one from the right-hand of g of h of x? Well, we can do the same exercise. As we approach one from the right, it looks like h is approaching two from below, from values less than two. And so if we are approaching two from below, because remember, whatever h is outputting is the input into g. So if the thing that we're inputting into g is approaching two from below, that means that g once again is going to be approaching negative two. So this is a really, really, really interesting case where the limit of g of x as x approaches two does not exist. But because on h of x, when we approach from both the left and the right-hand side, h is approaching two from below. We just have to think about the left-handed limit as we approach two from below or from the left on g. Because in both situations, we are approaching negative two."}, {"video_title": "Limits of composite functions external limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "And so if we are approaching two from below, because remember, whatever h is outputting is the input into g. So if the thing that we're inputting into g is approaching two from below, that means that g once again is going to be approaching negative two. So this is a really, really, really interesting case where the limit of g of x as x approaches two does not exist. But because on h of x, when we approach from both the left and the right-hand side, h is approaching two from below. We just have to think about the left-handed limit as we approach two from below or from the left on g. Because in both situations, we are approaching negative two. And so that is going to be our limit. When the left-handed and the right-hand limit are the same, that is going to be your limit. It is equal to negative two."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's what we're going to cover in this video. What you see here is a flow chart developed by the team at Khan Academy. And I'm essentially going to work through that flow chart. It looks a little bit complicated at first, but hopefully it'll make sense as we talk it through. So the goal is, hey, we want to find the limit of f of x as x approaches a. So what this is telling us to do is, well, the first thing, just try to substitute what happens when x equals a. Let's evaluate f of a."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks a little bit complicated at first, but hopefully it'll make sense as we talk it through. So the goal is, hey, we want to find the limit of f of x as x approaches a. So what this is telling us to do is, well, the first thing, just try to substitute what happens when x equals a. Let's evaluate f of a. And this flow chart says, well, if f of a is equal to a real number, it's saying we're done. But then there's this little caveat here, probably. And the reason why is that the limit is a different thing than the value of the function."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's evaluate f of a. And this flow chart says, well, if f of a is equal to a real number, it's saying we're done. But then there's this little caveat here, probably. And the reason why is that the limit is a different thing than the value of the function. Sometimes they happen to be the same. In fact, that's the definition of a continuous function, which we talk about in previous videos. But sometimes they aren't the same."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the reason why is that the limit is a different thing than the value of the function. Sometimes they happen to be the same. In fact, that's the definition of a continuous function, which we talk about in previous videos. But sometimes they aren't the same. This will not necessarily be true if you're dealing with some function that has a point discontinuity like that, or a jump discontinuity, or a function that looks like this. This would not necessarily be the case. But if at that point you're trying to find the limit towards the, if as you approach this point right over here, the function is continuous, it's behaving somewhat normally, then this is a good thing to keep in mind."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But sometimes they aren't the same. This will not necessarily be true if you're dealing with some function that has a point discontinuity like that, or a jump discontinuity, or a function that looks like this. This would not necessarily be the case. But if at that point you're trying to find the limit towards the, if as you approach this point right over here, the function is continuous, it's behaving somewhat normally, then this is a good thing to keep in mind. You could just say, hey, can I just evaluate the function at that, at that a over there? So in general, if you're dealing with pretty plain vanilla functions like an x squared, or if you're dealing with rational expressions like this, or trigonometric expressions, and if you're able to just evaluate the function, and it gives you a real number, you are probably done. If you're dealing with some type of a function that has all sorts of special cases, and it's piecewise defined, as we've seen in previous other videos, I would be a little bit more skeptical."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if at that point you're trying to find the limit towards the, if as you approach this point right over here, the function is continuous, it's behaving somewhat normally, then this is a good thing to keep in mind. You could just say, hey, can I just evaluate the function at that, at that a over there? So in general, if you're dealing with pretty plain vanilla functions like an x squared, or if you're dealing with rational expressions like this, or trigonometric expressions, and if you're able to just evaluate the function, and it gives you a real number, you are probably done. If you're dealing with some type of a function that has all sorts of special cases, and it's piecewise defined, as we've seen in previous other videos, I would be a little bit more skeptical. Or if you know visually around that point there's some type of jump, or some type of discontinuity, you've got to be a little bit more careful. But in general, this is a pretty good rule of thumb. If you're dealing with plain vanilla functions that are continuous, if you evaluate at x equals a, and you get a real number, that's probably going to be the limit."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you're dealing with some type of a function that has all sorts of special cases, and it's piecewise defined, as we've seen in previous other videos, I would be a little bit more skeptical. Or if you know visually around that point there's some type of jump, or some type of discontinuity, you've got to be a little bit more careful. But in general, this is a pretty good rule of thumb. If you're dealing with plain vanilla functions that are continuous, if you evaluate at x equals a, and you get a real number, that's probably going to be the limit. But now let's think about the other scenarios. What happens if you evaluate it, and you get some number divided by zero? Well, that case, you are probably dealing with a vertical asymptote."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you're dealing with plain vanilla functions that are continuous, if you evaluate at x equals a, and you get a real number, that's probably going to be the limit. But now let's think about the other scenarios. What happens if you evaluate it, and you get some number divided by zero? Well, that case, you are probably dealing with a vertical asymptote. And what do we mean by a vertical asymptote? Well, look at this example right over here, where you're just saying the limit, we'll do that in a darker color. So if we're talking about the limit as x approaches one, of one over x minus one."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that case, you are probably dealing with a vertical asymptote. And what do we mean by a vertical asymptote? Well, look at this example right over here, where you're just saying the limit, we'll do that in a darker color. So if we're talking about the limit as x approaches one, of one over x minus one. If you just try to evaluate this expression at x equals one, you would get one over one minus one, which is equal to one over zero, which says, okay, I'm falling into this vertical asymptote case. And at that point, if you wanted to just understand what was going on there, or even verify that it's a vertical asymptote, well, then you could try out some numbers, you could try to plot it. You could say, all right, I probably have a vertical asymptote here at x equals one."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we're talking about the limit as x approaches one, of one over x minus one. If you just try to evaluate this expression at x equals one, you would get one over one minus one, which is equal to one over zero, which says, okay, I'm falling into this vertical asymptote case. And at that point, if you wanted to just understand what was going on there, or even verify that it's a vertical asymptote, well, then you could try out some numbers, you could try to plot it. You could say, all right, I probably have a vertical asymptote here at x equals one. So that's my vertical asymptote. And you could try out some values. Well, let's see, if x is greater than one, the denominator is going to be positive."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say, all right, I probably have a vertical asymptote here at x equals one. So that's my vertical asymptote. And you could try out some values. Well, let's see, if x is greater than one, the denominator is going to be positive. And so my graph, and you would get this from trying out a bunch of values, might look something like this. And then for values less than negative one, or less than one, I should say, you're gonna get negative values. And so your graph might look something like that."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see, if x is greater than one, the denominator is going to be positive. And so my graph, and you would get this from trying out a bunch of values, might look something like this. And then for values less than negative one, or less than one, I should say, you're gonna get negative values. And so your graph might look something like that. And so you have this vertical asymptote. That's probably what you have. Now, there are cases, very special cases, where you won't necessarily have the vertical asymptote."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so your graph might look something like that. And so you have this vertical asymptote. That's probably what you have. Now, there are cases, very special cases, where you won't necessarily have the vertical asymptote. One example of that would be something like one over x minus x. This one here is actually undefined for any x you give it. So it would be very, you will not have a vertical asymptote."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, there are cases, very special cases, where you won't necessarily have the vertical asymptote. One example of that would be something like one over x minus x. This one here is actually undefined for any x you give it. So it would be very, you will not have a vertical asymptote. But this is a very special case. Most times, you do have a vertical asymptote there. But let's say we don't fall into either of those situations."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would be very, you will not have a vertical asymptote. But this is a very special case. Most times, you do have a vertical asymptote there. But let's say we don't fall into either of those situations. What if when we evaluate the function, we get zero over zero? And here is an example of that. Limit as x approaches negative one of this rational expression."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's say we don't fall into either of those situations. What if when we evaluate the function, we get zero over zero? And here is an example of that. Limit as x approaches negative one of this rational expression. And let's try to evaluate it. You get negative one squared, which is one, minus negative one, which is plus one, minus two, so you get zero in the numerator. And then in the denominator, you have negative one squared, which is one, minus two times negative one, so plus two minus three, which is equal to zero."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Limit as x approaches negative one of this rational expression. And let's try to evaluate it. You get negative one squared, which is one, minus negative one, which is plus one, minus two, so you get zero in the numerator. And then in the denominator, you have negative one squared, which is one, minus two times negative one, so plus two minus three, which is equal to zero. Now this is known as indeterminate form. And so on our flow chart, we then continue to the right side of it. And so here's a bunch of techniques for trying to tackle something in indeterminate form."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then in the denominator, you have negative one squared, which is one, minus two times negative one, so plus two minus three, which is equal to zero. Now this is known as indeterminate form. And so on our flow chart, we then continue to the right side of it. And so here's a bunch of techniques for trying to tackle something in indeterminate form. And likely in a few weeks, you will learn another technique that involves a little bit more calculus called L'Hopital's Rule that we don't tackle here because that involves calculus, while all of these techniques can be done with things before calculus, some algebraic techniques, and some trigonometric techniques. So the first thing that you might wanna try to do, especially if you're dealing with a rational expression like this and you're getting indeterminate form, is try to factor it. Try to see if you can simplify this expression."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so here's a bunch of techniques for trying to tackle something in indeterminate form. And likely in a few weeks, you will learn another technique that involves a little bit more calculus called L'Hopital's Rule that we don't tackle here because that involves calculus, while all of these techniques can be done with things before calculus, some algebraic techniques, and some trigonometric techniques. So the first thing that you might wanna try to do, especially if you're dealing with a rational expression like this and you're getting indeterminate form, is try to factor it. Try to see if you can simplify this expression. And this expression here, you can factor it. This is the same thing as x, let's see, x minus two times x plus one over, let's see, x, well, this would be x minus three times x plus one. If what I just did seems completely foreign to you, I encourage you to watch the videos on factoring polynomials or factoring quadratics."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Try to see if you can simplify this expression. And this expression here, you can factor it. This is the same thing as x, let's see, x minus two times x plus one over, let's see, x, well, this would be x minus three times x plus one. If what I just did seems completely foreign to you, I encourage you to watch the videos on factoring polynomials or factoring quadratics. And so you can see here, all right, look, if I make the, I can simplify this because as long as x does not equal negative one, these two things are going to cancel out. So I can say that this is going to be equal to x minus two over x minus three for x does not equal negative one. Sometimes people forget to do this part."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If what I just did seems completely foreign to you, I encourage you to watch the videos on factoring polynomials or factoring quadratics. And so you can see here, all right, look, if I make the, I can simplify this because as long as x does not equal negative one, these two things are going to cancel out. So I can say that this is going to be equal to x minus two over x minus three for x does not equal negative one. Sometimes people forget to do this part. This is if you're really being mathematically precise. This entire expression is the same as this one because this entire expression is still not defined at x equals negative one, although you can substitute x equals negative one here and now get a value. So if you substitute x equals negative one here, even if it's formally, if we're formally taking it away to be mathematically equivalent, this would be negative one minus two, which would be negative three, over negative one minus three, which would be negative four, which is equal to 3 4ths."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sometimes people forget to do this part. This is if you're really being mathematically precise. This entire expression is the same as this one because this entire expression is still not defined at x equals negative one, although you can substitute x equals negative one here and now get a value. So if you substitute x equals negative one here, even if it's formally, if we're formally taking it away to be mathematically equivalent, this would be negative one minus two, which would be negative three, over negative one minus three, which would be negative four, which is equal to 3 4ths. So if this condition wasn't here, you could just evaluate it straight up and this is a pretty plain vanilla function. Wouldn't expect to see anything crazy happening here. And if I can just evaluate it at x equals negative one, I feel pretty good, I feel pretty good."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you substitute x equals negative one here, even if it's formally, if we're formally taking it away to be mathematically equivalent, this would be negative one minus two, which would be negative three, over negative one minus three, which would be negative four, which is equal to 3 4ths. So if this condition wasn't here, you could just evaluate it straight up and this is a pretty plain vanilla function. Wouldn't expect to see anything crazy happening here. And if I can just evaluate it at x equals negative one, I feel pretty good, I feel pretty good. So once again, we're now going and factoring. We're able to factor. We evaluate, we simplify it."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if I can just evaluate it at x equals negative one, I feel pretty good, I feel pretty good. So once again, we're now going and factoring. We're able to factor. We evaluate, we simplify it. We evaluate the expression, the simplified expression now and now we were able to get a value. We were able to get 3 4ths and so we can feel pretty good that the limit here in this situation is 3 4ths. Now let's, and I would categorize what we've seen so far as the bulk of the limit exercises that you will likely encounter."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We evaluate, we simplify it. We evaluate the expression, the simplified expression now and now we were able to get a value. We were able to get 3 4ths and so we can feel pretty good that the limit here in this situation is 3 4ths. Now let's, and I would categorize what we've seen so far as the bulk of the limit exercises that you will likely encounter. Now the next two, I would call slightly fancier techniques. So if you get indeterminate form, especially you'll sometimes see it with radical expressions like this, rational radical expressions, you might want to multiply by a conjugate. So for example, in this situation right here, if you just tried to evaluate it at x equals four, you get the square root of four minus two over four minus four, which is zero over zero."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's, and I would categorize what we've seen so far as the bulk of the limit exercises that you will likely encounter. Now the next two, I would call slightly fancier techniques. So if you get indeterminate form, especially you'll sometimes see it with radical expressions like this, rational radical expressions, you might want to multiply by a conjugate. So for example, in this situation right here, if you just tried to evaluate it at x equals four, you get the square root of four minus two over four minus four, which is zero over zero. So it's that indeterminate form and the technique here, because we're seeing this radical and irrational expression, is hey, let's maybe, maybe we can somehow get rid of that radical or simplify it somehow. So let me rewrite it. Square root of x minus two over x minus four."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, in this situation right here, if you just tried to evaluate it at x equals four, you get the square root of four minus two over four minus four, which is zero over zero. So it's that indeterminate form and the technique here, because we're seeing this radical and irrational expression, is hey, let's maybe, maybe we can somehow get rid of that radical or simplify it somehow. So let me rewrite it. Square root of x minus two over x minus four. When we say conjugate, let's multiply it by the square root of x plus two over the square root of x plus two. Once again, it's the same expression over the same expression, so I'm not fundamentally changing its value. And so this is going to be equal to, well if I have a plus b times a minus b, I'm gonna get a difference of squares."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Square root of x minus two over x minus four. When we say conjugate, let's multiply it by the square root of x plus two over the square root of x plus two. Once again, it's the same expression over the same expression, so I'm not fundamentally changing its value. And so this is going to be equal to, well if I have a plus b times a minus b, I'm gonna get a difference of squares. So it's gonna be square root of x squared, which is, let me just write it, it's gonna be square root of x squared minus four over, well square root of x squared is just going to be x minus four. So let me rewrite it that way. So it's x minus four over x minus four times square root of x plus two."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to, well if I have a plus b times a minus b, I'm gonna get a difference of squares. So it's gonna be square root of x squared, which is, let me just write it, it's gonna be square root of x squared minus four over, well square root of x squared is just going to be x minus four. So let me rewrite it that way. So it's x minus four over x minus four times square root of x plus two. Square root of x plus two. Well this was useful because now I can cancel out x equals four, or x minus four, right over here. And once again, if I wanted it mathematically to be the exact same expression, I would say well, now this is going to be equal to one over the square root of x plus two for x does not equal four."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's x minus four over x minus four times square root of x plus two. Square root of x plus two. Well this was useful because now I can cancel out x equals four, or x minus four, right over here. And once again, if I wanted it mathematically to be the exact same expression, I would say well, now this is going to be equal to one over the square root of x plus two for x does not equal four. But we can definitely see what this function is approaching if we just now substitute x equals four into this simplified expression. And so that's just gonna be one over, so if we just substitute, if we just substitute x equals four here, you get one over square root of four plus two, which is equal to 1 4th. And once again, you can feel pretty good that this is going to be your limit."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, if I wanted it mathematically to be the exact same expression, I would say well, now this is going to be equal to one over the square root of x plus two for x does not equal four. But we can definitely see what this function is approaching if we just now substitute x equals four into this simplified expression. And so that's just gonna be one over, so if we just substitute, if we just substitute x equals four here, you get one over square root of four plus two, which is equal to 1 4th. And once again, you can feel pretty good that this is going to be your limit. We've gone back into the green zone. If you were to actually plot this original function, you would have a point discontinuity, you would have a gap at x equals four. But then when you do that simplification and factoring out that x minus, or canceling out that x minus four, that gap would disappear."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, you can feel pretty good that this is going to be your limit. We've gone back into the green zone. If you were to actually plot this original function, you would have a point discontinuity, you would have a gap at x equals four. But then when you do that simplification and factoring out that x minus, or canceling out that x minus four, that gap would disappear. And so that's essentially what you're doing. You're trying to find the limit as we approach that gap, which we got right there. Now this final one, this is dealing with trig identities."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then when you do that simplification and factoring out that x minus, or canceling out that x minus four, that gap would disappear. And so that's essentially what you're doing. You're trying to find the limit as we approach that gap, which we got right there. Now this final one, this is dealing with trig identities. And in order to do these, you have to be pretty adept at your trig identities. So if we're saying the limit as, let me do that in a darker color. So if we're saying the limit as x approaches zero of sine of x over sine of two x, well, sine of zero zero, sine of zero zero, you're gonna get zero over zero."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this final one, this is dealing with trig identities. And in order to do these, you have to be pretty adept at your trig identities. So if we're saying the limit as, let me do that in a darker color. So if we're saying the limit as x approaches zero of sine of x over sine of two x, well, sine of zero zero, sine of zero zero, you're gonna get zero over zero. Once again, indeterminate form, we fall into this category. And now you might recognize this is going to be equal to the limit as x approaches zero of sine of x. We can rewrite sine of two x as two sine x cosine x."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we're saying the limit as x approaches zero of sine of x over sine of two x, well, sine of zero zero, sine of zero zero, you're gonna get zero over zero. Once again, indeterminate form, we fall into this category. And now you might recognize this is going to be equal to the limit as x approaches zero of sine of x. We can rewrite sine of two x as two sine x cosine x. And then those two can cancel out for all x's not equaling, for all x's not equaling zero if you wanna be really mathematical precise. And so there would have been a gap there for sure on the original graph if you were to graph y equals this. But now for the limit purposes, you could say this limit is going to be the limit as x approaches zero of one over two cosine of x."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can rewrite sine of two x as two sine x cosine x. And then those two can cancel out for all x's not equaling, for all x's not equaling zero if you wanna be really mathematical precise. And so there would have been a gap there for sure on the original graph if you were to graph y equals this. But now for the limit purposes, you could say this limit is going to be the limit as x approaches zero of one over two cosine of x. And now we can go back to this green condition right over here, because we can evaluate this at x equals zero. It's gonna be one over two times cosine of zero. Cosine of zero is one."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now for the limit purposes, you could say this limit is going to be the limit as x approaches zero of one over two cosine of x. And now we can go back to this green condition right over here, because we can evaluate this at x equals zero. It's gonna be one over two times cosine of zero. Cosine of zero is one. So this is going to be equal to 1 1\u20442. Now in general, none of these techniques work. And you'll encounter a few other techniques further on once you learn more calculus."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of zero is one. So this is going to be equal to 1 1\u20442. Now in general, none of these techniques work. And you'll encounter a few other techniques further on once you learn more calculus. Then you fall on the baseline, approximation. And approximation, you can do it numerically. Try values really, really, really, really close to the number you're trying to find the limit on."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you'll encounter a few other techniques further on once you learn more calculus. Then you fall on the baseline, approximation. And approximation, you can do it numerically. Try values really, really, really, really close to the number you're trying to find the limit on. You know, if you're trying to find the limit as x approaches zero, try 0.00000000001. Try negative 0.0000001. If you're trying to find the limit as x approaches four, try 4.0000001."}, {"video_title": "Strategy in finding limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Try values really, really, really, really close to the number you're trying to find the limit on. You know, if you're trying to find the limit as x approaches zero, try 0.00000000001. Try negative 0.0000001. If you're trying to find the limit as x approaches four, try 4.0000001. Try 3.999999999999. And see what happens. But that's kind of the last ditch, the last ditch effort."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And just from an inspection, you can see that the function is not defined when x is equal to 3. You get 0 over 0. It's not defined. So to answer this question, let's try to rewrite the same exact function definition slightly differently. So let's say f of x is going to be equal to, and I'm going to think of two cases. I'm going to think of the case when x is greater than 3 and when x is less than 3. So when x is, we'll do this in two different colors actually."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So to answer this question, let's try to rewrite the same exact function definition slightly differently. So let's say f of x is going to be equal to, and I'm going to think of two cases. I'm going to think of the case when x is greater than 3 and when x is less than 3. So when x is, we'll do this in two different colors actually. When x, do it in green, that's not green. When x is greater than 3, what does this function simplify to? Well, whatever I get up here, I'm going to get a positive value up here."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So when x is, we'll do this in two different colors actually. When x, do it in green, that's not green. When x is greater than 3, what does this function simplify to? Well, whatever I get up here, I'm going to get a positive value up here. And then if I take the absolute value, it's going to be the exact same thing. So for x is greater than 3, this is going to be the exact same thing as x minus 3 over x minus 3. Because if x is greater than 3, the numerator is going to be positive."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Well, whatever I get up here, I'm going to get a positive value up here. And then if I take the absolute value, it's going to be the exact same thing. So for x is greater than 3, this is going to be the exact same thing as x minus 3 over x minus 3. Because if x is greater than 3, the numerator is going to be positive. You take the absolute value of that, you're not going to change its value. So you get this right over here. Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Because if x is greater than 3, the numerator is going to be positive. You take the absolute value of that, you're not going to change its value. So you get this right over here. Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3. Similarly, let's think about what happens when x is less than 3. When x is less than 3, well, x minus 3 is going to be a negative number. When you take the absolute value of that, you're essentially negating it."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3. Similarly, let's think about what happens when x is less than 3. When x is less than 3, well, x minus 3 is going to be a negative number. When you take the absolute value of that, you're essentially negating it. So it's going to be the negative of x minus 3 over x minus 3. Or if you were to simplify these two things, for any value as long as x does not equal 3, this part right over here simplifies to 1, so you're left with a negative 1. Negative 1 for x is less than 3."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "When you take the absolute value of that, you're essentially negating it. So it's going to be the negative of x minus 3 over x minus 3. Or if you were to simplify these two things, for any value as long as x does not equal 3, this part right over here simplifies to 1, so you're left with a negative 1. Negative 1 for x is less than 3. I encourage you, if you don't believe what I just said, try it out with some numbers. Try out some numbers, 3.1, 3.001, 3.5, 4, 7, any number greater than 3, you're going to get 1. You're going to get the same thing divided by the same thing."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Negative 1 for x is less than 3. I encourage you, if you don't believe what I just said, try it out with some numbers. Try out some numbers, 3.1, 3.001, 3.5, 4, 7, any number greater than 3, you're going to get 1. You're going to get the same thing divided by the same thing. And try values for x less than 3. You're going to get negative 1 no matter what you try. So let's visualize this function now."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "You're going to get the same thing divided by the same thing. And try values for x less than 3. You're going to get negative 1 no matter what you try. So let's visualize this function now. So now let me draw some axes. That's my x-axis. And then this is my f of x-axis."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So let's visualize this function now. So now let me draw some axes. That's my x-axis. And then this is my f of x-axis. So y is equal to f of x. And what we care about is x is equal to 3. So x is equal to 1, 2, 3, 4, 5."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And then this is my f of x-axis. So y is equal to f of x. And what we care about is x is equal to 3. So x is equal to 1, 2, 3, 4, 5. We could keep going. And let's say this is positive 1, 2. So that's y is equal to 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So x is equal to 1, 2, 3, 4, 5. We could keep going. And let's say this is positive 1, 2. So that's y is equal to 1. This is y is equal to negative 1 and negative 2, and we can keep going. So this way that we've rewritten the function is the exact same function as this. We've just written it a different way."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So that's y is equal to 1. This is y is equal to negative 1 and negative 2, and we can keep going. So this way that we've rewritten the function is the exact same function as this. We've just written it a different way. And so what we're saying is our function is undefined at 3. But if our x's are greater than 3, our function is equal to 1. So if our x is greater than 3, our function is equal to 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "We've just written it a different way. And so what we're saying is our function is undefined at 3. But if our x's are greater than 3, our function is equal to 1. So if our x is greater than 3, our function is equal to 1. So it looks like that. And it's undefined at 3. And if x is less than 3, our function is equal to negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So if our x is greater than 3, our function is equal to 1. So it looks like that. And it's undefined at 3. And if x is less than 3, our function is equal to negative 1. So it looks like that. Let me do it in that same color. It looks like this."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And if x is less than 3, our function is equal to negative 1. So it looks like that. Let me do it in that same color. It looks like this. Once again, it's undefined at 3. So it looks like that. So now let's try to answer our question."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "It looks like this. Once again, it's undefined at 3. So it looks like that. So now let's try to answer our question. What is the limit as x approaches 3? Well, let's think about the limit as x approaches 3 from the negative direction, from values less than 3. So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So now let's try to answer our question. What is the limit as x approaches 3? Well, let's think about the limit as x approaches 3 from the negative direction, from values less than 3. So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction. And all this notation here, I wrote this negative as a superscript right after the 3 says. So let's think about the limit as we're approaching from the left. So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction. And all this notation here, I wrote this negative as a superscript right after the 3 says. So let's think about the limit as we're approaching from the left. So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1. We go to 1, f of x is equal to negative 1. We go to 2, f of x is equal to negative 1. If you go to 2.999999, f of x is equal to negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1. We go to 1, f of x is equal to negative 1. We go to 2, f of x is equal to negative 1. If you go to 2.999999, f of x is equal to negative 1. So it looks like it is approaching negative 1 if you approach from the left-hand side. Now let's think about the limit of f of x as x approaches 3 from the positive direction, from values greater than 3. So here we see when x is equal to 5, f of x is equal to 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "If you go to 2.999999, f of x is equal to negative 1. So it looks like it is approaching negative 1 if you approach from the left-hand side. Now let's think about the limit of f of x as x approaches 3 from the positive direction, from values greater than 3. So here we see when x is equal to 5, f of x is equal to 1. When x is equal to 4, f of x is equal to 1. When x is equal to 3.0000001, f of x is equal to 1. So it seems to be approaching positive 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So here we see when x is equal to 5, f of x is equal to 1. When x is equal to 4, f of x is equal to 1. When x is equal to 3.0000001, f of x is equal to 1. So it seems to be approaching positive 1. So now we have something strange. We seem to be approaching a different value when we approach from the left than when we approach from the right. And if we're approaching two different values, then the limit does not exist."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So it seems to be approaching positive 1. So now we have something strange. We seem to be approaching a different value when we approach from the left than when we approach from the right. And if we're approaching two different values, then the limit does not exist. So this limit right over here does not exist. Or another way of saying it, the limit of a function f of x as x approaches some value c is equal to L if and only if the limit of f of x as x approaches c from the negative direction is equal to the limit of f of x as x approaches c from the positive direction, which is equal to L. This did not happen here. The limit when we approached from the left was negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And if we're approaching two different values, then the limit does not exist. So this limit right over here does not exist. Or another way of saying it, the limit of a function f of x as x approaches some value c is equal to L if and only if the limit of f of x as x approaches c from the negative direction is equal to the limit of f of x as x approaches c from the positive direction, which is equal to L. This did not happen here. The limit when we approached from the left was negative 1. The limit when we approached from the right was positive 1. So we did not get the same limits when we approached from either side. So the limit does not exist in this case."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm going to do it over various intervals. So first, let's think about the area under the curve between x is equal to 0 and x is equal to pi over 2. So we're talking about this area right over here. Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And all this is is kind of reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles. And then you take an infinite number of those infinitely thin rectangles. And that's kind of what this notation is trying to depict."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And all this is is kind of reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles. And then you take an infinite number of those infinitely thin rectangles. And that's kind of what this notation is trying to depict. But we know how to do this already. The second fundamental theorem of calculus helps us. We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's kind of what this notation is trying to depict. But we know how to do this already. The second fundamental theorem of calculus helps us. We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2. And from that, subtract it evaluated at 0. So what's the antiderivative of cosine of x? Or what's an antiderivative?"}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2. And from that, subtract it evaluated at 0. So what's the antiderivative of cosine of x? Or what's an antiderivative? Well, we know if we take the derivative, let me write this up here. We know that if we take the derivative of sine of x, we get cosine of x. So the antiderivative of cosine of x is sine of x."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or what's an antiderivative? Well, we know if we take the derivative, let me write this up here. We know that if we take the derivative of sine of x, we get cosine of x. So the antiderivative of cosine of x is sine of x. Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the antiderivative of cosine of x is sine of x. Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x. Because the derivative of a constant is 0. This could be pi. This could be 5."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x. Because the derivative of a constant is 0. This could be pi. This could be 5. This could be a million. This could be a googol. This could be any crazy number."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "This could be 5. This could be a million. This could be a googol. This could be any crazy number. But the derivative of this is still going to be cosine x. So when I say that we have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, because in this case, the constant is 0."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "This could be any crazy number. But the derivative of this is still going to be cosine x. So when I say that we have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, because in this case, the constant is 0. So let's evaluate. So one way that we can denote this, the antiderivative of cosine of x, or an antiderivative of cosine of x, is sine of x. And we're going to evaluate it at pi over 2."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine of x is probably the simplest, because in this case, the constant is 0. So let's evaluate. So one way that we can denote this, the antiderivative of cosine of x, or an antiderivative of cosine of x, is sine of x. And we're going to evaluate it at pi over 2. And from that, subtract it, evaluate it at 0. So this is going to be equal to sine of pi over 2 minus sine of 0, which is equal to sine of pi over 2 is 1. Sine of 0 is 0."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to evaluate it at pi over 2. And from that, subtract it, evaluate it at 0. So this is going to be equal to sine of pi over 2 minus sine of 0, which is equal to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. Now let's do something interesting."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. Now let's do something interesting. Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's do something interesting. Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area. We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're talking about this area. We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi over 2. So this is going to be equal to sine of 3 pi over 2 minus sine of pi over 2. What's sine of 3 pi over 2?"}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi over 2. So this is going to be equal to sine of 3 pi over 2 minus sine of pi over 2. What's sine of 3 pi over 2? If you visualize the unit circle real fast, 3 pi over 2 is going all the way 3 4ths around the unit circle. So it's right over there. So the sine is the y-coordinate on that unit circle."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's sine of 3 pi over 2? If you visualize the unit circle real fast, 3 pi over 2 is going all the way 3 4ths around the unit circle. So it's right over there. So the sine is the y-coordinate on that unit circle. So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2, pi over 2 is just going straight up like that."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the sine is the y-coordinate on that unit circle. So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2, pi over 2 is just going straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here, sine of pi over 2, pi over 2 is just going straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here. We've got a negative area. Now, how does that make sense?"}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here. We've got a negative area. Now, how does that make sense? We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is below the x-axis."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, how does that make sense? We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think that we have an area of 2, but it's all below the x-axis in this case. And so it is signed as negative 2. The actual area is 2, but since it's below the x-axis, we get a negative right over here."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think that we have an area of 2, but it's all below the x-axis in this case. And so it is signed as negative 2. The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let's do one more interesting one. Let's find the definite integral from 0 to 3 pi over 2 of cosine of x dx. Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let's do one more interesting one. Let's find the definite integral from 0 to 3 pi over 2 of cosine of x dx. Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2. And what do you think is going to happen? Well, let's evaluate it. It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2. And what do you think is going to happen? Well, let's evaluate it. It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under all this orange area that I just outlined is clearly not negative, or any area isn't a negative number. And the area isn't even 1."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under all this orange area that I just outlined is clearly not negative, or any area isn't a negative number. And the area isn't even 1. But what just happened here? Well, we saw in the first case that this first area was 1. The area of this first region right over here is 1."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the area isn't even 1. But what just happened here? Well, we saw in the first case that this first area was 1. The area of this first region right over here is 1. And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1. Or another way to say it, your net area is negative 1."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "The area of this first region right over here is 1. And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1. Or another way to say it, your net area is negative 1. So it's taking the one region above the x-axis and then subtracting the two below it. So what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus, it's essentially finding the net area above the x-axis. And if we get a negative number, that means that the net area is that actually most of the area is below the x-axis."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or another way to say it, your net area is negative 1. So it's taking the one region above the x-axis and then subtracting the two below it. So what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus, it's essentially finding the net area above the x-axis. And if we get a negative number, that means that the net area is that actually most of the area is below the x-axis. If we get 0, then that means it all nets out. And if you want to see a case at 0, take the integral from 0 all the way to 2 pi. And this will evaluate to 0 because you have an area of 1 and another area of 1."}, {"video_title": "Area between a curve and the x-axis negative area   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we get a negative number, that means that the net area is that actually most of the area is below the x-axis. If we get 0, then that means it all nets out. And if you want to see a case at 0, take the integral from 0 all the way to 2 pi. And this will evaluate to 0 because you have an area of 1 and another area of 1. But it nets out with this area of negative 2. Let's try that out. So if we go from 0 to 2 pi of cosine of x dx, this is going to be equal to sine of 2 pi minus sine of 0, which is equal to 0 minus 0, which is indeed equal to 0."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are now going to cover the famous or perhaps infamous potato problem from the 2017 AP Calculus exam. At time t equals zero, a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is 91 degrees Celsius at time t equals zero, and the internal temperature of the potato is greater than 27 degrees for all times t greater than zero. I would guess that the ambient room temperature is 27 degrees Celsius, and so that's why the temperature would approach this, but it would always stay a little bit greater than that as t gets larger and larger. The internal temperature of the potato at time t minutes can be modeled by the function h that satisfies the differential equation. dh dt, the derivative of our internal temperature with respect to time, is equal to negative 1 4th times r, the difference between our internal temperature and the ambient room temperature, where h of t is measured in degrees Celsius and h of zero is equal to 91. So before I even read part a, let's just make sense of what this differential equation is telling us, and let's see if it's consistent with our intuition."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I would guess that the ambient room temperature is 27 degrees Celsius, and so that's why the temperature would approach this, but it would always stay a little bit greater than that as t gets larger and larger. The internal temperature of the potato at time t minutes can be modeled by the function h that satisfies the differential equation. dh dt, the derivative of our internal temperature with respect to time, is equal to negative 1 4th times r, the difference between our internal temperature and the ambient room temperature, where h of t is measured in degrees Celsius and h of zero is equal to 91. So before I even read part a, let's just make sense of what this differential equation is telling us, and let's see if it's consistent with our intuition. So let me draw some axes here. So this is my y-axis, so that is my y-axis, and this right over here is my t-axis. Now if the ambient room temperature is 27 degrees Celsius, so I'll just draw there, that's what the room temperature is doing, and so we know at t equals zero, our potato is at 91 degrees, so let's see, that's 27, 91 might be right over there."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So before I even read part a, let's just make sense of what this differential equation is telling us, and let's see if it's consistent with our intuition. So let me draw some axes here. So this is my y-axis, so that is my y-axis, and this right over here is my t-axis. Now if the ambient room temperature is 27 degrees Celsius, so I'll just draw there, that's what the room temperature is doing, and so we know at t equals zero, our potato is at 91 degrees, so let's see, that's 27, 91 might be right over there. 91, this is all in degrees Celsius, and what you would expect intuitively is that it would start to cool, and when it's temperature, when there's a big difference between the potato and the room, maybe its rate of change is steeper than when there's a little difference. So you would expect the graph to look something like this. So you would expect it to look something like this, and then asymptote towards a temperature of 27 degrees Celsius."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now if the ambient room temperature is 27 degrees Celsius, so I'll just draw there, that's what the room temperature is doing, and so we know at t equals zero, our potato is at 91 degrees, so let's see, that's 27, 91 might be right over there. 91, this is all in degrees Celsius, and what you would expect intuitively is that it would start to cool, and when it's temperature, when there's a big difference between the potato and the room, maybe its rate of change is steeper than when there's a little difference. So you would expect the graph to look something like this. So you would expect it to look something like this, and then asymptote towards a temperature of 27 degrees Celsius. So this is what you would expect to see, and this differential equation is consistent with that, where the rate of change, notice, this is for all t greater than zero, this is going to be a negative value, because our potato is greater than 27 for t greater than zero so this part here is going to be positive, but then you multiply positive times negative 1 4th, you're constantly going to have a negative rate of change, which makes sense, the potato is cooling down. And it also makes sense that your rate of change is proportional to the difference between the temperature of the potato and the ambient room temperature. When there's a big difference, you expect a steeper rate of change, but then when there's less of a difference, the rate of change, you can imagine, becomes less and less and less negative as we asymptote towards the ambient temperature."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you would expect it to look something like this, and then asymptote towards a temperature of 27 degrees Celsius. So this is what you would expect to see, and this differential equation is consistent with that, where the rate of change, notice, this is for all t greater than zero, this is going to be a negative value, because our potato is greater than 27 for t greater than zero so this part here is going to be positive, but then you multiply positive times negative 1 4th, you're constantly going to have a negative rate of change, which makes sense, the potato is cooling down. And it also makes sense that your rate of change is proportional to the difference between the temperature of the potato and the ambient room temperature. When there's a big difference, you expect a steeper rate of change, but then when there's less of a difference, the rate of change, you can imagine, becomes less and less and less negative as we asymptote towards the ambient temperature. So with this out of the way, now let's tackle part A. Write an equation for the line tangent to the graph of h at t equals zero. Use this equation to approximate the internal temperature of the potato at time t equals three."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "When there's a big difference, you expect a steeper rate of change, but then when there's less of a difference, the rate of change, you can imagine, becomes less and less and less negative as we asymptote towards the ambient temperature. So with this out of the way, now let's tackle part A. Write an equation for the line tangent to the graph of h at t equals zero. Use this equation to approximate the internal temperature of the potato at time t equals three. So what are we going to do? Well, we're going to think about what's going on at time t equals zero, right over here. We want the equation of the tangent line, which might look something like this, at t equals zero."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Use this equation to approximate the internal temperature of the potato at time t equals three. So what are we going to do? Well, we're going to think about what's going on at time t equals zero, right over here. We want the equation of the tangent line, which might look something like this, at t equals zero. So this thing would be of the form y is equal to the slope of the equation of the tangent line. Well, it would be the derivative of our function at that point, so dh dt, times t, plus our y-intercept. Where does it intersect the y-axis here?"}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We want the equation of the tangent line, which might look something like this, at t equals zero. So this thing would be of the form y is equal to the slope of the equation of the tangent line. Well, it would be the derivative of our function at that point, so dh dt, times t, plus our y-intercept. Where does it intersect the y-axis here? Well, when t is equal to zero, the value of this equation is going to be 91, because it intersects our graph right at that point, that point zero comma 91, plus 91. So what is our derivative of h with respect to t at time t equals zero, right at this point right over here? Well, we just have to look at this."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Where does it intersect the y-axis here? Well, when t is equal to zero, the value of this equation is going to be 91, because it intersects our graph right at that point, that point zero comma 91, plus 91. So what is our derivative of h with respect to t at time t equals zero, right at this point right over here? Well, we just have to look at this. You could also write this as h prime of t right over here. So if we want to think about h prime of zero, that's going to be equal to negative 1 1\u2074 times h of zero minus 27. What is our initial temperature minus 27?"}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we just have to look at this. You could also write this as h prime of t right over here. So if we want to think about h prime of zero, that's going to be equal to negative 1 1\u2074 times h of zero minus 27. What is our initial temperature minus 27? This is, of course, 91 degrees. They tell us that multiple times. We've even drawn it a few times."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is our initial temperature minus 27? This is, of course, 91 degrees. They tell us that multiple times. We've even drawn it a few times. 91 minus 27 is 64. 64 times negative 1\u2074 is equal to negative 16. So this is negative 16 right over here."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've even drawn it a few times. 91 minus 27 is 64. 64 times negative 1\u2074 is equal to negative 16. So this is negative 16 right over here. So just like that, we have the equation for the line tangent to the graph of h at t is equal to zero. I'll write it one more time. It is, we've got a mini drum roll here, y is equal to negative 16t plus 91."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is negative 16 right over here. So just like that, we have the equation for the line tangent to the graph of h at t is equal to zero. I'll write it one more time. It is, we've got a mini drum roll here, y is equal to negative 16t plus 91. That's the equation of that tangent line right over there. And then they say, we want to use this equation to approximate the internal temperature of the potato at time t equals three. So let's say that this is time t equals three right over here."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is, we've got a mini drum roll here, y is equal to negative 16t plus 91. That's the equation of that tangent line right over there. And then they say, we want to use this equation to approximate the internal temperature of the potato at time t equals three. So let's say that this is time t equals three right over here. We want to approximate the temperature that this model describes right over here, but we're gonna do it using the line. So we're gonna evaluate the line at t equals three. So then we would get, let's see, negative 16 times three plus 91 is equal to, this is negative 48 plus 91 is equal to, what is that, 43."}, {"video_title": "2017 AP Calculus AB BC 4a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that this is time t equals three right over here. We want to approximate the temperature that this model describes right over here, but we're gonna do it using the line. So we're gonna evaluate the line at t equals three. So then we would get, let's see, negative 16 times three plus 91 is equal to, this is negative 48 plus 91 is equal to, what is that, 43. So this is equal to 43 degrees Celsius. So this right over here is the equation for the line tangent to the graph of h at t is equal to zero. And this right over here is our approximation using that equation of the tangent line of the internal temperature of the potato at time t is equal to three."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so like always, pause this video and see if you can figure it out. All right, now the way I'm going to tackle it is I'm gonna look at each of these graphs and try to think what would their derivatives look like? So for this first one, we can see our derivative right over here, the slope of our tangent line. It would be a little bit negative, and then it gets more and more and more negative. And as we approach this vertical asymptote right over here, it looks like it's approaching negative infinity. So the derivative would actually, over here, it would be a little bit less than zero, but then it would get more and more and more negative, and then it would approach negative infinity. So it would have a similar shape, general shape, to the graph itself, at least to the left of this vertical asymptote."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would be a little bit negative, and then it gets more and more and more negative. And as we approach this vertical asymptote right over here, it looks like it's approaching negative infinity. So the derivative would actually, over here, it would be a little bit less than zero, but then it would get more and more and more negative, and then it would approach negative infinity. So it would have a similar shape, general shape, to the graph itself, at least to the left of this vertical asymptote. Now what about to the right of the vertical asymptote? Right to the right of the vertical asymptote, it looks like the slope of the tangent line is very negative, it's very negative, but then it becomes less and less and less, less and less and less negative, and it looks like it is approaching, it is approaching zero. So on this side, the derivative starts out super negative, and then it looks like it is, the derivative is going to asymptote towards zero, something like that."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would have a similar shape, general shape, to the graph itself, at least to the left of this vertical asymptote. Now what about to the right of the vertical asymptote? Right to the right of the vertical asymptote, it looks like the slope of the tangent line is very negative, it's very negative, but then it becomes less and less and less, less and less and less negative, and it looks like it is approaching, it is approaching zero. So on this side, the derivative starts out super negative, and then it looks like it is, the derivative is going to asymptote towards zero, something like that. So based on what we just, it actually looks like, so based on what I just sketched, it just looks like this right graph is a good candidate for the derivative of this left graph. You might say, what's wrong with this blue graph? Well, this blue graph out here, notice it's positive."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on this side, the derivative starts out super negative, and then it looks like it is, the derivative is going to asymptote towards zero, something like that. So based on what we just, it actually looks like, so based on what I just sketched, it just looks like this right graph is a good candidate for the derivative of this left graph. You might say, what's wrong with this blue graph? Well, this blue graph out here, notice it's positive. So if this were the derivative of the left graph, then that means that the left graph would need a positive slope out here, but it doesn't have a positive slope. It's a slightly negative slope becoming super negative. And so right here, we're slightly negative, and then we become very negative."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this blue graph out here, notice it's positive. So if this were the derivative of the left graph, then that means that the left graph would need a positive slope out here, but it doesn't have a positive slope. It's a slightly negative slope becoming super negative. And so right here, we're slightly negative, and then we become very negative. And so maybe this is, let's call, let's call this F, and maybe this is F prime. This is F prime right over here. And now let's look at this middle graph."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so right here, we're slightly negative, and then we become very negative. And so maybe this is, let's call, let's call this F, and maybe this is F prime. This is F prime right over here. And now let's look at this middle graph. What would its derivative do? So over here, our slope is slightly negative, and then it becomes more and more and more and more and more negative. And so the derivative of this might look like it has to be slightly negative, but then it gets more and more and more and more negative as we approach that vertical asymptote."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's look at this middle graph. What would its derivative do? So over here, our slope is slightly negative, and then it becomes more and more and more and more and more negative. And so the derivative of this might look like it has to be slightly negative, but then it gets more and more and more and more negative as we approach that vertical asymptote. And in the right side of the vertical asymptote, our derivative is very positive here, and then it gets less and less and less and less and less and less positive. And so we start, our derivative would be very positive, and then it would get less and less and less and less positive. It looks like it might, the slope here might be asymptoting towards zero."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the derivative of this might look like it has to be slightly negative, but then it gets more and more and more and more negative as we approach that vertical asymptote. And in the right side of the vertical asymptote, our derivative is very positive here, and then it gets less and less and less and less and less and less positive. And so we start, our derivative would be very positive, and then it would get less and less and less and less positive. It looks like it might, the slope here might be asymptoting towards zero. So our graph might look something like that. Well, the left graph right here looks a lot like what I just sketched out as a candidate derivative for this blue graph, for this middle graph. And so I would say that this is f, then this is the derivative of that, which would make it f prime."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like it might, the slope here might be asymptoting towards zero. So our graph might look something like that. Well, the left graph right here looks a lot like what I just sketched out as a candidate derivative for this blue graph, for this middle graph. And so I would say that this is f, then this is the derivative of that, which would make it f prime. And then we already established that this right graph is the derivative of the left one. So if it's the derivative of f prime, it's not f prime itself, it's the second derivative. So I feel pretty good about that."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so I would say that this is f, then this is the derivative of that, which would make it f prime. And then we already established that this right graph is the derivative of the left one. So if it's the derivative of f prime, it's not f prime itself, it's the second derivative. So I feel pretty good about that. And just for good measure, we could think about what the derivative of this graph would look like. Here the slope is slightly negative, but then it gets more and more and more and more and more negative. So the derivative would have a similar shape here."}, {"video_title": "Connecting f, f', and f'' graphically (another example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I feel pretty good about that. And just for good measure, we could think about what the derivative of this graph would look like. Here the slope is slightly negative, but then it gets more and more and more and more and more negative. So the derivative would have a similar shape here. And then here, our derivative would be very positive, and it gets less and less and less and less and less positive. And so we start very positive, and then it gets less and less and less and less and less positive. And so as a general shape, it actually does look a lot like this first graph."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is introduce ourselves to the notion of a definite integral. And with indefinite integrals and derivatives, this is really one of the pillars of calculus. And as we'll see, they are all related, and we'll see that more and more in future videos, and we'll also get a better appreciation for even where the notation of a definite integral comes from. So let me draw some functions here. And we're actually gonna start thinking about areas under curves. So let me draw coordinate axes here. So that's my y-axis."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw some functions here. And we're actually gonna start thinking about areas under curves. So let me draw coordinate axes here. So that's my y-axis. This is my x-axis. Actually, I'm gonna do two cases. So this is my y-axis."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's my y-axis. This is my x-axis. Actually, I'm gonna do two cases. So this is my y-axis. This is my x-axis. And let's say I have some function here. So this is f of x, right over there."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is my y-axis. This is my x-axis. And let's say I have some function here. So this is f of x, right over there. And let's say that this is x equals a. And let me draw a line going straight up like that. And let's say that this is x equals b, just like that."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is f of x, right over there. And let's say that this is x equals a. And let me draw a line going straight up like that. And let's say that this is x equals b, just like that. And what we want to do is concern ourselves with the area under the graph, under the graph of y is equal to f of x, and above the x-axis, and between these two bounds, between x equals a and x equals b. So this area right over here. And you can already get an appreciation."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's say that this is x equals b, just like that. And what we want to do is concern ourselves with the area under the graph, under the graph of y is equal to f of x, and above the x-axis, and between these two bounds, between x equals a and x equals b. So this area right over here. And you can already get an appreciation. We're not used to finding areas where one of the boundaries, or as we'll see in the future, many of the boundaries could actually be curves. But that's one of the powers of the definite integral, and one of the powers of integral calculus. And so the notation for this area right over here would be the definite integral."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can already get an appreciation. We're not used to finding areas where one of the boundaries, or as we'll see in the future, many of the boundaries could actually be curves. But that's one of the powers of the definite integral, and one of the powers of integral calculus. And so the notation for this area right over here would be the definite integral. And so we're gonna have our lower bound at x equals a, so we'll write it there. We'll have our upper bound at x equals b, right over there. We're taking the area under the curve of f of x, f of x, and then dx."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the notation for this area right over here would be the definite integral. And so we're gonna have our lower bound at x equals a, so we'll write it there. We'll have our upper bound at x equals b, right over there. We're taking the area under the curve of f of x, f of x, and then dx. Now in the future, we're going to, especially once we start looking at Riemann sums, we'll get a better understanding of where this notation comes from. This actually comes from Leibniz, one of the founders of calculus. This is known as the summa symbol."}, {"video_title": "Definite integrals intro   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're taking the area under the curve of f of x, f of x, and then dx. Now in the future, we're going to, especially once we start looking at Riemann sums, we'll get a better understanding of where this notation comes from. This actually comes from Leibniz, one of the founders of calculus. This is known as the summa symbol. But for the sake of this video, you just need to know what this represents. This right over here, this represents the area under f of x between x equals a and x equals b. So this value and this expression should be the same."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's a closed interval. It also includes 1 and it includes 4. You can view this as the domain of our function as we have defined it. So given this, given this information, this function definition, what I would like you to do is come up with the absolute maximum value of f. Where f is defined right over here, where f is defined on this closed interval. I encourage you to pause this video and think about it on your own. So the extreme value theorem tells us, look, if we've got some closed interval, I'm going to speak in generalities here, so let's say that's our x-axis. And let's say we have some function that's defined on a closed interval."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So given this, given this information, this function definition, what I would like you to do is come up with the absolute maximum value of f. Where f is defined right over here, where f is defined on this closed interval. I encourage you to pause this video and think about it on your own. So the extreme value theorem tells us, look, if we've got some closed interval, I'm going to speak in generalities here, so let's say that's our x-axis. And let's say we have some function that's defined on a closed interval. We have a couple of different scenarios for what that function might look like on that closed interval. So we might hit a maximum point. We might hit a maximum point at the beginning of the interval, something like that."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's say we have some function that's defined on a closed interval. We have a couple of different scenarios for what that function might look like on that closed interval. So we might hit a maximum point. We might hit a maximum point at the beginning of the interval, something like that. We might hit an absolute maximum point at the end of the interval. So it might look something like this. So that's at the end of the interval."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "We might hit a maximum point at the beginning of the interval, something like that. We might hit an absolute maximum point at the end of the interval. So it might look something like this. So that's at the end of the interval. Or we might hit an absolute maximum point someplace in between. And that could look something like this. It could look like this."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's at the end of the interval. Or we might hit an absolute maximum point someplace in between. And that could look something like this. It could look like this. And at this maximum point, the slope of the tangent line is zero. So here, the derivative is zero. Or we could have a maximum point someplace in between that looks like this."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "It could look like this. And at this maximum point, the slope of the tangent line is zero. So here, the derivative is zero. Or we could have a maximum point someplace in between that looks like this. And if it looks like this, then here the derivative would be undefined. So we have a couple of different tangent lines that you could place right over there. So what we need to do is let's test the different endpoints."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or we could have a maximum point someplace in between that looks like this. And if it looks like this, then here the derivative would be undefined. So we have a couple of different tangent lines that you could place right over there. So what we need to do is let's test the different endpoints. Let's test the function at the beginning. Let's test the function at the end of the interval. And then let's see if there's any points where the derivative is either zero or the derivative is undefined."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we need to do is let's test the different endpoints. Let's test the function at the beginning. Let's test the function at the end of the interval. And then let's see if there's any points where the derivative is either zero or the derivative is undefined. And these points where the derivative is either zero or it's undefined, we've seen them before, we call these, of course, critical numbers. And this would be either in either case, actually if we assume that that's happening at the same number, we would call that a critical number. So those are the different candidates."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let's see if there's any points where the derivative is either zero or the derivative is undefined. And these points where the derivative is either zero or it's undefined, we've seen them before, we call these, of course, critical numbers. And this would be either in either case, actually if we assume that that's happening at the same number, we would call that a critical number. So those are the different candidates. Now you could have a critical number in between where, say, the slope is zero, say something like this, but it isn't the maximum or minimum. But what we can do is if we can find all the critical numbers, we can then test the function evaluated at the critical numbers and the function evaluated at the endpoints, and we can see which of those are the largest. All of those are the possible candidates for where f hits a maximum value."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So those are the different candidates. Now you could have a critical number in between where, say, the slope is zero, say something like this, but it isn't the maximum or minimum. But what we can do is if we can find all the critical numbers, we can then test the function evaluated at the critical numbers and the function evaluated at the endpoints, and we can see which of those are the largest. All of those are the possible candidates for where f hits a maximum value. So first we could think about, well, actually, let's find the critical numbers first since we have to do it. So let's take the derivative of f. f prime of x is going to be equal to the derivative of natural log of x is 1 over x. So it's going to be 8 over x minus 2x."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of those are the possible candidates for where f hits a maximum value. So first we could think about, well, actually, let's find the critical numbers first since we have to do it. So let's take the derivative of f. f prime of x is going to be equal to the derivative of natural log of x is 1 over x. So it's going to be 8 over x minus 2x. And let's set that equal to zero. So if we focus on this part right over here, we could add 2x to both sides, and we would get 8 over x is equal to 2x. Multiply both sides by x."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be 8 over x minus 2x. And let's set that equal to zero. So if we focus on this part right over here, we could add 2x to both sides, and we would get 8 over x is equal to 2x. Multiply both sides by x. We get 8 is equal to 2x squared. Divide both sides by 2. You get 4 is equal to x squared."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Multiply both sides by x. We get 8 is equal to 2x squared. Divide both sides by 2. You get 4 is equal to x squared. And if we were just purely solving this equation, we would get x is equal to plus or minus 2. Now we're saying that the function is only defined over this interval, so negative 2 isn't part of its domain, so we're only going to focus on x is equal to 2. This right over here is definitely a critical number."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "You get 4 is equal to x squared. And if we were just purely solving this equation, we would get x is equal to plus or minus 2. Now we're saying that the function is only defined over this interval, so negative 2 isn't part of its domain, so we're only going to focus on x is equal to 2. This right over here is definitely a critical number. Now have we found all of the critical numbers? Well, this is the only number other than negative 2, the only number in the interval that will make f prime of x equal to zero. What about where it's undefined?"}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here is definitely a critical number. Now have we found all of the critical numbers? Well, this is the only number other than negative 2, the only number in the interval that will make f prime of x equal to zero. What about where it's undefined? Well, f prime of x would be undefined. The only place it would be undefined is if you stuck a zero right over here in the denominator, but zero is not in the interval. So the only critical number in the interval is x equals 2."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "What about where it's undefined? Well, f prime of x would be undefined. The only place it would be undefined is if you stuck a zero right over here in the denominator, but zero is not in the interval. So the only critical number in the interval is x equals 2. Now we just have to test f at the different endpoints and at the critical number and see which of those is the highest. We're going to test f of 1, which is equal to 8 times the natural log of 1 minus 1 squared. We'll test f of 4, which is equal to 8 times the natural log of 4 minus 4 squared, which is, of course, 16."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the only critical number in the interval is x equals 2. Now we just have to test f at the different endpoints and at the critical number and see which of those is the highest. We're going to test f of 1, which is equal to 8 times the natural log of 1 minus 1 squared. We'll test f of 4, which is equal to 8 times the natural log of 4 minus 4 squared, which is, of course, 16. And we're going to test f of 2. So these are the endpoints, and this is a critical number, 8 times the natural log of 2 minus minus 2 squared. Now which of these is going to be the largest?"}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "We'll test f of 4, which is equal to 8 times the natural log of 4 minus 4 squared, which is, of course, 16. And we're going to test f of 2. So these are the endpoints, and this is a critical number, 8 times the natural log of 2 minus minus 2 squared. Now which of these is going to be the largest? And it might be tempting to get a calculator out, but actually let's see if we can get a little intuition here. So this is the natural log of 1 is 0. e to the 0 power is equal to 1, so 8 times 0 is 0. So this evaluates to negative 1."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now which of these is going to be the largest? And it might be tempting to get a calculator out, but actually let's see if we can get a little intuition here. So this is the natural log of 1 is 0. e to the 0 power is equal to 1, so 8 times 0 is 0. So this evaluates to negative 1. Now let's see, what does this evaluate to? The natural log of 4, e is 2.7, on and on and on. So this number is going to be between 1 and 2."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this evaluates to negative 1. Now let's see, what does this evaluate to? The natural log of 4, e is 2.7, on and on and on. So this number is going to be between 1 and 2. So it's going to be between 1 and 2. And it's actually going to be, well, between 1 and 2, you multiply that times 8, you're going to be between 8 and 16. And then you subtract 16, so that means you're going to be between 0 and negative 8."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this number is going to be between 1 and 2. So it's going to be between 1 and 2. And it's actually going to be, well, between 1 and 2, you multiply that times 8, you're going to be between 8 and 16. And then you subtract 16, so that means you're going to be between 0 and negative 8. So it's not clear, at least without using a calculator in this very rough way, which of these is large. Both of these are negative numbers, though. Now what about this, natural log of 2?"}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you subtract 16, so that means you're going to be between 0 and negative 8. So it's not clear, at least without using a calculator in this very rough way, which of these is large. Both of these are negative numbers, though. Now what about this, natural log of 2? Natural log of 2 is going to be some fraction. It's going to be more than 1 half. And since it's more than 1 half, this whole thing is going to be more than 4, which means this whole thing is going to be positive."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what about this, natural log of 2? Natural log of 2 is going to be some fraction. It's going to be more than 1 half. And since it's more than 1 half, this whole thing is going to be more than 4, which means this whole thing is going to be positive. So this is negative, this is negative, this is positive. And these are our only candidates for our maximum value. So I would go with this one."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And since it's more than 1 half, this whole thing is going to be more than 4, which means this whole thing is going to be positive. So this is negative, this is negative, this is positive. And these are our only candidates for our maximum value. So I would go with this one. Our maximum value happens when x is equal to 2, and that maximum value is 8 natural log of 2 minus 4. That is the absolute maximum value over the interval, I guess we could say over the domain that this function is defined. If we want to verify it with a calculator, we of course could."}, {"video_title": "Finding absolute extrema on a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I would go with this one. Our maximum value happens when x is equal to 2, and that maximum value is 8 natural log of 2 minus 4. That is the absolute maximum value over the interval, I guess we could say over the domain that this function is defined. If we want to verify it with a calculator, we of course could. So we already figured out this one, but let's see. f of 4, 8 natural log of 4 minus 16 is equal to negative 5. So this one is definitely not the maximum value."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Let's say that we have the differential equation dy dx, or the derivative of y with respect to x, is equal to negative x over y. Let's say we don't know how to find the solutions to this, but we at least want to get a sense of what the solutions might look like. And to do that, what we could do is we could look at a coordinate plane. So, let me draw some axes here. So, let me draw a relatively straight line. Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So, let me draw some axes here. So, let me draw a relatively straight line. Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question. So, let's actually do that. Let me set up a little table here."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question. So, let's actually do that. Let me set up a little table here. Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Let me set up a little table here. Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going. What about when x is one and y is one?"}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going. What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero. So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Well then, it's negative one over zero. So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark. Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So, I'll put that as a question mark. Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here. So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one negative one?"}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Well, you would have a slope of negative one here. So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one negative one? Well, now it's negative one over negative one. Your slope is now one. So, one negative one."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "What about if you had one negative one? Well, now it's negative one over negative one. Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So, one negative one. If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two. That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "We could even do two negative two. That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "You'd have a slope of negative one right over here. And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head. I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "I'm now just doing them in my head. I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Here, your slope, what if it was negative one, one. It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So, negative negative one is one over one. So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this. This would be a solution."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this. This would be a solution. So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do something, maybe it would have to do something like this."}, {"video_title": "Creating a slope field   First order differential equations   Khan Academy.mp3", "Sentence": "This would be a solution. So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do something, maybe it would have to do something like this. Or, if it was, if this were a point on the function over here, it would have to do something like this. And once again, I'm doing this based on what the slope field is telling me. So, this field that I'm creating, where I'm taking, I'm sampling a bunch of points and I'm visualizing the slope with the line segment, once again, this is called a slope, slope field."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You've already spent a lot of your mathematical lives talking about functions. The basic idea is give a valid input into a function, so a member of that function's domain, and then the function is going to tell you for that input what is going to be the corresponding output. And we call that corresponding output f of x. So for example, there's many ways of defining functions. You could say something like f of x is equal to x squared. So that means that whatever x, whatever you input into the function, the output is going to be that input squared. You could have something defined like this."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, there's many ways of defining functions. You could say something like f of x is equal to x squared. So that means that whatever x, whatever you input into the function, the output is going to be that input squared. You could have something defined like this. F of x is equal to x squared if x odd, and you could say it's equal to x to the third otherwise. So if it's an odd integer, it's an odd integer, you just square it, but otherwise for any other real number, you take it to the third power. This is a valid way of defining a function."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could have something defined like this. F of x is equal to x squared if x odd, and you could say it's equal to x to the third otherwise. So if it's an odd integer, it's an odd integer, you just square it, but otherwise for any other real number, you take it to the third power. This is a valid way of defining a function. What we're going to do in this video is explore a new way, or a potentially new way for you, of defining a function, and that's by using a definite integral, but it's the same general idea. So what we have graphed here, this is the t-axis, this is the y-axis, and we have the graph of the function f, or you could view this as the graph of y is equal to f of t. Now what I want, and this is another way of representing what outputs you might get for a given input here if t is one, f of t is five. If t is four, f of t is three."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is a valid way of defining a function. What we're going to do in this video is explore a new way, or a potentially new way for you, of defining a function, and that's by using a definite integral, but it's the same general idea. So what we have graphed here, this is the t-axis, this is the y-axis, and we have the graph of the function f, or you could view this as the graph of y is equal to f of t. Now what I want, and this is another way of representing what outputs you might get for a given input here if t is one, f of t is five. If t is four, f of t is three. But I'm now going to define a new function based on a definite integral of f of t. Let's define our new function. Let's say g, let's call it g of x. Let's make it equal to the definite integral from negative two to x of f of t dt."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If t is four, f of t is three. But I'm now going to define a new function based on a definite integral of f of t. Let's define our new function. Let's say g, let's call it g of x. Let's make it equal to the definite integral from negative two to x of f of t dt. Now pause this video, really take a look at it. This might look really fancy, but what's happening here is given an input x, g of x is going to be based on what the definite integral here would be for that x. And so we can set up a little table here to think about some potential values."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's make it equal to the definite integral from negative two to x of f of t dt. Now pause this video, really take a look at it. This might look really fancy, but what's happening here is given an input x, g of x is going to be based on what the definite integral here would be for that x. And so we can set up a little table here to think about some potential values. So let's say x, and let's say g of x right over here. So if x is one, what is g of x going to be equal to? All right, so g of one is going to be equal to the definite integral going from negative two."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can set up a little table here to think about some potential values. So let's say x, and let's say g of x right over here. So if x is one, what is g of x going to be equal to? All right, so g of one is going to be equal to the definite integral going from negative two. Now x is going to be equal to one in this situation. That's what we're inputting into the function. So one is our upper bound of f of t dt."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so g of one is going to be equal to the definite integral going from negative two. Now x is going to be equal to one in this situation. That's what we're inputting into the function. So one is our upper bound of f of t dt. And what is that equal to? Well, that's going to be the area under the curve and above the t-axis between t equals negative two and t is equal to one. So it's gonna be this area here."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one is our upper bound of f of t dt. And what is that equal to? Well, that's going to be the area under the curve and above the t-axis between t equals negative two and t is equal to one. So it's gonna be this area here. And since it's on a grid, we can actually figure this out. We can actually break this up into two sections. This rectangular section is three wide and five high, so it has an area of 15 square units."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's gonna be this area here. And since it's on a grid, we can actually figure this out. We can actually break this up into two sections. This rectangular section is three wide and five high, so it has an area of 15 square units. And this little triangular section up here is two wide and one high, two times one times 1 1\u20442, area of a triangle. This is going to be another one. So that area is going to be equal to 16."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This rectangular section is three wide and five high, so it has an area of 15 square units. And this little triangular section up here is two wide and one high, two times one times 1 1\u20442, area of a triangle. This is going to be another one. So that area is going to be equal to 16. What if x is equal to two? What is g of two going to be equal to? Pause this video and try to figure that out."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that area is going to be equal to 16. What if x is equal to two? What is g of two going to be equal to? Pause this video and try to figure that out. Well, g of two is going to be equal to the definite integral from negative two. And now our upper bound's going to be our input into the function to two of f of t dt. So that's going to be going from here all the way now to here."}, {"video_title": "Functions defined by definite integrals (accumulation functions)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause this video and try to figure that out. Well, g of two is going to be equal to the definite integral from negative two. And now our upper bound's going to be our input into the function to two of f of t dt. So that's going to be going from here all the way now to here. And so it's the area we just calculated. It's all of this stuff, which we figured out was 16 square units, plus another one, two, three, four, five square units. So 16 plus five, this is going to be equal to 21."}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "In this video, we're gonna try to understand limits of composite functions, or at least a way of thinking about limits of composite functions. And in particular, we're gonna think about the case where we're trying to find the limit as x approaches a of f of g of x. And we're going to see under certain circumstances, this is going to be equal to f of the limit, the limit as x approaches a of g of x. And what are those circumstances you are asking? Well, this is going to be true if and only if two things are true. First of all, this limit needs to exist. So the limit as x approaches a of g of x needs to exist."}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "And what are those circumstances you are asking? Well, this is going to be true if and only if two things are true. First of all, this limit needs to exist. So the limit as x approaches a of g of x needs to exist. So that needs to exist. And then on top of that, the function f needs to be continuous at this point, and f continuous at L. So let's look at some examples and see if we can apply this idea, or see if we can't apply it. So here I have two functions that are graphically represented right over here."}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "So the limit as x approaches a of g of x needs to exist. So that needs to exist. And then on top of that, the function f needs to be continuous at this point, and f continuous at L. So let's look at some examples and see if we can apply this idea, or see if we can't apply it. So here I have two functions that are graphically represented right over here. Let me make sure I have enough space for them. And what we see on the left-hand side is our function f, and what we see on the right-hand side is our function g. So first, let's figure out what is the limit as x approaches negative three of f of g of x. Pause this video and see, first of all, does this theorem apply?"}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "So here I have two functions that are graphically represented right over here. Let me make sure I have enough space for them. And what we see on the left-hand side is our function f, and what we see on the right-hand side is our function g. So first, let's figure out what is the limit as x approaches negative three of f of g of x. Pause this video and see, first of all, does this theorem apply? And if it does apply, what is this limit? So the first thing we need to see is does this theorem apply? So first of all, if we were to find the limit as x approaches negative three of g of x, what is that?"}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "Pause this video and see, first of all, does this theorem apply? And if it does apply, what is this limit? So the first thing we need to see is does this theorem apply? So first of all, if we were to find the limit as x approaches negative three of g of x, what is that? Well, when we're approaching negative three from the right, it looks like our function is actually at three. And it looks like when we're approaching negative three from the left, it looks like our function is at three. So it looks like this limit is three, even though the value g of negative three is negative two, but it's a point discontinuity."}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "So first of all, if we were to find the limit as x approaches negative three of g of x, what is that? Well, when we're approaching negative three from the right, it looks like our function is actually at three. And it looks like when we're approaching negative three from the left, it looks like our function is at three. So it looks like this limit is three, even though the value g of negative three is negative two, but it's a point discontinuity. As we approach it from either side, the value of the function is at three. So this thing is going to be three. So it exists, so we meet that first condition."}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "So it looks like this limit is three, even though the value g of negative three is negative two, but it's a point discontinuity. As we approach it from either side, the value of the function is at three. So this thing is going to be three. So it exists, so we meet that first condition. And then the second question is, is our function f continuous at this limit, continuous at three? So when x equals three, yeah, it looks like at that point, our function is definitely continuous. And so we could say that this limit is going to be the same thing as this equals f of the limit as x approaches negative three of g of x, close the parentheses."}, {"video_title": "Theorem for limits of composite functions   Limits and contiuity   AP Calculus   Khan Academy.mp3", "Sentence": "So it exists, so we meet that first condition. And then the second question is, is our function f continuous at this limit, continuous at three? So when x equals three, yeah, it looks like at that point, our function is definitely continuous. And so we could say that this limit is going to be the same thing as this equals f of the limit as x approaches negative three of g of x, close the parentheses. And we know that this is equal to three. And we know that f of three is going to be equal to negative one. So this met the conditions for this theorem, and we were able to use the theorem to actually solve this limit."}, {"video_title": "2015 AP Calculus AB 5c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is true if and only if F prime prime of x goes from, goes from positive, positive to negative, negative, or vice versa, or vice versa. So where do we see F prime prime of x going from positive to negative? Well that's going to be true, that's going to be true if and only if F prime of x goes from being increasing to decreasing, or vice versa, F prime goes from increasing, increasing to decreasing, decreasing, or vice versa. I'm using a lot of vice versa here. So now let's, and I wanted to think of it in terms of F prime because we have the graph of F prime. So F prime goes from increasing to decreasing, or vice versa, or we could go from decreasing to increasing. Well let's think about it."}, {"video_title": "2015 AP Calculus AB 5c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm using a lot of vice versa here. So now let's, and I wanted to think of it in terms of F prime because we have the graph of F prime. So F prime goes from increasing to decreasing, or vice versa, or we could go from decreasing to increasing. Well let's think about it. Let's see, over here, F prime is, F prime is decreasing, decreasing, decreasing, decreasing, and then it increases. So we have a point of inflection right over here, right when F prime of x is zero. So, and that's because F prime is differentiable, so the derivative is definitely, the derivative is zero right at that point of inflection right over here, so if that happens at x equals negative one and over here, then F prime starts increasing, but then it, right at x equals one, then it starts decreasing."}, {"video_title": "2015 AP Calculus AB 5c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well let's think about it. Let's see, over here, F prime is, F prime is decreasing, decreasing, decreasing, decreasing, and then it increases. So we have a point of inflection right over here, right when F prime of x is zero. So, and that's because F prime is differentiable, so the derivative is definitely, the derivative is zero right at that point of inflection right over here, so if that happens at x equals negative one and over here, then F prime starts increasing, but then it, right at x equals one, then it starts decreasing. So at x equals one, we have another point of inflection, and that's where we have that zero, that the zero, a tangent line with slope zero, and then we're decreasing, decreasing, decreasing, decreasing, decreasing, increasing. All right, so this is going to be another point of inflection, x equals three. So these are our three points of inflection."}, {"video_title": "2015 AP Calculus AB 5c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, and that's because F prime is differentiable, so the derivative is definitely, the derivative is zero right at that point of inflection right over here, so if that happens at x equals negative one and over here, then F prime starts increasing, but then it, right at x equals one, then it starts decreasing. So at x equals one, we have another point of inflection, and that's where we have that zero, that the zero, a tangent line with slope zero, and then we're decreasing, decreasing, decreasing, decreasing, decreasing, increasing. All right, so this is going to be another point of inflection, x equals three. So these are our three points of inflection. So this happens, so this happens, happens at x equals negative one, x equals one, and x equals three. So these three points, and on our graph of F prime, where we see F prime goes from decreasing to increasing, or increasing to decreasing, or decreasing to increasing. All right, all right."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we want to do without having to graph g, we want to figure out at what x values does g have a relative maximum. And just to remind us what's going on at a relative maximum, so let me draw a hypothetical function right over here. So a relative maximum is going to happen, so you can visually inspect this, okay, that looks like a relative maximum, this is kind of top of a mountain or top of a hill. These all look like relative maxima. And what's in common? Well, the graph, the function, is going from increasing to decreasing at each of those points. It's going from increasing to decreasing."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "These all look like relative maxima. And what's in common? Well, the graph, the function, is going from increasing to decreasing at each of those points. It's going from increasing to decreasing. Increasing to decreasing at either of the points. Or you could say that the first derivative is going from positive to negative. So if you look at this interval right over here, g prime is greater than zero, and then over the next interval, when you're decreasing, g prime would be less than zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going from increasing to decreasing. Increasing to decreasing at either of the points. Or you could say that the first derivative is going from positive to negative. So if you look at this interval right over here, g prime is greater than zero, and then over the next interval, when you're decreasing, g prime would be less than zero. So what we really need to think about is when does g prime, so let me see, relative, we care about relative maximum point, and so that's essentially asking when does g prime go when does g prime go from positive to negative? From, I wrote frore, from g prime greater than zero to g prime less than zero. And the values that we could look at, or the points, are our critical points."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you look at this interval right over here, g prime is greater than zero, and then over the next interval, when you're decreasing, g prime would be less than zero. So what we really need to think about is when does g prime, so let me see, relative, we care about relative maximum point, and so that's essentially asking when does g prime go when does g prime go from positive to negative? From, I wrote frore, from g prime greater than zero to g prime less than zero. And the values that we could look at, or the points, are our critical points. And critical points are where g prime is either zero or it is undefined. So let's think about it. Where is g prime of x equal to zero?"}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the values that we could look at, or the points, are our critical points. And critical points are where g prime is either zero or it is undefined. So let's think about it. Where is g prime of x equal to zero? G prime of x is equal to zero when, let's just take g prime of x. We're gonna leverage the power rule right here. Four x to the third power, four x to the third minus five x to the fourth, minus five x to the fourth is equal to zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "Where is g prime of x equal to zero? G prime of x is equal to zero when, let's just take g prime of x. We're gonna leverage the power rule right here. Four x to the third power, four x to the third minus five x to the fourth, minus five x to the fourth is equal to zero. Let's see, we can factor out an x to the third. We have x to the third times four minus five x is equal to zero. So this is going to happen when x is equal to zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "Four x to the third power, four x to the third minus five x to the fourth, minus five x to the fourth is equal to zero. Let's see, we can factor out an x to the third. We have x to the third times four minus five x is equal to zero. So this is going to happen when x is equal to zero. That I could, let me not skip steps. So this is going to happen when x to the third is equal to zero, or four minus five x is equal to zero. For x to the third equaling zero, well that's only gonna happen when x is equal to zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to happen when x is equal to zero. That I could, let me not skip steps. So this is going to happen when x to the third is equal to zero, or four minus five x is equal to zero. For x to the third equaling zero, well that's only gonna happen when x is equal to zero. And four minus five x equaling zero, we'll add five x to both sides. You get four is equal to five x. Divide both sides by five, you get four fifths is equal to x. So here, these are the two places where our derivative is equal to zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "For x to the third equaling zero, well that's only gonna happen when x is equal to zero. And four minus five x equaling zero, we'll add five x to both sides. You get four is equal to five x. Divide both sides by five, you get four fifths is equal to x. So here, these are the two places where our derivative is equal to zero. Now are there any places where our derivative is undefined? Well, our function right over here is just a straight up polynomial. Our derivative is another polynomial."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here, these are the two places where our derivative is equal to zero. Now are there any places where our derivative is undefined? Well, our function right over here is just a straight up polynomial. Our derivative is another polynomial. It is defined for all real numbers. So these are our two critical, our critical points, or we could even say critical values. So let's think about what g prime is doing on either side of these critical values."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our derivative is another polynomial. It is defined for all real numbers. So these are our two critical, our critical points, or we could even say critical values. So let's think about what g prime is doing on either side of these critical values. And I'll draw a little number line here to help us visualize this. And so, so there we go, a little bit of a number line. And let's see, we care about zero, and we care about four fifths."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what g prime is doing on either side of these critical values. And I'll draw a little number line here to help us visualize this. And so, so there we go, a little bit of a number line. And let's see, we care about zero, and we care about four fifths. So let's say this is negative one, this is zero, this is one. And so we have one critical point at, let me do this in magenta. We have one critical point here at x equals zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we care about zero, and we care about four fifths. So let's say this is negative one, this is zero, this is one. And so we have one critical point at, let me do this in magenta. We have one critical point here at x equals zero. And then we have another critical point, I will do this at x equals four fifths. So four fifths is right around there. So that is four fifths."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have one critical point here at x equals zero. And then we have another critical point, I will do this at x equals four fifths. So four fifths is right around there. So that is four fifths. And let's just think about what g, what g prime is doing in these intervals. And these critical points are the only places where g prime might switch sides, switch signs. So let's first think about this, let me pick some colors I haven't used yet."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is four fifths. And let's just think about what g, what g prime is doing in these intervals. And these critical points are the only places where g prime might switch sides, switch signs. So let's first think about this, let me pick some colors I haven't used yet. So let's think about the interval from negative infinity to zero. So this is the open interval from negative infinity to zero. And we could just plug in a value, we could, let's try negative one."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's first think about this, let me pick some colors I haven't used yet. So let's think about the interval from negative infinity to zero. So this is the open interval from negative infinity to zero. And we could just plug in a value, we could, let's try negative one. Negative one is pretty straightforward to evaluate. So let's see, you have four, you're gonna have four times negative one to the third power so that's gonna be four times negative one, minus five times negative one to the fourth power. So that's just gonna be one."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could just plug in a value, we could, let's try negative one. Negative one is pretty straightforward to evaluate. So let's see, you have four, you're gonna have four times negative one to the third power so that's gonna be four times negative one, minus five times negative one to the fourth power. So that's just gonna be one. So let's see, this is going to be negative four minus five, which is negative nine. So right over here, g prime is equal to negative nine. And so we know over this whole interval, since it's to the left of this critical point, we know that g prime of x is less than zero."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's just gonna be one. So let's see, this is going to be negative four minus five, which is negative nine. So right over here, g prime is equal to negative nine. And so we know over this whole interval, since it's to the left of this critical point, we know that g prime of x is less than zero. And so our function itself is decreasing over this interval. And so we know we need to go from increasing to decreasing. So you can already say, well we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we know over this whole interval, since it's to the left of this critical point, we know that g prime of x is less than zero. And so our function itself is decreasing over this interval. And so we know we need to go from increasing to decreasing. So you can already say, well we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it. But anyway, let's just think about what's happening in the other intervals. So in the interval between zero and 4 5ths, so that interval right over there, so it's between zero and 4 5ths. Well let's just sample a number there."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you can already say, well we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it. But anyway, let's just think about what's happening in the other intervals. So in the interval between zero and 4 5ths, so that interval right over there, so it's between zero and 4 5ths. Well let's just sample a number there. Let's say the number, I don't know, 1 1 2 might be really straightforward. So we can evaluate g prime of 1 1 2. G prime of 1 1 2 is equal to four times 1 1 2 to the third power."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well let's just sample a number there. Let's say the number, I don't know, 1 1 2 might be really straightforward. So we can evaluate g prime of 1 1 2. G prime of 1 1 2 is equal to four times 1 1 2 to the third power. 1 1 2 to the third power is 1 8th. So it's 4 8ths, or it's just 1 1 2, minus five times 1 1 2 to the fourth. So that's 5 1 6ths."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "G prime of 1 1 2 is equal to four times 1 1 2 to the third power. 1 1 2 to the third power is 1 8th. So it's 4 8ths, or it's just 1 1 2, minus five times 1 1 2 to the fourth. So that's 5 1 6ths. Minus 5 1 6ths. And so this is equal to 8 1 6ths minus 5 1 6ths, which is equal to 3 1 6ths, but the important thing is it's equal to a positive value. So in this blue interval right over there, and actually let me put 4 5ths in a different color so we see that it's not part of that interval."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's 5 1 6ths. Minus 5 1 6ths. And so this is equal to 8 1 6ths minus 5 1 6ths, which is equal to 3 1 6ths, but the important thing is it's equal to a positive value. So in this blue interval right over there, and actually let me put 4 5ths in a different color so we see that it's not part of that interval. So in this light blue interval right here between zero and 4 5ths, g prime of x is greater than zero, so we know our function is increasing. And so let's see what's happening to the right of this. And the easiest value to try out would just be 1."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in this blue interval right over there, and actually let me put 4 5ths in a different color so we see that it's not part of that interval. So in this light blue interval right here between zero and 4 5ths, g prime of x is greater than zero, so we know our function is increasing. And so let's see what's happening to the right of this. And the easiest value to try out would just be 1. So let's try out x equals 1. It's in that interval. So when x equals 1, I'll just write g prime of 1 is equal to 4 minus 5, which is equal to negative 1."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the easiest value to try out would just be 1. So let's try out x equals 1. It's in that interval. So when x equals 1, I'll just write g prime of 1 is equal to 4 minus 5, which is equal to negative 1. So g prime of x is less than zero. g prime of x is less than zero. So our function, so we could say g is increasing here, it is decreasing, oh sorry, let me be careful."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x equals 1, I'll just write g prime of 1 is equal to 4 minus 5, which is equal to negative 1. So g prime of x is less than zero. g prime of x is less than zero. So our function, so we could say g is increasing here, it is decreasing, oh sorry, let me be careful. G is decreasing here, the function itself is decreasing because our derivative is negative, then our function is increasing here because our derivative is positive, and then our function is decreasing here. So at what critical point are we going from increasing to decreasing? Well we're doing that at x equals 4 5ths."}, {"video_title": "Worked example finding relative extrema   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our function, so we could say g is increasing here, it is decreasing, oh sorry, let me be careful. G is decreasing here, the function itself is decreasing because our derivative is negative, then our function is increasing here because our derivative is positive, and then our function is decreasing here. So at what critical point are we going from increasing to decreasing? Well we're doing that at x equals 4 5ths. So we have a relative maximum at x equals 4 5ths. If they said, well where do we have a relative minimum point? Well that's going to happen at x equals zero."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "Let's say that we have the function capital F of x, which we're going to define as the definite integral from one to sine of x, so that's an interesting upper bound right over there, of two t minus one, and of course, dt. And what we are curious about is trying to figure out what is F prime of x going to be equal to? So pause this video and see if you can figure that out. All right, so some of you might have been a little bit challenged by this notion of, hey, instead of an x on this upper bound, I now have a sine of x. If it was just an x, I could have used the fundamental theorem of calculus. Just to review that, if I had a function, let me call it h of x, if I have h of x that was defined as the definite integral from one to x of two t minus one dt, we know from the fundamental theorem of calculus that h prime of x would be simply this inner function with the t replaced by the x. It would just be two x minus one."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "All right, so some of you might have been a little bit challenged by this notion of, hey, instead of an x on this upper bound, I now have a sine of x. If it was just an x, I could have used the fundamental theorem of calculus. Just to review that, if I had a function, let me call it h of x, if I have h of x that was defined as the definite integral from one to x of two t minus one dt, we know from the fundamental theorem of calculus that h prime of x would be simply this inner function with the t replaced by the x. It would just be two x minus one. Pretty straightforward. But this one isn't quite as straightforward. Instead of having an x up here, our upper bound is a sine of x."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "It would just be two x minus one. Pretty straightforward. But this one isn't quite as straightforward. Instead of having an x up here, our upper bound is a sine of x. So one way to think about it is if we were to define g of x as being equal to sine of x, equal to sine of x, our capital F of x can be expressed as capital F of x is the same thing as h of, h of, instead of an x, everywhere we see an x, we're replacing it with a sine of x. So it's h of g of x, g of x. You can see the g of x right over there."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "Instead of having an x up here, our upper bound is a sine of x. So one way to think about it is if we were to define g of x as being equal to sine of x, equal to sine of x, our capital F of x can be expressed as capital F of x is the same thing as h of, h of, instead of an x, everywhere we see an x, we're replacing it with a sine of x. So it's h of g of x, g of x. You can see the g of x right over there. So you replace x with g of x for where in this expression, you get h of g of x, and that is capital F of x. Now why am I doing all of that? Well, this might start making you think about the chain rule, because if this is true, then that means that capital F prime of x is going to be equal to h prime of g of x, h prime of g of x, times g prime of x."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "You can see the g of x right over there. So you replace x with g of x for where in this expression, you get h of g of x, and that is capital F of x. Now why am I doing all of that? Well, this might start making you think about the chain rule, because if this is true, then that means that capital F prime of x is going to be equal to h prime of g of x, h prime of g of x, times g prime of x. And so what would that be? Well, we already know what h prime of x is. So, let me do this in another color."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "Well, this might start making you think about the chain rule, because if this is true, then that means that capital F prime of x is going to be equal to h prime of g of x, h prime of g of x, times g prime of x. And so what would that be? Well, we already know what h prime of x is. So, let me do this in another color. This part right over here is going to be equal to, everywhere we see an x here, we'll replace with a g of x. So it's going to be two, two times sine of x, two sine of x, and then minus one, minus one. This is this right over here."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule   AP\u00ae\ufe0e Calculus   Khan Academy.mp3", "Sentence": "So, let me do this in another color. This part right over here is going to be equal to, everywhere we see an x here, we'll replace with a g of x. So it's going to be two, two times sine of x, two sine of x, and then minus one, minus one. This is this right over here. And then what's g prime of x? G prime of x, well, g prime of x is just, of course, the derivative of sine of x is cosine of x, is cosine of x. So this part right over here is going to be cosine of x."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've already thought about what a definite integral means. If I'm taking the definite integral from a to b of f of x dx, I can just view that as the area below my function f, so if this is my y axis, this is my x axis, and y is equal to f of x, so something like that, y is equal to f of x, and if this is a and if this is b, I could just view this expression as being equal to this area, but what if my function was not above the x axis? What if it was below the x axis? So these are going to be equivalent. Let's say, let me just draw that scenario, so let me draw a scenario where it's my x axis, that is my y axis, and let's say I have, let's say I have a function that looks like that, so that is y is equal to g of x, and let's say that this right over here is a, and this right over here is b, and let's say that this area right over here is equal to five. Well, if I were to ask you what is the definite integral from a to b of g of x dx, what do you think it is going to be? Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x axis."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "So these are going to be equivalent. Let's say, let me just draw that scenario, so let me draw a scenario where it's my x axis, that is my y axis, and let's say I have, let's say I have a function that looks like that, so that is y is equal to g of x, and let's say that this right over here is a, and this right over here is b, and let's say that this area right over here is equal to five. Well, if I were to ask you what is the definite integral from a to b of g of x dx, what do you think it is going to be? Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x axis. You might be tempted to say, hey, this is just going to be equal to five, but you have to be very careful, because if you're looking at the area above your curve and below your x axis versus below your curve and above the x axis, this definite integral is actually going to be the negative of the area. Now, we'll see later on why this will work out nicely with a whole set of integration properties, but if you want to get some intuition for it, let's just think about velocity versus time graphs. So if I, in my horizontal axis, that is time."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x axis. You might be tempted to say, hey, this is just going to be equal to five, but you have to be very careful, because if you're looking at the area above your curve and below your x axis versus below your curve and above the x axis, this definite integral is actually going to be the negative of the area. Now, we'll see later on why this will work out nicely with a whole set of integration properties, but if you want to get some intuition for it, let's just think about velocity versus time graphs. So if I, in my horizontal axis, that is time. My vertical axis, this is velocity, and velocity is going to be measured in meters per second. Time is going to be measured in seconds. Time is measured in seconds, and actually I'm gonna do two scenarios here."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I, in my horizontal axis, that is time. My vertical axis, this is velocity, and velocity is going to be measured in meters per second. Time is going to be measured in seconds. Time is measured in seconds, and actually I'm gonna do two scenarios here. So let's say that I have a first velocity time graph. Let's just call it v one of t, which is equal to three, and it would be three meters per second. So one, two, three."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "Time is measured in seconds, and actually I'm gonna do two scenarios here. So let's say that I have a first velocity time graph. Let's just call it v one of t, which is equal to three, and it would be three meters per second. So one, two, three. So it would look like that. That is v one of t. And if I were to look at the definite integral going from time equals one to time equals five of v sub one of t dt, what would this be equal to? Well, here my function is above my t axis, so I'll just go from one to five, which will be around there, and I could just think about the area here, and this area is pretty easy to calculate."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one, two, three. So it would look like that. That is v one of t. And if I were to look at the definite integral going from time equals one to time equals five of v sub one of t dt, what would this be equal to? Well, here my function is above my t axis, so I'll just go from one to five, which will be around there, and I could just think about the area here, and this area is pretty easy to calculate. It's going to be three meters per second times four seconds. That's my change in time. And so this is going to be 12 meters."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, here my function is above my t axis, so I'll just go from one to five, which will be around there, and I could just think about the area here, and this area is pretty easy to calculate. It's going to be three meters per second times four seconds. That's my change in time. And so this is going to be 12 meters. And so this is going to be equal to 12. And one way to conceptualize this is this gives us our change in position. If my velocity is three meters per second, and since it's positive, you can conceptualize that as it's going to the right at three meters per second, what is my change in position?"}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be 12 meters. And so this is going to be equal to 12. And one way to conceptualize this is this gives us our change in position. If my velocity is three meters per second, and since it's positive, you can conceptualize that as it's going to the right at three meters per second, what is my change in position? Well, I would have gone 12 meters to the right. And you don't need calculus to figure that out. Three meters per second times four seconds would be 12 meters."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "If my velocity is three meters per second, and since it's positive, you can conceptualize that as it's going to the right at three meters per second, what is my change in position? Well, I would have gone 12 meters to the right. And you don't need calculus to figure that out. Three meters per second times four seconds would be 12 meters. But what if it were the other way around? What if I had another velocity function? Let's call that v sub two of t that is equal to negative two meters per second."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "Three meters per second times four seconds would be 12 meters. But what if it were the other way around? What if I had another velocity function? Let's call that v sub two of t that is equal to negative two meters per second. And it's just a constant negative two meters per second. So this is v sub two of t right over here. What would or what should the definite integral from one to five of v sub two of t be dt be equal to?"}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's call that v sub two of t that is equal to negative two meters per second. And it's just a constant negative two meters per second. So this is v sub two of t right over here. What would or what should the definite integral from one to five of v sub two of t be dt be equal to? Well, it should be equal to my change in position. But if my velocity is negative, that means I'm moving to the left. That means my change in position should be to the left as opposed to to the right."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "What would or what should the definite integral from one to five of v sub two of t be dt be equal to? Well, it should be equal to my change in position. But if my velocity is negative, that means I'm moving to the left. That means my change in position should be to the left as opposed to to the right. And so we can just look at this area right over here. Well, if you just look at it as the rectangle, it's gonna be two times four, which is equal to eight. But you have to be very careful."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "That means my change in position should be to the left as opposed to to the right. And so we can just look at this area right over here. Well, if you just look at it as the rectangle, it's gonna be two times four, which is equal to eight. But you have to be very careful. Since it is below my horizontal axis and above my function, this is going to be negative. And this should make a lot of sense. If I'm going two meters per second to the left for four seconds, or another way to think about it, if I'm going negative two meters per second for four seconds then my change in position is going to be negative eight meters."}, {"video_title": "Negative definite integrals   Integration and accumulation of change   AP Calculus AB   Khan Academy.mp3", "Sentence": "But you have to be very careful. Since it is below my horizontal axis and above my function, this is going to be negative. And this should make a lot of sense. If I'm going two meters per second to the left for four seconds, or another way to think about it, if I'm going negative two meters per second for four seconds then my change in position is going to be negative eight meters. I would have moved eight meters to the left if we say the convention is negative means to the left. So the big takeaway is if it's below your function and above the horizontal axis, the definite integral, and if your a is less than b, then your definite integral is going to be positive. If your a is less than b, but your function over that interval is below the horizontal axis, then your definite integral is going to be negative."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what I want to think about is what is the limit of f of x as x approaches infinity? And there are several ways that you could do this. You could actually try to plug in larger and larger numbers for x and see if it seems to be approaching some value. Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is as x gets very, very large, let's just focus on the numerator, as x gets very, very large, this term right over here in the numerator, 4x to the 5th, is going to become much, much more significant than any of these other things. Something squaring gets large, but something being raised to the 5th power gets raised that much, much faster."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is as x gets very, very large, let's just focus on the numerator, as x gets very, very large, this term right over here in the numerator, 4x to the 5th, is going to become much, much more significant than any of these other things. Something squaring gets large, but something being raised to the 5th power gets raised that much, much faster. Similarly, in the denominator, this term right over here, the highest degree term, 6x to the 5th, is going to grow much, much, much faster than any of these other terms. Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the 5th power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the 5th over 6x to the 5th for very large x."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Something squaring gets large, but something being raised to the 5th power gets raised that much, much faster. Similarly, in the denominator, this term right over here, the highest degree term, 6x to the 5th, is going to grow much, much, much faster than any of these other terms. Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the 5th power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the 5th over 6x to the 5th for very large x. Or we could say as x approaches infinity. Now what could this be simplified to? Well, you have x to the 5th divided by x to the 5th."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as x gets very, very, very large, this thing is going to approximate 4x to the 5th over 6x to the 5th for very large x. Or we could say as x approaches infinity. Now what could this be simplified to? Well, you have x to the 5th divided by x to the 5th. These are going to grow together. So you can think of them as canceling out, and so you are left with 2 thirds. So what you could say is the limit of f of x as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much, and so it's going to approach 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you have x to the 5th divided by x to the 5th. These are going to grow together. So you can think of them as canceling out, and so you are left with 2 thirds. So what you could say is the limit of f of x as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much, and so it's going to approach 2 thirds. Now let's look at the graph and see if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2 thirds. So let's look at the graph."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what you could say is the limit of f of x as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much, and so it's going to approach 2 thirds. Now let's look at the graph and see if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2 thirds. So let's look at the graph. So right here is the graph. Got it from Wolfram Alpha. We see indeed as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at the graph. So right here is the graph. Got it from Wolfram Alpha. We see indeed as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2 thirds. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We see indeed as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2 thirds. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2 thirds. So let me draw it as neatly as I can. So this right over here is y is equal to 2 thirds. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have a horizontal asymptote right at 2 thirds. So let me draw it as neatly as I can. So this right over here is y is equal to 2 thirds. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2 thirds. When we just look at the graph here, it seems like the same thing is happening from the bottom direction when x approaches negative infinity. So we could say the limit of f of x as x approaches negative infinity, that also looks like it's 2 thirds. We can use the exact same logic."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2 thirds. When we just look at the graph here, it seems like the same thing is happening from the bottom direction when x approaches negative infinity. So we could say the limit of f of x as x approaches negative infinity, that also looks like it's 2 thirds. We can use the exact same logic. When x becomes a very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the 5th and the 6x to the 5th. So this is true for very large x's. It's also true for very negative x's."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can use the exact same logic. When x becomes a very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the 5th and the 6x to the 5th. So this is true for very large x's. It's also true for very negative x's. So we could also say as x approaches negative infinity, this is also true. Then the x to the 5th over the x to the 5th is going to cancel out. These are the dominant terms."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's also true for very negative x's. So we could also say as x approaches negative infinity, this is also true. Then the x to the 5th over the x to the 5th is going to cancel out. These are the dominant terms. We're going to get it equaling 2 thirds. Once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "These are the dominant terms. We're going to get it equaling 2 thirds. Once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2 thirds. Whether we take the limit of f of x as x approaches infinity, we get 2 thirds. And the limit of f of x as x approaches negative infinity is 2 thirds. In general, whenever you do this, you just have to think about what terms are going to dominate the rest and focus on those."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the rate of change of the area, a of t, so our area's going to be a function of t, what is the rate of change of the area of the triangle at that instant? And so what we're going to do in this exercise, instead of going straight and trying to solve it, what we need to do here is to identify the various units of different expressions, and then try to think about what information is given and what's not, and then that will actually equip us to actually solve this rate of change problem. So let's just do this first part. Let's match each expression with its units, and like always, pause the video and see if you can do it on your own. All right, so the first one is b prime of t. So this is the rate of change of which the base is changing with respect to time. So if we think about it, b of t, that is the base, that is going to be in meters. So this is going to be in meters."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's match each expression with its units, and like always, pause the video and see if you can do it on your own. All right, so the first one is b prime of t. So this is the rate of change of which the base is changing with respect to time. So if we think about it, b of t, that is the base, that is going to be in meters. So this is going to be in meters. If we say b prime of t, this is going to be how much our base is changing with respect to time. So this is going to be meters per, and they give us right over here, they say it's decreasing at a rate of 13 meters per hour. So the units here are meters per hour."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be in meters. If we say b prime of t, this is going to be how much our base is changing with respect to time. So this is going to be meters per, and they give us right over here, they say it's decreasing at a rate of 13 meters per hour. So the units here are meters per hour. And so b prime of t, that is going to be in meters per hour. A at time t sub zero. Remember, a is the area of our triangle, and we're measuring everything in meters, as you can tell from the information they've given us."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the units here are meters per hour. And so b prime of t, that is going to be in meters per hour. A at time t sub zero. Remember, a is the area of our triangle, and we're measuring everything in meters, as you can tell from the information they've given us. And so area is going to be in square units, and so it's going to be in square meters. Now the height at time t sub zero. Well, both the base and the height, those are lengths, they're gonna be measured in meters."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, a is the area of our triangle, and we're measuring everything in meters, as you can tell from the information they've given us. And so area is going to be in square units, and so it's going to be in square meters. Now the height at time t sub zero. Well, both the base and the height, those are lengths, they're gonna be measured in meters. And so our height at time t sub zero is going to be in meters. And then here we have the rate of change of our area with respect to time. So our area, we already know, is in meters squared."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, both the base and the height, those are lengths, they're gonna be measured in meters. And so our height at time t sub zero is going to be in meters. And then here we have the rate of change of our area with respect to time. So our area, we already know, is in meters squared. But we wanna know, this here, this is going to be the rate of change of our area with respect to time. So it's going to be an amount of area per unit time, and time here, we're using hours, as you can see from some of the information they've given us. So this is going to be area per unit time, or meters squared per hour."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our area, we already know, is in meters squared. But we wanna know, this here, this is going to be the rate of change of our area with respect to time. So it's going to be an amount of area per unit time, and time here, we're using hours, as you can see from some of the information they've given us. So this is going to be area per unit time, or meters squared per hour. So it's going to be right over here. So it's area per unit time, and the length we're using in this is meters, and time is hours. All right."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be area per unit time, or meters squared per hour. So it's going to be right over here. So it's area per unit time, and the length we're using in this is meters, and time is hours. All right. Now they say, match each expression with its given value. So what is the base of the triangle at time t sub zero? Do they give that to us?"}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right. Now they say, match each expression with its given value. So what is the base of the triangle at time t sub zero? Do they give that to us? Well, let's see. They say at a certain time, at a certain instant, t sub zero, the base, I'm gonna underline this in a different color, the base at a certain instant, t sub zero, the base is five meters. So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Do they give that to us? Well, let's see. They say at a certain time, at a certain instant, t sub zero, the base, I'm gonna underline this in a different color, the base at a certain instant, t sub zero, the base is five meters. So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters. So this is five meters right over here. Now what about the rate of change of the base with respect to time? Do they tell us that?"}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters. So this is five meters right over here. Now what about the rate of change of the base with respect to time? Do they tell us that? Well, look right over here. That's actually the first piece of information they gave us. So the base, B of t of a triangle, is decreasing at a rate of 13 meters per hour."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Do they tell us that? Well, look right over here. That's actually the first piece of information they gave us. So the base, B of t of a triangle, is decreasing at a rate of 13 meters per hour. So the rate of change of the base, that is B prime of t, which is equal to dB dt, and they tell us that that is, it's decreasing at a rate of 13 meters per hour. So that would be negative 13 meters per hour. And so the rate of change of the base with respect to time is going to be negative 13."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the base, B of t of a triangle, is decreasing at a rate of 13 meters per hour. So the rate of change of the base, that is B prime of t, which is equal to dB dt, and they tell us that that is, it's decreasing at a rate of 13 meters per hour. So that would be negative 13 meters per hour. And so the rate of change of the base with respect to time is going to be negative 13. They gave us that. Now A prime of t, this is the rate of change of the area at time t sub zero. Did they give us this?"}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the rate of change of the base with respect to time is going to be negative 13. They gave us that. Now A prime of t, this is the rate of change of the area at time t sub zero. Did they give us this? Well, they ask us that. What is the rate of change of the area, A of t of the triangle at that instant? So this is what we actually need to figure out, but they haven't given it to us, otherwise there's no problem to solve."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Did they give us this? Well, they ask us that. What is the rate of change of the area, A of t of the triangle at that instant? So this is what we actually need to figure out, but they haven't given it to us, otherwise there's no problem to solve. So this one right over here is not given. This is what we are trying to solve for. And then finally we have the first derivative of the height with respect to time."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is what we actually need to figure out, but they haven't given it to us, otherwise there's no problem to solve. So this one right over here is not given. This is what we are trying to solve for. And then finally we have the first derivative of the height with respect to time. So you could view this as dh dt. What is this going to be? Do they give it to us?"}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally we have the first derivative of the height with respect to time. So you could view this as dh dt. What is this going to be? Do they give it to us? Well, look right over here. They say the height of the triangle is increasing at a rate of six meters per hour. So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t, and they're telling us that it is increasing at six meters per hour."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Do they give it to us? Well, look right over here. They say the height of the triangle is increasing at a rate of six meters per hour. So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t, and they're telling us that it is increasing at six meters per hour. So it's gonna be positive six meters per hour. So they did indeed give us that. Now why is all of this a useful exercise to go through?"}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t, and they're telling us that it is increasing at six meters per hour. So it's gonna be positive six meters per hour. So they did indeed give us that. Now why is all of this a useful exercise to go through? Well, now we are really ready to solve the question because in general, if we're talking about any triangle, we know that area is equal to 1 1 2 base times height. Now in this situation, area and our base and our height, they're all going to be functions of t. So we could write a of t is equal to 1 1 2 times b of t times h of t. And if we wanna find the rate of change of our area at that instant, and the instant that they're talking about is at time t sub zero, well then what we would wanna do is take the derivative of both sides with respect to t. So the derivative on the left-hand side with respect to t would be a prime of t, and then on the right-hand side, it would be 1 1 2 times, and we would actually use a combination of, well, it's really just the product rule right over here, the derivative of the first function with respect to t, so it's b prime of t times the second function. This is just the product rule here, plus the first function, b of t, times the derivative of the second function with respect to time."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now why is all of this a useful exercise to go through? Well, now we are really ready to solve the question because in general, if we're talking about any triangle, we know that area is equal to 1 1 2 base times height. Now in this situation, area and our base and our height, they're all going to be functions of t. So we could write a of t is equal to 1 1 2 times b of t times h of t. And if we wanna find the rate of change of our area at that instant, and the instant that they're talking about is at time t sub zero, well then what we would wanna do is take the derivative of both sides with respect to t. So the derivative on the left-hand side with respect to t would be a prime of t, and then on the right-hand side, it would be 1 1 2 times, and we would actually use a combination of, well, it's really just the product rule right over here, the derivative of the first function with respect to t, so it's b prime of t times the second function. This is just the product rule here, plus the first function, b of t, times the derivative of the second function with respect to time. And we need to figure out not just the general expression, they want us to know what the rate of change of the area, so a prime of t at that instant, at t sub zero. So what we wanna figure out, we wanna figure out a prime at time t sub zero. Well, that's just going to be equal to 1 1 2 times b prime of t sub zero times h of t sub zero plus b of t sub zero times h prime of t sub zero."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is just the product rule here, plus the first function, b of t, times the derivative of the second function with respect to time. And we need to figure out not just the general expression, they want us to know what the rate of change of the area, so a prime of t at that instant, at t sub zero. So what we wanna figure out, we wanna figure out a prime at time t sub zero. Well, that's just going to be equal to 1 1 2 times b prime of t sub zero times h of t sub zero plus b of t sub zero times h prime of t sub zero. Now, this might seem daunting, except they've given us a lot of this information. What is b prime of t sub zero? Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour, so they gave us this."}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's just going to be equal to 1 1 2 times b prime of t sub zero times h of t sub zero plus b of t sub zero times h prime of t sub zero. Now, this might seem daunting, except they've given us a lot of this information. What is b prime of t sub zero? Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour, so they gave us this. And h, what is the height at time t sub zero? Well, they tell us right over here. At a certain instant, the base is five meters and the height is one meter, so they give us both b and h at t sub zero, so they gave us this, they gave us this, and what is the rate of change of the height at time t sub zero?"}, {"video_title": "Analyzing related rates problems expressions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour, so they gave us this. And h, what is the height at time t sub zero? Well, they tell us right over here. At a certain instant, the base is five meters and the height is one meter, so they give us both b and h at t sub zero, so they gave us this, they gave us this, and what is the rate of change of the height at time t sub zero? Well, they tell us the height of the triangle is increasing at a rate of six meters per hour. So they tell us that as well. All of that stuff is given, and so you just have to plug it in to figure out what is the rate of change of the area at t sub zero, at that instant."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And like always, I encourage you to pause the video and see if you can work through this on your own. So let's look at this first statement. So this first statement says both the limit of g of x as x approaches six from the right-hand side and the limit as x approaches six from the left-hand side of g of x exist. All right, so let's first think about the limit of g of x as x approaches six from the right-hand side, as we approach six from values greater than six. So if we look over here, we could say, okay, when x is equal to nine, g of nine is right over there. G of eight is right over here. G of seven is right over here."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so let's first think about the limit of g of x as x approaches six from the right-hand side, as we approach six from values greater than six. So if we look over here, we could say, okay, when x is equal to nine, g of nine is right over there. G of eight is right over here. G of seven is right over here. It looks like it's between negative three and negative four. G of 6.5 looks like it's a little bit, it's still between negative three and negative four, but it's closer to negative three. G of 6.1 is even closer to negative three."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of seven is right over here. It looks like it's between negative three and negative four. G of 6.5 looks like it's a little bit, it's still between negative three and negative four, but it's closer to negative three. G of 6.1 is even closer to negative three. G of 6.01 is even closer to negative three. So it looks like the limit from the right-hand side does exist. So it looks like this one exists."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of 6.1 is even closer to negative three. G of 6.01 is even closer to negative three. So it looks like the limit from the right-hand side does exist. So it looks like this one exists. Now let's see, and I'm just looking at it graphically, and that's all they can expect you to do in an exercise like this. Now let's think about the limit as x approaches six from the left-hand side. So I could start anywhere, but let's say when x is equal to three, g of three is a little more than one."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks like this one exists. Now let's see, and I'm just looking at it graphically, and that's all they can expect you to do in an exercise like this. Now let's think about the limit as x approaches six from the left-hand side. So I could start anywhere, but let's say when x is equal to three, g of three is a little more than one. G of four looks like it's a little bit less than two. G of five looks like it's close to three. G of 5.5 looks like it's between five and six."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I could start anywhere, but let's say when x is equal to three, g of three is a little more than one. G of four looks like it's a little bit less than two. G of five looks like it's close to three. G of 5.5 looks like it's between five and six. G of 5.75 looks like it's approaching nine. And as we get closer and closer, as x gets closer and closer to six from below, from values to the left of six, it looks like we're unbounded. We are approaching infinity."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of 5.5 looks like it's between five and six. G of 5.75 looks like it's approaching nine. And as we get closer and closer, as x gets closer and closer to six from below, from values to the left of six, it looks like we're unbounded. We are approaching infinity. And so technically, we would say this limit does not exist. So this one does not exist. So I won't check this one off."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are approaching infinity. And so technically, we would say this limit does not exist. So this one does not exist. So I won't check this one off. Some people will say the limit is approaching infinity, but that technically is, infinity is not a value that you can say it is approaching in the classical, formal definition of a limit. So for these purposes, we would just say this does not exist. Now let's see."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I won't check this one off. Some people will say the limit is approaching infinity, but that technically is, infinity is not a value that you can say it is approaching in the classical, formal definition of a limit. So for these purposes, we would just say this does not exist. Now let's see. They say the limit as x approaches six of g of x exists. Well, the only way that the limit exists is if both the left and the right limits exist, and they approach the same thing. Well, we don't even, our limit as x approaches six from the negative side, or from the left-hand side, I guess I could say, does not even exist."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's see. They say the limit as x approaches six of g of x exists. Well, the only way that the limit exists is if both the left and the right limits exist, and they approach the same thing. Well, we don't even, our limit as x approaches six from the negative side, or from the left-hand side, I guess I could say, does not even exist. So this cannot be true. So that's not gonna be true. The first one's not gonna be true."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we don't even, our limit as x approaches six from the negative side, or from the left-hand side, I guess I could say, does not even exist. So this cannot be true. So that's not gonna be true. The first one's not gonna be true. G is defined at x equals six. So at x equals six, it doesn't look like g is defined. And looking at this graph, I can't tell you what g of six should be."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The first one's not gonna be true. G is defined at x equals six. So at x equals six, it doesn't look like g is defined. And looking at this graph, I can't tell you what g of six should be. We have an open circle over here, so g of six is not equal to negative three, and this goes up to infinity, and we have a vertical asymptote actually drawn right over here at x equals six. So g is not defined at x equals six. So I'll rule that one out."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And looking at this graph, I can't tell you what g of six should be. We have an open circle over here, so g of six is not equal to negative three, and this goes up to infinity, and we have a vertical asymptote actually drawn right over here at x equals six. So g is not defined at x equals six. So I'll rule that one out. G is continuous at x equals six. Well, you can see that it goes up to infinity, then it jumps down, back down here, then continues. So just, when you just think about it in common sense language, it looks very discontinuous."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll rule that one out. G is continuous at x equals six. Well, you can see that it goes up to infinity, then it jumps down, back down here, then continues. So just, when you just think about it in common sense language, it looks very discontinuous. And if you wanna think about it more formally, in order for something to be continuous, the limit needs to exist at that value. The function needs to be defined at that value, and the value of the function needs to be equal to the value of the limit. And neither of these, the first two conditions aren't true, and so these can't even equal each other, because neither of these exist."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So just, when you just think about it in common sense language, it looks very discontinuous. And if you wanna think about it more formally, in order for something to be continuous, the limit needs to exist at that value. The function needs to be defined at that value, and the value of the function needs to be equal to the value of the limit. And neither of these, the first two conditions aren't true, and so these can't even equal each other, because neither of these exist. So this is not continuous at x equals six, and so the only thing I could check here is none of the above. Let's do another one of these. So the first statement, both the right hand and the left hand limit exist as x approaches three."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And neither of these, the first two conditions aren't true, and so these can't even equal each other, because neither of these exist. So this is not continuous at x equals six, and so the only thing I could check here is none of the above. Let's do another one of these. So the first statement, both the right hand and the left hand limit exist as x approaches three. So let's think about it. So x equals three is where we have this little discontinuity here, this jump discontinuity. So let's approach, let's go from the positive, from values larger than three."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first statement, both the right hand and the left hand limit exist as x approaches three. So let's think about it. So x equals three is where we have this little discontinuity here, this jump discontinuity. So let's approach, let's go from the positive, from values larger than three. So when x is equal to five, g of five is a little bit more negative than negative three. G of four is between negative two and negative three. G of 3.5 is getting a little bit closer to negative two."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's approach, let's go from the positive, from values larger than three. So when x is equal to five, g of five is a little bit more negative than negative three. G of four is between negative two and negative three. G of 3.5 is getting a little bit closer to negative two. G of 3.1, it's getting even closer, closer to negative two. G of 3.01 is even closer to negative two. So it looks like this limit right over here, oh, I'm circling the wrong one."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of 3.5 is getting a little bit closer to negative two. G of 3.1, it's getting even closer, closer to negative two. G of 3.01 is even closer to negative two. So it looks like this limit right over here, oh, I'm circling the wrong one. It looks like this limit exists, and in fact, it looks like it is approaching negative two. So this right over here is equal to negative two, the limit of g of x as x approaches three from the right hand side. And now let's think about it from the left hand side."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks like this limit right over here, oh, I'm circling the wrong one. It looks like this limit exists, and in fact, it looks like it is approaching negative two. So this right over here is equal to negative two, the limit of g of x as x approaches three from the right hand side. And now let's think about it from the left hand side. So we can start, I can start here. G of one looks like it's a little bit greater than negative one. G of two, it's less than one."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's think about it from the left hand side. So we can start, I can start here. G of one looks like it's a little bit greater than negative one. G of two, it's less than one. G of 2.5 is between one and two. G of 2.9, looks like it's a little bit less than two. G of 2.99 is getting even closer to two."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of two, it's less than one. G of 2.5 is between one and two. G of 2.9, looks like it's a little bit less than two. G of 2.99 is getting even closer to two. G of 2.999999 would be even closer to, so it looks like this thing right over here is approaching two. So both of these limits, the limit from the right and the limit from the left, exist. The limit of g of x as x approaches three exists."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of 2.99 is getting even closer to two. G of 2.999999 would be even closer to, so it looks like this thing right over here is approaching two. So both of these limits, the limit from the right and the limit from the left, exist. The limit of g of x as x approaches three exists. So these are the one-sided limit. This is the actual limit. Now in order for this to exist, both the right and left-handed limits need to exist, and they need to approach the same value."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The limit of g of x as x approaches three exists. So these are the one-sided limit. This is the actual limit. Now in order for this to exist, both the right and left-handed limits need to exist, and they need to approach the same value. Well this first statement, we saw that both of these exist, but they aren't approaching the same value. From the left, we are, sorry, from the right, we are approaching negative two, and from the left, we are approaching two. So this limit does not exist."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now in order for this to exist, both the right and left-handed limits need to exist, and they need to approach the same value. Well this first statement, we saw that both of these exist, but they aren't approaching the same value. From the left, we are, sorry, from the right, we are approaching negative two, and from the left, we are approaching two. So this limit does not exist. So I will not check that out, or I will not check that box. G is defined at x equals three. Well when x equals three, we see a solid dot right over there."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this limit does not exist. So I will not check that out, or I will not check that box. G is defined at x equals three. Well when x equals three, we see a solid dot right over there. And so it is indeed defined. It is indeed defined there. G is continuous at x equals three."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well when x equals three, we see a solid dot right over there. And so it is indeed defined. It is indeed defined there. G is continuous at x equals three. Well in order for g to be continuous at x equals three, the limit must exist there. It must be defined there, and the value of the function there needs to be equal to the value of the limit. Well the function is defined there, but the limit doesn't exist there."}, {"video_title": "Worked example Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G is continuous at x equals three. Well in order for g to be continuous at x equals three, the limit must exist there. It must be defined there, and the value of the function there needs to be equal to the value of the limit. Well the function is defined there, but the limit doesn't exist there. So it cannot be continuous. It cannot be continuous there. So I would cross that out."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are being asked, what is the smallest, this is a little typo here, what is the smallest possible sum of squares of two numbers if their product is negative 16? So let's say that these two numbers are x and y, x and y. So how could we define the sum of the squares of the two numbers? So I'll just call that the sum of the squares, s for sum of the squares, and it would just be equal to x squared plus y squared. And this is what we wanna minimize. We want to minimize, minimize s. Now, right now, s is expressed as a function of x and y. We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable, and lucky for us, they give us another piece of information."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll just call that the sum of the squares, s for sum of the squares, and it would just be equal to x squared plus y squared. And this is what we wanna minimize. We want to minimize, minimize s. Now, right now, s is expressed as a function of x and y. We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable, and lucky for us, they give us another piece of information. Their product is negative 16. So x times y is equal to negative 16. So let's say we wanted this expression right over here only in terms of x."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable, and lucky for us, they give us another piece of information. Their product is negative 16. So x times y is equal to negative 16. So let's say we wanted this expression right over here only in terms of x. Well, then we could solve, we can figure out what y is in terms of x, and then substitute. So let's do that right over here. If we divide both sides by x, we get y is equal to negative 16 over x."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say we wanted this expression right over here only in terms of x. Well, then we could solve, we can figure out what y is in terms of x, and then substitute. So let's do that right over here. If we divide both sides by x, we get y is equal to negative 16 over x. And so let's replace our y in this expression with negative 16 over x. So then we would get our sum of squares as a function of x is going to be equal to x squared plus y squared. Y is negative 16 over x."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we divide both sides by x, we get y is equal to negative 16 over x. And so let's replace our y in this expression with negative 16 over x. So then we would get our sum of squares as a function of x is going to be equal to x squared plus y squared. Y is negative 16 over x. Negative 16 over x. And then that's what we will now square. So this is equal to x squared plus, what is this?"}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Y is negative 16 over x. Negative 16 over x. And then that's what we will now square. So this is equal to x squared plus, what is this? 256, 256 over x squared, or we could write that as 256, 256 x to the negative 2 power. That is the sum of our squares that we now want to minimize. Well, to minimize this, we would wanna look at the critical points of this, which is where the derivative is either zero or undefined, and see whether those critical points are possibly a minimum or a maximum point."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to x squared plus, what is this? 256, 256 over x squared, or we could write that as 256, 256 x to the negative 2 power. That is the sum of our squares that we now want to minimize. Well, to minimize this, we would wanna look at the critical points of this, which is where the derivative is either zero or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones, if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative, s prime, let me do this in a different color, s prime of x, I'll do it right over here, actually."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to minimize this, we would wanna look at the critical points of this, which is where the derivative is either zero or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones, if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative, s prime, let me do this in a different color, s prime of x, I'll do it right over here, actually. The derivative, s prime of x, with respect to x, is going to be equal to 2x times negative 2 times 5, 2x plus 256 times negative 2, so that's minus 512 x to the negative 3 power. X to the negative 3 power. Now, this is going to be undefined, this is going to be undefined when x is equal to zero, but if x is equal to zero, then y is undefined, so this whole thing breaks down."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative, s prime, let me do this in a different color, s prime of x, I'll do it right over here, actually. The derivative, s prime of x, with respect to x, is going to be equal to 2x times negative 2 times 5, 2x plus 256 times negative 2, so that's minus 512 x to the negative 3 power. X to the negative 3 power. Now, this is going to be undefined, this is going to be undefined when x is equal to zero, but if x is equal to zero, then y is undefined, so this whole thing breaks down. So that isn't a useful critical point, x equals zero. So let's think about any other ones. Well, it's defined everywhere else, so let's think about where the derivative is equal to zero."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, this is going to be undefined, this is going to be undefined when x is equal to zero, but if x is equal to zero, then y is undefined, so this whole thing breaks down. So that isn't a useful critical point, x equals zero. So let's think about any other ones. Well, it's defined everywhere else, so let's think about where the derivative is equal to zero. So when does this thing equal zero? So when does 2x minus 512 x to the negative 3 equal zero? Well, we can add 512 x to the negative 3 to both sides, so you get 2x is equal to 512 x to the negative 3rd power."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's defined everywhere else, so let's think about where the derivative is equal to zero. So when does this thing equal zero? So when does 2x minus 512 x to the negative 3 equal zero? Well, we can add 512 x to the negative 3 to both sides, so you get 2x is equal to 512 x to the negative 3rd power. We can multiply both sides times x to the 3rd power, multiply both sides times x to the 3rd, so all of the x's go away on the right-hand side. So you get 2x to the 4th is equal to 512. We can divide both sides by 2, and you get x to the 4th power is equal to 256."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can add 512 x to the negative 3 to both sides, so you get 2x is equal to 512 x to the negative 3rd power. We can multiply both sides times x to the 3rd power, multiply both sides times x to the 3rd, so all of the x's go away on the right-hand side. So you get 2x to the 4th is equal to 512. We can divide both sides by 2, and you get x to the 4th power is equal to 256. And so what is the 4th root of 256? Well, we could take the square root of both sides just to help us here. So let's see, if we take, so it's going to be x squared is going to be equal to, 256 is 16 squared, so this is 16, this is going to be x squared is equal to 16, or x is equal to 4."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can divide both sides by 2, and you get x to the 4th power is equal to 256. And so what is the 4th root of 256? Well, we could take the square root of both sides just to help us here. So let's see, if we take, so it's going to be x squared is going to be equal to, 256 is 16 squared, so this is 16, this is going to be x squared is equal to 16, or x is equal to 4. Now, that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here, but let's make sure it's a minimum value, and to do that, we can just do our second derivative test. So let's figure out, let's take the second derivative, s prime prime of x, and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2, and then we're going to have negative 3 times negative 512, so I'll just write that as plus 3 times 512, that's going to be 1536, is that right?"}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, if we take, so it's going to be x squared is going to be equal to, 256 is 16 squared, so this is 16, this is going to be x squared is equal to 16, or x is equal to 4. Now, that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here, but let's make sure it's a minimum value, and to do that, we can just do our second derivative test. So let's figure out, let's take the second derivative, s prime prime of x, and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2, and then we're going to have negative 3 times negative 512, so I'll just write that as plus 3 times 512, that's going to be 1536, is that right? Yeah, 3 times 500 is 1500, 3 times 12 is 36, x to the negative 4 power. So this thing is going to be, this thing right over here is actually going to be positive for any x, this thing is going to be positive for any x. x to the negative 4, even if it's a negative x value, that's going to be positive, everything else is positive, this thing is always positive. So we are always in a concave upwards situation."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So s prime prime of x is going to be equal to 2, and then we're going to have negative 3 times negative 512, so I'll just write that as plus 3 times 512, that's going to be 1536, is that right? Yeah, 3 times 500 is 1500, 3 times 12 is 36, x to the negative 4 power. So this thing is going to be, this thing right over here is actually going to be positive for any x, this thing is going to be positive for any x. x to the negative 4, even if it's a negative x value, that's going to be positive, everything else is positive, this thing is always positive. So we are always in a concave upwards situation. Concave upwards means that our graph might look something like that. Actually, I don't want to draw a little squiggle, it might look something like that, and you see the reason why the second derivative implies concave upwards, a second derivative positive means that our derivative is constantly increasing, so the derivative is constantly increasing, it's negative, less negative, even less negative, let me do that in a different color. You see it's negative, less negative, even less negative, zero, positive, more positive, so it's increasing over the entire place."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we are always in a concave upwards situation. Concave upwards means that our graph might look something like that. Actually, I don't want to draw a little squiggle, it might look something like that, and you see the reason why the second derivative implies concave upwards, a second derivative positive means that our derivative is constantly increasing, so the derivative is constantly increasing, it's negative, less negative, even less negative, let me do that in a different color. You see it's negative, less negative, even less negative, zero, positive, more positive, so it's increasing over the entire place. So if you have a critical point where the derivative is equal to zero, so the slope is equal to zero, and it's concave upwards, you see pretty clearly that we have minimized the function. So what is y going to be equal to? We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares, we could just put it back into this, but just for fun, we see that y would be negative 16 over x, so y would be equal to negative 4, and we could just figure out now what our sum of squares is."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You see it's negative, less negative, even less negative, zero, positive, more positive, so it's increasing over the entire place. So if you have a critical point where the derivative is equal to zero, so the slope is equal to zero, and it's concave upwards, you see pretty clearly that we have minimized the function. So what is y going to be equal to? We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares, we could just put it back into this, but just for fun, we see that y would be negative 16 over x, so y would be equal to negative 4, and we could just figure out now what our sum of squares is. Our minimum sum of squares is going to be equal to, it's going to be equal to four squared, which is 16, plus negative four squared, plus another 16, which is equal to 32. Now I know some of you might be thinking, hey, I could have done this without calculus. I could have just tried out numbers whose product is negative 16, and I probably would have tried out four and negative four in not too much time, and then I would have been able to maybe figure out that it's lower than if I did two and negative eight, or negative two and eight, or one and 16, and that's true."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares, we could just put it back into this, but just for fun, we see that y would be negative 16 over x, so y would be equal to negative 4, and we could just figure out now what our sum of squares is. Our minimum sum of squares is going to be equal to, it's going to be equal to four squared, which is 16, plus negative four squared, plus another 16, which is equal to 32. Now I know some of you might be thinking, hey, I could have done this without calculus. I could have just tried out numbers whose product is negative 16, and I probably would have tried out four and negative four in not too much time, and then I would have been able to maybe figure out that it's lower than if I did two and negative eight, or negative two and eight, or one and 16, and that's true. You probably would have been able to do that, but you still wouldn't have been able to feel good that that was a minimum value, because you wouldn't have tried out 4.01 or 4.0011. In fact, you couldn't have tried out all of the possible values. Remember, we didn't say that this is only integers."}, {"video_title": "Optimization sum of squares   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could have just tried out numbers whose product is negative 16, and I probably would have tried out four and negative four in not too much time, and then I would have been able to maybe figure out that it's lower than if I did two and negative eight, or negative two and eight, or one and 16, and that's true. You probably would have been able to do that, but you still wouldn't have been able to feel good that that was a minimum value, because you wouldn't have tried out 4.01 or 4.0011. In fact, you couldn't have tried out all of the possible values. Remember, we didn't say that this is only integers. It just happened to be that our values worked out to be, just worked out to be integers in this situation. You can imagine what would happen if the problem wasn't if their product is negative 16, but what if their product is negative 17, or what if their product is negative 16.5, or what if their product was pi squared? Then you wouldn't be able to try everything else out, and you would have to resort to doing what we did in this video."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You know the spot on the ground that is directly below the hot air balloon. Let's say it took off from that point. It's just been going straight up ever since. And you know, you've measured it out, that you're 500 meters away from there. So you know that you are 500 meters away from that. And you're also able to measure the angle between the horizontal and the hot air balloon. You could do that with, I don't know, I'm not exactly a surveyor, but I guess a viewfinder or something like that."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you know, you've measured it out, that you're 500 meters away from there. So you know that you are 500 meters away from that. And you're also able to measure the angle between the horizontal and the hot air balloon. You could do that with, I don't know, I'm not exactly a surveyor, but I guess a viewfinder or something like that. So you're able to, and I'm not sure if that's the right tool, but there are tools that you can measure the angles between the horizontal and something that's not on the horizontal. So you know that this angle right over here is pi over four radians, or 45 degrees. We're gonna keep it pi over four, because when you take derivatives of trig functions, you assume that you're dealing with radians."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could do that with, I don't know, I'm not exactly a surveyor, but I guess a viewfinder or something like that. So you're able to, and I'm not sure if that's the right tool, but there are tools that you can measure the angles between the horizontal and something that's not on the horizontal. So you know that this angle right over here is pi over four radians, or 45 degrees. We're gonna keep it pi over four, because when you take derivatives of trig functions, you assume that you're dealing with radians. So right over here, this is pi over four radians. And you also are able to measure the rate at which this angle is changing. So this is changing, changing, changing at 0.2 radians per minute."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna keep it pi over four, because when you take derivatives of trig functions, you assume that you're dealing with radians. So right over here, this is pi over four radians. And you also are able to measure the rate at which this angle is changing. So this is changing, changing, changing at 0.2 radians per minute. Now, my question to you, or the question that you're trying to figure out as you watch this hot air balloon is how fast is it rising right now? How fast is it rising just as the angle between the horizontal and kind of the line between you and the hot air balloon is pi over four radians, and that angle is changing at 0.2 radians per minute? So let's think about what we know and what we're trying to figure out."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is changing, changing, changing at 0.2 radians per minute. Now, my question to you, or the question that you're trying to figure out as you watch this hot air balloon is how fast is it rising right now? How fast is it rising just as the angle between the horizontal and kind of the line between you and the hot air balloon is pi over four radians, and that angle is changing at 0.2 radians per minute? So let's think about what we know and what we're trying to figure out. So we know a couple of things. We know that theta is equal to pi over four. If we call theta the angle right over here."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what we know and what we're trying to figure out. So we know a couple of things. We know that theta is equal to pi over four. If we call theta the angle right over here. So this is theta. We also know the rate at which theta is changing. We know d theta, let me do this in yellow."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we call theta the angle right over here. So this is theta. We also know the rate at which theta is changing. We know d theta, let me do this in yellow. We know d theta, dt, is equal to 0.2 radians per minute. Now, what are we trying to figure out? Well, we're trying to figure out the rate at which the height of the balloon is changing."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know d theta, let me do this in yellow. We know d theta, dt, is equal to 0.2 radians per minute. Now, what are we trying to figure out? Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here, this distance right over here, h, what we wanna figure out is dh dt. That's what we don't know. So what we'd wanna come up with is a relationship between dh dt, d theta dt, and maybe theta if we need it."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here, this distance right over here, h, what we wanna figure out is dh dt. That's what we don't know. So what we'd wanna come up with is a relationship between dh dt, d theta dt, and maybe theta if we need it. Or another way to think about it, if we can come up with a relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h? Well, it's a little bit of trigonometry."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we'd wanna come up with is a relationship between dh dt, d theta dt, and maybe theta if we need it. Or another way to think about it, if we can come up with a relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h? Well, it's a little bit of trigonometry. We know, we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent, that's the definition of tangent. So let's write that down."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's a little bit of trigonometry. We know, we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent, that's the definition of tangent. So let's write that down. So we know that the tangent, the tangent of theta, tangent of theta is equal to the opposite side, the opposite side is equal to h, is equal to h over the adjacent side, which we know is going to be a fixed 500, is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between theta, d theta, and d h, or d theta, d t, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of this, of both sides of this, with respect to t implicitly. So let's do that."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's write that down. So we know that the tangent, the tangent of theta, tangent of theta is equal to the opposite side, the opposite side is equal to h, is equal to h over the adjacent side, which we know is going to be a fixed 500, is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between theta, d theta, and d h, or d theta, d t, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of this, of both sides of this, with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now, let's take the derivative with respect to t. So d, d t, I'm gonna take the derivative with respect to t on the left, we're gonna take the derivative with respect to t on the right. So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's write it like that. And now, let's take the derivative with respect to t. So d, d t, I'm gonna take the derivative with respect to t on the left, we're gonna take the derivative with respect to t on the right. So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta d t. Once again this is just d, the derivative, the, sorry, the derivative of the tangent. The tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative of theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we'd want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta d t. Once again this is just d, the derivative, the, sorry, the derivative of the tangent. The tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative of theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we'd want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. Fair enough, so this is the left-hand side, and then the right-hand side becomes, well, it's just going to be 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're taking the derivative with respect to t, not just the derivative with respect to theta. Fair enough, so this is the left-hand side, and then the right-hand side becomes, well, it's just going to be 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that. Let's do that right over here. So we get secant squared of theta."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that. Let's do that right over here. So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4 times d theta dt. Well, that is just 0.2."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4 times d theta dt. Well, that is just 0.2. So times 0.2. And then this is going to be equal to 1 over 500. And we want to make sure, since this is in radians per minute, we're going to get meters per minute."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that is just 0.2. So times 0.2. And then this is going to be equal to 1 over 500. And we want to make sure, since this is in radians per minute, we're going to get meters per minute. And this is meters right over here. We're going to get meters per minute right over here. We just want to make sure we know what our units are doing."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we want to make sure, since this is in radians per minute, we're going to get meters per minute. And this is meters right over here. We're going to get meters per minute right over here. We just want to make sure we know what our units are doing. I haven't written the units here to save some space. But we get 1 over 500 times dh dt. So if we want to solve for dh dt, you can multiply both sides by 500."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We just want to make sure we know what our units are doing. I haven't written the units here to save some space. But we get 1 over 500 times dh dt. So if we want to solve for dh dt, you can multiply both sides by 500. And you get the rate at which our height is changing is equal to 500 times secant squared of pi over 4. That is 1 over cosine squared of pi over 4. Cosine of pi over 4."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we want to solve for dh dt, you can multiply both sides by 500. And you get the rate at which our height is changing is equal to 500 times secant squared of pi over 4. That is 1 over cosine squared of pi over 4. Cosine of pi over 4. Let me write this over here. Cosine of pi over 4 is square root of 2 over 2. Cosine squared of pi over 4 is going to be equal to 2 over 4, which is equal to 1 half."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of pi over 4. Let me write this over here. Cosine of pi over 4 is square root of 2 over 2. Cosine squared of pi over 4 is going to be equal to 2 over 4, which is equal to 1 half. And so secant squared of pi over 4 is just 1 over that is equal to 2. So this is going to be equal to, let me rewrite this. So the secant squared of pi over 4, let me erase this right over here."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine squared of pi over 4 is going to be equal to 2 over 4, which is equal to 1 half. And so secant squared of pi over 4 is just 1 over that is equal to 2. So this is going to be equal to, let me rewrite this. So the secant squared of pi over 4, let me erase this right over here. Secant squared of pi over 4, all of this business right over here, simplifies to 2. So times 2 times 0.2. So what is this going to be?"}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the secant squared of pi over 4, let me erase this right over here. Secant squared of pi over 4, all of this business right over here, simplifies to 2. So times 2 times 0.2. So what is this going to be? This is going to be 500 times 0.4. So this is equal to 500 times 0.4, which is equal to, let me make sure I get this right. This would be with two 0's and one behind the decimal."}, {"video_title": "Related rates balloon   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is this going to be? This is going to be 500 times 0.4. So this is equal to 500 times 0.4, which is equal to, let me make sure I get this right. This would be with two 0's and one behind the decimal. Yep, there you go. It would be 200. So that rate at which our height is changing with respect to time right at that moment is 200 meters per minute."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is y is equal to one over x squared. This is y is equal to one over x. And we explored what's the limit as x approaches zero in either of those scenarios. And in this left scenario, we saw as x becomes less and less negative, as it approaches zero from the left-hand side, our, the value of one over x squared is unbounded in the positive direction. And the same thing happens as we approach x from the right. As we become less and less positive, but we are still positive, the value of one over x squared becomes unbounded in the positive direction. So in that video, we just said, hey, one could say that this limit is unbounded."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in this left scenario, we saw as x becomes less and less negative, as it approaches zero from the left-hand side, our, the value of one over x squared is unbounded in the positive direction. And the same thing happens as we approach x from the right. As we become less and less positive, but we are still positive, the value of one over x squared becomes unbounded in the positive direction. So in that video, we just said, hey, one could say that this limit is unbounded. But what we're going to do in this video is introduce new notation. Instead of just saying it's unbounded, we could say, hey, from both the left and the right, it looks like we're going to positive infinity. So we can introduce this notation of saying, hey, this is going to infinity, which you will sometimes see used."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in that video, we just said, hey, one could say that this limit is unbounded. But what we're going to do in this video is introduce new notation. Instead of just saying it's unbounded, we could say, hey, from both the left and the right, it looks like we're going to positive infinity. So we can introduce this notation of saying, hey, this is going to infinity, which you will sometimes see used. Some people would call this unbounded. Some people say it does not exist because it's not approaching some finite value. While some people will use this notation of the limit going to infinity."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can introduce this notation of saying, hey, this is going to infinity, which you will sometimes see used. Some people would call this unbounded. Some people say it does not exist because it's not approaching some finite value. While some people will use this notation of the limit going to infinity. But what about this scenario? Can we use our new notation here? Well, when we approach zero from the left, it looks like we're unbounded in the negative direction."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "While some people will use this notation of the limit going to infinity. But what about this scenario? Can we use our new notation here? Well, when we approach zero from the left, it looks like we're unbounded in the negative direction. And when we approach zero from the right, we're unbounded in the positive direction. So here, you still could not say that the limit is approaching infinity because from the right, it's approaching infinity, but from the left, it's approaching negative infinity. So you would still say that this does not exist."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, when we approach zero from the left, it looks like we're unbounded in the negative direction. And when we approach zero from the right, we're unbounded in the positive direction. So here, you still could not say that the limit is approaching infinity because from the right, it's approaching infinity, but from the left, it's approaching negative infinity. So you would still say that this does not exist. You could do one-sided limits here, which if you're not familiar with, I encourage you to review it on Khan Academy. If you said the limit of one over x as x approaches zero from the left-hand side from values less than zero, well, then you would look at this right over here and say, well, look, it looks like we're going unbounded in the negative direction. So you would say this is equal to negative infinity."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you would still say that this does not exist. You could do one-sided limits here, which if you're not familiar with, I encourage you to review it on Khan Academy. If you said the limit of one over x as x approaches zero from the left-hand side from values less than zero, well, then you would look at this right over here and say, well, look, it looks like we're going unbounded in the negative direction. So you would say this is equal to negative infinity. And of course, if you said the limit as x approaches zero from the right of one over x, well, here, you're unbounded in the positive direction, so that's going to be equal to positive infinity. Let's do an example problem from Khan Academy based on this idea and this notation. So here it says, consider graphs A, B, and C. The dashed lines represent asymptotes."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you would say this is equal to negative infinity. And of course, if you said the limit as x approaches zero from the right of one over x, well, here, you're unbounded in the positive direction, so that's going to be equal to positive infinity. Let's do an example problem from Khan Academy based on this idea and this notation. So here it says, consider graphs A, B, and C. The dashed lines represent asymptotes. Which of the graphs agree with this statement that the limit as x approaches one of h of x is equal to infinity? Pause this video and see if you can figure it out. All right, let's go through each of these."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here it says, consider graphs A, B, and C. The dashed lines represent asymptotes. Which of the graphs agree with this statement that the limit as x approaches one of h of x is equal to infinity? Pause this video and see if you can figure it out. All right, let's go through each of these. So we wanna think about what happens at x equals one, so that's right over here on graph A. So as we approach x equals one, so let me write this. So the limit, let me do this for the different graphs."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's go through each of these. So we wanna think about what happens at x equals one, so that's right over here on graph A. So as we approach x equals one, so let me write this. So the limit, let me do this for the different graphs. So for graph A, the limit as x approaches one from the left, that looks like it's unbounded in the positive direction, that equals infinity. And the limit as x approaches one from the right, well, that looks like it's going to negative infinity. That equals negative infinity."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the limit, let me do this for the different graphs. So for graph A, the limit as x approaches one from the left, that looks like it's unbounded in the positive direction, that equals infinity. And the limit as x approaches one from the right, well, that looks like it's going to negative infinity. That equals negative infinity. And since these are going in two different directions, you wouldn't be able to say that the limit as x approaches one from both directions is equal to infinity, so I would rule this one out. Now let's look at choice B. What's the limit as x approaches one from the left?"}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That equals negative infinity. And since these are going in two different directions, you wouldn't be able to say that the limit as x approaches one from both directions is equal to infinity, so I would rule this one out. Now let's look at choice B. What's the limit as x approaches one from the left? And of course, these are of h of x. Gotta write that down. So of h of x, right over here. Well, as we approach from the left, we are going to, looks like we're going to positive infinity and it looks like the limit of h of x as we approach one from the right is also going to positive infinity."}, {"video_title": "Introduction to infinite limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's the limit as x approaches one from the left? And of course, these are of h of x. Gotta write that down. So of h of x, right over here. Well, as we approach from the left, we are going to, looks like we're going to positive infinity and it looks like the limit of h of x as we approach one from the right is also going to positive infinity. And so since we're approaching, you could say the same direction of infinity, you could say this for B. So B meets the constraints, but let's just check C to make sure. Well, you can see very clearly, x equals one, that as we approach it from the left, we go to negative infinity and as we approach from the right, we go to positive infinity so this, once again, would not be approaching the same infinity, so you would rule this one out as well."}, {"video_title": "Unbounded limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause this video and see if you can figure that out. Well, when you try to figure it out, you immediately see something interesting happening at x equals zero. The closer we get to zero from the left, you take one over x squared, it just gets larger and larger and larger. It doesn't approach some finite value. It's unbounded, it has no bound. And the same thing is happening as we approach from the right. As we get values closer and closer to zero from the right, we get larger and larger values for one over x squared without bound."}, {"video_title": "Unbounded limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It doesn't approach some finite value. It's unbounded, it has no bound. And the same thing is happening as we approach from the right. As we get values closer and closer to zero from the right, we get larger and larger values for one over x squared without bound. So terminology that folks will sometimes use where they're both going in the same direction but it's unbounded is they'll say this limit is unbounded. In some context, you might hear teachers say that this limit does not exist or and it definitely does not exist if you're thinking about approaching a finite value. In future videos, we'll start to introduce ideas of infinity and notations around limits and infinity where we can get a little bit more specific about what type of limit this is."}, {"video_title": "Unbounded limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "As we get values closer and closer to zero from the right, we get larger and larger values for one over x squared without bound. So terminology that folks will sometimes use where they're both going in the same direction but it's unbounded is they'll say this limit is unbounded. In some context, you might hear teachers say that this limit does not exist or and it definitely does not exist if you're thinking about approaching a finite value. In future videos, we'll start to introduce ideas of infinity and notations around limits and infinity where we can get a little bit more specific about what type of limit this is. But with that out of the way, let's look at another scenario. This right over here, you might recognize, is the graph of y is equal to one over x. So I'm gonna ask you the same question."}, {"video_title": "Unbounded limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "In future videos, we'll start to introduce ideas of infinity and notations around limits and infinity where we can get a little bit more specific about what type of limit this is. But with that out of the way, let's look at another scenario. This right over here, you might recognize, is the graph of y is equal to one over x. So I'm gonna ask you the same question. Pause this video and think about what's the limit of one over x as x approaches zero. Pause this video and figure it out. All right, so here when we approach from the left, we get more and more and more negative values while when we approach from the right, we're getting more and more positive values."}, {"video_title": "Unbounded limits   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm gonna ask you the same question. Pause this video and think about what's the limit of one over x as x approaches zero. Pause this video and figure it out. All right, so here when we approach from the left, we get more and more and more negative values while when we approach from the right, we're getting more and more positive values. So in this situation where we're not getting unbounded in the same direction, the previous example, we were being unbounded in the positive direction but here from the left, we're getting unbounded in the negative direction while from the right, we're getting unbounded in the positive direction. And so when you're thinking about the limit as you approach a point, if it's not even approaching the same value or even the same direction, you would just clearly say that this limit does not exist. Does not exist."}, {"video_title": "Worked examples Definite integral properties 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're given the graph of f of x and of y equals f of x, and the area between f of x and the x-axis over different intervals. Well, when you look at this, you actually don't even have to look at this graph over here because in general, if I have the definite integral of any function, f of x dx, from, let's say, a to the same value, from one value to the same value, this is always going to be equal to zero. We're going from three to three. We could be going from negative pi to negative pi. It's always going to be zero. One way to think about it, we're starting and stopping here at three, so we're not capturing any area. Let's do another one."}, {"video_title": "Worked examples Definite integral properties 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could be going from negative pi to negative pi. It's always going to be zero. One way to think about it, we're starting and stopping here at three, so we're not capturing any area. Let's do another one. So here, we want to find the definite integral from seven to four of f of x dx. So we want to go from seven to four. So we want to go from seven to four."}, {"video_title": "Worked examples Definite integral properties 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do another one. So here, we want to find the definite integral from seven to four of f of x dx. So we want to go from seven to four. So we want to go from seven to four. Now, you might be tempted to say, okay, well, look, the area between f of x and x is two, so maybe this thing is two. But the key realization is this area only applies when you have the lower bound as the lower bound and the higher value as the higher bound. So the integral from four to seven of f of x dx, this thing, this thing is equal to two."}, {"video_title": "Worked examples Definite integral properties 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to go from seven to four. Now, you might be tempted to say, okay, well, look, the area between f of x and x is two, so maybe this thing is two. But the key realization is this area only applies when you have the lower bound as the lower bound and the higher value as the higher bound. So the integral from four to seven of f of x dx, this thing, this thing is equal to two. This thing is depicting that area right over there. So what about this, where we've switched it? Instead of going from four to seven, we're going from seven to four."}, {"video_title": "Worked examples Definite integral properties 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the integral from four to seven of f of x dx, this thing, this thing is equal to two. This thing is depicting that area right over there. So what about this, where we've switched it? Instead of going from four to seven, we're going from seven to four. Well, the key realization is is if you switch the bounds, and this is a key definite integral property, that's going to give you the negative value. So this is going to be equal to the negative of the integral from four to seven of f of x dx. And so this is going to be negative, and we just figured out the integral from four to seven of f of x dx."}, {"video_title": "Worked examples Definite integral properties 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of going from four to seven, we're going from seven to four. Well, the key realization is is if you switch the bounds, and this is a key definite integral property, that's going to give you the negative value. So this is going to be equal to the negative of the integral from four to seven of f of x dx. And so this is going to be negative, and we just figured out the integral from four to seven of f of x dx. Well, now that is this area. F of x is above the x-axis. It's a positive area."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you have a horizontal tangent right over, a horizontal tangent right over there. Let me draw that a little bit neater. Right over there, a horizontal tangent right over there, and a horizontal tangent right over there. All right, the areas of the regions bounded by the x-axis in the graph of f prime on the intervals negative two to one, closed intervals from negative two to one, so this region right over here, and the region from one to four, so this region right over there, they tell us the areas are nine and 12 respectively. So that area is nine, and that area is 12. So part a, find all x-coordinates at which f has a relative maximum. Give a reason for your answer."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, the areas of the regions bounded by the x-axis in the graph of f prime on the intervals negative two to one, closed intervals from negative two to one, so this region right over here, and the region from one to four, so this region right over there, they tell us the areas are nine and 12 respectively. So that area is nine, and that area is 12. So part a, find all x-coordinates at which f has a relative maximum. Give a reason for your answer. All x-coordinates at which f has a relative maximum. So you might say, oh, look, this looks like a relative maximum over here, but this isn't f, this is the graph of f prime. So let's think about what needs to, we don't have a graph of f in front of us."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Give a reason for your answer. All x-coordinates at which f has a relative maximum. So you might say, oh, look, this looks like a relative maximum over here, but this isn't f, this is the graph of f prime. So let's think about what needs to, we don't have a graph of f in front of us. So let's think about what needs to be true for f to have a relative maximum at a point. So let's, we are probably familiar with what relative maxima look like. They look like a little lump like that."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what needs to, we don't have a graph of f in front of us. So let's think about what needs to be true for f to have a relative maximum at a point. So let's, we are probably familiar with what relative maxima look like. They look like a little lump like that. They could also actually look like that, but since this is a differentiable function over the interval, we're probably not dealing with a relative maximum that looks like that. And so what do we know about a relative maximum point? So let's say, let's say that's our relative maximum."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "They look like a little lump like that. They could also actually look like that, but since this is a differentiable function over the interval, we're probably not dealing with a relative maximum that looks like that. And so what do we know about a relative maximum point? So let's say, let's say that's our relative maximum. Well, as we approach our relative maximum for values for, as we have x-values that are approaching the x-value of our relative maximum point, as we approach it from values below that x-value, we see that we have a positive slope. Our function needs to be increasing. So over here, over here, we see f is increasing, going into the relative maximum point."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say, let's say that's our relative maximum. Well, as we approach our relative maximum for values for, as we have x-values that are approaching the x-value of our relative maximum point, as we approach it from values below that x-value, we see that we have a positive slope. Our function needs to be increasing. So over here, over here, we see f is increasing, going into the relative maximum point. f is increasing, which means that the derivative of f, the derivative of f must be greater than zero. And then after we pass that maximum point, after we pass that maximum point, we see that our function needs to be decreasing. Let me just add another color."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here, over here, we see f is increasing, going into the relative maximum point. f is increasing, which means that the derivative of f, the derivative of f must be greater than zero. And then after we pass that maximum point, after we pass that maximum point, we see that our function needs to be decreasing. Let me just add another color. We see that our function is decreasing right over here. So f decreasing, decreasing, which means that f-prime of x needs to be less than zero. So our relative maximum point should be, should happen at an x-value."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just add another color. We see that our function is decreasing right over here. So f decreasing, decreasing, which means that f-prime of x needs to be less than zero. So our relative maximum point should be, should happen at an x-value. It should happen at an x-value where our first derivative transitions from being greater than zero to being less than zero. So what x-values, let me say this. So we have f has relative, let me just write it shorthand, relative maximum at x-values where f-prime transitions, transitions, transitions from positive, positive, to negative, to, let me write this a little bit neater, to negative, to negative."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our relative maximum point should be, should happen at an x-value. It should happen at an x-value where our first derivative transitions from being greater than zero to being less than zero. So what x-values, let me say this. So we have f has relative, let me just write it shorthand, relative maximum at x-values where f-prime transitions, transitions, transitions from positive, positive, to negative, to, let me write this a little bit neater, to negative, to negative. And where do we see f-prime transitioning from positive to negative? Well, over here we see that only happening once. We see right here f-prime is positive, positive, positive, and then it goes negative, negative, negative."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have f has relative, let me just write it shorthand, relative maximum at x-values where f-prime transitions, transitions, transitions from positive, positive, to negative, to, let me write this a little bit neater, to negative, to negative. And where do we see f-prime transitioning from positive to negative? Well, over here we see that only happening once. We see right here f-prime is positive, positive, positive, and then it goes negative, negative, negative. So we see f-prime is positive over here. And then right when we hit x equals negative two, f-prime becomes negative. f-prime becomes negative."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We see right here f-prime is positive, positive, positive, and then it goes negative, negative, negative. So we see f-prime is positive over here. And then right when we hit x equals negative two, f-prime becomes negative. f-prime becomes negative. So we know that the function itself, not f-prime, f must be increasing here because f-prime is positive. And then our function f is decreasing here because f-prime is negative. And so this happens at x equals two."}, {"video_title": "2015 AP Calculus AB 5a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "f-prime becomes negative. So we know that the function itself, not f-prime, f must be increasing here because f-prime is positive. And then our function f is decreasing here because f-prime is negative. And so this happens at x equals two. So let me write that down. This happens at x equals two. This happens, happens at x equals two."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "For zero is less than or equal to t is less than or equal to 10, Bob's velocity is modeled by b of t is equal to t to the third minus 60 squared plus 300, where t is measured in minutes and b of t is measured in meters per minute. Find Bob's acceleration at time t equals 5. Well, acceleration, this is a velocity function right over here, so the acceleration is going to be the derivative of the velocity function with respect to time. What is the rate of change of velocity with respect to time? That's acceleration. So we really just want to evaluate. This Bob's acceleration at t equals 5, that's going to be b prime of 5."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the rate of change of velocity with respect to time? That's acceleration. So we really just want to evaluate. This Bob's acceleration at t equals 5, that's going to be b prime of 5. So let's first figure out what b prime of t is. b prime of t is equal to, we'll take the derivative here, it's pretty straightforward, just use the power rule. So it's going to be 3t squared minus 12t, 2 times negative 6 is 12, or negative 12, and then the derivative of 300, 300 doesn't change with respect to time, so it's just a zero."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This Bob's acceleration at t equals 5, that's going to be b prime of 5. So let's first figure out what b prime of t is. b prime of t is equal to, we'll take the derivative here, it's pretty straightforward, just use the power rule. So it's going to be 3t squared minus 12t, 2 times negative 6 is 12, or negative 12, and then the derivative of 300, 300 doesn't change with respect to time, so it's just a zero. And so b prime of 5 is going to be equal to 3 times 5 squared minus 12 times 5, which is equal to 75 minus 60, which is equal to 15. And the units here, this is acceleration, so this is going to be, his velocity was in meters per minute, and so this is going to be meters per minute per minute, because remember, time is in terms of minutes. So we could write as meters, let me write it out, meters per minute per minute, which is the same thing as meters per minute, meters per minute squared."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be 3t squared minus 12t, 2 times negative 6 is 12, or negative 12, and then the derivative of 300, 300 doesn't change with respect to time, so it's just a zero. And so b prime of 5 is going to be equal to 3 times 5 squared minus 12 times 5, which is equal to 75 minus 60, which is equal to 15. And the units here, this is acceleration, so this is going to be, his velocity was in meters per minute, and so this is going to be meters per minute per minute, because remember, time is in terms of minutes. So we could write as meters, let me write it out, meters per minute per minute, which is the same thing as meters per minute, meters per minute squared. Alright, let's do the next part. Based on the model, based on the model b from part c, find Bob's average velocity during the interval from zero is less than or equal to t, is less than or equal to 10. And if the notion of average velocity or average value of a function is completely foreign to you, I encourage you to watch the videos on Khan Academy on finding the average of a function."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could write as meters, let me write it out, meters per minute per minute, which is the same thing as meters per minute, meters per minute squared. Alright, let's do the next part. Based on the model, based on the model b from part c, find Bob's average velocity during the interval from zero is less than or equal to t, is less than or equal to 10. And if the notion of average velocity or average value of a function is completely foreign to you, I encourage you to watch the videos on Khan Academy on finding the average of a function. But straight, just to kind of cut to the chase, the average velocity, the average velocity is going to be the area under the velocity curve divided by our change in time. So the area under the velocity curve from t equals zero to t equals 10 of b of t, of b of t, dt, divided by our change in time. So it's going to be divided by, well, you're going from zero to 10, so 10 minus zero is going to be equal to 10."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if the notion of average velocity or average value of a function is completely foreign to you, I encourage you to watch the videos on Khan Academy on finding the average of a function. But straight, just to kind of cut to the chase, the average velocity, the average velocity is going to be the area under the velocity curve divided by our change in time. So the area under the velocity curve from t equals zero to t equals 10 of b of t, of b of t, dt, divided by our change in time. So it's going to be divided by, well, you're going from zero to 10, so 10 minus zero is going to be equal to 10. And if you wanted the intuition here, it's like, well, if you know the area of something, and if you wanted to find its average height, you could just divide by its width, and that's what we're doing here. If we know the area of something, we want to figure out its average height, and so you divide by its width. That's, I guess, a very high level intuition for where this expression came from."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be divided by, well, you're going from zero to 10, so 10 minus zero is going to be equal to 10. And if you wanted the intuition here, it's like, well, if you know the area of something, and if you wanted to find its average height, you could just divide by its width, and that's what we're doing here. If we know the area of something, we want to figure out its average height, and so you divide by its width. That's, I guess, a very high level intuition for where this expression came from. And so this is going to be equal to 1 tenth times the integral from zero to 10, and b of t is t to the third power minus 6t squared plus 300 dt. And so this is going to be equal to 1 over 10. Take the antiderivative here, so this is going to be t to the fourth over 4, t to the fourth over 4."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's, I guess, a very high level intuition for where this expression came from. And so this is going to be equal to 1 tenth times the integral from zero to 10, and b of t is t to the third power minus 6t squared plus 300 dt. And so this is going to be equal to 1 over 10. Take the antiderivative here, so this is going to be t to the fourth over 4, t to the fourth over 4. And then this is going to be, if we increase the exponent here by 1, it's t to the third, and then you divide by 3. So it's negative 6 divided by 3 is negative 2t to the third, and then plus 300t, 300t, and I'm going to evaluate it. I am going to evaluate it at 10, and subtract from that, and evaluate it at 0."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Take the antiderivative here, so this is going to be t to the fourth over 4, t to the fourth over 4. And then this is going to be, if we increase the exponent here by 1, it's t to the third, and then you divide by 3. So it's negative 6 divided by 3 is negative 2t to the third, and then plus 300t, 300t, and I'm going to evaluate it. I am going to evaluate it at 10, and subtract from that, and evaluate it at 0. And so this is going to be equal to 1 tenth, that same 1 tenth there. And when you evaluate all of this at 10, what are we going to get? Let's see, 10 to the fourth power is 10,000 divided by 4 is 2,500."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I am going to evaluate it at 10, and subtract from that, and evaluate it at 0. And so this is going to be equal to 1 tenth, that same 1 tenth there. And when you evaluate all of this at 10, what are we going to get? Let's see, 10 to the fourth power is 10,000 divided by 4 is 2,500. And then minus 2 times 10 to the third, so it's 2 times 1,000. So minus 2,000. And then 300 times 10, well that's plus 3,000."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see, 10 to the fourth power is 10,000 divided by 4 is 2,500. And then minus 2 times 10 to the third, so it's 2 times 1,000. So minus 2,000. And then 300 times 10, well that's plus 3,000. And then you subtract all of this evaluated at 0, which is just going to be 0. So this is going to be equal to 2,500 minus 2,000 is 500, plus 3,000. This all simplifies to 3,500, or 3,500."}, {"video_title": "2015 AP Calculus AB BC 3cd   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then 300 times 10, well that's plus 3,000. And then you subtract all of this evaluated at 0, which is just going to be 0. So this is going to be equal to 2,500 minus 2,000 is 500, plus 3,000. This all simplifies to 3,500, or 3,500. And then you divide it by 10. This is going to be 350, and it's an average velocity, meters per 350, meters per minute. And we are done."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can visualize that. So this is x equaling one right over here. This is the value of the function when x is equal to one, right over there. And then the tangent line looks something like, will look something like, oh no, I can do a better job than that. It's going to look something like that. And what we want to do is find the equation, the equation of that line. And if you are inspired, I encourage you to be, pause the video and try to work it out."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the tangent line looks something like, will look something like, oh no, I can do a better job than that. It's going to look something like that. And what we want to do is find the equation, the equation of that line. And if you are inspired, I encourage you to be, pause the video and try to work it out. Well, the way that we could do this is if we find the derivative at x equals one, the derivative is the slope of the tangent line. And so we'll know the slope of the tangent line, and we know that it contains that point, and then we can use that to find the equation of the tangent line. So let's actually just, let's just, so we want the equation of the tangent line when x is equal to one."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you are inspired, I encourage you to be, pause the video and try to work it out. Well, the way that we could do this is if we find the derivative at x equals one, the derivative is the slope of the tangent line. And so we'll know the slope of the tangent line, and we know that it contains that point, and then we can use that to find the equation of the tangent line. So let's actually just, let's just, so we want the equation of the tangent line when x is equal to one. So let's just first of all evaluate f of one. So f of one is equal to one to the third power, which is one, minus six times one squared, so it's just minus six, and then plus one, plus one, minus five. So this is equal to what?"}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's actually just, let's just, so we want the equation of the tangent line when x is equal to one. So let's just first of all evaluate f of one. So f of one is equal to one to the third power, which is one, minus six times one squared, so it's just minus six, and then plus one, plus one, minus five. So this is equal to what? Two minus 11, which is equal to negative nine. And that looks about right. That looks like about negative nine right over there."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to what? Two minus 11, which is equal to negative nine. And that looks about right. That looks like about negative nine right over there. The scales are different on the y and the x-axis. And so that is f of one. It is negative nine."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That looks like about negative nine right over there. The scales are different on the y and the x-axis. And so that is f of one. It is negative nine. Did I do that right? This is negative five, negative nine, yep, negative nine. And now let's evaluate what the derivative is at one."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is negative nine. Did I do that right? This is negative five, negative nine, yep, negative nine. And now let's evaluate what the derivative is at one. So what is f prime of x? f prime of x. Well here it's just a polynomial."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's evaluate what the derivative is at one. So what is f prime of x? f prime of x. Well here it's just a polynomial. Take the derivative of x to the third. Well we apply the power rule. We bring the three out front."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well here it's just a polynomial. Take the derivative of x to the third. Well we apply the power rule. We bring the three out front. So you get three x to the, and then we go one less than three to get the second power. And then you have minus six x squared. So you bring the two times the six to get 12."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We bring the three out front. So you get three x to the, and then we go one less than three to get the second power. And then you have minus six x squared. So you bring the two times the six to get 12. So minus 12 x to the, well two minus one is one power, so that's the same thing as 12x. And then plus the derivative of x is just one. That's just going to be one."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you bring the two times the six to get 12. So minus 12 x to the, well two minus one is one power, so that's the same thing as 12x. And then plus the derivative of x is just one. That's just going to be one. And if you view this as x to the first power, we're just bringing the one out front and decrementing the one. So we have one times x to the zero power, which is just one. And then the derivative of a constant here is just going to be zero."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's just going to be one. And if you view this as x to the first power, we're just bringing the one out front and decrementing the one. So we have one times x to the zero power, which is just one. And then the derivative of a constant here is just going to be zero. So this is our derivative of f, and if we want to evaluate it at one, f prime of one is going to be three times one squared, which is just three, minus 12 times one, so it's just minus 12, and then we have plus one. So this is three minus 12 is negative nine, and negative nine plus one is equal to negative eight. So we know the slope right over here is a slope of negative eight."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the derivative of a constant here is just going to be zero. So this is our derivative of f, and if we want to evaluate it at one, f prime of one is going to be three times one squared, which is just three, minus 12 times one, so it's just minus 12, and then we have plus one. So this is three minus 12 is negative nine, and negative nine plus one is equal to negative eight. So we know the slope right over here is a slope of negative eight. We know a point on that line, it contains the point one comma negative nine, so we could use that information to find the equation of the line. The line, just to remind ourselves, has the form y is equal to mx plus b, where m is the slope, so we know that y is going to be equal to negative 8x plus b and now we can substitute the x and y value that we know sits on that line to solve for b. So we know that y is equal to negative nine, let me just write this here, y is equal to negative nine when x is equal to one."}, {"video_title": "Tangents of polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know the slope right over here is a slope of negative eight. We know a point on that line, it contains the point one comma negative nine, so we could use that information to find the equation of the line. The line, just to remind ourselves, has the form y is equal to mx plus b, where m is the slope, so we know that y is going to be equal to negative 8x plus b and now we can substitute the x and y value that we know sits on that line to solve for b. So we know that y is equal to negative nine, let me just write this here, y is equal to negative nine when x is equal to one. And so we get negative nine is equal to negative eight times one, so negative eight plus b. Well, let's see, we could add eight to both sides and we get negative one is equal to b. So we're done."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So let's start with a little bit of a geometric or trigonometric construction that I have here. So this white circle, this is a unit circle. Let me label it as such. So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function?"}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle?"}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta? Well, cosine of zero is just one, and it's a continuous function, so this is just going to be one. So let's see."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "In the last video, we tried to come up with a somewhat rigorous definition of what a limit is, where we say when you say that the limit of f of x as x approaches c is equal to l, you're really saying, and this is a somewhat rigorous definition, that you can get f of x as close as you want to l by making x sufficiently close to c. So let's see if we can put a little bit meat on it. So instead of saying as close as you want, let's call that some positive number epsilon. So I'm just going to use the Greek letter epsilon right over there. So it really turns into a game. So you tell me how close you want. So this is the game. You tell me how close you want f of x to be to l. And you do this by giving me a positive number that we call epsilon, which is really how close you want f of x to be to l. So you give a positive number epsilon."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So it really turns into a game. So you tell me how close you want. So this is the game. You tell me how close you want f of x to be to l. And you do this by giving me a positive number that we call epsilon, which is really how close you want f of x to be to l. So you give a positive number epsilon. And epsilon is how close do you want to be. How close. So for example, if epsilon is 0.01, that says that you want f of x to be within 0.01 of epsilon."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "You tell me how close you want f of x to be to l. And you do this by giving me a positive number that we call epsilon, which is really how close you want f of x to be to l. So you give a positive number epsilon. And epsilon is how close do you want to be. How close. So for example, if epsilon is 0.01, that says that you want f of x to be within 0.01 of epsilon. And so what I then do is I say, well, OK, you've given me that epsilon. I'm going to find you. I will find you another number, find another positive number, another number which we'll call delta, the lowercase delta, the Greek letter delta, such that, so I'll say where, if x is within delta of c, then f of x will be within epsilon of our limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So for example, if epsilon is 0.01, that says that you want f of x to be within 0.01 of epsilon. And so what I then do is I say, well, OK, you've given me that epsilon. I'm going to find you. I will find you another number, find another positive number, another number which we'll call delta, the lowercase delta, the Greek letter delta, such that, so I'll say where, if x is within delta of c, then f of x will be within epsilon of our limit. So let's see if these are really saying the same thing. And this yellow definition right over here, we said you can get f of x as close as you want to l by making x sufficiently close to c. This second definition, which I kind of made as a little bit more of a game, is doing the same thing. Someone is saying how close they want f of x to be to l. And the burden is then to find a delta where as long as x is within delta of c, then f of x will be within epsilon of the limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "I will find you another number, find another positive number, another number which we'll call delta, the lowercase delta, the Greek letter delta, such that, so I'll say where, if x is within delta of c, then f of x will be within epsilon of our limit. So let's see if these are really saying the same thing. And this yellow definition right over here, we said you can get f of x as close as you want to l by making x sufficiently close to c. This second definition, which I kind of made as a little bit more of a game, is doing the same thing. Someone is saying how close they want f of x to be to l. And the burden is then to find a delta where as long as x is within delta of c, then f of x will be within epsilon of the limit. So that is doing it. It's saying, look, if we were constraining x in such a way that if x is in that range to c, then f of x will be as close as you want. So let's make this a little bit clearer by diagramming right over here."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "Someone is saying how close they want f of x to be to l. And the burden is then to find a delta where as long as x is within delta of c, then f of x will be within epsilon of the limit. So that is doing it. It's saying, look, if we were constraining x in such a way that if x is in that range to c, then f of x will be as close as you want. So let's make this a little bit clearer by diagramming right over here. You show up and you say, well, I want f of x to be within epsilon of our limit. So this right over here, this point right over here is our limit plus epsilon. And this right over here might be our limit minus."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So let's make this a little bit clearer by diagramming right over here. You show up and you say, well, I want f of x to be within epsilon of our limit. So this right over here, this point right over here is our limit plus epsilon. And this right over here might be our limit minus. This right over here is the limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "And this right over here might be our limit minus. This right over here is the limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit. And I can do that by defining a range around c. And really, I could visually look at this boundary. But I could even go narrower than boundary. I can go right over here."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "I think I can get your f of x within this range of our limit. And I can do that by defining a range around c. And really, I could visually look at this boundary. But I could even go narrower than boundary. I can go right over here. Says, OK, I meet your challenge. I will find another number, delta. So this right over here is c plus delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "I can go right over here. Says, OK, I meet your challenge. I will find another number, delta. So this right over here is c plus delta. This right over here is c minus. Let me write this down. Is c minus delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So this right over here is c plus delta. This right over here is c minus. Let me write this down. Is c minus delta. So I'll find you some delta so that if you take any x in the range c minus delta to c plus delta, and maybe the function's not even defined at c. So we think of ones that maybe aren't c, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range l plus epsilon or l minus epsilon."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "Is c minus delta. So I'll find you some delta so that if you take any x in the range c minus delta to c plus delta, and maybe the function's not even defined at c. So we think of ones that maybe aren't c, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range l plus epsilon or l minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon. Then I will find you a delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "They're going to be within the range l plus epsilon or l minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon. Then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here, but it's the exact same thing. So you give me, or let me write it this way."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "Then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here, but it's the exact same thing. So you give me, or let me write it this way. Given an epsilon greater than 0, we can find, so that's kind of the first part of the game, we can find a delta greater than 0 such that if x is within delta of c. So what's another way of saying that x is within delta of c? Well, one way you could say, well, what's the distance between x and c is going to be less than delta. This statement is true for any x that's within delta of c. The difference between the two is going to be less than delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So you give me, or let me write it this way. Given an epsilon greater than 0, we can find, so that's kind of the first part of the game, we can find a delta greater than 0 such that if x is within delta of c. So what's another way of saying that x is within delta of c? Well, one way you could say, well, what's the distance between x and c is going to be less than delta. This statement is true for any x that's within delta of c. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between c minus delta and c plus delta, and these are the x's that satisfy that right over here, then the distance between your f of x and your limit, and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is if the limit truly does exist, it truly is L, is if you give me any positive number epsilon, it could be a super, super small one, we can find a delta. So we can define a range around c so that if we take any x value that is within delta of c, that's all this statement is saying, that the distance between x and c is less than delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "This statement is true for any x that's within delta of c. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between c minus delta and c plus delta, and these are the x's that satisfy that right over here, then the distance between your f of x and your limit, and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is if the limit truly does exist, it truly is L, is if you give me any positive number epsilon, it could be a super, super small one, we can find a delta. So we can define a range around c so that if we take any x value that is within delta of c, that's all this statement is saying, that the distance between x and c is less than delta. So it's within delta c. So that's these points right over here. That f of those x's, the function evaluated at those x's is going to be within the range that you are specifying. It's going to be within epsilon of our limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So we can define a range around c so that if we take any x value that is within delta of c, that's all this statement is saying, that the distance between x and c is less than delta. So it's within delta c. So that's these points right over here. That f of those x's, the function evaluated at those x's is going to be within the range that you are specifying. It's going to be within epsilon of our limit. The f of x, the difference between f of x and your limit will be less than epsilon. Your f of x is going to sit someplace over there. So that's all the epsilon delta definition is telling us."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "With initial condition, at time zero, w is equal to 1400. And the units in this problem, it was 1400 tons, they told us, 1400 tons. So, it might seem strange to see a differential equations problem on an AP exam. You say, well, you know, this isn't fair. This isn't a differential equations class. And the thing to keep in mind is, if on an AP exam you see differential equations, they're really asking you to solve a differential equations that can be solved using the skills that you would learn in an AP class. And the main differential or class of differential equations that you could solve are the ones, are separable differential equations."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "You say, well, you know, this isn't fair. This isn't a differential equations class. And the thing to keep in mind is, if on an AP exam you see differential equations, they're really asking you to solve a differential equations that can be solved using the skills that you would learn in an AP class. And the main differential or class of differential equations that you could solve are the ones, are separable differential equations. The ones where you can separate the independent and the dependent variables. And if you don't have any idea of what I'm talking about, I'm about to show you. So, let me rewrite this differential equation."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the main differential or class of differential equations that you could solve are the ones, are separable differential equations. The ones where you can separate the independent and the dependent variables. And if you don't have any idea of what I'm talking about, I'm about to show you. So, let me rewrite this differential equation. We have dw dt is equal to one over 25 times w minus 300. Now, this differential equation, we have w here, and we have this dt, but we don't have t anywhere else. So, what I wanna do is I wanna get the parts that involve w onto the left-hand side."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, let me rewrite this differential equation. We have dw dt is equal to one over 25 times w minus 300. Now, this differential equation, we have w here, and we have this dt, but we don't have t anywhere else. So, what I wanna do is I wanna get the parts that involve w onto the left-hand side. And I wanna get anything that deals with dt on the right-hand side. So, let's divide both sides of this equation by w minus 300. So, we would have one over w minus 300, dw dt, the derivative of w with respect to t, is equal to, well, I divided this side by w minus 300, so it's equal to one over 25."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, what I wanna do is I wanna get the parts that involve w onto the left-hand side. And I wanna get anything that deals with dt on the right-hand side. So, let's divide both sides of this equation by w minus 300. So, we would have one over w minus 300, dw dt, the derivative of w with respect to t, is equal to, well, I divided this side by w minus 300, so it's equal to one over 25. Now, I wanna get the dt's on this side, so let's multiply both sides by dt. View the differential as just a very, very small change in t. And so, we get, and I'm a little bit imprecise with my notation here, but this is one way that you can think about it when you are dealing with differential equations like this, separable differential equations. You have one over w minus 300, dw, let me make sure that the dw doesn't look like it's in the denominator, dw, is equal to one over 25 dt."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, we would have one over w minus 300, dw dt, the derivative of w with respect to t, is equal to, well, I divided this side by w minus 300, so it's equal to one over 25. Now, I wanna get the dt's on this side, so let's multiply both sides by dt. View the differential as just a very, very small change in t. And so, we get, and I'm a little bit imprecise with my notation here, but this is one way that you can think about it when you are dealing with differential equations like this, separable differential equations. You have one over w minus 300, dw, let me make sure that the dw doesn't look like it's in the denominator, dw, is equal to one over 25 dt. And now, we can integrate both sides, which is essentially just saying that we're taking, since this is equal to this, that each little increment of w times this is equal to each little increment of dt times one over 25. We can then say, well, if we sum up all of the increments of w times this quantity, if we sum up the infinite number of infinitesimally small increments, that would be the same thing as summing up all of these characters, because each of these characters is really the same as each of these characters. So, let's do that sum."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have one over w minus 300, dw, let me make sure that the dw doesn't look like it's in the denominator, dw, is equal to one over 25 dt. And now, we can integrate both sides, which is essentially just saying that we're taking, since this is equal to this, that each little increment of w times this is equal to each little increment of dt times one over 25. We can then say, well, if we sum up all of the increments of w times this quantity, if we sum up the infinite number of infinitesimally small increments, that would be the same thing as summing up all of these characters, because each of these characters is really the same as each of these characters. So, let's do that sum. We're essentially just taking, we're just integrating both sides of this. And here, you might be able to do this in your head, or we could do the formal u substitution, just to make it clear what we're doing. If you said u is equal to w minus 300, then what is du?"}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, let's do that sum. We're essentially just taking, we're just integrating both sides of this. And here, you might be able to do this in your head, or we could do the formal u substitution, just to make it clear what we're doing. If you said u is equal to w minus 300, then what is du? If you wanted the differential, so if you wanna take the derivative of u with respect to w, you would say du with respect to w is equal to what? It is equal to one. The derivative of this with respect to w is one."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you said u is equal to w minus 300, then what is du? If you wanted the differential, so if you wanna take the derivative of u with respect to w, you would say du with respect to w is equal to what? It is equal to one. The derivative of this with respect to w is one. Or, if you multiply both sides times dw, you have du is equal to dw. And so, this left-hand side right over here, this integral could be rewritten as the integral of one over u du. We define this as u, and we just saw that du is the same thing as dw."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of this with respect to w is one. Or, if you multiply both sides times dw, you have du is equal to dw. And so, this left-hand side right over here, this integral could be rewritten as the integral of one over u du. We define this as u, and we just saw that du is the same thing as dw. So, this could be rewritten as du. And this is one of the basic integrals, so hopefully we've learned. And this is equal to the natural log of the absolute value of u."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "We define this as u, and we just saw that du is the same thing as dw. So, this could be rewritten as du. And this is one of the basic integrals, so hopefully we've learned. And this is equal to the natural log of the absolute value of u. Or, if you unwind the substitution, this is equal to the natural log of the absolute value of w minus 300. So, let me write this down. So, the left-hand side right over here, and you might be able to do that in your head."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is equal to the natural log of the absolute value of u. Or, if you unwind the substitution, this is equal to the natural log of the absolute value of w minus 300. So, let me write this down. So, the left-hand side right over here, and you might be able to do that in your head. You say, hey, look, you have w minus 300. Its derivative is one, so it's essentially there. So, I can just take this as if this was one over, I can just take the antiderivative here and treat this whole thing as kind of a variable."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, the left-hand side right over here, and you might be able to do that in your head. You say, hey, look, you have w minus 300. Its derivative is one, so it's essentially there. So, I can just take this as if this was one over, I can just take the antiderivative here and treat this whole thing as kind of a variable. So, it would just be the natural log of this entire thing, or the absolute value of that entire thing. So, the integral of the left-hand side is the natural log of w minus 300, of the absolute value of w minus 300. And the derivative of the right-hand side, and this is a little bit more straightforward, with respect to t, or sorry, the antiderivative with respect to t on the right-hand side is one over 25t, or we could say t over 25."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, I can just take this as if this was one over, I can just take the antiderivative here and treat this whole thing as kind of a variable. So, it would just be the natural log of this entire thing, or the absolute value of that entire thing. So, the integral of the left-hand side is the natural log of w minus 300, of the absolute value of w minus 300. And the derivative of the right-hand side, and this is a little bit more straightforward, with respect to t, or sorry, the antiderivative with respect to t on the right-hand side is one over 25t, or we could say t over 25. We'll write it as t over 25. And then, these are both indefinite integrals. You could have a constant on both sides or one side, but we could just put it on one side right over here."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the derivative of the right-hand side, and this is a little bit more straightforward, with respect to t, or sorry, the antiderivative with respect to t on the right-hand side is one over 25t, or we could say t over 25. We'll write it as t over 25. And then, these are both indefinite integrals. You could have a constant on both sides or one side, but we could just put it on one side right over here. You could have put plus some constant, you could have added some, you know, C1 over here, and then you could have had C2 over here, but then you could have subtracted C1 from both sides, and then just have a C3 constant over here. But to simplify it, you can just say, look, this is going to be equal to this plus some constant. You could put the constant on that side as well."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could have a constant on both sides or one side, but we could just put it on one side right over here. You could have put plus some constant, you could have added some, you know, C1 over here, and then you could have had C2 over here, but then you could have subtracted C1 from both sides, and then just have a C3 constant over here. But to simplify it, you can just say, look, this is going to be equal to this plus some constant. You could put the constant on that side as well. It doesn't matter, all the constants could kind of merge into one. Now, this is, we've gotten pretty far. We want to solve the particular solution."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could put the constant on that side as well. It doesn't matter, all the constants could kind of merge into one. Now, this is, we've gotten pretty far. We want to solve the particular solution. Right now, we only have this constant, and we also haven't written W as a function of t just right yet. It's kind of a, they're implicitly expressed, this is more of a relationship than a function right now. So the first thing we can do is try to solve for C. We're told in the problem, we're told in the problem that W of zero is equal to, W of zero is equal to 1,400 tons."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "We want to solve the particular solution. Right now, we only have this constant, and we also haven't written W as a function of t just right yet. It's kind of a, they're implicitly expressed, this is more of a relationship than a function right now. So the first thing we can do is try to solve for C. We're told in the problem, we're told in the problem that W of zero is equal to, W of zero is equal to 1,400 tons. They told us that that's one of our initial conditions. So we know that the natural log, that when t is equal to zero, W is equal to 1,400. So 1,400, W is equal to 1,400, when t is equal to zero, is equal to zero plus C. Or another way to think about it is, the natural log of the absolute value of 1,100 is equal to C. And obviously this is a positive value, so we can just say that C is equal to the natural log of 1,100."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first thing we can do is try to solve for C. We're told in the problem, we're told in the problem that W of zero is equal to, W of zero is equal to 1,400 tons. They told us that that's one of our initial conditions. So we know that the natural log, that when t is equal to zero, W is equal to 1,400. So 1,400, W is equal to 1,400, when t is equal to zero, is equal to zero plus C. Or another way to think about it is, the natural log of the absolute value of 1,100 is equal to C. And obviously this is a positive value, so we can just say that C is equal to the natural log of 1,100. We can drop the absolute value signs, because the absolute value of 1,100 is 1,100. So now we can rewrite this part right over here. Let me do it in yellow, let me start, that's not yellow, let me start right over here."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 1,400, W is equal to 1,400, when t is equal to zero, is equal to zero plus C. Or another way to think about it is, the natural log of the absolute value of 1,100 is equal to C. And obviously this is a positive value, so we can just say that C is equal to the natural log of 1,100. We can drop the absolute value signs, because the absolute value of 1,100 is 1,100. So now we can rewrite this part right over here. Let me do it in yellow, let me start, that's not yellow, let me start right over here. So we could write the natural log of the absolute value of W minus 300. And actually, we know that W is an increasing function. W starts at 1,400 and only goes up from there as t gets larger and larger."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do it in yellow, let me start, that's not yellow, let me start right over here. So we could write the natural log of the absolute value of W minus 300. And actually, we know that W is an increasing function. W starts at 1,400 and only goes up from there as t gets larger and larger. So if we're starting at 1,400 and only getting larger and larger values for W, this expression right here is never going to get negative for positive t values, and that's all we care about. And so we can actually drop the absolute value sign. This will always be a positive value."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "W starts at 1,400 and only goes up from there as t gets larger and larger. So if we're starting at 1,400 and only getting larger and larger values for W, this expression right here is never going to get negative for positive t values, and that's all we care about. And so we can actually drop the absolute value sign. This will always be a positive value. So we could say the absolute value of W minus 300 is equal to t over 25, and we've just solved for our constant, plus the natural log of 1,100. And now we can just solve this explicitly for W. We can raise e to both sides, this value is equal to this value, the left-hand side is equal to the right-hand side. So e raised to this power is the same thing as e raised to that power."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "This will always be a positive value. So we could say the absolute value of W minus 300 is equal to t over 25, and we've just solved for our constant, plus the natural log of 1,100. And now we can just solve this explicitly for W. We can raise e to both sides, this value is equal to this value, the left-hand side is equal to the right-hand side. So e raised to this power is the same thing as e raised to that power. So we have e to the natural log of W minus 300 is going to be equal to e to the t over 25 plus the natural log of 1,100. Let me take this, let me actually, just to get the real estate, let me delete that. All right, now I can keep working right below that."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So e raised to this power is the same thing as e raised to that power. So we have e to the natural log of W minus 300 is going to be equal to e to the t over 25 plus the natural log of 1,100. Let me take this, let me actually, just to get the real estate, let me delete that. All right, now I can keep working right below that. And so e to the natural log of something is just going to be equal to that something. The natural log is what power do I have to raise e to to get this thing? Well, I'm raising e to that power, so I'm going to get this thing."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now I can keep working right below that. And so e to the natural log of something is just going to be equal to that something. The natural log is what power do I have to raise e to to get this thing? Well, I'm raising e to that power, so I'm going to get this thing. So the left-hand side just becomes W minus 300. And the right-hand side, we could rewrite this as e to the t over 25 times e to the natural log of 1,100. Right, we have the same base, different exponents."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I'm raising e to that power, so I'm going to get this thing. So the left-hand side just becomes W minus 300. And the right-hand side, we could rewrite this as e to the t over 25 times e to the natural log of 1,100. Right, we have the same base, different exponents. If you wanted to simplify this, you would just add the exponents, which is exactly what we did up here. So these two things are equivalent. E to the natural log of 1,100, that is 1,100."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right, we have the same base, different exponents. If you wanted to simplify this, you would just add the exponents, which is exactly what we did up here. So these two things are equivalent. E to the natural log of 1,100, that is 1,100. The natural log of 1,100 is a power that you raise e to to get to 1,100. So if you raise e to that power, you get 1,100. So that is 1,100."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "E to the natural log of 1,100, that is 1,100. The natural log of 1,100 is a power that you raise e to to get to 1,100. So if you raise e to that power, you get 1,100. So that is 1,100. So we get W minus 300 is equal to 1,100, 1,100 e to the t over 25. And then you just add 300 to both sides. W is equal to 1,100 e to the t over 25 plus 300."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is 1,100. So we get W minus 300 is equal to 1,100, 1,100 e to the t over 25. And then you just add 300 to both sides. W is equal to 1,100 e to the t over 25 plus 300. And we're done. We have solved the differential equation. We found the particular solution."}, {"video_title": "2011 Calculus AB free response #5c.   AP Calculus AB   Khan Academy.mp3", "Sentence": "W is equal to 1,100 e to the t over 25 plus 300. And we're done. We have solved the differential equation. We found the particular solution. We had to solve for c to find that particular solution. But we're done, and we did it using just basic integration. We didn't have to use any of the fancy differential equation solving tools that you might learn in a more advanced class."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Differentiability. So let's think about that first. It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero. Well just add f of C to both sides and what do you get? Well you get the limit as x approaches C of f of x is equal to f of C. And this is the definition of continuity, the limit of my function as x approaches C is equal to the function, is equal to the value of the function at C. This is, this means that our function is continuous. Continuous at C. So just a reminder, we started assuming f differentiable at C, we use that fact to evaluate this limit right over here, which we got to be equal to zero."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "I've drawn a crazy-looking function here in yellow. And what I want to think about is when this function takes on maximum values and minimum values. And for the sake of this video, we can assume that the graph of this function just keeps getting lower and lower and lower as x becomes more and more negative, and lower and lower and lower as x goes beyond the interval that I've depicted right over here. So what is the maximum value that this function takes on? Well, we can eyeball that. It looks like it's at this point. It looks like it's at that point right over there."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is the maximum value that this function takes on? Well, we can eyeball that. It looks like it's at this point. It looks like it's at that point right over there. So we would call this a global maximum. The function never takes on a value larger than this. So we could say that we have a global maximum at the point x naught, because f of x naught is greater than or equal to f of x for any other x in the domain."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like it's at that point right over there. So we would call this a global maximum. The function never takes on a value larger than this. So we could say that we have a global maximum at the point x naught, because f of x naught is greater than or equal to f of x for any other x in the domain. And that's pretty obvious when you look at it like this. Now, do we have a global minimum point the way that I've drawn it? Well, no."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say that we have a global maximum at the point x naught, because f of x naught is greater than or equal to f of x for any other x in the domain. And that's pretty obvious when you look at it like this. Now, do we have a global minimum point the way that I've drawn it? Well, no. This function can take on arbitrarily negative values. It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, no. This function can take on arbitrarily negative values. It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity. So we have, let me write this down, we have no global minimum. Now, let me ask you a question. Do we have local minima or local maximum?"}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It approaches negative infinity as x approaches positive infinity. So we have, let me write this down, we have no global minimum. Now, let me ask you a question. Do we have local minima or local maximum? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum. So do we have a local minima here or a local minimum here?"}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Do we have local minima or local maximum? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum. So do we have a local minima here or a local minimum here? Well, a local minimum, you could imagine, means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum. And I'm not giving you a very rigorous definition here, but one way to think about it is we can say that we have a local minimum point at x1 is if we have a region around x1 where f of x1 is less than an f of x for any x in this region right over here."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So do we have a local minima here or a local minimum here? Well, a local minimum, you could imagine, means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum. And I'm not giving you a very rigorous definition here, but one way to think about it is we can say that we have a local minimum point at x1 is if we have a region around x1 where f of x1 is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it right over there. Now, do we have any other local minima?"}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm not giving you a very rigorous definition here, but one way to think about it is we can say that we have a local minimum point at x1 is if we have a region around x1 where f of x1 is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it right over there. Now, do we have any other local minima? Well, it doesn't look like we do. Now, what about local maxima? Well, this one right over here, let me do it in purple."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, do we have any other local minima? Well, it doesn't look like we do. Now, what about local maxima? Well, this one right over here, let me do it in purple. I don't want to get people confused, actually. Let me do it in this color. This point right over here looks like a local maximum."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this one right over here, let me do it in purple. I don't want to get people confused, actually. Let me do it in this color. This point right over here looks like a local maximum. Local, not lox. That would have to deal with salmon. Local maximum right over there."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "This point right over here looks like a local maximum. Local, not lox. That would have to deal with salmon. Local maximum right over there. So we could say at the point x2, we have a local maximum point at x2 because f of x2 is larger than f of x for any x around a neighborhood around x2. I'm not being very rigorous, but you can see it just by looking at it. So that's fair enough."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Local maximum right over there. So we could say at the point x2, we have a local maximum point at x2 because f of x2 is larger than f of x for any x around a neighborhood around x2. I'm not being very rigorous, but you can see it just by looking at it. So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now, how can we identify those if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now, how can we identify those if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points. So at this first point right over here, if I were to try to visualize the tangent line, let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's look at the derivative at each of these points. So at this first point right over here, if I were to try to visualize the tangent line, let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0. So we would say that f prime of x0 is equal to 0. So the slope of the tangent line at this point is 0. What about over here?"}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope here is 0. So we would say that f prime of x0 is equal to 0. So the slope of the tangent line at this point is 0. What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0. What about over here?"}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0. What about over here? Well, here, the tangent line is actually not well-defined. We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "What about over here? Well, here, the tangent line is actually not well-defined. We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting, and once again, I'm not rigorously proving it to you. I just want you to get the intuition here."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting, and once again, I'm not rigorously proving it to you. I just want you to get the intuition here. We see that if we have some type of an extrema, and we're not talking about when x is at an endpoint of an interval. Just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is, let's say we have an interval from there."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "I just want you to get the intuition here. We see that if we have some type of an extrema, and we're not talking about when x is at an endpoint of an interval. Just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is, let's say we have an interval from there. So let's say a function starts right over there and then keeps going. This would be a maximum point, but it would be an endpoint. We're not talking about endpoints right now."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're saying, let's say that the function is, let's say we have an interval from there. So let's say a function starts right over there and then keeps going. This would be a maximum point, but it would be an endpoint. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that or points like this. We're talking about the points in between."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that or points like this. We're talking about the points in between. So if you have a point inside of an interval, it's going to be a minimum or maximum, and we see the intuition here. If you have non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or maximum point at some point x is equal to a and x isn't the endpoint of some interval, this tells you something interesting, or at least we have the intuition."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're talking about the points in between. So if you have a point inside of an interval, it's going to be a minimum or maximum, and we see the intuition here. If you have non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or maximum point at some point x is equal to a and x isn't the endpoint of some interval, this tells you something interesting, or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0, or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you know that you have a minimum or maximum point at some point x is equal to a and x isn't the endpoint of some interval, this tells you something interesting, or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0, or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0. Derivative is 0. Derivative is undefined. And we have a word for these points where the derivative is either 0 or the derivative is undefined."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative is 0. Derivative is 0. Derivative is undefined. And we have a word for these points where the derivative is either 0 or the derivative is undefined. We call them critical points. So for the sake of this function, the critical points are, if we could include x sub 0, we could include x sub 1, at x sub 0 and x sub 1 the derivative is 0, and x sub 2 where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we have a word for these points where the derivative is either 0 or the derivative is undefined. We call them critical points. So for the sake of this function, the critical points are, if we could include x sub 0, we could include x sub 1, at x sub 0 and x sub 1 the derivative is 0, and x sub 2 where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point where the derivative is 0 or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point where the derivative is 0 or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. If we look at the tangent line right over here, if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of a critical point, x sub 3 would also be a critical point."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to think about that, let's imagine this point right over here. So let's call this x sub 3. If we look at the tangent line right over here, if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of a critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or a maximum point, that's not an endpoint. It's definitely going to be a critical point."}, {"video_title": "Critical points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So based on our definition of a critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or a maximum point, that's not an endpoint. It's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear, at all of these points, we're at a minimum or maximum point. This, we're at a critical point."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let f of x be equal to negative one over x minus one squared. Select the correct description of the one-sided limits of f at x equals one. And so we can see we have a bunch of choices where we're approaching x from the right-hand side and we're approaching x from the left-hand side. And we're trying to figure out do we get unbounded on either of those in the positive, towards positive infinity or negative infinity. And there's a couple of ways to tackle it. The most straightforward, let's just consider each of these separately. So we could think about the limit of f of x as x approaches one from the positive direction and limit of f of x as x approaches one from the left-hand side."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're trying to figure out do we get unbounded on either of those in the positive, towards positive infinity or negative infinity. And there's a couple of ways to tackle it. The most straightforward, let's just consider each of these separately. So we could think about the limit of f of x as x approaches one from the positive direction and limit of f of x as x approaches one from the left-hand side. This is from the right-hand side. This is from the left-hand side. So I'm just gonna make a table and try out some values as we approach one as we approach one from the different sides."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could think about the limit of f of x as x approaches one from the positive direction and limit of f of x as x approaches one from the left-hand side. This is from the right-hand side. This is from the left-hand side. So I'm just gonna make a table and try out some values as we approach one as we approach one from the different sides. X, f of x, and I'll do the same thing over here. So we are going to have our x and have our f of x. And if we approach one from the right-hand side here, that'd be approaching one from above."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm just gonna make a table and try out some values as we approach one as we approach one from the different sides. X, f of x, and I'll do the same thing over here. So we are going to have our x and have our f of x. And if we approach one from the right-hand side here, that'd be approaching one from above. So we could try 1.1. We could try 1.01. Now f of 1.1 is negative one over 1.1 minus one squared."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we approach one from the right-hand side here, that'd be approaching one from above. So we could try 1.1. We could try 1.01. Now f of 1.1 is negative one over 1.1 minus one squared. So see, this denominator here is going to be.1 squared. So this is going to be 0.01. And so this is going to be negative 100."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now f of 1.1 is negative one over 1.1 minus one squared. So see, this denominator here is going to be.1 squared. So this is going to be 0.01. And so this is going to be negative 100. So let me just write that down. That's going to be negative 100. So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be negative 100. So let me just write that down. That's going to be negative 100. So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared. Well, in this denominator, this is going to be point, this is the same thing as 0.01 squared, which is the same thing as 0.0001, 1 10,000th. And so the negative 1 1 10,000th is going to be negative 10,000. So let's just write that down."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared. Well, in this denominator, this is going to be point, this is the same thing as 0.01 squared, which is the same thing as 0.0001, 1 10,000th. And so the negative 1 1 10,000th is going to be negative 10,000. So let's just write that down. Negative 10,000. And so this looks like, as we get closer, because notice, as I'm going here, I am approaching one from the positive direction. I'm getting closer and closer to one from above, and I'm going unbounded towards negative infinity."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just write that down. Negative 10,000. And so this looks like, as we get closer, because notice, as I'm going here, I am approaching one from the positive direction. I'm getting closer and closer to one from above, and I'm going unbounded towards negative infinity. So this looks like it is negative infinity. Now we could do the same thing from the left-hand side. I could do 0.9."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm getting closer and closer to one from above, and I'm going unbounded towards negative infinity. So this looks like it is negative infinity. Now we could do the same thing from the left-hand side. I could do 0.9. I could do 0.99. 0.99. Now 0.9 is actually also going to get me negative 100, because 0.9 minus one is going to be negative 0.1."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could do 0.9. I could do 0.99. 0.99. Now 0.9 is actually also going to get me negative 100, because 0.9 minus one is going to be negative 0.1. But then when you square it, the negative goes away, so you get 0.01, and then one divided by that is 100, but then you have the negative, so this is also negative 100. And if you don't follow those calculations, I'll do it, let me do it one more time, just so you see it clearly. This is going to be negative one over, and I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from below, from the left-hand side."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now 0.9 is actually also going to get me negative 100, because 0.9 minus one is going to be negative 0.1. But then when you square it, the negative goes away, so you get 0.01, and then one divided by that is 100, but then you have the negative, so this is also negative 100. And if you don't follow those calculations, I'll do it, let me do it one more time, just so you see it clearly. This is going to be negative one over, and I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from below, from the left-hand side. So this is going to be 0.99 minus one squared. Well, 0.99 minus one is going to be negative 100th, so this is going to be negative 0.01 squared. When you square it, the negative goes away, and you're left with one 10,000th, so this is going to be 0.0001, and so when you evaluate this, you get 10,000."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be negative one over, and I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from below, from the left-hand side. So this is going to be 0.99 minus one squared. Well, 0.99 minus one is going to be negative 100th, so this is going to be negative 0.01 squared. When you square it, the negative goes away, and you're left with one 10,000th, so this is going to be 0.0001, and so when you evaluate this, you get 10,000. So that, or sorry, you get negative 10,000. So in either case, regardless of which direction we approach from, we are approaching negative infinity. So that is this choice right over here."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you square it, the negative goes away, and you're left with one 10,000th, so this is going to be 0.0001, and so when you evaluate this, you get 10,000. So that, or sorry, you get negative 10,000. So in either case, regardless of which direction we approach from, we are approaching negative infinity. So that is this choice right over here. Now there's other ways you could have tackled this. If you just look at kind of the structure of this expression here, the numerator is a constant, so that's clearly always going to be positive. Let's ignore this negative for the time being."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is this choice right over here. Now there's other ways you could have tackled this. If you just look at kind of the structure of this expression here, the numerator is a constant, so that's clearly always going to be positive. Let's ignore this negative for the time being. That negative's out front. This numerator, this one, is always going to be positive. Down here, we're taking the, at x equals one, well, this becomes zero, and the whole expression becomes undefined, but as we approach one, x minus one could be positive or negative, as we see over here, but then when we square it, this is going to become positive as well."}, {"video_title": "Analyzing unbounded limits rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's ignore this negative for the time being. That negative's out front. This numerator, this one, is always going to be positive. Down here, we're taking the, at x equals one, well, this becomes zero, and the whole expression becomes undefined, but as we approach one, x minus one could be positive or negative, as we see over here, but then when we square it, this is going to become positive as well. So the denominator's going to be positive for any x other than one, so positive divided by positive is going to be positive, but then you have a negative out front. So this thing is going to be negative for any x other than one, and it's actually not defined at x equals one. And so you could, from that, you could deduce, well, okay, then we can only go to negative infinity."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I have these brackets here. So it also includes a and b in the interval. So let me graph this just so we get a sense of what I'm talking about. So that's my vertical axis. This is my horizontal axis. I'm going to label my horizontal axis t so that we can save x for later. I can still make this y right over there."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's my vertical axis. This is my horizontal axis. I'm going to label my horizontal axis t so that we can save x for later. I can still make this y right over there. And let me graph. This right over here is the graph of y is equal to f of t. Now, our lower endpoint is a. So that's a right over there."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I can still make this y right over there. And let me graph. This right over here is the graph of y is equal to f of t. Now, our lower endpoint is a. So that's a right over there. Our upper boundary is b. Let me make that clear. And actually, just to show that we're including that endpoint, let me make them bold lines, filled in lines."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's a right over there. Our upper boundary is b. Let me make that clear. And actually, just to show that we're including that endpoint, let me make them bold lines, filled in lines. So lower boundary a, upper boundary b. We're just saying, and I've drawn it this way, that f is continuous on that. Now, let's define some new function."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually, just to show that we're including that endpoint, let me make them bold lines, filled in lines. So lower boundary a, upper boundary b. We're just saying, and I've drawn it this way, that f is continuous on that. Now, let's define some new function. That's the area under the curve between a and some point that's in our interval. Let me pick this right over here, x. So let's define some new function to capture the area under the curve between a and x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, let's define some new function. That's the area under the curve between a and some point that's in our interval. Let me pick this right over here, x. So let's define some new function to capture the area under the curve between a and x. Well, how do we denote the area under the curve between two endpoints? Well, we just use our definite integral. That's our Riemann integral."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's define some new function to capture the area under the curve between a and x. Well, how do we denote the area under the curve between two endpoints? Well, we just use our definite integral. That's our Riemann integral. It's really that right now, before we come up with the conclusion of this video, it really just represents the area under the curve between two endpoints. So this right over here, we can say, is the definite integral from a to x of f of t dt. Now, this right over here is going to be a function of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's our Riemann integral. It's really that right now, before we come up with the conclusion of this video, it really just represents the area under the curve between two endpoints. So this right over here, we can say, is the definite integral from a to x of f of t dt. Now, this right over here is going to be a function of x. And let me make it clear. Where x is in the interval between a and b. This thing right over here is going to be another function of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, this right over here is going to be a function of x. And let me make it clear. Where x is in the interval between a and b. This thing right over here is going to be another function of x. This value is going to depend on what x we actually choose. So let's define this as a function of x. So I'm going to say that this is equal to uppercase f of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This thing right over here is going to be another function of x. This value is going to depend on what x we actually choose. So let's define this as a function of x. So I'm going to say that this is equal to uppercase f of x. So all fair and good. Uppercase f of x is a function, if you give me an x value that's between a and b, it'll tell you the area under lowercase f of t between a and x. Now, the cool part."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm going to say that this is equal to uppercase f of x. So all fair and good. Uppercase f of x is a function, if you give me an x value that's between a and b, it'll tell you the area under lowercase f of t between a and x. Now, the cool part. The fundamental theorem of calculus. The fundamental theorem of calculus tells us, let me write this down, because this is a big deal. Fundamental theorem of calculus tells us that if we were to take the derivative of our capital F, so the derivative, let me make sure I have enough space here."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, the cool part. The fundamental theorem of calculus. The fundamental theorem of calculus tells us, let me write this down, because this is a big deal. Fundamental theorem of calculus tells us that if we were to take the derivative of our capital F, so the derivative, let me make sure I have enough space here. So if I were to take the derivative of capital F with respect to x, which is the same thing as taking the derivative of this with respect to x, which is equal to the derivative of all of this business. Let me copy this. So copy and then paste."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Fundamental theorem of calculus tells us that if we were to take the derivative of our capital F, so the derivative, let me make sure I have enough space here. So if I were to take the derivative of capital F with respect to x, which is the same thing as taking the derivative of this with respect to x, which is equal to the derivative of all of this business. Let me copy this. So copy and then paste. Which is the same thing. I've defined capital F as this stuff. So if I'm taking the derivative of the left-hand side, the same thing as taking the derivative of the right-hand side, the fundamental theorem of calculus tells us that this is going to be equal to f, lowercase f of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So copy and then paste. Which is the same thing. I've defined capital F as this stuff. So if I'm taking the derivative of the left-hand side, the same thing as taking the derivative of the right-hand side, the fundamental theorem of calculus tells us that this is going to be equal to f, lowercase f of x. Now, why is this a big deal? Why does it get such an important title as the fundamental theorem of calculus? Well, it tells us that for any continuous function f, if I define a function that is the area under the curve between a and x right over here, that the derivative of that function is going to be f. So let me make it clear."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I'm taking the derivative of the left-hand side, the same thing as taking the derivative of the right-hand side, the fundamental theorem of calculus tells us that this is going to be equal to f, lowercase f of x. Now, why is this a big deal? Why does it get such an important title as the fundamental theorem of calculus? Well, it tells us that for any continuous function f, if I define a function that is the area under the curve between a and x right over here, that the derivative of that function is going to be f. So let me make it clear. Every continuous function, every continuous f has an antiderivative capital F of x. That by itself is a cool thing. But the other really cool thing, or I guess these are somewhat related, is remember, coming into this, all we did, we just viewed the definite integral as symbolizing the area under the curve between two points."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it tells us that for any continuous function f, if I define a function that is the area under the curve between a and x right over here, that the derivative of that function is going to be f. So let me make it clear. Every continuous function, every continuous f has an antiderivative capital F of x. That by itself is a cool thing. But the other really cool thing, or I guess these are somewhat related, is remember, coming into this, all we did, we just viewed the definite integral as symbolizing the area under the curve between two points. That's where that Riemann definition of integration comes from. But now we see a connection between that and derivatives. When you're taking the definite integral, one way of thinking, especially if you're taking a definite integral between a lower boundary and an x, one way to think about it is you're essentially taking an antiderivative."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the other really cool thing, or I guess these are somewhat related, is remember, coming into this, all we did, we just viewed the definite integral as symbolizing the area under the curve between two points. That's where that Riemann definition of integration comes from. But now we see a connection between that and derivatives. When you're taking the definite integral, one way of thinking, especially if you're taking a definite integral between a lower boundary and an x, one way to think about it is you're essentially taking an antiderivative. So we now see a connection. This is why it is the fundamental theorem of calculus. It connects differential calculus and integral calculus."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you're taking the definite integral, one way of thinking, especially if you're taking a definite integral between a lower boundary and an x, one way to think about it is you're essentially taking an antiderivative. So we now see a connection. This is why it is the fundamental theorem of calculus. It connects differential calculus and integral calculus. Connection between derivatives, or maybe I should say antiderivatives, and integration, which before this video we just viewed integration as area under curve. Now that we see it has a connection to derivatives. Well, how would you actually use the fundamental theorem of calculus?"}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It connects differential calculus and integral calculus. Connection between derivatives, or maybe I should say antiderivatives, and integration, which before this video we just viewed integration as area under curve. Now that we see it has a connection to derivatives. Well, how would you actually use the fundamental theorem of calculus? Well, maybe in the context of a calculus class. And we'll do the intuition for why this happens or why this is true and maybe a proof in later videos. But how would you actually apply this right over here?"}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, how would you actually use the fundamental theorem of calculus? Well, maybe in the context of a calculus class. And we'll do the intuition for why this happens or why this is true and maybe a proof in later videos. But how would you actually apply this right over here? Well, let's say that someone told you that they want to find the derivative. Let me just in a new color just to show this as an example. Let's say someone wanted to find the derivative with respect to x of the integral from, I don't know, I'll pick some random number here."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But how would you actually apply this right over here? Well, let's say that someone told you that they want to find the derivative. Let me just in a new color just to show this as an example. Let's say someone wanted to find the derivative with respect to x of the integral from, I don't know, I'll pick some random number here. So pi to x of, I'll put something crazy here, cosine squared of t over the natural log of t minus the square root of t dt. So they want you to take the derivative with respect to x of this crazy thing. Remember, this thing in the parentheses is a function of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say someone wanted to find the derivative with respect to x of the integral from, I don't know, I'll pick some random number here. So pi to x of, I'll put something crazy here, cosine squared of t over the natural log of t minus the square root of t dt. So they want you to take the derivative with respect to x of this crazy thing. Remember, this thing in the parentheses is a function of x. It's value. It's going to have a value that is dependent on x. You give it a different x, it's going to have a different value."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, this thing in the parentheses is a function of x. It's value. It's going to have a value that is dependent on x. You give it a different x, it's going to have a different value. So what's the derivative of this with respect to x? Well, the fundamental theorem of calculus tells us it can be very simple. We essentially, and you can even pattern match up here, and we'll get more intuition of why this is true in future videos."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You give it a different x, it's going to have a different value. So what's the derivative of this with respect to x? Well, the fundamental theorem of calculus tells us it can be very simple. We essentially, and you can even pattern match up here, and we'll get more intuition of why this is true in future videos. But essentially, everywhere where you see this right over here is an f of t. Everywhere you see a t, replace it with an x, and it becomes an f of x. So this is going to be equal to cosine squared of x over the natural log of x minus the square root of x. You take the derivative of the indefinite integral where the upper boundary is x right over here, it just becomes whatever you were taking the integral of, that as a function instead of t, that is now a function of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We essentially, and you can even pattern match up here, and we'll get more intuition of why this is true in future videos. But essentially, everywhere where you see this right over here is an f of t. Everywhere you see a t, replace it with an x, and it becomes an f of x. So this is going to be equal to cosine squared of x over the natural log of x minus the square root of x. You take the derivative of the indefinite integral where the upper boundary is x right over here, it just becomes whatever you were taking the integral of, that as a function instead of t, that is now a function of x. So it can really simplify sometimes taking a derivative. And sometimes you'll see on exams these trick problems where you have this really hairy thing that you take a definite integral of and then take the derivative. And you just have to remember the fundamental theorem of calculus, the thing that ties it all together, connects derivatives and integration, that you can just simplify it by realizing that this is just going to be, instead of a function of lowercase f of t, it's going to be lowercase f of x."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You take the derivative of the indefinite integral where the upper boundary is x right over here, it just becomes whatever you were taking the integral of, that as a function instead of t, that is now a function of x. So it can really simplify sometimes taking a derivative. And sometimes you'll see on exams these trick problems where you have this really hairy thing that you take a definite integral of and then take the derivative. And you just have to remember the fundamental theorem of calculus, the thing that ties it all together, connects derivatives and integration, that you can just simplify it by realizing that this is just going to be, instead of a function of lowercase f of t, it's going to be lowercase f of x. Let me make it clear. In this example right over here, this right over here was lowercase f of t, and now it became lowercase f of x. This right over here was our a."}, {"video_title": "Fundamental theorem of calculus (Part 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you just have to remember the fundamental theorem of calculus, the thing that ties it all together, connects derivatives and integration, that you can just simplify it by realizing that this is just going to be, instead of a function of lowercase f of t, it's going to be lowercase f of x. Let me make it clear. In this example right over here, this right over here was lowercase f of t, and now it became lowercase f of x. This right over here was our a. And notice, it doesn't matter what the lower boundary of a actually is. You don't have anything on the right-hand side that is in some way dependent on a. Anyway, hope you enjoyed that."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you've never heard of the term composite functions or if the first few minutes of this video look unfamiliar to you, I encourage you to watch the algebra videos on composite functions on Khan Academy. The goal of this one is to really be a little bit of a practice before we get into some skills that are necessary in calculus, in particular, the chain rule. So let's just review what a composite function is. So let's say that I have, let's say that I have f of x, f of x being equal to one plus x. And let's say that we have g of x is equal to, let's say g of x is equal to cosine of x. So what would f of g of x be? F of g of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that I have, let's say that I have f of x, f of x being equal to one plus x. And let's say that we have g of x is equal to, let's say g of x is equal to cosine of x. So what would f of g of x be? F of g of x. And I encourage you to pause this video and try to work it out on your own. Well, one way to think about it is the input into f of x is no longer x, it is g of x. So everywhere where we see an x in the definition of f of x, we would replace with a g of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "F of g of x. And I encourage you to pause this video and try to work it out on your own. Well, one way to think about it is the input into f of x is no longer x, it is g of x. So everywhere where we see an x in the definition of f of x, we would replace with a g of x. So this is going to be equal to, this is going to be equal to one plus, instead of the input being x, the input is g of x, so the output is one plus g of x. And g of x, of course, is cosine of x. So instead of writing g of x there, I could write cosine of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So everywhere where we see an x in the definition of f of x, we would replace with a g of x. So this is going to be equal to, this is going to be equal to one plus, instead of the input being x, the input is g of x, so the output is one plus g of x. And g of x, of course, is cosine of x. So instead of writing g of x there, I could write cosine of x. And one way to visualize this is I'm putting my x into g of x first. So x goes into the function g, and it outputs g of x. And then we're gonna take that output, g of x, and then input it into f of x, or input it into the function f, I should say."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So instead of writing g of x there, I could write cosine of x. And one way to visualize this is I'm putting my x into g of x first. So x goes into the function g, and it outputs g of x. And then we're gonna take that output, g of x, and then input it into f of x, or input it into the function f, I should say. We input into the function f, and that is going to output f of whatever the input was, and the input is g of x, g of x. So now with that review out of the way, let's see if we can go the other way around. Let's see if we can look at some type of a function definition and say, hey, can we express that as a composition of other functions?"}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're gonna take that output, g of x, and then input it into f of x, or input it into the function f, I should say. We input into the function f, and that is going to output f of whatever the input was, and the input is g of x, g of x. So now with that review out of the way, let's see if we can go the other way around. Let's see if we can look at some type of a function definition and say, hey, can we express that as a composition of other functions? So let's start with, let's say that I have a g of x is equal to cosine of sine of x plus one. And I also wanna state, there's oftentimes more than one way to compose a, or to construct a function based on the composition of other ones. But with that said, pause this video and say, hey, can I express g of x as a composition of two other functions?"}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if we can look at some type of a function definition and say, hey, can we express that as a composition of other functions? So let's start with, let's say that I have a g of x is equal to cosine of sine of x plus one. And I also wanna state, there's oftentimes more than one way to compose a, or to construct a function based on the composition of other ones. But with that said, pause this video and say, hey, can I express g of x as a composition of two other functions? Let's say an f and an h of x. So there's a couple of ways that you could think about it. You could say, all right, let's see, I have this sine of x right over here."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But with that said, pause this video and say, hey, can I express g of x as a composition of two other functions? Let's say an f and an h of x. So there's a couple of ways that you could think about it. You could say, all right, let's see, I have this sine of x right over here. So what if I called that an f of x? So let's say I call that, well, actually, let me use a different variable so we don't get confused here. Let me call this u of x, the sine of x right over there."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say, all right, let's see, I have this sine of x right over here. So what if I called that an f of x? So let's say I call that, well, actually, let me use a different variable so we don't get confused here. Let me call this u of x, the sine of x right over there. So this would be cosine of u of x plus one. And so if we then divided, if we then defined another function as v of x being equal to cosine of whatever its input is plus one, well, then this looks like the composition of v and u of x. Instead of v of x, if we did v of u of x, then this would be cosine of u of x plus one."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me call this u of x, the sine of x right over there. So this would be cosine of u of x plus one. And so if we then divided, if we then defined another function as v of x being equal to cosine of whatever its input is plus one, well, then this looks like the composition of v and u of x. Instead of v of x, if we did v of u of x, then this would be cosine of u of x plus one. Let me write that down. So if we wrote v of u of x, which is sine of x, if we did v of u of x, that is going to be equal to cosine of, instead of an x plus one, it's going to be a u of x plus one. And u of x, let me write this here, u of x is equal to sine of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of v of x, if we did v of u of x, then this would be cosine of u of x plus one. Let me write that down. So if we wrote v of u of x, which is sine of x, if we did v of u of x, that is going to be equal to cosine of, instead of an x plus one, it's going to be a u of x plus one. And u of x, let me write this here, u of x is equal to sine of x. That's how we've set this up. So we can either write cosine of u of x plus one or cosine of sine of x plus one, which was exactly what we had before. And so this function g of x, if we say u of x is equal to sine of x, if we say u of x is equal to sine of x and v of x is equal to cosine of x plus one, then we could write g of x as the composition of these two functions."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And u of x, let me write this here, u of x is equal to sine of x. That's how we've set this up. So we can either write cosine of u of x plus one or cosine of sine of x plus one, which was exactly what we had before. And so this function g of x, if we say u of x is equal to sine of x, if we say u of x is equal to sine of x and v of x is equal to cosine of x plus one, then we could write g of x as the composition of these two functions. Now, you could even make it a composition of three functions. We could keep u of x to be equal to sine of x. We could define, let's say, a w of x to be equal to x plus one."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this function g of x, if we say u of x is equal to sine of x, if we say u of x is equal to sine of x and v of x is equal to cosine of x plus one, then we could write g of x as the composition of these two functions. Now, you could even make it a composition of three functions. We could keep u of x to be equal to sine of x. We could define, let's say, a w of x to be equal to x plus one. And so then, let's think about it. W of x, w of u of x, I should say, w of u, let me do the same color, w of u of x is going to be equal to, now my input is no longer x, it's a u of x, so it's going to be a u of x plus one or just sine of x plus one. So that's sine of x plus one."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could define, let's say, a w of x to be equal to x plus one. And so then, let's think about it. W of x, w of u of x, I should say, w of u, let me do the same color, w of u of x is going to be equal to, now my input is no longer x, it's a u of x, so it's going to be a u of x plus one or just sine of x plus one. So that's sine of x plus one. And then if we define a third function, let's say, let's see, I'll call it, let's call it h. I'm running out of variables. Well, there are a lot of letters left. So if I say h of x is just equal to the cosine of whatever I input, so it's equal to the cosine of x, well then h of w of u of x is gonna be g of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's sine of x plus one. And then if we define a third function, let's say, let's see, I'll call it, let's call it h. I'm running out of variables. Well, there are a lot of letters left. So if I say h of x is just equal to the cosine of whatever I input, so it's equal to the cosine of x, well then h of w of u of x is gonna be g of x. Let me write that down. H of w of u of x, u of x, is going to be equal to, remember, h of x takes the cosine of whatever its input is. So it's gonna take the cosine."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I say h of x is just equal to the cosine of whatever I input, so it's equal to the cosine of x, well then h of w of u of x is gonna be g of x. Let me write that down. H of w of u of x, u of x, is going to be equal to, remember, h of x takes the cosine of whatever its input is. So it's gonna take the cosine. Now, its input is w of u of x. We already figured out w of u of x is going to be this business. So it's going to be sine of x plus one."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's gonna take the cosine. Now, its input is w of u of x. We already figured out w of u of x is going to be this business. So it's going to be sine of x plus one. Where the u of x is sine of x, but then we inputted that into w, so we got sine of x plus one. And then we inputted that into h to get cosine of that, which is our original expression, which is equal to g of x. So the whole point here is to appreciate how to recognize compositions of functions."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be sine of x plus one. Where the u of x is sine of x, but then we inputted that into w, so we got sine of x plus one. And then we inputted that into h to get cosine of that, which is our original expression, which is equal to g of x. So the whole point here is to appreciate how to recognize compositions of functions. Now, I wanna stress, it's not always going to be a composition of a function. For example, if I have some function, let me just clear this out. If I had some function f of x is equal to cosine of x times sine of x, it would be hard to express this as a composition of functions, but I can represent it as the product of functions."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the whole point here is to appreciate how to recognize compositions of functions. Now, I wanna stress, it's not always going to be a composition of a function. For example, if I have some function, let me just clear this out. If I had some function f of x is equal to cosine of x times sine of x, it would be hard to express this as a composition of functions, but I can represent it as the product of functions. For example, I could say cosine of x, I could say u of x is equal to cosine of x, and I could say v of x, let me use a different color, I could say v of x is equal to sine of x. And so here, f of x wouldn't be the composition of u and v, it would be the product. F of x is equal to u of x times v of x, v of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I had some function f of x is equal to cosine of x times sine of x, it would be hard to express this as a composition of functions, but I can represent it as the product of functions. For example, I could say cosine of x, I could say u of x is equal to cosine of x, and I could say v of x, let me use a different color, I could say v of x is equal to sine of x. And so here, f of x wouldn't be the composition of u and v, it would be the product. F of x is equal to u of x times v of x, v of x. If we were to take the composition, if we were to say u of v of x, pause the video, think about what that is, and that's a little bit of a review. Well, this is going to be, I take u of x, takes the cosine of whatever is input, and now the input is v of x, which would be sine of x. Sine of x."}, {"video_title": "Identifying composite functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "F of x is equal to u of x times v of x, v of x. If we were to take the composition, if we were to say u of v of x, pause the video, think about what that is, and that's a little bit of a review. Well, this is going to be, I take u of x, takes the cosine of whatever is input, and now the input is v of x, which would be sine of x. Sine of x. And then if you did v of u of x, well, that'd be the other way around. It would be sine of cosine of x. But anyway, this is, once again, just to help us recognize, hey, do I have, when I look at an expression or a function definition, am I looking at products of functions?"}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we've got a Riemann sum. We're gonna take the limit as n approaches infinity. And the goal of this video is to see if we can rewrite this as a definite integral. I encourage you to pause the video and see if you can work through it on your own. So let's remind ourselves how a definite integral can relate to a Riemann sum. So if I have the definite integral from a to b of f of x, f of x dx, we have seen in other videos, this is going to be the limit as n approaches infinity of the sum, capital sigma, going from i equals one to n. And so we're essentially going to sum the areas of a bunch of rectangles, where the width of each of those rectangles, we can write as a delta x. So your width is going to be delta x of each of those rectangles."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "I encourage you to pause the video and see if you can work through it on your own. So let's remind ourselves how a definite integral can relate to a Riemann sum. So if I have the definite integral from a to b of f of x, f of x dx, we have seen in other videos, this is going to be the limit as n approaches infinity of the sum, capital sigma, going from i equals one to n. And so we're essentially going to sum the areas of a bunch of rectangles, where the width of each of those rectangles, we can write as a delta x. So your width is going to be delta x of each of those rectangles. And then your height is going to be the value of the function evaluated someplace in that delta x. If we're doing a right Riemann sum, we would do the right end of that rectangle or of that subinterval. And so we would start at our lower bound, a, and we would add as many delta x's as our index specifies."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "So your width is going to be delta x of each of those rectangles. And then your height is going to be the value of the function evaluated someplace in that delta x. If we're doing a right Riemann sum, we would do the right end of that rectangle or of that subinterval. And so we would start at our lower bound, a, and we would add as many delta x's as our index specifies. So if i is equal to one, we add one delta x. So we'd be at the right of the first rectangle. If i is equal to two, we add two delta x's."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we would start at our lower bound, a, and we would add as many delta x's as our index specifies. So if i is equal to one, we add one delta x. So we'd be at the right of the first rectangle. If i is equal to two, we add two delta x's. So this is going to be delta x times our index. So this is the general form that we have seen before. And so one possibility, you could even do a little bit of pattern matching right over here."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "If i is equal to two, we add two delta x's. So this is going to be delta x times our index. So this is the general form that we have seen before. And so one possibility, you could even do a little bit of pattern matching right over here. Our function looks like the natural log function. So that looks like our f of x. It's the natural log function."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so one possibility, you could even do a little bit of pattern matching right over here. Our function looks like the natural log function. So that looks like our f of x. It's the natural log function. So I could write that. So f of x looks like the natural log of x. What else do we see?"}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's the natural log function. So I could write that. So f of x looks like the natural log of x. What else do we see? Well, a, that looks like two. A is equal to two. What would our delta x be?"}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "What else do we see? Well, a, that looks like two. A is equal to two. What would our delta x be? Well, you can see this right over here, this thing that we're multiplying that just is divided by n, and it's not multiplying by an i. So this looks like our delta x. And this right over here looks like delta x times i."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "What would our delta x be? Well, you can see this right over here, this thing that we're multiplying that just is divided by n, and it's not multiplying by an i. So this looks like our delta x. And this right over here looks like delta x times i. So it looks like our delta x is equal to five over n. So what can we tell so far? Well, we could say that, okay, this thing up here, the original thing, is going to be equal to the definite integral. We know our lower bound is going from two to, we haven't figured out our upper bound yet."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this right over here looks like delta x times i. So it looks like our delta x is equal to five over n. So what can we tell so far? Well, we could say that, okay, this thing up here, the original thing, is going to be equal to the definite integral. We know our lower bound is going from two to, we haven't figured out our upper bound yet. We haven't figured out our b yet. But our function is the natural log of x, and then I will just write a dx here. So in order to complete writing this definite integral, I need to be able to write the upper bound."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know our lower bound is going from two to, we haven't figured out our upper bound yet. We haven't figured out our b yet. But our function is the natural log of x, and then I will just write a dx here. So in order to complete writing this definite integral, I need to be able to write the upper bound. And the way to figure out the upper bound is by looking at our delta x. Because the way that we would figure out a delta x for this Riemann sum here, we would say that delta x is equal to the difference between our bounds divided by how many sections we want to divide it in, divided by n. So it's equal to b minus a, b minus a over n. Over n. And so you can pattern match here. If this is delta x is equal to b minus a over n, let me write this down."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in order to complete writing this definite integral, I need to be able to write the upper bound. And the way to figure out the upper bound is by looking at our delta x. Because the way that we would figure out a delta x for this Riemann sum here, we would say that delta x is equal to the difference between our bounds divided by how many sections we want to divide it in, divided by n. So it's equal to b minus a, b minus a over n. Over n. And so you can pattern match here. If this is delta x is equal to b minus a over n, let me write this down. So this is going to be equal to b, b minus our a, which is two. All of that over n. So b minus two is equal to five, which would make b equal to seven. B is equal to seven."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "If this is delta x is equal to b minus a over n, let me write this down. So this is going to be equal to b, b minus our a, which is two. All of that over n. So b minus two is equal to five, which would make b equal to seven. B is equal to seven. So there you have it. We have our original limit, our Riemann limit, or limit of our Riemann sum being rewritten as a definite integral. And once again, I want to emphasize why this makes sense."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "B is equal to seven. So there you have it. We have our original limit, our Riemann limit, or limit of our Riemann sum being rewritten as a definite integral. And once again, I want to emphasize why this makes sense. If I wanted to draw this, it would look something like this. I'm gonna try to hand draw the natural log function. It looks something like this."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, I want to emphasize why this makes sense. If I wanted to draw this, it would look something like this. I'm gonna try to hand draw the natural log function. It looks something like this. This right over here would be one. And so let's say this is two. And so we're going from two to seven."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks something like this. This right over here would be one. And so let's say this is two. And so we're going from two to seven. This isn't exactly right. So our definite integral is concerned with the area under the curve from two until seven. And so this Riemann sum you could view as an approximation when n isn't approaching infinity."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we're going from two to seven. This isn't exactly right. So our definite integral is concerned with the area under the curve from two until seven. And so this Riemann sum you could view as an approximation when n isn't approaching infinity. But what you're saying is, look, when i is equal to one, your first one is going to be of width five over n. So this is essentially saying our difference between two and seven, we're taking that distance five, dividing it into n rectangles. And so this first one is going to have a width of five over n. And then what's the height gonna be? Well, it's a right Riemann sum."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this Riemann sum you could view as an approximation when n isn't approaching infinity. But what you're saying is, look, when i is equal to one, your first one is going to be of width five over n. So this is essentially saying our difference between two and seven, we're taking that distance five, dividing it into n rectangles. And so this first one is going to have a width of five over n. And then what's the height gonna be? Well, it's a right Riemann sum. So we're using the value of the function right over there, right at two plus five over n. So this value right over here, this is the natural log, the natural log of two plus five over n. And since this is the first rectangle, times one. Times one. And we could keep going."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's a right Riemann sum. So we're using the value of the function right over there, right at two plus five over n. So this value right over here, this is the natural log, the natural log of two plus five over n. And since this is the first rectangle, times one. Times one. And we could keep going. This one right over here, the width is the same, five over n. But what's the height? Well, the height here, this height right over here is gonna be the natural log of two plus five over n times two. Times two."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could keep going. This one right over here, the width is the same, five over n. But what's the height? Well, the height here, this height right over here is gonna be the natural log of two plus five over n times two. Times two. This is for i is equal to two. This is i is equal to one. And so hopefully you are seeing that this makes sense."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "Times two. This is for i is equal to two. This is i is equal to one. And so hopefully you are seeing that this makes sense. The area of this first rectangle is going to be natural log of two plus five over n times one. Times five over n. The second one over here, natural log of two plus five over n times two. Times five over n. And so this is calculating the sum of the areas of these rectangles."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "In this video, we will cover the power rule, which really simplifies our life when it comes to taking derivatives, especially derivatives of polynomials. You are probably already familiar with the definition of a derivative. The limit is delta x approaches 0 of f of x plus delta x minus f of x, all of that over delta x. And it really just comes out of trying to find the slope of a tangent line at any given point. But we're going to see what the power rule is. It simplifies our life. We won't have to take these sometimes complicated limits."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it really just comes out of trying to find the slope of a tangent line at any given point. But we're going to see what the power rule is. It simplifies our life. We won't have to take these sometimes complicated limits. And we're not going to prove it in this video. But we'll hopefully get a sense of how to use it. And in future videos, we'll get a sense of why it makes sense and even prove it."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We won't have to take these sometimes complicated limits. And we're not going to prove it in this video. But we'll hopefully get a sense of how to use it. And in future videos, we'll get a sense of why it makes sense and even prove it. So the power rule just tells us that if I have some function, f of x, and it's equal to some power of x, so x to the n power, where n does not equal 0. So n can be anything. It can be positive and negative."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in future videos, we'll get a sense of why it makes sense and even prove it. So the power rule just tells us that if I have some function, f of x, and it's equal to some power of x, so x to the n power, where n does not equal 0. So n can be anything. It can be positive and negative. It does not have to be an integer. The power rule tells us that the derivative of this, f prime of x, is just going to be equal to n. So you're literally bringing this out front. n times x."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It can be positive and negative. It does not have to be an integer. The power rule tells us that the derivative of this, f prime of x, is just going to be equal to n. So you're literally bringing this out front. n times x. And then you just decrement the power. Times x to the n minus 1 power. So let's do a couple of examples just to make sure that that actually makes sense."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "n times x. And then you just decrement the power. Times x to the n minus 1 power. So let's do a couple of examples just to make sure that that actually makes sense. So let's ask ourselves, well, let's say that f of x was equal to x squared. Based on the power rule, what is f prime of x going to be equal to? Well, in this situation, our n is 2."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do a couple of examples just to make sure that that actually makes sense. So let's ask ourselves, well, let's say that f of x was equal to x squared. Based on the power rule, what is f prime of x going to be equal to? Well, in this situation, our n is 2. So we bring the 2 out front. 2 times x to the 2 minus 1 power. So that's going to be 2 times x to the first power, which is just equal to 2x."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, in this situation, our n is 2. So we bring the 2 out front. 2 times x to the 2 minus 1 power. So that's going to be 2 times x to the first power, which is just equal to 2x. That was pretty straightforward. Let's think about the situation where, let's say, we have g of x is equal to x to the third power. What is g prime of x going to be in this scenario?"}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's going to be 2 times x to the first power, which is just equal to 2x. That was pretty straightforward. Let's think about the situation where, let's say, we have g of x is equal to x to the third power. What is g prime of x going to be in this scenario? Well, n is 3. So we just literally pattern match here. You're probably finding this shockingly straightforward."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is g prime of x going to be in this scenario? Well, n is 3. So we just literally pattern match here. You're probably finding this shockingly straightforward. So this is going to be 3 times x to the 3 minus 1 power. Or this is going to be equal to 3x squared. And we're done."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're probably finding this shockingly straightforward. So this is going to be 3 times x to the 3 minus 1 power. Or this is going to be equal to 3x squared. And we're done. In the next video, we'll think about whether this actually makes sense. Let's do one more example just to show it doesn't have to necessarily apply to only these kind of positive integers. We could have a scenario where maybe we have h of x. h of x is equal to x to the negative 100 power."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're done. In the next video, we'll think about whether this actually makes sense. Let's do one more example just to show it doesn't have to necessarily apply to only these kind of positive integers. We could have a scenario where maybe we have h of x. h of x is equal to x to the negative 100 power. The power rule tells us that h prime of x would be equal to what? Well, n is negative 100. So it's negative 100x to the negative 100 minus 1, which is equal to negative 100x to the negative 101."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could have a scenario where maybe we have h of x. h of x is equal to x to the negative 100 power. The power rule tells us that h prime of x would be equal to what? Well, n is negative 100. So it's negative 100x to the negative 100 minus 1, which is equal to negative 100x to the negative 101. Let's do one more. Let's say we had z of x. z of x is equal to x to the 2.571 power. And we are concerned with what is z prime of x."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's negative 100x to the negative 100 minus 1, which is equal to negative 100x to the negative 101. Let's do one more. Let's say we had z of x. z of x is equal to x to the 2.571 power. And we are concerned with what is z prime of x. Well, once again, power rule simplifies our life. n is 2.571. So it's going to be 2.571 times x to the 2.571 minus 1 power."}, {"video_title": "Power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we are concerned with what is z prime of x. Well, once again, power rule simplifies our life. n is 2.571. So it's going to be 2.571 times x to the 2.571 minus 1 power. So it's going to be equal to, let me make sure I'm not falling off the bottom of the page, 2.571 times x to the 1.571 power. Hopefully you enjoyed that. In the next few videos, we will not only expose you to more properties of derivatives."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is try to find the derivative with respect to x of x squared sine of x, all of that to the third power. And what's going to be interesting is that there's multiple ways to tackle it, and I encourage you to pause the video and see if you can work through it on your own. So there's actually multiple techniques. One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to x of something to the third power. So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to x of something to the third power. So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. This is just an application of the chain rule. Now, the second part, what would that be? The second part here, this is another color, in orange."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. This is just an application of the chain rule. Now, the second part, what would that be? The second part here, this is another color, in orange. Well, here I would apply the product rule. I have the product of two expressions, so I would take the derivative of, let me write this down. So this is gonna be the product rule, product rule."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The second part here, this is another color, in orange. Well, here I would apply the product rule. I have the product of two expressions, so I would take the derivative of, let me write this down. So this is gonna be the product rule, product rule. I would take the derivative of the first expression, so x, derivative of x squared is two x, let me write a little bit to the right. This is gonna be two x times the second expression, sine of x, plus the first expression, x squared, times the derivative of the second one, cosine of x. That's just the product rule as applied to this part right over here."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is gonna be the product rule, product rule. I would take the derivative of the first expression, so x, derivative of x squared is two x, let me write a little bit to the right. This is gonna be two x times the second expression, sine of x, plus the first expression, x squared, times the derivative of the second one, cosine of x. That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that. So all of this I could rewrite as, let's see. This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that. So all of this I could rewrite as, let's see. This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So x squared squared is x to the fourth, and then sine of x squared is sine squared of x, and then all of that is being multiplied by that. And if we want, we can algebraically simplify, we can distribute everything out, in which case, what would we get? Well, let's see."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So x squared squared is x to the fourth, and then sine of x squared is sine squared of x, and then all of that is being multiplied by that. And if we want, we can algebraically simplify, we can distribute everything out, in which case, what would we get? Well, let's see. Three times two is six. X to the fourth times x is x to the fifth. Sine squared of x times sine of x is sine of x to the third power."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see. Three times two is six. X to the fourth times x is x to the fifth. Sine squared of x times sine of x is sine of x to the third power. And then, let's see, three, so plus three, x to the fourth times x squared is x to the sixth power, and then I have sine squared of x, sine squared of x, cosine of x. So there you have it. That's one strategy, chain rule first and then product rule."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine squared of x times sine of x is sine of x to the third power. And then, let's see, three, so plus three, x to the fourth times x squared is x to the sixth power, and then I have sine squared of x, sine squared of x, cosine of x. So there you have it. That's one strategy, chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's one strategy, chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have, if I'm taking the product things to some exponent, well, that's the same thing of each of them raised to the exponent and then the product of the two. Now, how would we tackle this?"}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have, if I'm taking the product things to some exponent, well, that's the same thing of each of them raised to the exponent and then the product of the two. Now, how would we tackle this? Well, I, here, I would do the product rule first. So let's do that. So let's do the product rule."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, how would we tackle this? Well, I, here, I would do the product rule first. So let's do that. So let's do the product rule. So we're gonna take the derivative of the first expression. So derivative of x to the sixth is six x to the fifth times the second expression, sine to the third of x, or sine of x to the third power, plus the first x to the sixth times the derivative of the second, and I'm just gonna write that, d dx of sine of x to the third power. Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do the product rule. So we're gonna take the derivative of the first expression. So derivative of x to the sixth is six x to the fifth times the second expression, sine to the third of x, or sine of x to the third power, plus the first x to the sixth times the derivative of the second, and I'm just gonna write that, d dx of sine of x to the third power. Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. And so what is this going to be? Well, I have the derivative of something to the third power, so that's going to be three times that something squared times the derivative of that something. So in this case, the something is sine of x, and the derivative of sine of x is cosine of x."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. And so what is this going to be? Well, I have the derivative of something to the third power, so that's going to be three times that something squared times the derivative of that something. So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. And then I have all of this business over here. I have six x to the fifth, sine to the third, or sine of x to the third power, plus x to the sixth. And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. And then I have all of this business over here. I have six x to the fifth, sine to the third, or sine of x to the third power, plus x to the sixth. And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. This term is exactly equivalent to this term, the way it's written. And then this is exactly, if you multiply three times x to the sixth, sine of x squared cosine of x. So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. This term is exactly equivalent to this term, the way it's written. And then this is exactly, if you multiply three times x to the sixth, sine of x squared cosine of x. So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. But the point here is that there's multiple strategies. You could use the chain rule first and then the product rule or you could use the product rule first and then the chain rule. In this case, you could debate which one is faster."}, {"video_title": "Applying the chain rule and product rule   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. But the point here is that there's multiple strategies. You could use the chain rule first and then the product rule or you could use the product rule first and then the chain rule. In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster. But sometimes, but these two are pretty close, but sometimes it'll be more clear than not which one is preferable. You really wanna minimize the amount of hairiness, the number of steps, the chances for careless mistakes you might have."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to assume we know that the limit as x approaches zero of sine of x over x, that this is equal to one. And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done. Using that fact and a little bit of trig identities and a little bit of algebraic manipulation, we were able to show that our original limit, the limit as x approaches zero of one minus cosine of x over x is equal to zero. I encourage you to graph it. You will see that that makes sense from a graphical point of view as well."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is less than or equal to zero, f is one minus two sine of x. When x is greater than zero, f is e to the negative four x. Show that f is continuous at x equals zero. So for something to be continuous at x equals zero, let's think about what has to happen. So if I have a function here, so that is my x-axis, and let's say this is my y-axis, and we care about what happens at x equals zero. So x equals zero there, and let's say that this is our function. Let's say that this is our function."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for something to be continuous at x equals zero, let's think about what has to happen. So if I have a function here, so that is my x-axis, and let's say this is my y-axis, and we care about what happens at x equals zero. So x equals zero there, and let's say that this is our function. Let's say that this is our function. So maybe our function maybe looks something like this. I don't know what, well, this one in particular probably looks something like this. This one in particular, this one in particular might look, who knows, and then it might look something like that."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that this is our function. So maybe our function maybe looks something like this. I don't know what, well, this one in particular probably looks something like this. This one in particular, this one in particular might look, who knows, and then it might look something like that. And the particulars aren't that important. We just have to think about what they're asking. In order for it to be continuous here, the limit as we approach from the left, the limit as we approach from the left, so the limit as we approach zero from the left, should be equal to the value of the function at zero."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "This one in particular, this one in particular might look, who knows, and then it might look something like that. And the particulars aren't that important. We just have to think about what they're asking. In order for it to be continuous here, the limit as we approach from the left, the limit as we approach from the left, so the limit as we approach zero from the left, should be equal to the value of the function at zero. So this should, so the limit of f of x should be equal to f of zero, which should be equal to the limit as we approach from the right, which should be equal to the limit as we approach zero from the right. So that should be equal to the limit as x approaches 0 from the right of f of x. And the reason why this matters is if this wasn't the case, if f of 0 wasn't the same as the limit, then we might have a gap right over there."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "In order for it to be continuous here, the limit as we approach from the left, the limit as we approach from the left, so the limit as we approach zero from the left, should be equal to the value of the function at zero. So this should, so the limit of f of x should be equal to f of zero, which should be equal to the limit as we approach from the right, which should be equal to the limit as we approach zero from the right. So that should be equal to the limit as x approaches 0 from the right of f of x. And the reason why this matters is if this wasn't the case, if f of 0 wasn't the same as the limit, then we might have a gap right over there. So you could have limits. So you could have a situation like this, where you have a gap right over there. And then it looks something like that."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the reason why this matters is if this wasn't the case, if f of 0 wasn't the same as the limit, then we might have a gap right over there. So you could have limits. So you could have a situation like this, where you have a gap right over there. And then it looks something like that. So the two limits from the left and the right both exist. And the limit at that point would exist. But if the function itself does not equal that value there, if it equals something else, then the function would not be continuous."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it looks something like that. So the two limits from the left and the right both exist. And the limit at that point would exist. But if the function itself does not equal that value there, if it equals something else, then the function would not be continuous. So that's why the limit has to be equal to the value of the function in order for it to be continuous. So let's think about whether all of these things equal each other. First of all, let's just think about the value of the function there."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if the function itself does not equal that value there, if it equals something else, then the function would not be continuous. So that's why the limit has to be equal to the value of the function in order for it to be continuous. So let's think about whether all of these things equal each other. First of all, let's just think about the value of the function there. So remember, we're doing part a. f of 0 is equal to, we're going to use this first case, because that's the case for x is less than or equal to 0. So f of 0 is going to be equal to 1 minus 2 sine of 0. Well, sine of 0 is 0."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "First of all, let's just think about the value of the function there. So remember, we're doing part a. f of 0 is equal to, we're going to use this first case, because that's the case for x is less than or equal to 0. So f of 0 is going to be equal to 1 minus 2 sine of 0. Well, sine of 0 is 0. 2 times 0 is 0. So this whole thing is 0. 1 minus 0 is 1."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, sine of 0 is 0. 2 times 0 is 0. So this whole thing is 0. 1 minus 0 is 1. Fair enough. Now let's think about the limit as x approaches 0 from the left-hand side. So as we approach 0 from the left-hand side of f of x."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "1 minus 0 is 1. Fair enough. Now let's think about the limit as x approaches 0 from the left-hand side. So as we approach 0 from the left-hand side of f of x. So as we approach 0 from the left-hand side, we're dealing with values of x that are less than 0. So once again, we're dealing with this first case up here. So this is the limit as x approaches 0 from the left-hand side of 1 minus 2 sine of x."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as we approach 0 from the left-hand side of f of x. So as we approach 0 from the left-hand side, we're dealing with values of x that are less than 0. So once again, we're dealing with this first case up here. So this is the limit as x approaches 0 from the left-hand side of 1 minus 2 sine of x. Now sine of x is continuous. It's a continuous function. So this is going to be the same thing as 1 minus 2 sine of 0, which we've already figured out, which is exactly equal to 1."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the limit as x approaches 0 from the left-hand side of 1 minus 2 sine of x. Now sine of x is continuous. It's a continuous function. So this is going to be the same thing as 1 minus 2 sine of 0, which we've already figured out, which is exactly equal to 1. So this is equal to 1. So the value of the limit as we approach from the left is the same as the value of the function. Now let's do it as we approach from the right, as we approach from values of x greater than 0."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be the same thing as 1 minus 2 sine of 0, which we've already figured out, which is exactly equal to 1. So this is equal to 1. So the value of the limit as we approach from the left is the same as the value of the function. Now let's do it as we approach from the right, as we approach from values of x greater than 0. So let's think about the limit as we approach 0 from the right of f of x. So here, we're dealing with values of x that are larger than 0. So we're dealing with this case right over here."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's do it as we approach from the right, as we approach from values of x greater than 0. So let's think about the limit as we approach 0 from the right of f of x. So here, we're dealing with values of x that are larger than 0. So we're dealing with this case right over here. So that's going to be the limit as x approaches 0 from the right of e to the negative 4x. And for the x's that we care about, or actually in general, this is a continuous function. This is a continuous function."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're dealing with this case right over here. So that's going to be the limit as x approaches 0 from the right of e to the negative 4x. And for the x's that we care about, or actually in general, this is a continuous function. This is a continuous function. So this is going to be the same thing as e to the negative 4 times 0, which is just e to the 0, which is just 1. So this, once again, is equal to 1. So the function is equal to 1 at that point."}, {"video_title": "2011 Calculus AB free response #6a   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is a continuous function. So this is going to be the same thing as e to the negative 4 times 0, which is just e to the 0, which is just 1. So this, once again, is equal to 1. So the function is equal to 1 at that point. The limit as we approach from the left is equal to 1. And the limit as we approach from the right is equal to 1. So the function is continuous there."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So a continuous function, let's say that's my y-axis, that is my x-axis, a function is going to be continuous over some interval if it just doesn't have any jumps or discontinuities or gaps over that interval. So if it's connected, and it for sure has to be defined over that interval without any gaps. So for example, a continuous function could look something like this. This function, let me make that line a little bit thicker. So this function right over here is continuous. It is connected over this interval, the interval that we can see. Now examples of discontinuous functions over interval or non-continuous functions, well, they would have gaps of some kind."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This function, let me make that line a little bit thicker. So this function right over here is continuous. It is connected over this interval, the interval that we can see. Now examples of discontinuous functions over interval or non-continuous functions, well, they would have gaps of some kind. They could have some type of an asymptotic discontinuity. So something like that, that makes it discontinuous. They could have a jump discontinuity, something like that."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now examples of discontinuous functions over interval or non-continuous functions, well, they would have gaps of some kind. They could have some type of an asymptotic discontinuity. So something like that, that makes it discontinuous. They could have a jump discontinuity, something like that. They could just have a gap where they're not defined. So they could have a gap where they're not defined. Or maybe they actually are defined there, but it's removable discontinuity."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "They could have a jump discontinuity, something like that. They could just have a gap where they're not defined. So they could have a gap where they're not defined. Or maybe they actually are defined there, but it's removable discontinuity. So all of these are examples of discontinuous functions. Now if you want the more mathy understanding of that, and we've looked at this before, we say that a function, f, is continuous at some value x equals a if and only if, draw my little two-way arrows here, say if and only if the limit of f of x as x approaches a is equal to the value of the function at a. So once again, in order to be continuous there, you at least have to be defined there."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or maybe they actually are defined there, but it's removable discontinuity. So all of these are examples of discontinuous functions. Now if you want the more mathy understanding of that, and we've looked at this before, we say that a function, f, is continuous at some value x equals a if and only if, draw my little two-way arrows here, say if and only if the limit of f of x as x approaches a is equal to the value of the function at a. So once again, in order to be continuous there, you at least have to be defined there. Now when you look at these, the one thing that jumps out at me, in order to be continuous for all real numbers, you have to be defined for all real numbers, and g of x is not defined for all real numbers. It's not defined for negative values of x. And so we would rule this one out."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, in order to be continuous there, you at least have to be defined there. Now when you look at these, the one thing that jumps out at me, in order to be continuous for all real numbers, you have to be defined for all real numbers, and g of x is not defined for all real numbers. It's not defined for negative values of x. And so we would rule this one out. So let's think about f of x equals e to the x. It is defined for all real numbers, and as we'll see, most of the common functions that you've learned in math, they don't have these strange jumps or gaps or discontinuities. Some of them do, functions like one over x and things like that, but things like e to the x, it doesn't have any of those."}, {"video_title": "Functions continuous on all real numbers   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we would rule this one out. So let's think about f of x equals e to the x. It is defined for all real numbers, and as we'll see, most of the common functions that you've learned in math, they don't have these strange jumps or gaps or discontinuities. Some of them do, functions like one over x and things like that, but things like e to the x, it doesn't have any of those. We could graph e to the x. E to the x looks something like this. It's defined for all real numbers. There's no jumps or gaps of any kind, and so this f of x is continuous for all real numbers, and f only."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "We've already seen scenarios where we start with a differential equation and then we generate a slope field that describes the solutions to the differential equation and then we use that to visualize those solutions. What I want to do in this video is do an exercise that takes us the other way. Start with a slope field and figure out which differential equation is the slope field describing the solutions for. And so I encourage you to look at each of these options and think about which of these differential equations is being described by this slope field. I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And so I encourage you to look at each of these options and think about which of these differential equations is being described by this slope field. I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here?"}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "I could have picked any other. I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "It looks like this slope right over here could be two. This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation."}, {"video_title": "Differential equation from slope field   First order differential equations   Khan Academy.mp3", "Sentence": "And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation. Y is equal to, y is equal to negative x, whoops, y is equal to negative x minus one. And you can verify that. If y is equal to negative x minus one, then the x and negative x cancel out, you're just left with dy dx is equal to negative one, which is exactly what's being described by this slope field."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Continuous, not differentiable, differentiable, not continuous, both continuous and differentiable, neither continuous nor differentiable. Now one of these we can knock out right from the get-go. You cannot, in order to be differentiable, you need to be continuous there, so you cannot have differentiable but not continuous, so let's just rule that one out. And now let's think about continuity. So let's first think about continuity. And frankly, if it isn't continuous, then it's not going to be differentiable. So let's think about it a little bit."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's think about continuity. So let's first think about continuity. And frankly, if it isn't continuous, then it's not going to be differentiable. So let's think about it a little bit. So in order to be continuous, f of, let me use a darker color, f of three needs to be equal to the limit of f of x as x approaches three. Now what is f of three? Well, let's see, we fall into this case right over here, because x is equal to three."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about it a little bit. So in order to be continuous, f of, let me use a darker color, f of three needs to be equal to the limit of f of x as x approaches three. Now what is f of three? Well, let's see, we fall into this case right over here, because x is equal to three. So six times three is 18, minus nine is nine, so this is nine. So the limit of f of x as x approaches three needs to be equal to nine. So let's first think about the limit as we approach from the left-hand side, the limit as x approaches three, as x approaches three from the left-hand side of f of x."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see, we fall into this case right over here, because x is equal to three. So six times three is 18, minus nine is nine, so this is nine. So the limit of f of x as x approaches three needs to be equal to nine. So let's first think about the limit as we approach from the left-hand side, the limit as x approaches three, as x approaches three from the left-hand side of f of x. Well, when x is less than three, we fall into this case. So f of x is just going to be equal to x squared. And so this is defined and continuous for all real numbers, so we can just substitute the three in there."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's first think about the limit as we approach from the left-hand side, the limit as x approaches three, as x approaches three from the left-hand side of f of x. Well, when x is less than three, we fall into this case. So f of x is just going to be equal to x squared. And so this is defined and continuous for all real numbers, so we can just substitute the three in there. So this is going to be equal to nine. Now what's the limit of, as we approach three from the right-hand side of f of x? Well, as we approach from the right, this one right over here is, f of x is equal to six x minus nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is defined and continuous for all real numbers, so we can just substitute the three in there. So this is going to be equal to nine. Now what's the limit of, as we approach three from the right-hand side of f of x? Well, as we approach from the right, this one right over here is, f of x is equal to six x minus nine. So we just write six x minus nine. And once again, six x minus nine is defined and continuous for all real numbers, so we can just pop a three in there, and you get 18 minus nine. Well, this is also equal to nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, as we approach from the right, this one right over here is, f of x is equal to six x minus nine. So we just write six x minus nine. And once again, six x minus nine is defined and continuous for all real numbers, so we can just pop a three in there, and you get 18 minus nine. Well, this is also equal to nine. So the right and left-hand, the left and right-hand limits both equal nine, which is equal to the value of the function there, so it is definitely continuous. So we can rule out, we can rule out this choice right over there. And now let's think about differentiability."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is also equal to nine. So the right and left-hand, the left and right-hand limits both equal nine, which is equal to the value of the function there, so it is definitely continuous. So we can rule out, we can rule out this choice right over there. And now let's think about differentiability. So in order to be differentiable, I'll just, differentiable, in order to be differentiable, the limit as x approaches three of f of x minus f of three over x minus three needs to exist. So let's see if we can evaluate this. So first of all, we know what f of three is."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's think about differentiability. So in order to be differentiable, I'll just, differentiable, in order to be differentiable, the limit as x approaches three of f of x minus f of three over x minus three needs to exist. So let's see if we can evaluate this. So first of all, we know what f of three is. F of three, we've already evaluated this. This is going to be nine. And let's see if we can evaluate this limit, or let's see what the limit is as we approach from the left-hand side and the right-hand side."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So first of all, we know what f of three is. F of three, we've already evaluated this. This is going to be nine. And let's see if we can evaluate this limit, or let's see what the limit is as we approach from the left-hand side and the right-hand side. And if they're approaching the same thing, then we know that this, and that same thing that they're approaching is the limit. So let's first think about the limit as x approaches three from the left-hand side. So it's over x minus three, and we have f of x minus nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see if we can evaluate this limit, or let's see what the limit is as we approach from the left-hand side and the right-hand side. And if they're approaching the same thing, then we know that this, and that same thing that they're approaching is the limit. So let's first think about the limit as x approaches three from the left-hand side. So it's over x minus three, and we have f of x minus nine. But as we approach from the left-hand side, this is f of x. As x is less than three, f of x is equal to x squared. So this would be, instead of f of x minus nine, I'll write x squared minus nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's over x minus three, and we have f of x minus nine. But as we approach from the left-hand side, this is f of x. As x is less than three, f of x is equal to x squared. So this would be, instead of f of x minus nine, I'll write x squared minus nine. And x squared minus nine, this is a difference of squares, so this is x plus three times x minus three. X plus three times x minus three. And so these would cancel out."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this would be, instead of f of x minus nine, I'll write x squared minus nine. And x squared minus nine, this is a difference of squares, so this is x plus three times x minus three. X plus three times x minus three. And so these would cancel out. We can say that this equivalent to x plus three, as long as x does not equal three, and that's okay, because we're approaching from the left. And as we approach from the left, well, x plus three is defined for all real numbers, it's continuous for all real numbers, so we can just substitute the three in there. So we would get a six."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so these would cancel out. We can say that this equivalent to x plus three, as long as x does not equal three, and that's okay, because we're approaching from the left. And as we approach from the left, well, x plus three is defined for all real numbers, it's continuous for all real numbers, so we can just substitute the three in there. So we would get a six. So now let's try to evaluate the limit as we approach from the right-hand side. So once again, it's f of x, but as we approach from the right-hand side, f of x is six x minus nine. That's our f of x."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would get a six. So now let's try to evaluate the limit as we approach from the right-hand side. So once again, it's f of x, but as we approach from the right-hand side, f of x is six x minus nine. That's our f of x. And then we have minus f of three, which is nine. So it's six x minus 18. Six x minus 18."}, {"video_title": "Differentiability at a point algebraic (function is differentiable)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's our f of x. And then we have minus f of three, which is nine. So it's six x minus 18. Six x minus 18. Well, that's the same thing as six times x minus three. And as we approach from the right, well, that's just going to be equal to six. So it looks like our derivative exists there, and it is equal to the limit as x approaches three of all of this business is equal to six, because the limit as we approach from the left and the right is also equal to six."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's the base of our, this purple, this, I guess, mauve or purple color is the base of it. But then it's kind of popping out of our screen. What I've drawn here in blue, you could view this as kind of the top ridge of the figure. And if you were to take cross-sections of the figure, and that's what this yellow line is, if you were to take cross-sections of this figure that are vertical, that are, I should say, perpendicular to the x-axis, those cross-sections are going to be isosceles-right triangles. So this cross-section is going to look like this if you were to look at, if you were to flatten it out. So over here it's sitting, it's popping out of your page or out of your screen, but if you were to actually flatten it out, the cross-section would look like this. It's going to be an isosceles-right triangle where the hypotenuse of the isosceles-right triangle sits along the base."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you were to take cross-sections of the figure, and that's what this yellow line is, if you were to take cross-sections of this figure that are vertical, that are, I should say, perpendicular to the x-axis, those cross-sections are going to be isosceles-right triangles. So this cross-section is going to look like this if you were to look at, if you were to flatten it out. So over here it's sitting, it's popping out of your page or out of your screen, but if you were to actually flatten it out, the cross-section would look like this. It's going to be an isosceles-right triangle where the hypotenuse of the isosceles-right triangle sits along the base. So it's isosceles, so that's equal to that. It's a right triangle. And then this distance between that point and this point is the same as the distance between f of x and g of x for this x value right over there."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be an isosceles-right triangle where the hypotenuse of the isosceles-right triangle sits along the base. So it's isosceles, so that's equal to that. It's a right triangle. And then this distance between that point and this point is the same as the distance between f of x and g of x for this x value right over there. And obviously that changes as we change our x value. And to help us visualize the shape here, I've kind of drawn a picture of our coordinate plane. If we view it as an angle, if we're kind of above it, you can kind of start to see how this figure would look."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this distance between that point and this point is the same as the distance between f of x and g of x for this x value right over there. And obviously that changes as we change our x value. And to help us visualize the shape here, I've kind of drawn a picture of our coordinate plane. If we view it as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base of it. I've drawn the base of it right over there. Maybe I should, to make it clear, let me make it like this."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we view it as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base of it. I've drawn the base of it right over there. Maybe I should, to make it clear, let me make it like this. Let me shade it in kind of parallel to these cross-sections. So I've drawn the base right over there. There's two other sides."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe I should, to make it clear, let me make it like this. Let me shade it in kind of parallel to these cross-sections. So I've drawn the base right over there. There's two other sides. There's the side that's on, at least the way I've drawn it here, I guess you could view it as its top side or the left side right over there. And over on this picture, that would be this when we're looking at it from above. And then you have this other side."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "There's two other sides. There's the side that's on, at least the way I've drawn it here, I guess you could view it as its top side or the left side right over there. And over on this picture, that would be this when we're looking at it from above. And then you have this other side. I guess on this view, this would be called kind of the right side. And over here, this is kind of the, when you view it over here, this is the bottom side. So the whole reason why I've set this up and we're attempting to visualize this figure is I want to see if you can come up with a definite integral that describes the volume of this figure that kind of almost looks like a football if you cut it in half or a rugby ball, but it's skewed a little bit as well."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you have this other side. I guess on this view, this would be called kind of the right side. And over here, this is kind of the, when you view it over here, this is the bottom side. So the whole reason why I've set this up and we're attempting to visualize this figure is I want to see if you can come up with a definite integral that describes the volume of this figure that kind of almost looks like a football if you cut it in half or a rugby ball, but it's skewed a little bit as well. But what's an expression, a definite integral that expresses the volume of this? And I encourage you to use the fact that it intersects at the point, these functions intersect at the point 0, 0 and c, d. So can you come up with some expression, a definite integral in terms of zeros and c's and d's and f's and g's that describe the volume of this figure? So I'm assuming you've paused the video and had a go at it, so let's think about it."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the whole reason why I've set this up and we're attempting to visualize this figure is I want to see if you can come up with a definite integral that describes the volume of this figure that kind of almost looks like a football if you cut it in half or a rugby ball, but it's skewed a little bit as well. But what's an expression, a definite integral that expresses the volume of this? And I encourage you to use the fact that it intersects at the point, these functions intersect at the point 0, 0 and c, d. So can you come up with some expression, a definite integral in terms of zeros and c's and d's and f's and g's that describe the volume of this figure? So I'm assuming you've paused the video and had a go at it, so let's think about it. So if we want to find the volume, one way to think about it is we could take, we could take the volume of, we could approximate the volume as the volume of these individual triangles. So that would be the area of each of these triangles times some very small depth, some very small depth, so let me just shade it in to show the depth, some very small depth, which we could call, we could call dx. So once again, we could find the volume of each of these by finding the area, the cross-sectional area there, and then multiplying that times a little bit, a little dx, a little dx, which would give us 3, so this is a little dx, which would give us 3-dimensional, that's our dx, I could write that a little bit neater, dx to give us a little bit of 3-dimensional depth."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm assuming you've paused the video and had a go at it, so let's think about it. So if we want to find the volume, one way to think about it is we could take, we could take the volume of, we could approximate the volume as the volume of these individual triangles. So that would be the area of each of these triangles times some very small depth, some very small depth, so let me just shade it in to show the depth, some very small depth, which we could call, we could call dx. So once again, we could find the volume of each of these by finding the area, the cross-sectional area there, and then multiplying that times a little bit, a little dx, a little dx, which would give us 3, so this is a little dx, which would give us 3-dimensional, that's our dx, I could write that a little bit neater, dx to give us a little bit of 3-dimensional depth. So how could we, what is the volume of one of these figures going to look like? Well, if we say, let me just call this height h, and we know that h is going to be f of x minus g of x, that's this distance right over here, so let's call that h. We know that h is going to be, and maybe I should say h of x, because it is going to be a function of x, h of x is going to be f of x, f of x minus g of x, minus g of x. But given an h, what is going to be the area of this triangle?"}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, we could find the volume of each of these by finding the area, the cross-sectional area there, and then multiplying that times a little bit, a little dx, a little dx, which would give us 3, so this is a little dx, which would give us 3-dimensional, that's our dx, I could write that a little bit neater, dx to give us a little bit of 3-dimensional depth. So how could we, what is the volume of one of these figures going to look like? Well, if we say, let me just call this height h, and we know that h is going to be f of x minus g of x, that's this distance right over here, so let's call that h. We know that h is going to be, and maybe I should say h of x, because it is going to be a function of x, h of x is going to be f of x, f of x minus g of x, minus g of x. But given an h, what is going to be the area of this triangle? Well, this is a 45, 45, 90 triangle. If this is 90, then this is going to have to be 45 degrees, that's going to have to be 45 degrees, and we know that the sides of a 45, 45, 90 triangle are square root of 2 times the hypotenuse. So this is going to be square root of 2, sorry, square root of 2 over 2 times the hypotenuse."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "But given an h, what is going to be the area of this triangle? Well, this is a 45, 45, 90 triangle. If this is 90, then this is going to have to be 45 degrees, that's going to have to be 45 degrees, and we know that the sides of a 45, 45, 90 triangle are square root of 2 times the hypotenuse. So this is going to be square root of 2, sorry, square root of 2 over 2 times the hypotenuse. Square root of 2 over 2 times the hypotenuse. And you can get that straight from the Pythagorean theorem. If this side, let's say it's length a, then this side has length a, so you're going to have a squared plus a squared is equal to the hypotenuse squared, or 2a squared is equal to the hypotenuse squared, or that a squared is equal to the hypotenuse squared over 2, or that a is equal to h over the square root of 2, which is the same thing as the square root of 2h over 2."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be square root of 2, sorry, square root of 2 over 2 times the hypotenuse. Square root of 2 over 2 times the hypotenuse. And you can get that straight from the Pythagorean theorem. If this side, let's say it's length a, then this side has length a, so you're going to have a squared plus a squared is equal to the hypotenuse squared, or 2a squared is equal to the hypotenuse squared, or that a squared is equal to the hypotenuse squared over 2, or that a is equal to h over the square root of 2, which is the same thing as the square root of 2h over 2. I just rationalized the denominator, multiplied both the numerator and the denominator by square root of 2. So that's where I got this from. And so what is the area going to be?"}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "If this side, let's say it's length a, then this side has length a, so you're going to have a squared plus a squared is equal to the hypotenuse squared, or 2a squared is equal to the hypotenuse squared, or that a squared is equal to the hypotenuse squared over 2, or that a is equal to h over the square root of 2, which is the same thing as the square root of 2h over 2. I just rationalized the denominator, multiplied both the numerator and the denominator by square root of 2. So that's where I got this from. And so what is the area going to be? Well, the area is just going to be your base times your height times 1 half. So let me write that down. So the area there, the area is just going to be the base, which is square root of 2 over 2 times our hypotenuse, times the height, which is square root of 2 over 2 times our hypotenuse, times 1 half."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what is the area going to be? Well, the area is just going to be your base times your height times 1 half. So let me write that down. So the area there, the area is just going to be the base, which is square root of 2 over 2 times our hypotenuse, times the height, which is square root of 2 over 2 times our hypotenuse, times 1 half. If we didn't do that 1 half, we'd be figuring out the area of this entire square. But this is obviously, we're concerned with the triangle. So what's this going to be?"}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area there, the area is just going to be the base, which is square root of 2 over 2 times our hypotenuse, times the height, which is square root of 2 over 2 times our hypotenuse, times 1 half. If we didn't do that 1 half, we'd be figuring out the area of this entire square. But this is obviously, we're concerned with the triangle. So what's this going to be? This is going to be square root of 2 over 2 times square root of 2 over 2 is going to be 1 half, and then you're going to have another 1 half. So it's 1 fourth h squared. Did I do that right?"}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's this going to be? This is going to be square root of 2 over 2 times square root of 2 over 2 is going to be 1 half, and then you're going to have another 1 half. So it's 1 fourth h squared. Did I do that right? This is, yes, it's going to be 2 over 4, which is 1 half, and then times another 1 half is 1 fourth h squared is the area. Now what's going to be the volume of each of these triangles? Well, the volume of each of these triangles right over here, the volume is just going to be our area times dx, or it's just going to be 1 fourth h squared times our depth, dx."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "Did I do that right? This is, yes, it's going to be 2 over 4, which is 1 half, and then times another 1 half is 1 fourth h squared is the area. Now what's going to be the volume of each of these triangles? Well, the volume of each of these triangles right over here, the volume is just going to be our area times dx, or it's just going to be 1 fourth h squared times our depth, dx. And so if we just integrated a bunch of these from our x equals 0 all the way to x equals c, we essentially have our volume of the entire figure. So how could we write that? So we want to write, we want to essentially define the volume of the figure, and we're kind of in the home stretch here."}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the volume of each of these triangles right over here, the volume is just going to be our area times dx, or it's just going to be 1 fourth h squared times our depth, dx. And so if we just integrated a bunch of these from our x equals 0 all the way to x equals c, we essentially have our volume of the entire figure. So how could we write that? So we want to write, we want to essentially define the volume of the figure, and we're kind of in the home stretch here. Actually, let me write it right. This is volume of a section, volume of a cross section. So what's the volume of the entire thing going to be?"}, {"video_title": "Volume with cross sections triangle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to write, we want to essentially define the volume of the figure, and we're kind of in the home stretch here. Actually, let me write it right. This is volume of a section, volume of a cross section. So what's the volume of the entire thing going to be? Well, the volume of the figure is going to be the definite integral from x equals 0 to x equals c of 1 fourth h squared. 1 fourth, but we know that h is equal to f of x minus g of x. So instead of h, I'm going to write f of x minus g of x squared dx."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, part C. At what time t, where t, where zero is less than or equal to t is less than or equal to eight, is the amount of water in the pipe at a minimum? Justify your answer. All right, well let's define a function w that represents the amount of water in the pipe at any time t, and then we can figure out how to figure out the minimum value for w over that interval. So let's just say w of t, so that's the amount of water in the pipe at any given time t, it's going to be equal to, well we start with 30 cubic feet of water in the pipe at t equals zero, they tell us that right up here. So we start with that much, and then we are going to add or take away some amount. So then it's going to be plus, we're going to sum up from time zero to time t, remember this is a function of t. Well let's sum up our net inflow times little small changes in time, and we'll sum up all those small changes in time from zero to t, and that'll give us our aggregate inflow, our aggregate net inflow. So what is our net inflow?"}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just say w of t, so that's the amount of water in the pipe at any given time t, it's going to be equal to, well we start with 30 cubic feet of water in the pipe at t equals zero, they tell us that right up here. So we start with that much, and then we are going to add or take away some amount. So then it's going to be plus, we're going to sum up from time zero to time t, remember this is a function of t. Well let's sum up our net inflow times little small changes in time, and we'll sum up all those small changes in time from zero to t, and that'll give us our aggregate inflow, our aggregate net inflow. So what is our net inflow? Well you have the function r, which says the rate at which the water is entering, and your function d, which is the rate at which water is exiting. And instead of using t as our variables here, I'm going to use x, since t is already one of our variable, is one of our boundaries of integration. And we can use any variable we want, it's really just a bit of a placeholder variable for the integration."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is our net inflow? Well you have the function r, which says the rate at which the water is entering, and your function d, which is the rate at which water is exiting. And instead of using t as our variables here, I'm going to use x, since t is already one of our variable, is one of our boundaries of integration. And we can use any variable we want, it's really just a bit of a placeholder variable for the integration. So our net inflow is r, I'll say r of x, minus d of x, and then dx. Once again, this is our net rate of inflow, this is our inflow minus our outflow, so our net rate of inflow times small changes, you could view this as time, although it's x, and then we sum up from zero to time is equal to t. So this is how much water we are going to have in the pipe at any time t. So let's now think about at what point does w hit a minimum in this interval? And there's three possibilities where w could hit a minimum point."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can use any variable we want, it's really just a bit of a placeholder variable for the integration. So our net inflow is r, I'll say r of x, minus d of x, and then dx. Once again, this is our net rate of inflow, this is our inflow minus our outflow, so our net rate of inflow times small changes, you could view this as time, although it's x, and then we sum up from zero to time is equal to t. So this is how much water we are going to have in the pipe at any time t. So let's now think about at what point does w hit a minimum in this interval? And there's three possibilities where w could hit a minimum point. It could hit a minimum right at the beginning, at w of zero, it could hit it at the end of our interval, at w of eight, at t equals eight, or it could be someplace in between where w is at a local maximum point, in which case the derivative of w will be equal to zero. So let's first evaluate w at the end points. So w of zero, well we know that, that's the bounds are gonna be zero to zero, they even gave that to us in the problem, is the initial state is we have 30 cubic feet of water in the pipe already."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there's three possibilities where w could hit a minimum point. It could hit a minimum right at the beginning, at w of zero, it could hit it at the end of our interval, at w of eight, at t equals eight, or it could be someplace in between where w is at a local maximum point, in which case the derivative of w will be equal to zero. So let's first evaluate w at the end points. So w of zero, well we know that, that's the bounds are gonna be zero to zero, they even gave that to us in the problem, is the initial state is we have 30 cubic feet of water in the pipe already. And now what is w of eight? Well this is going to be 30 plus the definite integral from zero to eight of r of x minus d of x minus d of x dx. And lucky for us, we're allowed to use a calculator, so let's use a calculator to evaluate this."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So w of zero, well we know that, that's the bounds are gonna be zero to zero, they even gave that to us in the problem, is the initial state is we have 30 cubic feet of water in the pipe already. And now what is w of eight? Well this is going to be 30 plus the definite integral from zero to eight of r of x minus d of x minus d of x dx. And lucky for us, we're allowed to use a calculator, so let's use a calculator to evaluate this. And so let me get my calculator out, and actually let me go see the definitions of r of x and d of x right over here so I can type them in. And what I'm gonna do, and there's a bunch of ways, the more advanced you get with a calculator that might save you some time if you're on the AP test, I'm gonna actually define a function that is r of x minus d of x. So everywhere I see a t, I'm gonna substitute it, I'm gonna use an x instead."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And lucky for us, we're allowed to use a calculator, so let's use a calculator to evaluate this. And so let me get my calculator out, and actually let me go see the definitions of r of x and d of x right over here so I can type them in. And what I'm gonna do, and there's a bunch of ways, the more advanced you get with a calculator that might save you some time if you're on the AP test, I'm gonna actually define a function that is r of x minus d of x. So everywhere I see a t, I'm gonna substitute it, I'm gonna use an x instead. And then I could use that later on to integrate it or to solve it in some way. So let's do that. So y one is gonna be 20 times sine of x squared divided by 35, close parentheses, minus, open parentheses, you have negative 0.04 times t to the third power, or sorry, so I'll write times x, x to the third power plus 0.4 times x squared plus 0.96 times x, and then I can close these parentheses."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So everywhere I see a t, I'm gonna substitute it, I'm gonna use an x instead. And then I could use that later on to integrate it or to solve it in some way. So let's do that. So y one is gonna be 20 times sine of x squared divided by 35, close parentheses, minus, open parentheses, you have negative 0.04 times t to the third power, or sorry, so I'll write times x, x to the third power plus 0.4 times x squared plus 0.96 times x, and then I can close these parentheses. So does that make sense? 20 times sine of x squared divided by 35 minus negative 0.04 x to the third plus 0.4 x squared plus 0.96 times x. All right, there you go."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So y one is gonna be 20 times sine of x squared divided by 35, close parentheses, minus, open parentheses, you have negative 0.04 times t to the third power, or sorry, so I'll write times x, x to the third power plus 0.4 times x squared plus 0.96 times x, and then I can close these parentheses. So does that make sense? 20 times sine of x squared divided by 35 minus negative 0.04 x to the third plus 0.4 x squared plus 0.96 times x. All right, there you go. I have defined y one. And now I can use that to evaluate different things. So let me now go back here."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, there you go. I have defined y one. And now I can use that to evaluate different things. So let me now go back here. And so let me evaluate what this is going to be. This is going to be 30 plus the definite integral. I'll go to the math functions."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me now go back here. And so let me evaluate what this is going to be. This is going to be 30 plus the definite integral. I'll go to the math functions. You scroll down a little bit, you have, this is the function for definite integral, function integral. And the function is y one. And I can go to vars, go to my y vars, it's gonna be a function variable."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll go to the math functions. You scroll down a little bit, you have, this is the function for definite integral, function integral. And the function is y one. And I can go to vars, go to my y vars, it's gonna be a function variable. So y one is what I select. I could have typed it in, but this will hopefully save time and I can reuse y one. And my variable of integration is x."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I can go to vars, go to my y vars, it's gonna be a function variable. So y one is what I select. I could have typed it in, but this will hopefully save time and I can reuse y one. And my variable of integration is x. And I'm going from zero to eight. From my lower bound is zero, my upper bound is eight. And let's see, did I type everything in?"}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And my variable of integration is x. And I'm going from zero to eight. From my lower bound is zero, my upper bound is eight. And let's see, did I type everything in? All right. And then let it munch a little bit. And there you have it."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, did I type everything in? All right. And then let it munch a little bit. And there you have it. It's approximately 48.544 cubic feet of water at t equals eight. So let me write that, 48.544. So this is approximately, approximately 48.544 cubic feet."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there you have it. It's approximately 48.544 cubic feet of water at t equals eight. So let me write that, 48.544. So this is approximately, approximately 48.544 cubic feet. And now let's see if there's any point in between where w hits a local maximum point or a local minimum point, I should say. I think I said a local maximum earlier on. I should say a local minimum point."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is approximately, approximately 48.544 cubic feet. And now let's see if there's any point in between where w hits a local maximum point or a local minimum point, I should say. I think I said a local maximum earlier on. I should say a local minimum point. And so if it's at a local minimum or maximum point, really, the derivative of w is going to be zero. So let's see, at what t do we get a zero derivative? So w prime of t. Well, the derivative of a constant with respect to t is zero."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I should say a local minimum point. And so if it's at a local minimum or maximum point, really, the derivative of w is going to be zero. So let's see, at what t do we get a zero derivative? So w prime of t. Well, the derivative of a constant with respect to t is zero. And the derivative of this with respect to t, and this comes straight out of the fundamental theorem of calculus, this is going to be, this is going to be r of t minus d of t. Where once again, this is a function. We have t as the upper bound. And so we have whatever here that we were integrating, but it's now going to be a function of t. And so in order for this to be equal to zero, we need to figure out when does r of t minus d of t equal zero."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So w prime of t. Well, the derivative of a constant with respect to t is zero. And the derivative of this with respect to t, and this comes straight out of the fundamental theorem of calculus, this is going to be, this is going to be r of t minus d of t. Where once again, this is a function. We have t as the upper bound. And so we have whatever here that we were integrating, but it's now going to be a function of t. And so in order for this to be equal to zero, we need to figure out when does r of t minus d of t equal zero. And lucky for us, we've already typed in r of t minus d of t. We've defined that as y sub one on our calculator. So let's go back here. And now I can use the solver."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we have whatever here that we were integrating, but it's now going to be a function of t. And so in order for this to be equal to zero, we need to figure out when does r of t minus d of t equal zero. And lucky for us, we've already typed in r of t minus d of t. We've defined that as y sub one on our calculator. So let's go back here. And now I can use the solver. So go to math, and then let's see if I scroll up. Let's see, I get to the solver. I could have scrolled down as well."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now I can use the solver. So go to math, and then let's see if I scroll up. Let's see, I get to the solver. I could have scrolled down as well. It's right below definite integral. So I go to the solver, and I say the equation zero equals, and now I could just go to y one. So I go to my y variables function."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could have scrolled down as well. It's right below definite integral. So I go to the solver, and I say the equation zero equals, and now I could just go to y one. So I go to my y variables function. I select y one. So my equation is equal to y one. And I press enter."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I go to my y variables function. I select y one. So my equation is equal to y one. And I press enter. And now I can put an initial guess for what x value is going to solve that equation. It's really a t value, but I'm using x as a variable here. And now I do alpha, and I click solve here."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I press enter. And now I can put an initial guess for what x value is going to solve that equation. It's really a t value, but I'm using x as a variable here. And now I do alpha, and I click solve here. It's the little blue you see right above the enter. And it gets to, well zero is one of them. Let's see if I can get another one."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now I do alpha, and I click solve here. It's the little blue you see right above the enter. And it gets to, well zero is one of them. Let's see if I can get another one. So let's see if I can get, see if I start at two. Alpha, solve, let it munch on that a little bit. Okay, so this is actually within the interval."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if I can get another one. So let's see if I can get, see if I start at two. Alpha, solve, let it munch on that a little bit. Okay, so this is actually within the interval. So approximately 3.272. Within 3.272. So t is approximately 3.272."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Okay, so this is actually within the interval. So approximately 3.272. Within 3.272. So t is approximately 3.272. That's where we have a local minimum point. But now we have to evaluate w there. So we have to evaluate w at 3.272 to figure out is it truly lower than w of zero."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So t is approximately 3.272. That's where we have a local minimum point. But now we have to evaluate w there. So we have to evaluate w at 3.272 to figure out is it truly lower than w of zero. W of eight is even higher than w of zero. So how do we do that? Well luckily, we can go back."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have to evaluate w at 3.272 to figure out is it truly lower than w of zero. W of eight is even higher than w of zero. So how do we do that? Well luckily, we can go back. So second, quit. And that should be stored under the variable x. Yep, it's right there. So then we can say, well let's calculate."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well luckily, we can go back. So second, quit. And that should be stored under the variable x. Yep, it's right there. So then we can say, well let's calculate. We want to calculate the function evaluated when t is 3.272. So our function is 30, 30, plus the definite integral. Go to the math."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So then we can say, well let's calculate. We want to calculate the function evaluated when t is 3.272. So our function is 30, 30, plus the definite integral. Go to the math. So definite integral. And once again, we have y1. So let me go to my y variables function."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Go to the math. So definite integral. And once again, we have y1. So let me go to my y variables function. Y1, we already defined that is r of x minus d of x. Our variable of integration is x. Our lower bound is zero."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me go to my y variables function. Y1, we already defined that is r of x minus d of x. Our variable of integration is x. Our lower bound is zero. Zero. And our upper bound is 3.272, which we stored in the variable x. So we can actually, it's a little confusing."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our lower bound is zero. Zero. And our upper bound is 3.272, which we stored in the variable x. So we can actually, it's a little confusing. This is saying what's our variable of integration. This is the actual value for our upper bound. And so let's munch on it a little bit."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can actually, it's a little confusing. This is saying what's our variable of integration. This is the actual value for our upper bound. And so let's munch on it a little bit. And we get 27.965. So this is approximately equal to 27.965. So now we're ready to answer, at what time t is the amount of water in the pipe at a minimum?"}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's munch on it a little bit. And we get 27.965. So this is approximately equal to 27.965. So now we're ready to answer, at what time t is the amount of water in the pipe at a minimum? We can see that at time 3.272, we have less water in the pipe than either right when we started or right at the end of our interval at t equals eight. So at what time t? We say t is equal to 3.272."}, {"video_title": "2015 AP Calculus AB BC 1c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we're ready to answer, at what time t is the amount of water in the pipe at a minimum? We can see that at time 3.272, we have less water in the pipe than either right when we started or right at the end of our interval at t equals eight. So at what time t? We say t is equal to 3.272. We could write w of 3.272 is less than w of zero, which is less than w of eight. So this indeed, just by knowing that the derivative there is zero, it could be a minimum or a maximum, but the fact that it's lower than both of the endpoints, this tells us that hey, this is a minimum, minimum, minimum point right over there. Or we could say that t equals 3.272 is where we hit our minimum, minimum value."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "And once again, it looks like you might be able to use the fundamental theorem of calculus. The big giveaway is that you're taking the derivative of a definite integral that gives you a function of x. But here I have x on both the upper and the lower boundary. And the fundamental theorem of calculus, at least from what we've seen, is when we have x's only on the upper boundary. And then, of course, it's an x squared, but we've seen examples of that already when we use the chain rule to do it. But how can we break this up and put this in a form that's a little bit closer to what we're familiar with when we apply the fundamental theorem of calculus? And to realize that, we really just have to attempt to graph what this is representing."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "And the fundamental theorem of calculus, at least from what we've seen, is when we have x's only on the upper boundary. And then, of course, it's an x squared, but we've seen examples of that already when we use the chain rule to do it. But how can we break this up and put this in a form that's a little bit closer to what we're familiar with when we apply the fundamental theorem of calculus? And to realize that, we really just have to attempt to graph what this is representing. So let's say that this is our lowercase f of x, or I should say f of t. So let's call this lowercase f of t. And let's graph it over the interval between x and x squared. So let's say this is my y-axis. This is my t-axis."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "And to realize that, we really just have to attempt to graph what this is representing. So let's say that this is our lowercase f of x, or I should say f of t. So let's call this lowercase f of t. And let's graph it over the interval between x and x squared. So let's say this is my y-axis. This is my t-axis. And let's say that this right over here is y is equal to f of t. I'm drawing it generally. I don't know what this exactly looks like. And we're going to talk about the interval between x and x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "This is my t-axis. And let's say that this right over here is y is equal to f of t. I'm drawing it generally. I don't know what this exactly looks like. And we're going to talk about the interval between x and x squared. So if we're going to talk about the interval between x, which is right over here, it's the lower bound. So x and x squared. It's the lower bound, at least for this definite integral."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "And we're going to talk about the interval between x and x squared. So if we're going to talk about the interval between x, which is right over here, it's the lower bound. So x and x squared. It's the lower bound, at least for this definite integral. We don't know for sure. It depends on what x you choose on which one is actually smaller. But let's just say that for the sake of visualizing, we'll draw x right over here."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "It's the lower bound, at least for this definite integral. We don't know for sure. It depends on what x you choose on which one is actually smaller. But let's just say that for the sake of visualizing, we'll draw x right over here. And we will draw x squared right over here. So this whole expression, this entire definite integral, is essentially asking for, is essentially representing this entire area. The entire area under the curve."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "But let's just say that for the sake of visualizing, we'll draw x right over here. And we will draw x squared right over here. So this whole expression, this entire definite integral, is essentially asking for, is essentially representing this entire area. The entire area under the curve. But what we could do is introduce a constant that's someplace in between x and x squared. Let's say that constant is c. And break this area into two different areas with c as the divider. So that same exact whole area, we can now write it as two separate integrals."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "The entire area under the curve. But what we could do is introduce a constant that's someplace in between x and x squared. Let's say that constant is c. And break this area into two different areas with c as the divider. So that same exact whole area, we can now write it as two separate integrals. So one integral that represents this area right over here. And then another integral that represents this area right over there. And we just say c is some constant between x and x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "So that same exact whole area, we can now write it as two separate integrals. So one integral that represents this area right over here. And then another integral that represents this area right over there. And we just say c is some constant between x and x squared. Well, how can we denote this area in purple? Well, that's going to be, so this thing is going to be equal to the sum of these two areas. The purple area we can show is the definite integral from x to c of our function of t. Cosine t over t dt."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "And we just say c is some constant between x and x squared. Well, how can we denote this area in purple? Well, that's going to be, so this thing is going to be equal to the sum of these two areas. The purple area we can show is the definite integral from x to c of our function of t. Cosine t over t dt. And then to that, we're going to add the green area. And then we'll get the original area. So the green area, for the green area, our lower bound of integration is now our constant c. And our upper bound of integration is x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "The purple area we can show is the definite integral from x to c of our function of t. Cosine t over t dt. And then to that, we're going to add the green area. And then we'll get the original area. So the green area, for the green area, our lower bound of integration is now our constant c. And our upper bound of integration is x squared. It's going to be of cosine t over t dt. And this is a form where if we know how to apply the chain rule, we can apply the fundamental theorem of calculus. And this is almost in a form."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "So the green area, for the green area, our lower bound of integration is now our constant c. And our upper bound of integration is x squared. It's going to be of cosine t over t dt. And this is a form where if we know how to apply the chain rule, we can apply the fundamental theorem of calculus. And this is almost in a form. We're used to seeing it where the x is the upper bound. And we already know what happens. We can swap these two bounds, but it will just be the negative of that integral."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "And this is almost in a form. We're used to seeing it where the x is the upper bound. And we already know what happens. We can swap these two bounds, but it will just be the negative of that integral. So this is going to be equal to, let me rewrite it, the negative of the definite integral from c to x of cosine t over t dt. And then we have plus the definite integral that goes from c to x squared of cosine t over t dt. So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "We can swap these two bounds, but it will just be the negative of that integral. So this is going to be equal to, let me rewrite it, the negative of the definite integral from c to x of cosine t over t dt. And then we have plus the definite integral that goes from c to x squared of cosine t over t dt. So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus. So if we want to find f prime of x, well, applying the derivative operator over here, we're going to have a negative out front. It's going to be equal to negative cosine x over x. Once again, just the fundamental theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus. So if we want to find f prime of x, well, applying the derivative operator over here, we're going to have a negative out front. It's going to be equal to negative cosine x over x. Once again, just the fundamental theorem of calculus. And then plus, we're first going to take the derivative of this thing with respect to x squared. And that's going to give you cosine of x squared over x squared. Wherever you saw a t, you replace it with an x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "Once again, just the fundamental theorem of calculus. And then plus, we're first going to take the derivative of this thing with respect to x squared. And that's going to give you cosine of x squared over x squared. Wherever you saw a t, you replace it with an x squared. And then you're going to multiply that times the derivative of x squared with respect to x. So that's just going to be derivative of x squared with respect to x is just 2x. And we're done."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds   Khan Academy.mp3", "Sentence": "Wherever you saw a t, you replace it with an x squared. And then you're going to multiply that times the derivative of x squared with respect to x. So that's just going to be derivative of x squared with respect to x is just 2x. And we're done. We just need to simplify this thing. All of this is going to be equal to negative cosine x over x plus, well, this is going to cancel out with just one of those, plus 2 cosine x squared over x. And I guess we could simplify it even more as being equal to, and we can swap these."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I want to do in this video is think about the idea of an average value of a function over some closed interval. So what do I mean by that? And how can we think about what average value of a function even means? So let's say that's my y-axis. And let's say that this is my, this right over here is my x-axis. And let me draw a function here. So let's say the function looks something like that."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that's my y-axis. And let's say that this is my, this right over here is my x-axis. And let me draw a function here. So let's say the function looks something like that. That's the graph of y is equal to, y is equal to f of x. And now let's think about a closed interval. So we're gonna think about the closed interval between a and b, including a and b."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say the function looks something like that. That's the graph of y is equal to, y is equal to f of x. And now let's think about a closed interval. So we're gonna think about the closed interval between a and b, including a and b. That's what makes it closed. We're including our endpoints. So we're gonna think about this interval right over here."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna think about the closed interval between a and b, including a and b. That's what makes it closed. We're including our endpoints. So we're gonna think about this interval right over here. So between x is equal to a and x is equal to b, what is the average value of this function? One way to think about it is what is the average height of this function? So how could we, what would that mean?"}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna think about this interval right over here. So between x is equal to a and x is equal to b, what is the average value of this function? One way to think about it is what is the average height of this function? So how could we, what would that mean? Well, one way to think about it, it would be some height so that if we multiply it times the width of this interval, we'll get the area under the curve. So another way to think about it is the area under the curve right over here, the area, actually let me do this in a different color. So the area under this curve right over here, I'll shade it in yellow."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how could we, what would that mean? Well, one way to think about it, it would be some height so that if we multiply it times the width of this interval, we'll get the area under the curve. So another way to think about it is the area under the curve right over here, the area, actually let me do this in a different color. So the area under this curve right over here, I'll shade it in yellow. We already know that we can express this as the definite integral from a to b of f of x dx. The average value of our function over this closed interval a, b, let me write that, over the closed interval between a and b, including a and b, we could think about it as some height, some height, let me do this in a new color, some value of our function, some height, let me think about it, maybe some height right over here, so that if we multiply this height times this width, we're gonna get the area of a rectangle. The rectangle would be the area of this rectangle right over here, and that rectangle is going to have the same area as the area under the curve, which is a reasonable way if you kind of remember how when you even think about finding the area, or one way to think about the area of a trapezoid, you can multiply, if you have a trapezoid, if you have a trapezoid like this, you have a trapezoid like this, this is, you could kind of turn 90 degrees, but you multiply the height times the average width of the trapezoid, and then that will give you its area, so this would be the average width, which in a trapezoid like this would just be halfway between, this function is not linear, so it's not necessarily going to be halfway in between, but it's that same idea."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area under this curve right over here, I'll shade it in yellow. We already know that we can express this as the definite integral from a to b of f of x dx. The average value of our function over this closed interval a, b, let me write that, over the closed interval between a and b, including a and b, we could think about it as some height, some height, let me do this in a new color, some value of our function, some height, let me think about it, maybe some height right over here, so that if we multiply this height times this width, we're gonna get the area of a rectangle. The rectangle would be the area of this rectangle right over here, and that rectangle is going to have the same area as the area under the curve, which is a reasonable way if you kind of remember how when you even think about finding the area, or one way to think about the area of a trapezoid, you can multiply, if you have a trapezoid, if you have a trapezoid like this, you have a trapezoid like this, this is, you could kind of turn 90 degrees, but you multiply the height times the average width of the trapezoid, and then that will give you its area, so this would be the average width, which in a trapezoid like this would just be halfway between, this function is not linear, so it's not necessarily going to be halfway in between, but it's that same idea. So how could we use this idea, where this is right over here, this height, this height right over here, we could call this, we could call this the function's average, the function's average. How could we use all of this to come up with a formula for the average of a function over this closed interval? Well, let's just express in math what we've already said."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "The rectangle would be the area of this rectangle right over here, and that rectangle is going to have the same area as the area under the curve, which is a reasonable way if you kind of remember how when you even think about finding the area, or one way to think about the area of a trapezoid, you can multiply, if you have a trapezoid, if you have a trapezoid like this, you have a trapezoid like this, this is, you could kind of turn 90 degrees, but you multiply the height times the average width of the trapezoid, and then that will give you its area, so this would be the average width, which in a trapezoid like this would just be halfway between, this function is not linear, so it's not necessarily going to be halfway in between, but it's that same idea. So how could we use this idea, where this is right over here, this height, this height right over here, we could call this, we could call this the function's average, the function's average. How could we use all of this to come up with a formula for the average of a function over this closed interval? Well, let's just express in math what we've already said. We already said that this function average should be some height, so let's say the function average, so that's a height, and if I multiply it times the width of this interval, so this width right over here, this width right over here is just going to be the larger value minus the smaller value, so that's going to be b minus a, so the average value of the function times the width of the interval should give us an area that is equivalent to the area under the curve, so it should be equal to the definite integral from a to b of f of x dx, and so if we just, if we knew all of this other stuff, we could solve for the function's average. The function's average, if we divide both sides by b minus a, the function's average is going to be equal to, just dividing both sides by b minus a, you're going to get one over b minus a times the definite integral, the definite integral from a to b of f of x dx, or another way to think about it, you're going to figure out what the area under the curve is over that interval. You're going to divide that by the width, and then you're going to have the function's average."}, {"video_title": "Average value over a closed interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's just express in math what we've already said. We already said that this function average should be some height, so let's say the function average, so that's a height, and if I multiply it times the width of this interval, so this width right over here, this width right over here is just going to be the larger value minus the smaller value, so that's going to be b minus a, so the average value of the function times the width of the interval should give us an area that is equivalent to the area under the curve, so it should be equal to the definite integral from a to b of f of x dx, and so if we just, if we knew all of this other stuff, we could solve for the function's average. The function's average, if we divide both sides by b minus a, the function's average is going to be equal to, just dividing both sides by b minus a, you're going to get one over b minus a times the definite integral, the definite integral from a to b of f of x dx, or another way to think about it, you're going to figure out what the area under the curve is over that interval. You're going to divide that by the width, and then you're going to have the function's average. One way to think about it, you're going to have the average height, and once again, I'd like to remind you that because you shouldn't just sit there and try to memorize this thing. Just get a conceptual understanding of what this is really just trying to say. Area under the curve divided by the width, well, that's just going to give you the average height."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then they say there's no point c between in that interval for which f prime of c is equal to the average rate of change. Explain why the statement does not contradict the mean value theorem. Fascinating. All right, let's do this first part first. The average rate of change of f on the interval. Sounds like a very fancy thing, but the average rate of change on the interval is really just the slope of the line that connects the endpoints of the interval. So this right over here, those are the endpoints, and let's figure out the slope of that line."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's do this first part first. The average rate of change of f on the interval. Sounds like a very fancy thing, but the average rate of change on the interval is really just the slope of the line that connects the endpoints of the interval. So this right over here, those are the endpoints, and let's figure out the slope of that line. So going from this point to that point, our change in x, which is in a color that you're likely to see, our change in x over here, our change in x is equal to seven. And you could get that by taking three minus negative four. You could literally just count."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here, those are the endpoints, and let's figure out the slope of that line. So going from this point to that point, our change in x, which is in a color that you're likely to see, our change in x over here, our change in x is equal to seven. And you could get that by taking three minus negative four. You could literally just count. One, two, three, four, five, six, seven. That's our change in x, and our change in y, when we run over seven, when we go seven to the right, our change in y is equal to negative two. We went from negative one to negative three is equal to negative two."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could literally just count. One, two, three, four, five, six, seven. That's our change in x, and our change in y, when we run over seven, when we go seven to the right, our change in y is equal to negative two. We went from negative one to negative three is equal to negative two. So slope, which is change in y over change in x, rise over run, is equal to seven over negative two, or, oh, actually, let me, no, I did it the other way around. Change in y is negative two over seven. So negative two sevens."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "We went from negative one to negative three is equal to negative two. So slope, which is change in y over change in x, rise over run, is equal to seven over negative two, or, oh, actually, let me, no, I did it the other way around. Change in y is negative two over seven. So negative two sevens. And that's all it is. Now, you could kind of think of it in kind of the more fancy sense, but you're gonna get the exact same answer if you said, oh, well, look, the average rate of change of f on the interval, well, the rate of change of f is f prime, f prime of x. That is the rate of change of f at any point x."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative two sevens. And that's all it is. Now, you could kind of think of it in kind of the more fancy sense, but you're gonna get the exact same answer if you said, oh, well, look, the average rate of change of f on the interval, well, the rate of change of f is f prime, f prime of x. That is the rate of change of f at any point x. And so if you wanna find the average value of this over the interval, you would integrate it from our starting point, from negative four to three dx, and then you would divide it by your change in x. So then you would divide it by, you would have it over one over seven. Over one over seven."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is the rate of change of f at any point x. And so if you wanna find the average value of this over the interval, you would integrate it from our starting point, from negative four to three dx, and then you would divide it by your change in x. So then you would divide it by, you would have it over one over seven. Over one over seven. But then this part right over here is just going to be the same thing as f of three minus f of four. And then this over here, you have a seven in the denominator. So this is really just your change in x."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "Over one over seven. But then this part right over here is just going to be the same thing as f of three minus f of four. And then this over here, you have a seven in the denominator. So this is really just your change in x. This is your change in x, which it was by definition, or how we actually set up this average right over here. And this over here is really just your change in y. So it really is just the slope between the endpoints."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is really just your change in x. This is your change in x, which it was by definition, or how we actually set up this average right over here. And this over here is really just your change in y. So it really is just the slope between the endpoints. So we did the first part. Our average rate of change of f on the interval is negative two sevenths. And then let's think about the second part."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it really is just the slope between the endpoints. So we did the first part. Our average rate of change of f on the interval is negative two sevenths. And then let's think about the second part. They say there's no point c in that interval for which f prime of c is equal to the average rate of change. Explain why the statement does not contradict the mean value theorem. So the mean value theorem, just as a little bit of a review, as a little bit of review, it says that if we have some type of an interval, if you have an interval, so let me draw some axes right over here."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let's think about the second part. They say there's no point c in that interval for which f prime of c is equal to the average rate of change. Explain why the statement does not contradict the mean value theorem. So the mean value theorem, just as a little bit of a review, as a little bit of review, it says that if we have some type of an interval, if you have an interval, so let me draw some axes right over here. If you have an interval, so let me draw an interval like this. And over that interval, you have a differentiable function. You have a differentiable function."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the mean value theorem, just as a little bit of a review, as a little bit of review, it says that if we have some type of an interval, if you have an interval, so let me draw some axes right over here. If you have an interval, so let me draw an interval like this. And over that interval, you have a differentiable function. You have a differentiable function. So maybe my function looks like this. It says there's at least one point c on that interval where the derivative at that point c is equal to the average rate of change. So the way I've drawn it right over here, the average rate of change of this function, I'll do it in magenta, is this right over here."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have a differentiable function. So maybe my function looks like this. It says there's at least one point c on that interval where the derivative at that point c is equal to the average rate of change. So the way I've drawn it right over here, the average rate of change of this function, I'll do it in magenta, is this right over here. And the mean value theorem says is that if this is differentiable, there's at least one point c in this interval where I have the same slope, where the tangent line has the same slope, where the derivative is the same as the slope, as that average slope. And you can see here, it's probably right over here, you have one of those points. And actually, we probably have multiple of them."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the way I've drawn it right over here, the average rate of change of this function, I'll do it in magenta, is this right over here. And the mean value theorem says is that if this is differentiable, there's at least one point c in this interval where I have the same slope, where the tangent line has the same slope, where the derivative is the same as the slope, as that average slope. And you can see here, it's probably right over here, you have one of those points. And actually, we probably have multiple of them. We probably have another point right over here that's like that. And then another point there that's like that. And if you think about it, it's kind of intuitive that at some point, you know, here we're, we obviously have a larger slope, and over here we have a smaller slope."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually, we probably have multiple of them. We probably have another point right over here that's like that. And then another point there that's like that. And if you think about it, it's kind of intuitive that at some point, you know, here we're, we obviously have a larger slope, and over here we have a smaller slope. And since it's differentiable, our derivative is continuous. So at some point, at some point, the slope has to get to exactly what the average slope is. Now let's think about our little conundrum with this question right over here."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you think about it, it's kind of intuitive that at some point, you know, here we're, we obviously have a larger slope, and over here we have a smaller slope. And since it's differentiable, our derivative is continuous. So at some point, at some point, the slope has to get to exactly what the average slope is. Now let's think about our little conundrum with this question right over here. Why is there not a point c for which f prime of c is equal to the average rate of change? And you can even verify that for yourself because from negative four to zero, our slope is positive. We have a positive slope here."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's think about our little conundrum with this question right over here. Why is there not a point c for which f prime of c is equal to the average rate of change? And you can even verify that for yourself because from negative four to zero, our slope is positive. We have a positive slope here. And then the slope just jumps down to negative two. It just jumps down to negative two, which is a much more negative slope than this. So it never goes to negative two sevenths."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have a positive slope here. And then the slope just jumps down to negative two. It just jumps down to negative two, which is a much more negative slope than this. So it never goes to negative two sevenths. And the reason is is that this is not a differentiable function at x is equal to zero. It is not differentiable at x equals zero. Our slope jumps here."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it never goes to negative two sevenths. And the reason is is that this is not a differentiable function at x is equal to zero. It is not differentiable at x equals zero. Our slope jumps here. And because it's not differentiable, the mean value theorem doesn't apply. Now you can imagine if this was differentiable, if this did have a continuous derivative, then you would find a point. So if this looks something like this instead, I'll continue it over here."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our slope jumps here. And because it's not differentiable, the mean value theorem doesn't apply. Now you can imagine if this was differentiable, if this did have a continuous derivative, then you would find a point. So if this looks something like this instead, I'll continue it over here. If it looks something like this instead, where it had a continuous derivative, then there would be a point where the slope was the same as the average slope. Maybe it would have been right over there. So it's really because f is not differentiable over the entire interval."}, {"video_title": "2011 Calculus AB free response #4d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if this looks something like this instead, I'll continue it over here. If it looks something like this instead, where it had a continuous derivative, then there would be a point where the slope was the same as the average slope. Maybe it would have been right over there. So it's really because f is not differentiable over the entire interval. It does not have a continuous derivative. The derivative jumps from a positive value approaching zero here, and it just jumps straight down to negative two right over here. It doesn't go continuously through all the values in between."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to pause this video, and especially if you've seen the other videos on introducing the idea of the average value of a function, figure out what this is. What is the average value of our function f over this interval? So I'm assuming you've had a go at it. Let's just visualize what's going on, and then we can actually find the average. So that's my y-axis. This is my x-axis. Now over the interval between zero and three, so let's say that this is zero."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just visualize what's going on, and then we can actually find the average. So that's my y-axis. This is my x-axis. Now over the interval between zero and three, so let's say that this is zero. This is one, two, three. It's a closed interval. When x is zero, f of zero is going to be one."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now over the interval between zero and three, so let's say that this is zero. This is one, two, three. It's a closed interval. When x is zero, f of zero is going to be one. So we're gonna be right over here. F of one is two, so it's gonna be, so it's one, two, three. Actually, let me make my scale a little bit smaller on that."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is zero, f of zero is going to be one. So we're gonna be right over here. F of one is two, so it's gonna be, so it's one, two, three. Actually, let me make my scale a little bit smaller on that. I have to go all the way up to 10. So this is gonna be 10. This is gonna be five."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me make my scale a little bit smaller on that. I have to go all the way up to 10. So this is gonna be 10. This is gonna be five. And then one, two, three. Actually, let me, this is the hardest part, is making this even. So let's see, this is gonna be in the middle."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is gonna be five. And then one, two, three. Actually, let me, this is the hardest part, is making this even. So let's see, this is gonna be in the middle. Pretty good. And then let's see, in the middle, and then we have that. No, good enough."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, this is gonna be in the middle. Pretty good. And then let's see, in the middle, and then we have that. No, good enough. All right, so we're gonna be there, and we're gonna be there. I have, obviously, different scales for x and y-axes. Two squared plus one is five."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "No, good enough. All right, so we're gonna be there, and we're gonna be there. I have, obviously, different scales for x and y-axes. Two squared plus one is five. Three squared plus one is 10. So it's gonna look something like this. This is what our function is going to look like."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two squared plus one is five. Three squared plus one is 10. So it's gonna look something like this. This is what our function is going to look like. So that's the graph of y is equal to f of x. And we care about the average value on the interval, closed interval between zero and three. Between zero and three."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is what our function is going to look like. So that's the graph of y is equal to f of x. And we care about the average value on the interval, closed interval between zero and three. Between zero and three. So one way to think about it, you could apply the formula, but it's very important to think about what does that formula actually mean? And once again, you shouldn't memorize this formula because it actually kind of falls out out of what it actually means. So the average of our function is going to be, it's going to be equal to the definite integral over this interval."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "Between zero and three. So one way to think about it, you could apply the formula, but it's very important to think about what does that formula actually mean? And once again, you shouldn't memorize this formula because it actually kind of falls out out of what it actually means. So the average of our function is going to be, it's going to be equal to the definite integral over this interval. So essentially, the area under this curve. So it's going to be the definite integral from zero to three of f of x, which is x squared plus one, dx. And then we're gonna take this area."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the average of our function is going to be, it's going to be equal to the definite integral over this interval. So essentially, the area under this curve. So it's going to be the definite integral from zero to three of f of x, which is x squared plus one, dx. And then we're gonna take this area. We're gonna take this area right over here, and we're gonna divide it by the width of our interval to essentially come up with the average height or the average value of our function. So we're gonna divide it by b minus a, or three minus zero, which is just going to be three. And so now we just have to evaluate this."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're gonna take this area. We're gonna take this area right over here, and we're gonna divide it by the width of our interval to essentially come up with the average height or the average value of our function. So we're gonna divide it by b minus a, or three minus zero, which is just going to be three. And so now we just have to evaluate this. So this is going to be equal to 1 3rd times, see, the antiderivative of x squared is x to the third over three. Antiderivative of one is x, and we're going to evaluate it from zero to three. And so this is going to be equal to 1 3rd times, when we evaluate it at three, let me use another color here, when we evaluate it at three, it's going to be three to the third divided by three."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now we just have to evaluate this. So this is going to be equal to 1 3rd times, see, the antiderivative of x squared is x to the third over three. Antiderivative of one is x, and we're going to evaluate it from zero to three. And so this is going to be equal to 1 3rd times, when we evaluate it at three, let me use another color here, when we evaluate it at three, it's going to be three to the third divided by three. Well, that's just going to be 27 divided by three. That's nine plus three. And then when we evaluate it at zero, minus zero, minus zero."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to 1 3rd times, when we evaluate it at three, let me use another color here, when we evaluate it at three, it's going to be three to the third divided by three. Well, that's just going to be 27 divided by three. That's nine plus three. And then when we evaluate it at zero, minus zero, minus zero. So it's just minus, when you evaluate it at zero, it's just gonna be zero. And so we are left with, I'm gonna make the brackets that same color. This is going to be 1 3rd times 12, 1 3rd times 12, which is equal to four."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then when we evaluate it at zero, minus zero, minus zero. So it's just minus, when you evaluate it at zero, it's just gonna be zero. And so we are left with, I'm gonna make the brackets that same color. This is going to be 1 3rd times 12, 1 3rd times 12, which is equal to four. Which is equal to four. So this is the average value of our function. The average value of our function over this interval, over this interval, is equal, the average value of our function is equal to four."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be 1 3rd times 12, 1 3rd times 12, which is equal to four. Which is equal to four. So this is the average value of our function. The average value of our function over this interval, over this interval, is equal, the average value of our function is equal to four. And notice, our function actually hits that value at some point in the interval. At some point in the interval, something lower than two, but greater than one, we could maybe call that C. It looks like our function hits that value. And this is actually, this is actually a generally true thing."}, {"video_title": "Calculating average value of function over interval   AP Calculus AB   Khan Academy.mp3", "Sentence": "The average value of our function over this interval, over this interval, is equal, the average value of our function is equal to four. And notice, our function actually hits that value at some point in the interval. At some point in the interval, something lower than two, but greater than one, we could maybe call that C. It looks like our function hits that value. And this is actually, this is actually a generally true thing. This is a mean value theorem for integrals, and we'll go into more depth there. But you can see that this kind of does look like its average value. That if you imagine the box, if you multiply this height, this average value times this width, you would have this area right over here, and this area right over here is the same, this area that I'm highlighting in yellow right over here, is the same as the area under the curve."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that we have the function g of x. And it is equal to the definite integral from 19 to x of the cube root of t dt. And what I'm curious about finding or trying to figure out is, what is g prime of 27? What is that equal to? Pause this video and try to think about it. And I'll give you a little bit of a hint. Think about the second fundamental theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "What is that equal to? Pause this video and try to think about it. And I'll give you a little bit of a hint. Think about the second fundamental theorem of calculus. All right, now let's work on this together. So we wanna figure out what g prime, we could try to figure out what g prime of x is and then evaluate that at 27. And the best way that I can think about doing that is by taking the derivative of both sides of this equation."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "Think about the second fundamental theorem of calculus. All right, now let's work on this together. So we wanna figure out what g prime, we could try to figure out what g prime of x is and then evaluate that at 27. And the best way that I can think about doing that is by taking the derivative of both sides of this equation. So let's take the derivative of both sides of that equation. So the left-hand side, we'll take the derivative with respect to x of g of x, and the right-hand side, the derivative with respect to x of all of this business. Now the left-hand side is pretty straightforward."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "And the best way that I can think about doing that is by taking the derivative of both sides of this equation. So let's take the derivative of both sides of that equation. So the left-hand side, we'll take the derivative with respect to x of g of x, and the right-hand side, the derivative with respect to x of all of this business. Now the left-hand side is pretty straightforward. The derivative with respect to x of g of x, that's just going to be g prime of x. But what is the right-hand side going to be equal to? Well, that's where the second fundamental theorem of calculus is useful."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "Now the left-hand side is pretty straightforward. The derivative with respect to x of g of x, that's just going to be g prime of x. But what is the right-hand side going to be equal to? Well, that's where the second fundamental theorem of calculus is useful. I'll write it right over here. Second fundamental, I'll abbreviate a little bit, theorem, theorem of calculus. It tells us, let's say we have some function capital F of x, and it's equal to the definite integral from a, some constant a to x of lowercase f of t dt."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's where the second fundamental theorem of calculus is useful. I'll write it right over here. Second fundamental, I'll abbreviate a little bit, theorem, theorem of calculus. It tells us, let's say we have some function capital F of x, and it's equal to the definite integral from a, some constant a to x of lowercase f of t dt. The second fundamental theorem of calculus tells us that if our lowercase f, if lowercase f is continuous on the interval from a to x, so I'll write it this way, on the closed interval from a to x, then the derivative of our capital F of x, so capital F prime of x, is just going to be equal to our inner function f evaluated at x instead of t. It's going to become lowercase f of x. Now I know when you first saw this, you thought that, hey, this might be some cryptic thing that you might not use too often. But we're going to see that it's actually very, very useful."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "It tells us, let's say we have some function capital F of x, and it's equal to the definite integral from a, some constant a to x of lowercase f of t dt. The second fundamental theorem of calculus tells us that if our lowercase f, if lowercase f is continuous on the interval from a to x, so I'll write it this way, on the closed interval from a to x, then the derivative of our capital F of x, so capital F prime of x, is just going to be equal to our inner function f evaluated at x instead of t. It's going to become lowercase f of x. Now I know when you first saw this, you thought that, hey, this might be some cryptic thing that you might not use too often. But we're going to see that it's actually very, very useful. And even in the future, and some of you might already know, there's multiple ways to try to think about a definite integral like this, and you'll learn it in the future. But this can be extremely simplifying, especially if you have a hairy definite integral like this. And so this just tells us, hey, look, the derivative with respect to x of all of this business, first we have to check that our inner function, which would be analogous to our lowercase f here, is this continuous on the interval from 19 to x?"}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "But we're going to see that it's actually very, very useful. And even in the future, and some of you might already know, there's multiple ways to try to think about a definite integral like this, and you'll learn it in the future. But this can be extremely simplifying, especially if you have a hairy definite integral like this. And so this just tells us, hey, look, the derivative with respect to x of all of this business, first we have to check that our inner function, which would be analogous to our lowercase f here, is this continuous on the interval from 19 to x? Well, no matter what x is, this is going to be continuous over that interval because this is continuous for all x's. And so we meet this first condition, or our major condition. And so then we could just say, all right, then the derivative of all of this is just going to be this inner function replacing t with x."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "And so this just tells us, hey, look, the derivative with respect to x of all of this business, first we have to check that our inner function, which would be analogous to our lowercase f here, is this continuous on the interval from 19 to x? Well, no matter what x is, this is going to be continuous over that interval because this is continuous for all x's. And so we meet this first condition, or our major condition. And so then we could just say, all right, then the derivative of all of this is just going to be this inner function replacing t with x. So we're going to get the cube root. Instead of the cube root of t, you're gonna get the cube root of x. And so we can go back to our original question."}, {"video_title": "Finding derivative with fundamental theorem of calculus   AP\u00ae\ufe0e Calculus AB   Khan Academy.mp3", "Sentence": "And so then we could just say, all right, then the derivative of all of this is just going to be this inner function replacing t with x. So we're going to get the cube root. Instead of the cube root of t, you're gonna get the cube root of x. And so we can go back to our original question. What is g prime of 27 going to be equal to? Well, it's going to be equal to the cube root of 27, which is, of course, equal to three. And we're done."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what do we mean by that? And this is a very, what I just said is not that rigorous, or not rigorous at all, is that, well, think about the point right over here, let's say that's our C. If I can draw the graph at that point, the value of the function at that point without picking up my pencil, or my pen, then it's continuous there. So I could just start here, and I don't have to pick up my pencil, and there you go, I can draw, I can go through that point, so we could say that our function is continuous there. But if I had a function that looked somewhat different than that, if I had a function that looked like this, let's say that it is defined up until then, and then there's a bit of a jump, and then it goes like this, well, this would be very hard to draw, this function would be very hard to draw going through X equals C without picking up my pen. Let's see, my pen is touching the screen, touching the screen, touching the screen, how do I keep drawing this function without picking up my pen? I would have to pick it up, and then move back down here, and so that is an intuitive sense that we are not continuous in this case right over here. But let's actually come up with a formal definition for continuity, and then see if it feels intuitive for us."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if I had a function that looked somewhat different than that, if I had a function that looked like this, let's say that it is defined up until then, and then there's a bit of a jump, and then it goes like this, well, this would be very hard to draw, this function would be very hard to draw going through X equals C without picking up my pen. Let's see, my pen is touching the screen, touching the screen, touching the screen, how do I keep drawing this function without picking up my pen? I would have to pick it up, and then move back down here, and so that is an intuitive sense that we are not continuous in this case right over here. But let's actually come up with a formal definition for continuity, and then see if it feels intuitive for us. So the formal definition of continuity, let's start here, we'll start with continuity at a point. So we could say the function F is continuous, continuous at X equals C, continuous at X equals C, if and only if, I'll draw this two-way arrow to show if and only if, the two-sided limit of F of X as X approaches C is equal to F of C. So this seems very technical, but let's just think about what it's saying. It's saying, look, if the limit as we approach C from the left and the right of F of X, if that's actually the value of our function there, then we are continuous at that point."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's actually come up with a formal definition for continuity, and then see if it feels intuitive for us. So the formal definition of continuity, let's start here, we'll start with continuity at a point. So we could say the function F is continuous, continuous at X equals C, continuous at X equals C, if and only if, I'll draw this two-way arrow to show if and only if, the two-sided limit of F of X as X approaches C is equal to F of C. So this seems very technical, but let's just think about what it's saying. It's saying, look, if the limit as we approach C from the left and the right of F of X, if that's actually the value of our function there, then we are continuous at that point. So let's look at three examples. Let's look at one example where, by our picking up the pencil idea, it feels like we are continuous at a point, and then let's think about a couple of examples where it doesn't seem like we're continuous at a point, and see how this more rigorous definition applies. So let's say that my function, so let's say this right over here is, Y is equal to F of X, and we care about the behavior right over here when X is equal to C. This is my X axis, that's my Y axis."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's saying, look, if the limit as we approach C from the left and the right of F of X, if that's actually the value of our function there, then we are continuous at that point. So let's look at three examples. Let's look at one example where, by our picking up the pencil idea, it feels like we are continuous at a point, and then let's think about a couple of examples where it doesn't seem like we're continuous at a point, and see how this more rigorous definition applies. So let's say that my function, so let's say this right over here is, Y is equal to F of X, and we care about the behavior right over here when X is equal to C. This is my X axis, that's my Y axis. So we care about the behavior when X is equal to C. And so notice, from our first intuitive sense, I can definitely draw this function as we go through X equals C without picking up my pencil, so it feels continuous there. There's no jumps or discontinuities that we can tell. It just kind of keeps on going."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that my function, so let's say this right over here is, Y is equal to F of X, and we care about the behavior right over here when X is equal to C. This is my X axis, that's my Y axis. So we care about the behavior when X is equal to C. And so notice, from our first intuitive sense, I can definitely draw this function as we go through X equals C without picking up my pencil, so it feels continuous there. There's no jumps or discontinuities that we can tell. It just kind of keeps on going. It seems all connected is one way to think about it. But let's think about this definition. Well, the limit as X approaches C from the left, it is, as we approach from the left, it looks like it is approaching, it looks like it is approaching F of C. So this is the value F of C right over here."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It just kind of keeps on going. It seems all connected is one way to think about it. But let's think about this definition. Well, the limit as X approaches C from the left, it is, as we approach from the left, it looks like it is approaching, it looks like it is approaching F of C. So this is the value F of C right over here. And as we approach from the right, as we approach from the right, it also looks like it's approaching F of C. And we are defined right at X equals C, and it is the value that we are approaching from both the left or the right. So this seems good in this scenario. So now let's look at some scenarios that we would have to pick up the pencil as we draw the function through that point, through that, when X is equal to C. So let's look at a scenario."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the limit as X approaches C from the left, it is, as we approach from the left, it looks like it is approaching, it looks like it is approaching F of C. So this is the value F of C right over here. And as we approach from the right, as we approach from the right, it also looks like it's approaching F of C. And we are defined right at X equals C, and it is the value that we are approaching from both the left or the right. So this seems good in this scenario. So now let's look at some scenarios that we would have to pick up the pencil as we draw the function through that point, through that, when X is equal to C. So let's look at a scenario. Let's look at a scenario where we have what's often called a point discontinuity, although you don't have to know at this point, not no pun intended, the formal terminology for it. So let's say we have a function that, let's see, this is C. And let's say our function looks something like this. So we go like this, and at C, let's say it's equal to that."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now let's look at some scenarios that we would have to pick up the pencil as we draw the function through that point, through that, when X is equal to C. So let's look at a scenario. Let's look at a scenario where we have what's often called a point discontinuity, although you don't have to know at this point, not no pun intended, the formal terminology for it. So let's say we have a function that, let's see, this is C. And let's say our function looks something like this. So we go like this, and at C, let's say it's equal to that. So F of C is right over here. Right over here. F of C would be that value."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we go like this, and at C, let's say it's equal to that. So F of C is right over here. Right over here. F of C would be that value. But what's the limit as X approaches C? So the limit as X approaches C, and this would be a two-sided limit of F of X, well, this is, as we approach from the left, it looks like we are approaching this value right over here. And from the right, it looks like we are approaching that same value, and so we could call that L. And L is different than F of C. And so in this case, by our formal definition, we will not be continuous at, F will not be continuous for X, or at the point X, or when X is equal to C. And you can see that there."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "F of C would be that value. But what's the limit as X approaches C? So the limit as X approaches C, and this would be a two-sided limit of F of X, well, this is, as we approach from the left, it looks like we are approaching this value right over here. And from the right, it looks like we are approaching that same value, and so we could call that L. And L is different than F of C. And so in this case, by our formal definition, we will not be continuous at, F will not be continuous for X, or at the point X, or when X is equal to C. And you can see that there. If we tried to draw this, okay, my pencil's touching the paper, touching the paper, touching the paper. Uh-oh, if I needed to keep drawing this function, I'd have to pick up my pencil, move it over here, then pick it up again, and then jump right back down. But this rigorous definition is giving us the same conclusion."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And from the right, it looks like we are approaching that same value, and so we could call that L. And L is different than F of C. And so in this case, by our formal definition, we will not be continuous at, F will not be continuous for X, or at the point X, or when X is equal to C. And you can see that there. If we tried to draw this, okay, my pencil's touching the paper, touching the paper, touching the paper. Uh-oh, if I needed to keep drawing this function, I'd have to pick up my pencil, move it over here, then pick it up again, and then jump right back down. But this rigorous definition is giving us the same conclusion. The limit as we approach X equals C from the left and the right, it's a different value than F of C, and so this is not continuous. Not, not continuous. And let's think about another scenario."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this rigorous definition is giving us the same conclusion. The limit as we approach X equals C from the left and the right, it's a different value than F of C, and so this is not continuous. Not, not continuous. And let's think about another scenario. Let's think about a scenario, and actually, maybe let's think about a scenario where the limit, the two-sided limit doesn't even exist. So, there are my axes, X and Y, and let's say it's doing something like this. Let's say it's doing something like this, and then it does something like this and goes like that."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's think about another scenario. Let's think about a scenario, and actually, maybe let's think about a scenario where the limit, the two-sided limit doesn't even exist. So, there are my axes, X and Y, and let's say it's doing something like this. Let's say it's doing something like this, and then it does something like this and goes like that. And let's say that this right over here is our C. And so, let's see, this is F of C right over here. That is, let me draw it a little bit neater. That is F of C, and it does look like the limit as X approaches C from the left."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say it's doing something like this, and then it does something like this and goes like that. And let's say that this right over here is our C. And so, let's see, this is F of C right over here. That is, let me draw it a little bit neater. That is F of C, and it does look like the limit as X approaches C from the left. So, from values less than C, it does look like that is approaching F of C. But if we look at the limit as X approaches C from the right, that looks like it's approaching some other value. That looks like it's approaching this value right over here, let's call it L. That's approaching L, and L does not equal F of C. And so, in this situation, the two-sided limit doesn't even exist. We're approaching two different values when we approach from the left and from the right."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is F of C, and it does look like the limit as X approaches C from the left. So, from values less than C, it does look like that is approaching F of C. But if we look at the limit as X approaches C from the right, that looks like it's approaching some other value. That looks like it's approaching this value right over here, let's call it L. That's approaching L, and L does not equal F of C. And so, in this situation, the two-sided limit doesn't even exist. We're approaching two different values when we approach from the left and from the right. And so, the limit doesn't even exist at C. This is definitely not going to be continuous. And this matches up to our expectations with our little, do we have to pick up the pencil test? If I have to draw this, I can leave my pencil, it's on the paper, it's on the paper, it's on the paper, it's on the paper."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're approaching two different values when we approach from the left and from the right. And so, the limit doesn't even exist at C. This is definitely not going to be continuous. And this matches up to our expectations with our little, do we have to pick up the pencil test? If I have to draw this, I can leave my pencil, it's on the paper, it's on the paper, it's on the paper, it's on the paper. How am I going to continue to draw this function, this graph of the function, without picking up my pencil? Pick it up, put it back down, and then keep drawing it. So, once again, this right over here is not continuous, both intuitively by our pick up the pencil definition, and also by this more rigorous definition, where, in this case, the limit, the two-sided limit at X equals C doesn't even exist."}, {"video_title": "Continuity at a point   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I have to draw this, I can leave my pencil, it's on the paper, it's on the paper, it's on the paper, it's on the paper. How am I going to continue to draw this function, this graph of the function, without picking up my pencil? Pick it up, put it back down, and then keep drawing it. So, once again, this right over here is not continuous, both intuitively by our pick up the pencil definition, and also by this more rigorous definition, where, in this case, the limit, the two-sided limit at X equals C doesn't even exist. So, we're definitely not gonna be continuous. But even when the two-sided limit does exist, but the limit is a different value than the value of the function, that will also not be continuous. The only situation that it's going to be continuous is if the two-sided limit approaches the same value as the value of the function."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're told the function f is twice differentiable with f of two equals one, f prime of two is equal to four, and f prime prime of two is equal to three. What is the value of the approximation of f of 1.9 using the line tangent to the graph of f at x equals two? So pause this video and see if you can figure this out. This is an actual question from a past AP calculus exam. All right, now let's do this together. And if I was actually doing this on exam, I would just cut to the chase and I would figure out the equation of the tangent line at x equals two, go through the point two comma one, and then I would figure out, okay, when x is equal to 1.9, what is the value of y? And that would be my approximation."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is an actual question from a past AP calculus exam. All right, now let's do this together. And if I was actually doing this on exam, I would just cut to the chase and I would figure out the equation of the tangent line at x equals two, go through the point two comma one, and then I would figure out, okay, when x is equal to 1.9, what is the value of y? And that would be my approximation. But for the sake of learning and getting the intuition here, let's just make sure we understand what's happening. So let me graph this. So let's say that's my y-axis, and then this is my x-axis, and this is x equals one, this is x equals two, this is y equals one."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that would be my approximation. But for the sake of learning and getting the intuition here, let's just make sure we understand what's happening. So let me graph this. So let's say that's my y-axis, and then this is my x-axis, and this is x equals one, this is x equals two, this is y equals one. We know that the point two comma one is on the graph of y is equal to f of x. So we know that point right over there is there. And we also know the slope of the tangent line."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that's my y-axis, and then this is my x-axis, and this is x equals one, this is x equals two, this is y equals one. We know that the point two comma one is on the graph of y is equal to f of x. So we know that point right over there is there. And we also know the slope of the tangent line. The slope of the tangent line is four. So it's gonna look something like this. It's gonna probably even be a little steeper than that."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we also know the slope of the tangent line. The slope of the tangent line is four. So it's gonna look something like this. It's gonna probably even be a little steeper than that. The slope of the tangent line is gonna look something like that. And we don't know much more about it. We know the second derivative here."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's gonna probably even be a little steeper than that. The slope of the tangent line is gonna look something like that. And we don't know much more about it. We know the second derivative here. But what they're asking us to do is without knowing what the function actually looks like, the function might look something like this. Let me just draw something. So the function might look, might look something like this."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know the second derivative here. But what they're asking us to do is without knowing what the function actually looks like, the function might look something like this. Let me just draw something. So the function might look, might look something like this. We're trying to figure out what f of 1.9 is. So if x is 1.9, f of 1.9, if that's the way the function actually looked, might be this value right over here. But we don't know for sure because we don't know much more about the function."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the function might look, might look something like this. We're trying to figure out what f of 1.9 is. So if x is 1.9, f of 1.9, if that's the way the function actually looked, might be this value right over here. But we don't know for sure because we don't know much more about the function. But what they're suggesting for us to do is use this tangent line. If we know the equation of this tangent line here, we could say, well, what does that tangent line equal when x equals 1.9? When x equals 1.9, it equals that point right over there."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we don't know for sure because we don't know much more about the function. But what they're suggesting for us to do is use this tangent line. If we know the equation of this tangent line here, we could say, well, what does that tangent line equal when x equals 1.9? When x equals 1.9, it equals that point right over there. And then we could use that as our approximation for f of 1.9. Well, to do that, we need to know the equation of the tangent line. And we could do that in point-slope form."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x equals 1.9, it equals that point right over there. And then we could use that as our approximation for f of 1.9. Well, to do that, we need to know the equation of the tangent line. And we could do that in point-slope form. We would just have to say y minus the y value that we know is on that line, the point two comma one we know is on that line. So y minus one is going to be equal to the slope of our tangent line, which we know is going to be equal to four, times x minus the x value that corresponds to that y value, so x minus two. So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could do that in point-slope form. We would just have to say y minus the y value that we know is on that line, the point two comma one we know is on that line. So y minus one is going to be equal to the slope of our tangent line, which we know is going to be equal to four, times x minus the x value that corresponds to that y value, so x minus two. So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9. So we'd say y minus one is equal to four times 1.9 minus two. 1.9 minus two is negative 0.1. And let's see, four times negative 0.1, this all simplifies to negative 0.4."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9. So we'd say y minus one is equal to four times 1.9 minus two. 1.9 minus two is negative 0.1. And let's see, four times negative 0.1, this all simplifies to negative 0.4. Now you add one to both sides, you get y is equal to, if you add one here, you're gonna get 0.6. So this, I didn't draw it quite to scale, 0.6 might be something closer to right around there. But there you go, that is our approximation for f of 1.9, which is choice B."}, {"video_title": "Worked example Approximation with local linearity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, four times negative 0.1, this all simplifies to negative 0.4. Now you add one to both sides, you get y is equal to, if you add one here, you're gonna get 0.6. So this, I didn't draw it quite to scale, 0.6 might be something closer to right around there. But there you go, that is our approximation for f of 1.9, which is choice B. And we're done. And one interesting thing to note is we didn't have to use all the information they gave us. We did not have to use this information about the second derivative in order to solve the problem."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I've graphed y is equal to cosine of x in blue and y is equal to sine of x in red. We're not going to prove what the derivatives are, but we're gonna know what they are and get an intuitive sense, and in future videos we'll actually do a proof. So let's start with sine of x. So the derivative can be viewed as a slope of the tangent line. So for example, at this point right over here, it looks like the slope of our tangent line should be zero. So our derivative function should be zero at that x value. Similarly, over here, it looks like the derivative is zero."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative can be viewed as a slope of the tangent line. So for example, at this point right over here, it looks like the slope of our tangent line should be zero. So our derivative function should be zero at that x value. Similarly, over here, it looks like the derivative is zero. The slope of the tangent line would be zero. So whatever our derivative function is at that x value, it should be equal to zero. If we look right over here on sine of x, it looks like the slope of the tangent line would be pretty close to one."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Similarly, over here, it looks like the derivative is zero. The slope of the tangent line would be zero. So whatever our derivative function is at that x value, it should be equal to zero. If we look right over here on sine of x, it looks like the slope of the tangent line would be pretty close to one. If that is the case, then in our derivative function, when x is equal to zero, that derivative function should be equal to one. Similarly, over here, it looks like the slope of the tangent line is negative one, which tells us that the derivative function should be hitting the value of negative one at that x value. So you're probably seeing something interesting emerge."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we look right over here on sine of x, it looks like the slope of the tangent line would be pretty close to one. If that is the case, then in our derivative function, when x is equal to zero, that derivative function should be equal to one. Similarly, over here, it looks like the slope of the tangent line is negative one, which tells us that the derivative function should be hitting the value of negative one at that x value. So you're probably seeing something interesting emerge. Everywhere, while we're trying to plot the slope of the tangent line, it seems to coincide with y is equal to cosine of x. And it is indeed the case that the derivative of sine of x is equal to cosine of x. And you can see that it makes sense not just at the points we tried, but even in the trends."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you're probably seeing something interesting emerge. Everywhere, while we're trying to plot the slope of the tangent line, it seems to coincide with y is equal to cosine of x. And it is indeed the case that the derivative of sine of x is equal to cosine of x. And you can see that it makes sense not just at the points we tried, but even in the trends. If you look at sine of x here, the slope is one, but then it becomes less and less and less positive all the way until it becomes zero. Cosine of x, the value of the function, is one, and it becomes less and less positive all the way until it equals zero. And you could keep doing that type of analysis to feel good about it."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can see that it makes sense not just at the points we tried, but even in the trends. If you look at sine of x here, the slope is one, but then it becomes less and less and less positive all the way until it becomes zero. Cosine of x, the value of the function, is one, and it becomes less and less positive all the way until it equals zero. And you could keep doing that type of analysis to feel good about it. In another video, we're going to prove this more rigorously. So now let's think about cosine of x. So cosine of x, right over here, the slope of the tangent line looks like it is zero."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could keep doing that type of analysis to feel good about it. In another video, we're going to prove this more rigorously. So now let's think about cosine of x. So cosine of x, right over here, the slope of the tangent line looks like it is zero. And so its derivative function needs to be zero at that point. So hey, maybe it's sine of x. Let's keep trying this."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So cosine of x, right over here, the slope of the tangent line looks like it is zero. And so its derivative function needs to be zero at that point. So hey, maybe it's sine of x. Let's keep trying this. So over here, cosine of x, it looks like the slope of the tangent line is negative one. And so we would want the derivative to go through that point right over there. All right, this is starting to seem, it doesn't seem like the derivative of cosine of x could be sine of x."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's keep trying this. So over here, cosine of x, it looks like the slope of the tangent line is negative one. And so we would want the derivative to go through that point right over there. All right, this is starting to seem, it doesn't seem like the derivative of cosine of x could be sine of x. In fact, this is the opposite of what sine of x is doing. Sine of x is at one, not negative one at that point. But that's an interesting theory."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, this is starting to seem, it doesn't seem like the derivative of cosine of x could be sine of x. In fact, this is the opposite of what sine of x is doing. Sine of x is at one, not negative one at that point. But that's an interesting theory. Maybe the derivative of cosine of x is negative sine of x. So let's plot that. So this does seem to coincide."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that's an interesting theory. Maybe the derivative of cosine of x is negative sine of x. So let's plot that. So this does seem to coincide. The derivative of cosine of x here looks like negative one, the slope of the tangent line. And negative sine of this x value is negative one. Over here, the derivative of cosine of x looks like it is zero."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this does seem to coincide. The derivative of cosine of x here looks like negative one, the slope of the tangent line. And negative sine of this x value is negative one. Over here, the derivative of cosine of x looks like it is zero. And negative sine of x is indeed zero. So it actually turns out that it is the case, that the derivative of cosine of x is negative sine of x. So these are really good to know."}, {"video_title": "Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Over here, the derivative of cosine of x looks like it is zero. And negative sine of x is indeed zero. So it actually turns out that it is the case, that the derivative of cosine of x is negative sine of x. So these are really good to know. These are kind of fundamental trigonometric derivatives to know, we'll be able to derive other things for them. And hopefully this video gives you a good intuitive sense of why this is true. And in future videos, we will prove it rigorously."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here I have the graphs of the two functions and the enclosed regions r and s. So the first thing they want us to figure out is find the sum of the areas of region r and s. So the sum of those areas, you could think about it, we're gonna go from x equals zero right over here to x equals two. So we're gonna take the integral from x equals, let me write this down, area of r plus s is equal to, let me write that a little bit neater. So the area of r plus s is going to be equal to, let's see, we can take the integral from x equals zero to x equals two, x equals zero to x equals two. And what are we going to integrate? Well, we're going to integrate the difference between the two functions and really the absolute value of the difference. Remember, we want the sums. We don't wanna have negative area here."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what are we going to integrate? Well, we're going to integrate the difference between the two functions and really the absolute value of the difference. Remember, we want the sums. We don't wanna have negative area here. We want the sum of the areas of regions r and s. And at some point, you have g of x is above f of x, and at other points, f of x is above g of x. But if we take the absolute value, it doesn't matter which one we're subtracting from the other, we're just getting the absolute value of the difference. So let's take the absolute value of f of x minus g of x dx."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't wanna have negative area here. We want the sum of the areas of regions r and s. And at some point, you have g of x is above f of x, and at other points, f of x is above g of x. But if we take the absolute value, it doesn't matter which one we're subtracting from the other, we're just getting the absolute value of the difference. So let's take the absolute value of f of x minus g of x dx. So that's gonna be the sum of the areas, or we could say this is going to be the integral from zero to two of the absolute value, f of x, is one plus x plus e to the x squared minus two x minus g of x. So minus x to the fourth plus 6.5 x squared minus six x minus two. Take the absolute value, dx."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's take the absolute value of f of x minus g of x dx. So that's gonna be the sum of the areas, or we could say this is going to be the integral from zero to two of the absolute value, f of x, is one plus x plus e to the x squared minus two x minus g of x. So minus x to the fourth plus 6.5 x squared minus six x minus two. Take the absolute value, dx. Now this would be pretty hairy to solve if we did not have access to a calculator, but lucky for us on this part of the AP exam, we can use a graphing calculator. So let's do that to evaluate this definite integral here. And if you're wondering why I say minus two instead of plus two, remember we're subtracting, we're subtracting g of x, we're finding the difference between them."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Take the absolute value, dx. Now this would be pretty hairy to solve if we did not have access to a calculator, but lucky for us on this part of the AP exam, we can use a graphing calculator. So let's do that to evaluate this definite integral here. And if you're wondering why I say minus two instead of plus two, remember we're subtracting, we're subtracting g of x, we're finding the difference between them. So let's input this function into my calculator. And I'm gonna do the same thing that I did in part one, where I'm just gonna define, let me turn it on. All right, so I'm gonna actually clear that out, and I'm gonna define this whole expression as y one."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you're wondering why I say minus two instead of plus two, remember we're subtracting, we're subtracting g of x, we're finding the difference between them. So let's input this function into my calculator. And I'm gonna do the same thing that I did in part one, where I'm just gonna define, let me turn it on. All right, so I'm gonna actually clear that out, and I'm gonna define this whole expression as y one. So I am going to take the absolute value, so let me see where the absolute value, it's been a little while since I last used one of these. So, actually maybe it's math, math number, oh there you go, absolute value. So it's the absolute value of one plus x plus second e to the, and then we have x squared minus two x, and then close that parentheses, and then you have minus x to the fourth power plus 6.5 times x squared minus six times x minus two, and then we have to close the parentheses around the absolute value."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so I'm gonna actually clear that out, and I'm gonna define this whole expression as y one. So I am going to take the absolute value, so let me see where the absolute value, it's been a little while since I last used one of these. So, actually maybe it's math, math number, oh there you go, absolute value. So it's the absolute value of one plus x plus second e to the, and then we have x squared minus two x, and then close that parentheses, and then you have minus x to the fourth power plus 6.5 times x squared minus six times x minus two, and then we have to close the parentheses around the absolute value. All right, so we've inputted y one, and so now let's go over here, and now let's evaluate this definite integral. So we go to math, and we scroll down to definite for function integral, so click on that, and we're gonna use y one, so we go to variables, we go to the right to go to y variables, it's a function variable that we just defined, and so we select y one, that's what we just inputted. Our variable of integration is x, and our bounds of integration, well, we're gonna go from x equals zero to x equals two, so we go from zero to two, and then we let the calculator munch on it a little bit, and we get, it's taking some time to calculate, it's still munching on it."}, {"video_title": "2015 AP Calculus AB 2a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's the absolute value of one plus x plus second e to the, and then we have x squared minus two x, and then close that parentheses, and then you have minus x to the fourth power plus 6.5 times x squared minus six times x minus two, and then we have to close the parentheses around the absolute value. All right, so we've inputted y one, and so now let's go over here, and now let's evaluate this definite integral. So we go to math, and we scroll down to definite for function integral, so click on that, and we're gonna use y one, so we go to variables, we go to the right to go to y variables, it's a function variable that we just defined, and so we select y one, that's what we just inputted. Our variable of integration is x, and our bounds of integration, well, we're gonna go from x equals zero to x equals two, so we go from zero to two, and then we let the calculator munch on it a little bit, and we get, it's taking some time to calculate, it's still munching on it. Let's see, this is, it's taking a good bit of time. There you go. Alright, so it's approximately 2.00."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "In a previous video, we used the quotient rule in order to find the derivatives of tangent of x and cotangent of x. And what I want to do in this video is to keep going and find the derivatives of secant of x and cosecant of x. So let's start with secant of x. The derivative with respect to x of secant of x. Well, secant of x is the same thing as, so we're going to find the derivative with respect to x of, secant of x is the same thing as one over, one over the cosine of x. And that's just the definition of secant. And you could, there's multiple ways you could do this."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative with respect to x of secant of x. Well, secant of x is the same thing as, so we're going to find the derivative with respect to x of, secant of x is the same thing as one over, one over the cosine of x. And that's just the definition of secant. And you could, there's multiple ways you could do this. When you learn the chain rule, that actually might be a more natural thing to use to evaluate the derivative here. But we know the quotient rule, so we will apply the quotient rule here. And it's no coincidence that you get to the same answer."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could, there's multiple ways you could do this. When you learn the chain rule, that actually might be a more natural thing to use to evaluate the derivative here. But we know the quotient rule, so we will apply the quotient rule here. And it's no coincidence that you get to the same answer. The quotient rule actually can be derived based on the chain rule and the product rule. But I won't keep going into that. Let's just apply the quotient rule right over here."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's no coincidence that you get to the same answer. The quotient rule actually can be derived based on the chain rule and the product rule. But I won't keep going into that. Let's just apply the quotient rule right over here. So this derivative is going to be equal to, it's going to be equal to the derivative of the top. Well, what's the derivative of one with respect to x? Well, that's just zero times the function on the bottom."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just apply the quotient rule right over here. So this derivative is going to be equal to, it's going to be equal to the derivative of the top. Well, what's the derivative of one with respect to x? Well, that's just zero times the function on the bottom. So times cosine of x, cosine of x, minus, minus the function on the top, well, that's just one, times the derivative on the bottom. Well, the derivative of the bottom is, derivative of cosine of x is negative sine of x. So we could put the sine of x there, but it's negative sine of x, so you have a minus and there would be a negative, so we could just make that a positive."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's just zero times the function on the bottom. So times cosine of x, cosine of x, minus, minus the function on the top, well, that's just one, times the derivative on the bottom. Well, the derivative of the bottom is, derivative of cosine of x is negative sine of x. So we could put the sine of x there, but it's negative sine of x, so you have a minus and there would be a negative, so we could just make that a positive. And then all of that over the function on the bottom squared. So cosine of x squared. And so zero times cosine of x, that is just zero."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could put the sine of x there, but it's negative sine of x, so you have a minus and there would be a negative, so we could just make that a positive. And then all of that over the function on the bottom squared. So cosine of x squared. And so zero times cosine of x, that is just zero. And so all we are left with is sine of x over cosine of x squared. And there's multiple ways that you could rewrite this if you like. You could say that this is the same thing as sine of x over cosine of x times one over cosine of x."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so zero times cosine of x, that is just zero. And so all we are left with is sine of x over cosine of x squared. And there's multiple ways that you could rewrite this if you like. You could say that this is the same thing as sine of x over cosine of x times one over cosine of x. And of course, this is tangent of x times secant of x. Secant of x. So you could say derivative of secant x is sine of x over cosine squared of x, or it is tangent of x times secant of x. So now let's do cosecant."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say that this is the same thing as sine of x over cosine of x times one over cosine of x. And of course, this is tangent of x times secant of x. Secant of x. So you could say derivative of secant x is sine of x over cosine squared of x, or it is tangent of x times secant of x. So now let's do cosecant. So the derivative with respect to x of cosecant of x, well, that's the same thing as the derivative with respect to x of one over sine of x. Cosecant is one over sine of x. I remember that, because you think it's cosecant, maybe it's the reciprocal of cosine, but it's not, it's the opposite of what you would expect. Cosine's reciprocal isn't cosecant, it is secant. Once again, opposite of what you would expect."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now let's do cosecant. So the derivative with respect to x of cosecant of x, well, that's the same thing as the derivative with respect to x of one over sine of x. Cosecant is one over sine of x. I remember that, because you think it's cosecant, maybe it's the reciprocal of cosine, but it's not, it's the opposite of what you would expect. Cosine's reciprocal isn't cosecant, it is secant. Once again, opposite of what you would expect. That starts with an S, this starts with a C. That starts with a C, that starts with an S. Just the way that it happened to be defined. But anyway, let's just evaluate this. Once again, we'll do the quotient rule, but you could also do this using the chain rule."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, opposite of what you would expect. That starts with an S, this starts with a C. That starts with a C, that starts with an S. Just the way that it happened to be defined. But anyway, let's just evaluate this. Once again, we'll do the quotient rule, but you could also do this using the chain rule. So it's going to be the derivative of the expression on top, which is zero, times the expression on the bottom, which is sine of x, sine of x, minus the expression on top, which is just one, times the derivative of the expression on the bottom, which is cosine of x, all of that over the expression on the bottom squared. Sine squared of x, that's zero, so we get negative cosine of x over sine, over sine squared of x. So that's one way to think about it."}, {"video_title": "Derivatives of sec(x) and csc(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, we'll do the quotient rule, but you could also do this using the chain rule. So it's going to be the derivative of the expression on top, which is zero, times the expression on the bottom, which is sine of x, sine of x, minus the expression on top, which is just one, times the derivative of the expression on the bottom, which is cosine of x, all of that over the expression on the bottom squared. Sine squared of x, that's zero, so we get negative cosine of x over sine, over sine squared of x. So that's one way to think about it. Or, if you like, you could view this, the same thing we did over here, this is the same thing as negative cosine of x over sine of x, times one over sine of x. And this is negative cotangent of x, negative cotangent of x, times, let me write it this way, times one over sine of x is cosecant of x. Cosecant, cosecant of x. So whichever one you find more useful."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm not going to prove it here, but I think the conceptual underpinning here should be straightforward. So the theorem tells us that suppose f is a function continuous at every point of the interval, the closed interval, so we're including a and b, so it's continuous at every point of the interval a, b. And so let me just draw a couple of examples of what f could look like just based on these first lines. So suppose f is a function continuous at every point of the interval a, b. So let me draw some axes here. So that's my y-axis, and this is my x-axis. So one situation, if this is a, and this is b, f is continuous at every point of the closed interval a and b, so that means it's got to be, for sure, defined at every point as well as being continuous."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So suppose f is a function continuous at every point of the interval a, b. So let me draw some axes here. So that's my y-axis, and this is my x-axis. So one situation, if this is a, and this is b, f is continuous at every point of the closed interval a and b, so that means it's got to be, for sure, defined at every point as well as being continuous. To be continuous, you have to be defined at every point, and the limit of the function as you approach that point should be equal to the value of the function of that point. And so the function is definitely going to be defined at f of a, so it's definitely going to have an f of a right over here, this right over here is f of a. Maybe f of b is higher, although we can look at different cases."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one situation, if this is a, and this is b, f is continuous at every point of the closed interval a and b, so that means it's got to be, for sure, defined at every point as well as being continuous. To be continuous, you have to be defined at every point, and the limit of the function as you approach that point should be equal to the value of the function of that point. And so the function is definitely going to be defined at f of a, so it's definitely going to have an f of a right over here, this right over here is f of a. Maybe f of b is higher, although we can look at different cases. So that would be our f of b. And they tell us it is a continuous function. It is a continuous function."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe f of b is higher, although we can look at different cases. So that would be our f of b. And they tell us it is a continuous function. It is a continuous function. So if you're trying to imagine continuous functions, one way to think about it is, if we're continuous over an interval, we take the value of the function at one point of the interval, and if it's continuous, we need to be able to get to the value of the function at the other point of the interval without picking up our pencil. So I can do all sorts of things, and it still has to be a function, so I can't do something like that, but as long as I don't pick up my pencil, this is a continuous function. So there you go."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is a continuous function. So if you're trying to imagine continuous functions, one way to think about it is, if we're continuous over an interval, we take the value of the function at one point of the interval, and if it's continuous, we need to be able to get to the value of the function at the other point of the interval without picking up our pencil. So I can do all sorts of things, and it still has to be a function, so I can't do something like that, but as long as I don't pick up my pencil, this is a continuous function. So there you go. Somehow the graph, I had to pick up my pencil if I had to do something like this, so I got to pick up my pencil and do something like that, well, that's not continuous anymore. If I had to do something like this, and oops, pick up my pencil, not continuous anymore. If I had to do something like, whoop, okay, pick up my pencil, not continuous anymore."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So there you go. Somehow the graph, I had to pick up my pencil if I had to do something like this, so I got to pick up my pencil and do something like that, well, that's not continuous anymore. If I had to do something like this, and oops, pick up my pencil, not continuous anymore. If I had to do something like, whoop, okay, pick up my pencil, not continuous anymore. So this is what a functioning that is continuous what the closed interval AB looks like. I can draw some other examples. In fact, let me do that."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I had to do something like, whoop, okay, pick up my pencil, not continuous anymore. So this is what a functioning that is continuous what the closed interval AB looks like. I can draw some other examples. In fact, let me do that. So let me draw one, maybe where F of B is less than F of A. So that's my Y axis, and this is my X axis. And once again, A and B don't both have to be positive."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "In fact, let me do that. So let me draw one, maybe where F of B is less than F of A. So that's my Y axis, and this is my X axis. And once again, A and B don't both have to be positive. They could both be negative. One could be, A could be negative, B could be positive, and maybe in this situation, then F of A and F of B, it could also be positive or negative. But let's take a situation where this is F of A."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, A and B don't both have to be positive. They could both be negative. One could be, A could be negative, B could be positive, and maybe in this situation, then F of A and F of B, it could also be positive or negative. But let's take a situation where this is F of A. So that right over there is F of A. This right over here is F of B. F of B. And once again, we're saying F is a continuous function."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's take a situation where this is F of A. So that right over there is F of A. This right over here is F of B. F of B. And once again, we're saying F is a continuous function. So I should be able to go from F of A to F of B. F of B, draw a function without having to pick up my pencil. So it could do something like this. Actually, I don't wanna make it go vertical."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, we're saying F is a continuous function. So I should be able to go from F of A to F of B. F of B, draw a function without having to pick up my pencil. So it could do something like this. Actually, I don't wanna make it go vertical. It could go like this, and then go down, and then do something like that. So these are both cases, and I could draw an infinite number of cases, where F is a function continuous at every point of the interval, the closed interval, from A to B. Now, given that, there's two ways to state the conclusion for the Intermediate Value Theorem."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, I don't wanna make it go vertical. It could go like this, and then go down, and then do something like that. So these are both cases, and I could draw an infinite number of cases, where F is a function continuous at every point of the interval, the closed interval, from A to B. Now, given that, there's two ways to state the conclusion for the Intermediate Value Theorem. You'll see it written in one of these ways, or something close to one of these ways, and that's why I included both of these. So one way to say it is, well, if this first statement is true, then F will take on every value between F of A and F of B over the interval. And you see in both of these cases, every interval, sorry, every value between F of A and F of B, so every value here, is being taken on at some point."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, given that, there's two ways to state the conclusion for the Intermediate Value Theorem. You'll see it written in one of these ways, or something close to one of these ways, and that's why I included both of these. So one way to say it is, well, if this first statement is true, then F will take on every value between F of A and F of B over the interval. And you see in both of these cases, every interval, sorry, every value between F of A and F of B, so every value here, is being taken on at some point. You can pick some value. You could pick some value, an arbitrary value, L right over here. Well, look, L happened right over there."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you see in both of these cases, every interval, sorry, every value between F of A and F of B, so every value here, is being taken on at some point. You can pick some value. You could pick some value, an arbitrary value, L right over here. Well, look, L happened right over there. If you pick L, well, L happened right over there, and actually, it also happened there, and it also happened there. And this second bullet point describes the Intermediate Value Theorem more that way. For any L between the values of F of A and F of B, there exists a number C in the closed interval from A to B, for which F of C equals L. So there exists at least one C. So in this case, that would be our C. Over here, there's potential, there's multiple candidates for C. That could be a candidate for C. That could be a C. So we could say there exists at least one number, at least one number, I'll throw that in there, at least one number C in the interval for which this is true."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, look, L happened right over there. If you pick L, well, L happened right over there, and actually, it also happened there, and it also happened there. And this second bullet point describes the Intermediate Value Theorem more that way. For any L between the values of F of A and F of B, there exists a number C in the closed interval from A to B, for which F of C equals L. So there exists at least one C. So in this case, that would be our C. Over here, there's potential, there's multiple candidates for C. That could be a candidate for C. That could be a C. So we could say there exists at least one number, at least one number, I'll throw that in there, at least one number C in the interval for which this is true. And something that might amuse you for a few minutes is try to draw a function where this first statement is true but somehow the second statement is not true. So you say, okay, well, let's assume that there's an L where there isn't a C in the interval. Well, let me try to do that."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "For any L between the values of F of A and F of B, there exists a number C in the closed interval from A to B, for which F of C equals L. So there exists at least one C. So in this case, that would be our C. Over here, there's potential, there's multiple candidates for C. That could be a candidate for C. That could be a C. So we could say there exists at least one number, at least one number, I'll throw that in there, at least one number C in the interval for which this is true. And something that might amuse you for a few minutes is try to draw a function where this first statement is true but somehow the second statement is not true. So you say, okay, well, let's assume that there's an L where there isn't a C in the interval. Well, let me try to do that. I'll draw it big so that we can really see how obvious that we have to take on all of the values between F of A and F of B is. So let me draw a big axis this time. So that's my Y axis, and that is my X axis."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let me try to do that. I'll draw it big so that we can really see how obvious that we have to take on all of the values between F of A and F of B is. So let me draw a big axis this time. So that's my Y axis, and that is my X axis. And I'll just do the case where, just for simplicity, that is A and that is B. And let's say that this is F of A. So that is F of A."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's my Y axis, and that is my X axis. And I'll just do the case where, just for simplicity, that is A and that is B. And let's say that this is F of A. So that is F of A. And let's say that this is F of B. So the dotted line. All right, F of B."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is F of A. And let's say that this is F of B. So the dotted line. All right, F of B. And we assume that we have a continuous function here. So the graph, I could draw it from F of A to F of B, from this point to this point without picking up my pencil. From this coordinate, A comma F of A, to this coordinate, B comma F of B without picking up my pencil."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, F of B. And we assume that we have a continuous function here. So the graph, I could draw it from F of A to F of B, from this point to this point without picking up my pencil. From this coordinate, A comma F of A, to this coordinate, B comma F of B without picking up my pencil. Well, let's assume that there is some L that we don't take on. Let's say there's some value L right over here, and we never take on this value. This continuous function never takes on this value as we go from X equaling A to X equal B."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "From this coordinate, A comma F of A, to this coordinate, B comma F of B without picking up my pencil. Well, let's assume that there is some L that we don't take on. Let's say there's some value L right over here, and we never take on this value. This continuous function never takes on this value as we go from X equaling A to X equal B. Let's see if I can draw that. Let's see if I can get from here to here without ever essentially crossing this dotted line. Well, let's see, I could, ooh, maybe I'll avoid it a little bit."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "This continuous function never takes on this value as we go from X equaling A to X equal B. Let's see if I can draw that. Let's see if I can get from here to here without ever essentially crossing this dotted line. Well, let's see, I could, ooh, maybe I'll avoid it a little bit. I'll avoid, but gee, how am I going to get there? Well, without picking up my pencil. Well, I really need to cross that line."}, {"video_title": "Intermediate value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see, I could, ooh, maybe I'll avoid it a little bit. I'll avoid, but gee, how am I going to get there? Well, without picking up my pencil. Well, I really need to cross that line. All right, well, there you go. I found we took on the value L, and it happened at C, which is in that closed interval. So once again, I'm not giving you a proof here, but hopefully you have a good intuition that the intermediate value theorem is kind of common sense."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is introduce ourselves to the quotient rule. And we're not going to prove it in this video. In a future video, we can prove it using the product rule, and we'll see it has some similarities to the product rule. But here, we'll learn about what it is and how and where to actually apply it. So, for example, if I have some function, f of x, and it can be expressed as the quotient of two expressions. So let's say u of x of x over v of x, then the quotient rule tells us that f prime of x is going to be equal to, and this is going to look a little bit complicated, but once we apply it, you'll hopefully get a little bit more comfortable with it. It's going to be equal to the derivative of the numerator function, u prime of x, times the denominator function, v of x, minus the numerator function, u of x, do that in that blue color, u of x times the derivative of the denominator function, times v prime of x."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But here, we'll learn about what it is and how and where to actually apply it. So, for example, if I have some function, f of x, and it can be expressed as the quotient of two expressions. So let's say u of x of x over v of x, then the quotient rule tells us that f prime of x is going to be equal to, and this is going to look a little bit complicated, but once we apply it, you'll hopefully get a little bit more comfortable with it. It's going to be equal to the derivative of the numerator function, u prime of x, times the denominator function, v of x, minus the numerator function, u of x, do that in that blue color, u of x times the derivative of the denominator function, times v prime of x. And this already looks very similar to the product rule. If this was u of x times v of x, then this is what we would get when we took the derivative if this was a plus sign, but this is here a minus sign. But we're not done yet."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be equal to the derivative of the numerator function, u prime of x, times the denominator function, v of x, minus the numerator function, u of x, do that in that blue color, u of x times the derivative of the denominator function, times v prime of x. And this already looks very similar to the product rule. If this was u of x times v of x, then this is what we would get when we took the derivative if this was a plus sign, but this is here a minus sign. But we're not done yet. We would then divide by the denominator function squared, v of x squared. So let's actually apply this idea. So let's say that we have f of x is equal to x squared over cosine of x."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we're not done yet. We would then divide by the denominator function squared, v of x squared. So let's actually apply this idea. So let's say that we have f of x is equal to x squared over cosine of x. Well, what could be our u of x and what could be our v of x? Well, our u of x could be our x squared, so that is u of x and u prime of x would be equal to two x. And then this could be our v of x."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that we have f of x is equal to x squared over cosine of x. Well, what could be our u of x and what could be our v of x? Well, our u of x could be our x squared, so that is u of x and u prime of x would be equal to two x. And then this could be our v of x. So this is v of x and v prime of x. The derivative of cosine of x with respect to x is equal to negative sine of x. And then we just apply this."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this could be our v of x. So this is v of x and v prime of x. The derivative of cosine of x with respect to x is equal to negative sine of x. And then we just apply this. So based on that, f prime of x is going to be equal to the derivative of the numerator function, that's two x right over here, that's that there. So it's gonna be two x times the denominator function. V of x is just cosine of x times cosine of x minus the numerator function, which is just x squared, x squared, times the derivative of the denominator function."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we just apply this. So based on that, f prime of x is going to be equal to the derivative of the numerator function, that's two x right over here, that's that there. So it's gonna be two x times the denominator function. V of x is just cosine of x times cosine of x minus the numerator function, which is just x squared, x squared, times the derivative of the denominator function. The derivative of cosine of x is negative sine x. So negative sine of x. Negative sine of x."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "V of x is just cosine of x times cosine of x minus the numerator function, which is just x squared, x squared, times the derivative of the denominator function. The derivative of cosine of x is negative sine x. So negative sine of x. Negative sine of x. All of that over, all of that over, the denominator function squared. So that's cosine of x and I'm going to square it. I could write it, of course, like this."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative sine of x. All of that over, all of that over, the denominator function squared. So that's cosine of x and I'm going to square it. I could write it, of course, like this. Actually, let me write it like that just to make it a little bit clearer. And at this point, we just have to simplify. This is going to be equal to, let's see, we're gonna get two x times cosine of x, x cosine of x, negative times a negative is a positive, plus x squared, x squared times sine of x, sine of x, all of that over, cosine of x squared, which I could write like this as well."}, {"video_title": "Quotient rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could write it, of course, like this. Actually, let me write it like that just to make it a little bit clearer. And at this point, we just have to simplify. This is going to be equal to, let's see, we're gonna get two x times cosine of x, x cosine of x, negative times a negative is a positive, plus x squared, x squared times sine of x, sine of x, all of that over, cosine of x squared, which I could write like this as well. And we're done. You could try to simplify it. In fact, there's no obvious ways to simplify this any further."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I hope to do in this video is prove that if a function is differentiable at some point C, that it's also going to be continuous at that point C. But before we do the proof, let's just remind ourselves what differentiability means and what continuity means. So first, differentiability. Differentiability. So let's think about that first. It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about that first. It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our x-axis. And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope? Well, it's going to be change in y over change in x."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity. Con-ti-nuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here."}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero. Well just add f of C to both sides and what do you get?"}, {"video_title": "Proof Differentiability implies continuity   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero. Well just add f of C to both sides and what do you get? Well you get the limit as x approaches C of f of x is equal to f of C. And this is the definition of continuity, the limit of my function as x approaches C is equal to the function, is equal to the value of the function at C. This is, this means that our function is continuous. Continuous at C. So just a reminder, we started assuming f differentiable at C, we use that fact to evaluate this limit right over here, which we got to be equal to zero. And if that limit is equal to zero, then it just follows, just doing a little bit of algebra and using properties of limits, that the limit as x approaches C of f of x is equal to f of C. And that's our definition of being continuous, continuous at the point C. So hopefully that satisfies you."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It tells us what the corresponding f of x is. And they say what is the best estimate for f prime of four? So this is the derivative of our function f when x is equal to four. Or another way to think about it, what is the slope of the tangent line when x is equal to four for f of x? So what is the best estimate for f prime of four we can make based on this table? So let's just visualize what's going on before we even look at the choices. So let me draw some axes here."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or another way to think about it, what is the slope of the tangent line when x is equal to four for f of x? So what is the best estimate for f prime of four we can make based on this table? So let's just visualize what's going on before we even look at the choices. So let me draw some axes here. And let me plot these points. We know that these would sit on the curve of y is equal to f of x. When x is zero, f of x is 72."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw some axes here. And let me plot these points. We know that these would sit on the curve of y is equal to f of x. When x is zero, f of x is 72. So this is the point zero, 72. This is the point three, 95. Clearly two different scales on the x and y-axis."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is zero, f of x is 72. So this is the point zero, 72. This is the point three, 95. Clearly two different scales on the x and y-axis. This is the point five, 112. This is 677. This is 954."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Clearly two different scales on the x and y-axis. This is the point five, 112. This is 677. This is 954. Actually, let me write out the, this is one, two, three, four, five, six, seven, eight, nine, and 10. Now they want us to know, they wanna know what is the derivative of our function when f is equal to four. Well, they haven't told us even what the value of f is at four."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 954. Actually, let me write out the, this is one, two, three, four, five, six, seven, eight, nine, and 10. Now they want us to know, they wanna know what is the derivative of our function when f is equal to four. Well, they haven't told us even what the value of f is at four. We don't know what that point is. But what they're trying to do is, well, we're trying to make a best estimate. And using these points, we don't even know exactly what the curve looks like."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, they haven't told us even what the value of f is at four. We don't know what that point is. But what they're trying to do is, well, we're trying to make a best estimate. And using these points, we don't even know exactly what the curve looks like. It could look like all sorts of things. We could try to fit a reasonably smooth curve. The curve might look something like that."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And using these points, we don't even know exactly what the curve looks like. It could look like all sorts of things. We could try to fit a reasonably smooth curve. The curve might look something like that. But it might be wackier. It might do something like this. Well, let me try to do it."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "The curve might look something like that. But it might be wackier. It might do something like this. Well, let me try to do it. It might look something like this. So we don't know for sure. All we know is that it needs to go through those points because they've just sampled the function at those points."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let me try to do it. It might look something like this. So we don't know for sure. All we know is that it needs to go through those points because they've just sampled the function at those points. But let's just, for the sake of this exercise, let's assume the simplest, let's say it's a nice, smooth curve without too many twists and turns that goes through these points just like that. So what they're asking, okay, when x is equal to four, if this yellow curve were the actual curve, then what is the slope of the tangent line at that point? So we would be visualizing that."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "All we know is that it needs to go through those points because they've just sampled the function at those points. But let's just, for the sake of this exercise, let's assume the simplest, let's say it's a nice, smooth curve without too many twists and turns that goes through these points just like that. So what they're asking, okay, when x is equal to four, if this yellow curve were the actual curve, then what is the slope of the tangent line at that point? So we would be visualizing that. Now to be clear, this tangent line that I just drew, this would be for this version of our function that I did connecting these points. That does not have to be the actual function. We know that the actual function has to go through those points."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would be visualizing that. Now to be clear, this tangent line that I just drew, this would be for this version of our function that I did connecting these points. That does not have to be the actual function. We know that the actual function has to go through those points. But I'm just doing this for visualization purposes. One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate. We don't know if it's even gonna be a good estimate."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that the actual function has to go through those points. But I'm just doing this for visualization purposes. One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate. We don't know if it's even gonna be a good estimate. It's just going to be a best estimate. So what we generally do when we just have some data around a point is let's use the data points that are closest to that point and find slopes of secant lines pretty close around that point. And that's going to give us our best estimate for the slope of the tangent line."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know if it's even gonna be a good estimate. It's just going to be a best estimate. So what we generally do when we just have some data around a point is let's use the data points that are closest to that point and find slopes of secant lines pretty close around that point. And that's going to give us our best estimate for the slope of the tangent line. So what points do we have near F of four or near the point four comma F of four? Well, they give us what F is equal to when x is equal to three. They give us this point right over here."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's going to give us our best estimate for the slope of the tangent line. So what points do we have near F of four or near the point four comma F of four? Well, they give us what F is equal to when x is equal to three. They give us this point right over here. Let me do this in another color. So three comma 95, that is that right over there. And they also give us five comma 112."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "They give us this point right over here. Let me do this in another color. So three comma 95, that is that right over there. And they also give us five comma 112. That is that point right over there. And so what we could do, we could say, well, what is the average rate of change between these two points? Another way to think about it is what is the slope of the secant line between those two points?"}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they also give us five comma 112. That is that point right over there. And so what we could do, we could say, well, what is the average rate of change between these two points? Another way to think about it is what is the slope of the secant line between those two points? And that would be our best estimate for the slope of the tangent line at x equals four. Do we know that it's a good estimate? Do we know that it's even close?"}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Another way to think about it is what is the slope of the secant line between those two points? And that would be our best estimate for the slope of the tangent line at x equals four. Do we know that it's a good estimate? Do we know that it's even close? No, we don't know for sure, but that would be the best estimate. It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine. These are pretty close around four."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Do we know that it's even close? No, we don't know for sure, but that would be the best estimate. It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine. These are pretty close around four. And so let's do that. Let's find the average rate of change between when x goes from three to five. So we can see here our change in x, let me do this in a new color."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "These are pretty close around four. And so let's do that. Let's find the average rate of change between when x goes from three to five. So we can see here our change in x, let me do this in a new color. So our change in x here is equal to plus two, and I can draw that out. My change in x here is plus two, and my change in y is going to be, when my x increased by two, my change in y is plus, let's see, this is, if I add 10, I get to 105, and then I add another seven, so this is plus 17. So this is plus 17 right over here."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can see here our change in x, let me do this in a new color. So our change in x here is equal to plus two, and I can draw that out. My change in x here is plus two, and my change in y is going to be, when my x increased by two, my change in y is plus, let's see, this is, if I add 10, I get to 105, and then I add another seven, so this is plus 17. So this is plus 17 right over here. Plus 17. And so my change in y over change in x, change in y over my change in x, for this secant line between when x is equaling three and x is equaling five, is going to be equal to 17 over two. 17 over two, which is equal to 8.5."}, {"video_title": "Estimating derivatives   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is plus 17 right over here. Plus 17. And so my change in y over change in x, change in y over my change in x, for this secant line between when x is equaling three and x is equaling five, is going to be equal to 17 over two. 17 over two, which is equal to 8.5. So the slope of this green line here is 8.5, and that would be our best estimate for the slope of the tangent line when x equals four of the curve y is equal to f of x. And so lucky for us, the people who wrote this question had the exact same logic, and they did it right over there. So you wouldn't have to graph it the way I did."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Problem five, at the beginning of 2010, a landfill contained 1,400 tons of solid waste. The increasing function W, so I guess W constantly increases, the increasing function W models the total amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential equation, so the derivative of W with respect to time is equal to one over 25 times the quantity W minus 300. For the next 20 years, W is measured in tons, T is measured in years from the start of 2010. All right, let's get into this. So part A, use the tangent line to the graph of W at T equals zero to approximate the amount of solid waste that the landfill contains at the end of the first three months of 2010. So since time is in years, that would be three months would be 1 4th of a year, so T equals 1 4th."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "For the next 20 years, W is measured in tons, T is measured in years from the start of 2010. All right, let's get into this. So part A, use the tangent line to the graph of W at T equals zero to approximate the amount of solid waste that the landfill contains at the end of the first three months of 2010. So since time is in years, that would be three months would be 1 4th of a year, so T equals 1 4th. So it seems at first it's really daunting, they gave us a differential equation, we don't even know what W, what the actual function W is, how do we figure out its tangent line, but we just have to kind of think about what they're asking. So regardless of what W looks like, so let's think about it a little bit. So we don't know yet what W actually does look like, but they tell us it's an increasing function."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So since time is in years, that would be three months would be 1 4th of a year, so T equals 1 4th. So it seems at first it's really daunting, they gave us a differential equation, we don't even know what W, what the actual function W is, how do we figure out its tangent line, but we just have to kind of think about what they're asking. So regardless of what W looks like, so let's think about it a little bit. So we don't know yet what W actually does look like, but they tell us it's an increasing function. So at time zero, it has 1,400 tons in it, they tell us that right up here, and then it increases. We don't know what the function actually does look like, but let's say that that is W. So what they're saying in problem A, is they're saying use a tangent line to the graph at T equals zero. So there's, let me actually, let me draw W a little bit differently."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we don't know yet what W actually does look like, but they tell us it's an increasing function. So at time zero, it has 1,400 tons in it, they tell us that right up here, and then it increases. We don't know what the function actually does look like, but let's say that that is W. So what they're saying in problem A, is they're saying use a tangent line to the graph at T equals zero. So there's, let me actually, let me draw W a little bit differently. So W might look like this, W might look like this. So what they're saying is find the tangent line, find the slope of that tangent line at T equals zero. So there's some slope right over there."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So there's, let me actually, let me draw W a little bit differently. So W might look like this, W might look like this. So what they're saying is find the tangent line, find the slope of that tangent line at T equals zero. So there's some slope right over there. And then we can use to approximate, we can use this slope to kind of create a linear approximation of where we're gonna be a quarter of a year from then. So if you go, so although we don't know what W is just yet, we could take the slope of this line out to T is equal to, so this is our T axis, to T is equal to 1 4th, and wherever that line takes us out after a fourth of year, that will be at least a decent approximation. We're extrapolating forward from that first point and the current slope."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So there's some slope right over there. And then we can use to approximate, we can use this slope to kind of create a linear approximation of where we're gonna be a quarter of a year from then. So if you go, so although we don't know what W is just yet, we could take the slope of this line out to T is equal to, so this is our T axis, to T is equal to 1 4th, and wherever that line takes us out after a fourth of year, that will be at least a decent approximation. We're extrapolating forward from that first point and the current slope. So we really just have to figure out the slope of this line and just see where that line is at T is equal to 1 4th. And you say, wait, how do we know, how do we know what the derivative of W is at zero without knowing W? And that's where we can go straight to this differential equation."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're extrapolating forward from that first point and the current slope. So we really just have to figure out the slope of this line and just see where that line is at T is equal to 1 4th. And you say, wait, how do we know, how do we know what the derivative of W is at zero without knowing W? And that's where we can go straight to this differential equation. We can actually rewrite this differential equation using slightly different notation. This is the derivative of W with respect to T. We could write that as W prime of T is equal to one over 25, one over 25, times the function W, which is a function of T, minus 300. And when you look at it this way, it becomes a little bit clearer on how to figure out what the derivative of W is at zero."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's where we can go straight to this differential equation. We can actually rewrite this differential equation using slightly different notation. This is the derivative of W with respect to T. We could write that as W prime of T is equal to one over 25, one over 25, times the function W, which is a function of T, minus 300. And when you look at it this way, it becomes a little bit clearer on how to figure out what the derivative of W is at zero. We literally, the slope of this line, the slope of that line literally is just the derivative of W evaluated at zero. So let's literally take the derivative of W and evaluate it at zero. So we have W prime of zero is equal to one over 25 times W of zero, which we know."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when you look at it this way, it becomes a little bit clearer on how to figure out what the derivative of W is at zero. We literally, the slope of this line, the slope of that line literally is just the derivative of W evaluated at zero. So let's literally take the derivative of W and evaluate it at zero. So we have W prime of zero is equal to one over 25 times W of zero, which we know. We know that this is 1,400 tons of solid waste. This is how much waste there is at time zero, minus 300. So this right over here is 1,400."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have W prime of zero is equal to one over 25 times W of zero, which we know. We know that this is 1,400 tons of solid waste. This is how much waste there is at time zero, minus 300. So this right over here is 1,400. And so we have W prime at time equals zero. So our slope at time equals zero, our derivative at time equals zero, is equal to one over 25 times 1,400 minus 300 is 1,100. Minus 300 is 1,100."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is 1,400. And so we have W prime at time equals zero. So our slope at time equals zero, our derivative at time equals zero, is equal to one over 25 times 1,400 minus 300 is 1,100. Minus 300 is 1,100. And 25 goes into 1,100, it goes into 1,100 44, it would go into it 44 times, right? It goes four times into each 100. We have 1,100 here, so it'll go 44 times."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Minus 300 is 1,100. And 25 goes into 1,100, it goes into 1,100 44, it would go into it 44 times, right? It goes four times into each 100. We have 1,100 here, so it'll go 44 times. So it goes 44. So the slope here, the slope of this line, the slope of that line is 44. We could say M for slope, or actually let me just write down the word."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have 1,100 here, so it'll go 44 times. So it goes 44. So the slope here, the slope of this line, the slope of that line is 44. We could say M for slope, or actually let me just write down the word. The slope of this line is 44. And I just ate some peanuts or something, so my voice is a little dry, so bear with me. But the slope of this line is 24."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could say M for slope, or actually let me just write down the word. The slope of this line is 44. And I just ate some peanuts or something, so my voice is a little dry, so bear with me. But the slope of this line is 24. So how do we use that to find an approximation for the amount of waste that the landfill contains at the end of the first three months? Well, let me zoom in a little bit. And actually my fingers are all salty too, so maybe I don't have a proper grip on my pen, but I'll try my best."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the slope of this line is 24. So how do we use that to find an approximation for the amount of waste that the landfill contains at the end of the first three months? Well, let me zoom in a little bit. And actually my fingers are all salty too, so maybe I don't have a proper grip on my pen, but I'll try my best. So let's say, let me just zoom in a little bit more. So we're starting at 1,400 tons. So this is my W axis, this is my time axis."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually my fingers are all salty too, so maybe I don't have a proper grip on my pen, but I'll try my best. So let's say, let me just zoom in a little bit more. So we're starting at 1,400 tons. So this is my W axis, this is my time axis. We're starting at 1,400 tons, and we are increasing from there. They're telling us that it is an increasing function. So maybe it looks something like that."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is my W axis, this is my time axis. We're starting at 1,400 tons, and we are increasing from there. They're telling us that it is an increasing function. So maybe it looks something like that. And then our slope, our slope right over here, and I don't know, I haven't drawn W exactly that accurately. I'm just guessing what it might look like at this point. The tangent line has a slope of 44."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So maybe it looks something like that. And then our slope, our slope right over here, and I don't know, I haven't drawn W exactly that accurately. I'm just guessing what it might look like at this point. The tangent line has a slope of 44. So that is our tangent line. And what that's saying is if we go out one, if we go out one unit of time, which is one year, then we would have gone up 44 in tonnage. So if we use this line as approximation, after one year, this point would have 1,444 tons."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "The tangent line has a slope of 44. So that is our tangent line. And what that's saying is if we go out one, if we go out one unit of time, which is one year, then we would have gone up 44 in tonnage. So if we use this line as approximation, after one year, this point would have 1,444 tons. But we're not trying to approximate a year out, we're trying to approximate a fourth of a year out. So we're trying to approximate, so this would be half a year out, this is one fourth of a year out. We're trying to approximate, we're trying to approximate that point right over there."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we use this line as approximation, after one year, this point would have 1,444 tons. But we're not trying to approximate a year out, we're trying to approximate a fourth of a year out. So we're trying to approximate, so this would be half a year out, this is one fourth of a year out. We're trying to approximate, we're trying to approximate that point right over there. So it's going to be 1,400, this point right over here. And we could write down the equation of this line if we like. We could say this line, this line, so we'll call this the W approximation, because this isn't exactly our W function."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're trying to approximate, we're trying to approximate that point right over there. So it's going to be 1,400, this point right over here. And we could write down the equation of this line if we like. We could say this line, this line, so we'll call this the W approximation, because this isn't exactly our W function. This is equal to the slope of our line, 44 times time, plus our W intercept, we could say, or plus our initial condition, plus 1,400. So if you put time is equal to one fourth in there, you get it equal to 44, so let me say, let me write it this way. So our approximate W at time is equal to one fourth of a year is equal to 44 times one fourth, plus 1,400."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could say this line, this line, so we'll call this the W approximation, because this isn't exactly our W function. This is equal to the slope of our line, 44 times time, plus our W intercept, we could say, or plus our initial condition, plus 1,400. So if you put time is equal to one fourth in there, you get it equal to 44, so let me say, let me write it this way. So our approximate W at time is equal to one fourth of a year is equal to 44 times one fourth, plus 1,400. Running out of space and my salty hands are having trouble writing this properly. And so 44 times one fourth, or 44 divided by four, is 11 plus 1,400. And so you add them together, you get 1,411, or 1,411 tons."}, {"video_title": "2011 Calculus AB free response #5a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our approximate W at time is equal to one fourth of a year is equal to 44 times one fourth, plus 1,400. Running out of space and my salty hands are having trouble writing this properly. And so 44 times one fourth, or 44 divided by four, is 11 plus 1,400. And so you add them together, you get 1,411, or 1,411 tons. This is our approximation. We just took the slope from our starting point and used that slope as an approximation. It probably is not the exact amount of tonnage based on the actual function W, but it's an okay approximation."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I want to show you in this video is that implicit differentiation will give you the same result as, I guess we could say, explicit differentiation, when you can differentiate explicitly. So let's say that I have the relationship x times the square root of y is equal to 1. This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1 over x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we divide both sides by x, we get square root of y is equal to 1 over x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. We get dy dx is equal to negative 2 x to the negative 2 minus 1, x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is just an application of the chain rule. We get dy dx is equal to negative 2 x to the negative 2 minus 1, x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing. x times the square root of y and 1 right over there."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing. x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, we are just going to have to apply, well, actually we're going to apply both the product rule and the chain rule. The product rule tells us, so we have the product of two functions of x. You could view it that way."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, we are just going to have to apply, well, actually we're going to apply both the product rule and the chain rule. The product rule tells us, so we have the product of two functions of x. You could view it that way. So the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, plus x times the derivative with respect to x of the square root of y. Let me make it clear this bracket, of the square root of y. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could view it that way. So the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, plus x times the derivative with respect to x of the square root of y. Let me make it clear this bracket, of the square root of y. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. So we're just going to be left, this simplifies to 1, so we're just going to be left with the square root of y right over here."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. So we're just going to be left, this simplifies to 1, so we're just going to be left with the square root of y right over here. So we're just going to be, this is going to simplify to a square root of y. And what does this over here simplify to? Well, the derivative with respect to x of the square root of y, here we want to apply the chain rule, so let me make it clear."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're just going to be left, this simplifies to 1, so we're just going to be left with the square root of y right over here. So we're just going to be, this is going to simplify to a square root of y. And what does this over here simplify to? Well, the derivative with respect to x of the square root of y, here we want to apply the chain rule, so let me make it clear. So we have plus this x, plus this x, plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of the square root of something with respect to that something."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the derivative with respect to x of the square root of y, here we want to apply the chain rule, so let me make it clear. So we have plus this x, plus this x, plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of the square root of something with respect to that something. Well, the derivative of the square root of something with respect to that something or the derivative of something to the 1 half with respect to that something It's going to be 1 half times that something to the negative 1 half power. Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's going to be the derivative of the square root of something with respect to that something. Well, the derivative of the square root of something with respect to that something or the derivative of something to the 1 half with respect to that something It's going to be 1 half times that something to the negative 1 half power. Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1 half x to the negative 1 half. Now I'm just doing it with y's. But we're not done yet."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1 half x to the negative 1 half. Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is. That's actually what we're trying to solve for."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have to say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand side, on the right-hand side, we just have a 0."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's actually what we're trying to solve for. But to use the chain rule, we just have to say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand side, on the right-hand side, we just have a 0. And now once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over so I have, once again, more room to work with."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get this on the left-hand side, on the right-hand side, we just have a 0. And now once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over so I have, once again, more room to work with. Let me cut it, actually. And then let me paste it. Let me move it over right over here."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually, let me move all of this stuff over so I have, once again, more room to work with. Let me cut it, actually. And then let me paste it. Let me move it over right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me move it over right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator. And 1 half, I just put the 2 in the denominator there, times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator. And 1 half, I just put the 2 in the denominator there, times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here just to continue our simplification solving for dy dx."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "And 1 half, I just put the 2 in the denominator there, times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the reciprocal of this, is equal to 2 times the square root of y over x, over my yellow x, times the negative square root of y. Well, what's this going to simplify to?"}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's go back up here just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the reciprocal of this, is equal to 2 times the square root of y over x, over my yellow x, times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to the square root of y times the square root of y is just y. The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's this going to simplify to? This is going to be equal to the square root of y times the square root of y is just y. The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now, you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2x to the negative third power. x to the negative 3 power."}, {"video_title": "Showing explicit and implicit differentiation give same result   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you get negative 2y over x is equal to the derivative of y with respect to x. Now, you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2x to the negative third power. x to the negative 3 power. The key here is to realize that this thing right over here, we could solve explicitly in terms of, we could solve for y. So we could just make this substitution back here to see that these are the exact same thing. So if we make the substitution y is equal to 1 over x squared, you get dy dx, the derivative of y with respect to x, is equal to our negative 2, is equal to the negative 2 times 1 over x squared, and then all of that over x, which is equal to negative 2 over x to the third, which is exactly what we have over here."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "And it's a more constrained version of the general case we've been looking at, but it's still very powerful and very applicable. And the reason why we're going to go over this special case is because its proof is fairly straightforward and will give you an intuition for why L'Hospital's Rule works at all. So the special case of L'Hospital's Rule is a situation where f of a is equal to 0, f prime of a exists, g of a is equal to 0, g prime of a exists. If these constraints are met, then the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a. So it's very similar to the general case, it's a little bit more constrained. We're assuming that f prime of a exists, we're not just taking the limit now. We're assuming f prime of a and g prime of a actually exist."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "If these constraints are met, then the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a. So it's very similar to the general case, it's a little bit more constrained. We're assuming that f prime of a exists, we're not just taking the limit now. We're assuming f prime of a and g prime of a actually exist. But notice if we substitute a right over here, we get 0 over 0. But then if the derivatives exist, we can just evaluate the derivatives at a and then we get the limit. So this is very close to the general case of L'Hospital's Rule."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "We're assuming f prime of a and g prime of a actually exist. But notice if we substitute a right over here, we get 0 over 0. But then if the derivatives exist, we can just evaluate the derivatives at a and then we get the limit. So this is very close to the general case of L'Hospital's Rule. Now let's actually prove it. And to prove it, we're going to start with the right hand and then show that if we use the definition of derivatives, we get the left hand right over here. So let me do that."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is very close to the general case of L'Hospital's Rule. Now let's actually prove it. And to prove it, we're going to start with the right hand and then show that if we use the definition of derivatives, we get the left hand right over here. So let me do that. So I'll do it right over here. So f prime of a is equal to what by the definition of derivatives? Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me do that. So I'll do it right over here. So f prime of a is equal to what by the definition of derivatives? Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a. So this is literally just the slope between two points. So if you have your function f of x like this, this is the point a, f of a right over here. This right over here is the point x, f of x."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a. So this is literally just the slope between two points. So if you have your function f of x like this, this is the point a, f of a right over here. This right over here is the point x, f of x. This expression right over here is the slope between these two points. The change in our y value is f of x minus f of a. The change in our x value is x minus a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "This right over here is the point x, f of x. This expression right over here is the slope between these two points. The change in our y value is f of x minus f of a. The change in our x value is x minus a. So this expression is just the slope of this line. And we're just taking the line that connects these two points. That's the slope of it."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "The change in our x value is x minus a. So this expression is just the slope of this line. And we're just taking the line that connects these two points. That's the slope of it. I'll do that in white. The slope of the line that connects those two points. And we're taking the limit as x gets closer and closer and closer to a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "That's the slope of it. I'll do that in white. The slope of the line that connects those two points. And we're taking the limit as x gets closer and closer and closer to a. So this is just another way of writing the definition of the derivative. So that's fine. Let's do the same thing for g prime of a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "And we're taking the limit as x gets closer and closer and closer to a. So this is just another way of writing the definition of the derivative. So that's fine. Let's do the same thing for g prime of a. So f prime of a over g prime of a is going to be this business, which is in orange, f prime of a over g prime of a, which we can write as the limit as x approaches a of g of x minus g of a over x minus a. Well, in the numerator we're taking the limit as x approaches a. In the denominator we're taking the limit as x approaches a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's do the same thing for g prime of a. So f prime of a over g prime of a is going to be this business, which is in orange, f prime of a over g prime of a, which we can write as the limit as x approaches a of g of x minus g of a over x minus a. Well, in the numerator we're taking the limit as x approaches a. In the denominator we're taking the limit as x approaches a. So we can just rewrite this. This we can rewrite as the limit as x approaches a of all this business in orange, f of x minus f of a over x minus a over all the business in green, g of x minus g of a, all of that over x minus a. Now to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "In the denominator we're taking the limit as x approaches a. So we can just rewrite this. This we can rewrite as the limit as x approaches a of all this business in orange, f of x minus f of a over x minus a over all the business in green, g of x minus g of a, all of that over x minus a. Now to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a. So let's do that. Let's multiply by x minus a over x minus a. So the numerator x minus a and we're dividing by x minus a, those cancel out, and then these two cancel out, and we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a, and in the denominator we have g of x minus g of a. I think you see where this is going."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "Now to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a. So let's do that. Let's multiply by x minus a over x minus a. So the numerator x minus a and we're dividing by x minus a, those cancel out, and then these two cancel out, and we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a, and in the denominator we have g of x minus g of a. I think you see where this is going. What is f of a equal to? Well, we assumed f of a is equal to zero. That's why we're using L'Hospital's Rule from the get-go."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule   Differential Calculus   Khan Academy.mp3", "Sentence": "So the numerator x minus a and we're dividing by x minus a, those cancel out, and then these two cancel out, and we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a, and in the denominator we have g of x minus g of a. I think you see where this is going. What is f of a equal to? Well, we assumed f of a is equal to zero. That's why we're using L'Hospital's Rule from the get-go. f of a is equal to zero, g of a is equal to zero. f of a is equal to zero, g of a is equal to zero, and this simplifies to the limit as x approaches a of f prime of x, sorry, of f of x, we've got to be careful, of f of x over g of x. So we just showed that if f of a equals zero, g of a equals zero, and these two derivatives exist, then the derivatives evaluated a over each other are going to be equal to the limit as x approaches a of f of x over g of x, or the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if this is my y-axis, and this is, that's not that straight, this right over here is my x-axis. And let's say I have two functions. So I'm just gonna say it in general terms. So let's say I have a function right over here. So let's say it looks something like this. So that's one function. So this is y is equal to f of x."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say I have a function right over here. So let's say it looks something like this. So that's one function. So this is y is equal to f of x. And then I have another function that y is equal to g of x. So let's say it looks something like this. So y, the blue one right over here is y is equal to g of x."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is y is equal to f of x. And then I have another function that y is equal to g of x. So let's say it looks something like this. So y, the blue one right over here is y is equal to g of x. And like we did in the last video, I wanna think about the volume of the solid of revolution we will get if we essentially rotate the area between these two, and if we were to rotate it around the x-axis. So we're saying in very general terms, this could be anything, but in the way we've drawn it now, it would literally be that same truffle shape. It'd be a very similar truffle shape where on the outside it looks like a truffle, on the outside it looks like something like a truffle, and on the inside we have carved out a cone."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So y, the blue one right over here is y is equal to g of x. And like we did in the last video, I wanna think about the volume of the solid of revolution we will get if we essentially rotate the area between these two, and if we were to rotate it around the x-axis. So we're saying in very general terms, this could be anything, but in the way we've drawn it now, it would literally be that same truffle shape. It'd be a very similar truffle shape where on the outside it looks like a truffle, on the outside it looks like something like a truffle, and on the inside we have carved out a cone. Obviously this visualization is very specific to the way I've drawn these functions, but what we wanna do is generalize at least the mathematics of it. So how do we find a volume? Well, we could think of disks, but instead of thinking of disks, we're gonna think about washers now, which is essentially the exact same thing we did in the last video mathematically, but it's a slightly different way of conceptualizing it."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It'd be a very similar truffle shape where on the outside it looks like a truffle, on the outside it looks like something like a truffle, and on the inside we have carved out a cone. Obviously this visualization is very specific to the way I've drawn these functions, but what we wanna do is generalize at least the mathematics of it. So how do we find a volume? Well, we could think of disks, but instead of thinking of disks, we're gonna think about washers now, which is essentially the exact same thing we did in the last video mathematically, but it's a slightly different way of conceptualizing it. So imagine taking a little chunk between these two functions just like that. What is going to be the width of this chunk? Well, it's going to be equal to dx, and let's rotate that whole thing around the x-axis."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we could think of disks, but instead of thinking of disks, we're gonna think about washers now, which is essentially the exact same thing we did in the last video mathematically, but it's a slightly different way of conceptualizing it. So imagine taking a little chunk between these two functions just like that. What is going to be the width of this chunk? Well, it's going to be equal to dx, and let's rotate that whole thing around the x-axis. So if we rotate this thing around the x-axis, we end up with a washer. That's why we're gonna call this the washer method. It's really just kind of the disk method where you're gutting out the inside of a disk."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's going to be equal to dx, and let's rotate that whole thing around the x-axis. So if we rotate this thing around the x-axis, we end up with a washer. That's why we're gonna call this the washer method. It's really just kind of the disk method where you're gutting out the inside of a disk. So that's the inside of our washer, and then this is the outside of our washer. Outside of our washer looks something like that. Hopefully that makes sense."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's really just kind of the disk method where you're gutting out the inside of a disk. So that's the inside of our washer, and then this is the outside of our washer. Outside of our washer looks something like that. Hopefully that makes sense. And so the surface of our washer looks something like that. I know I could have drawn this a little bit better, but hopefully it serves the purpose so that we understand it. So the surface of our washer looks something like that, and it has depth of dx."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Hopefully that makes sense. And so the surface of our washer looks something like that. I know I could have drawn this a little bit better, but hopefully it serves the purpose so that we understand it. So the surface of our washer looks something like that, and it has depth of dx. It has depth dx. So let me see how well I can draw this. So depth dx, that's the side of this washer."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the surface of our washer looks something like that, and it has depth of dx. It has depth dx. So let me see how well I can draw this. So depth dx, that's the side of this washer. So a washer, you can imagine, is kind of a gutted out, it's a gutted out coin. It's a gutted out coin. So how do we find the volume?"}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So depth dx, that's the side of this washer. So a washer, you can imagine, is kind of a gutted out, it's a gutted out coin. It's a gutted out coin. So how do we find the volume? Well, once again, if we know the surface, if we know the area of the face of this washer, we can just multiply that times the depth. So it's gonna be the area of the face of the washer. So the area of the face of the washer, well, it could be the area if it wasn't gutted out, and what would that area be?"}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how do we find the volume? Well, once again, if we know the surface, if we know the area of the face of this washer, we can just multiply that times the depth. So it's gonna be the area of the face of the washer. So the area of the face of the washer, well, it could be the area if it wasn't gutted out, and what would that area be? Well, it would be pi times the overall, the outside radius squared. It would be pi times the outside radius squared. Well, what's the outside radius?"}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area of the face of the washer, well, it could be the area if it wasn't gutted out, and what would that area be? Well, it would be pi times the overall, the outside radius squared. It would be pi times the outside radius squared. Well, what's the outside radius? The radius that goes to the outside of the washer. Well, that's f of x. So it's gonna be f of x is the radius, and we're just going to square that."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's the outside radius? The radius that goes to the outside of the washer. Well, that's f of x. So it's gonna be f of x is the radius, and we're just going to square that. So that would give us, this expression right over here would give us the area of the entire face if it wasn't a washer, if it was a coin. But now we have to subtract out the inside. So what's the area of the inside?"}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's gonna be f of x is the radius, and we're just going to square that. So that would give us, this expression right over here would give us the area of the entire face if it wasn't a washer, if it was a coin. But now we have to subtract out the inside. So what's the area of the inside? What's the area of the inside? This part right over here. Well, we're gonna subtract it out."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's the area of the inside? What's the area of the inside? This part right over here. Well, we're gonna subtract it out. It's going to be pi times the radius of the inside squared. The radius of the inside squared. Well, what's the radius of the inside squared?"}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're gonna subtract it out. It's going to be pi times the radius of the inside squared. The radius of the inside squared. Well, what's the radius of the inside squared? Well, the inside, in this case, is g of x. It's going to be pi times g of x squared. That's the inner function, at least over the interval that we care about."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's the radius of the inside squared? Well, the inside, in this case, is g of x. It's going to be pi times g of x squared. That's the inner function, at least over the interval that we care about. So the area of this washer, we could just leave it like this, or we could factor out a pi. We could say it's equal to the area is equal to, if we factor out a pi, pi times f of x squared, f of x squared minus g of x squared. Minus, I don't, I have to write a parentheses there."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the inner function, at least over the interval that we care about. So the area of this washer, we could just leave it like this, or we could factor out a pi. We could say it's equal to the area is equal to, if we factor out a pi, pi times f of x squared, f of x squared minus g of x squared. Minus, I don't, I have to write a parentheses there. So we could write f of x squared minus g of x squared. And then if we want the volume, put that in that same yellow. If we want the volume of this thing, we just multiply it times the depth of each of those washers."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Minus, I don't, I have to write a parentheses there. So we could write f of x squared minus g of x squared. And then if we want the volume, put that in that same yellow. If we want the volume of this thing, we just multiply it times the depth of each of those washers. So the volume of each of these washers are going to be pi times, pi times f of x, f of x squared minus g of x squared, the outer function squared over our interval minus our inner function squared over the interval, and then times our depth. Times our depth. That'd be the volume of each of these washers."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we want the volume of this thing, we just multiply it times the depth of each of those washers. So the volume of each of these washers are going to be pi times, pi times f of x, f of x squared minus g of x squared, the outer function squared over our interval minus our inner function squared over the interval, and then times our depth. Times our depth. That'd be the volume of each of these washers. Now, for each, and that's going to be defined at a given x in our interval, but for each x at this interval, we can define a new washer. So there could be a washer out here and a washer out here. So we're going to take the sum of all of those washers and take the limit as we have smaller and smaller depths and we have an infinite number of infinitely thin washers."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "That'd be the volume of each of these washers. Now, for each, and that's going to be defined at a given x in our interval, but for each x at this interval, we can define a new washer. So there could be a washer out here and a washer out here. So we're going to take the sum of all of those washers and take the limit as we have smaller and smaller depths and we have an infinite number of infinitely thin washers. So we're going to take the integral over our interval from where these two things intersect, the interval that we care about. It doesn't have to be where they intersect, but in this case, that's what we'll do. So let's say x equals a to x equals b, although it could have been from, this could have been a, that could have been b, but this is our interval."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to take the sum of all of those washers and take the limit as we have smaller and smaller depths and we have an infinite number of infinitely thin washers. So we're going to take the integral over our interval from where these two things intersect, the interval that we care about. It doesn't have to be where they intersect, but in this case, that's what we'll do. So let's say x equals a to x equals b, although it could have been from, this could have been a, that could have been b, but this is our interval. We're saying in general terms from a to b. And this will give our volume. This right over here is the volume of each washer, and then we're summing up all of the washers and taking the limit as we have an infinite number of them."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say x equals a to x equals b, although it could have been from, this could have been a, that could have been b, but this is our interval. We're saying in general terms from a to b. And this will give our volume. This right over here is the volume of each washer, and then we're summing up all of the washers and taking the limit as we have an infinite number of them. So let's see if we apply this to the example in the last video, whether we get the exact same answer. Well, in the last video, y equals g of x was equal to x, and y is equal to f of x was equal to the square root of x. So let's evaluate that given what we just were able to derive."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here is the volume of each washer, and then we're summing up all of the washers and taking the limit as we have an infinite number of them. So let's see if we apply this to the example in the last video, whether we get the exact same answer. Well, in the last video, y equals g of x was equal to x, and y is equal to f of x was equal to the square root of x. So let's evaluate that given what we just were able to derive. So our volume, I'll do it up here. The volume is going to be the integral. What are the two intersection points?"}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's evaluate that given what we just were able to derive. So our volume, I'll do it up here. The volume is going to be the integral. What are the two intersection points? Well, over here, once again, we could have defined the interval someplace else, like between there and there, and we would have gotten a different shape, but the points that we care about, the way we visualize it is between x is equal to zero, x is equal to zero, and x is equal to one. That's where these two things intersect. We saw that in the last video."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "What are the two intersection points? Well, over here, once again, we could have defined the interval someplace else, like between there and there, and we would have gotten a different shape, but the points that we care about, the way we visualize it is between x is equal to zero, x is equal to zero, and x is equal to one. That's where these two things intersect. We saw that in the last video. Of pi, pi, times, what's f of x squared? f of x squared, square root of x squared is just x, minus g of x squared. G of x is x, that squared is x squared, and then we multiply times dx."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We saw that in the last video. Of pi, pi, times, what's f of x squared? f of x squared, square root of x squared is just x, minus g of x squared. G of x is x, that squared is x squared, and then we multiply times dx. So this is going to be equal to, we can factor out the pi, zero to one, x minus x squared dx, which is equal to pi times, let's see, the antiderivative of x is x squared over two, antiderivative of x squared is x squared over three, x to the third over three, and we're going to evaluate this from zero to one. So this is going to be equal to, I'm running out of my real estate a little bit, let me scroll over to the right a little bit. So this is going to be equal to pi times, well, when you evaluate this whole thing at one, you get, let's see, you get one half minus one third, one half minus one third, and then you subtract it, evaluate it at zero, but that's just going to be zero."}, {"video_title": "Generalizing the washer method   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of x is x, that squared is x squared, and then we multiply times dx. So this is going to be equal to, we can factor out the pi, zero to one, x minus x squared dx, which is equal to pi times, let's see, the antiderivative of x is x squared over two, antiderivative of x squared is x squared over three, x to the third over three, and we're going to evaluate this from zero to one. So this is going to be equal to, I'm running out of my real estate a little bit, let me scroll over to the right a little bit. So this is going to be equal to pi times, well, when you evaluate this whole thing at one, you get, let's see, you get one half minus one third, one half minus one third, and then you subtract it, evaluate it at zero, but that's just going to be zero. Zero squared over two minus zero to the third power over three, that's just all going to be zero. So when you subtract out zero, you're just left with this expression right over here. What's one half minus one third?"}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if we can give ourselves an intuitive understanding of the mean value theorem. And as we'll see, once you parse some of the mathematical lingo and notation, it's actually a quite intuitive theorem. And so let's just think about some function f. So let's say I have some function f, and we know a few things about this function. We know that it is continuous over the closed interval between x equals a and x is equal to b. And so when we put these brackets here, that just means closed interval. That means we're including, so when I put a bracket here, that means we're including the point a. And if I put the bracket on the right-hand side instead of a parenthesis, that means that we are including the point b."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that it is continuous over the closed interval between x equals a and x is equal to b. And so when we put these brackets here, that just means closed interval. That means we're including, so when I put a bracket here, that means we're including the point a. And if I put the bracket on the right-hand side instead of a parenthesis, that means that we are including the point b. And continuous just means we don't have any gaps or jumps in the function over this closed interval. Now let's also assume that it's differentiable over the open interval between a and b. So now we're saying, well, it's okay if it's not differentiable right at a, or if it's not differentiable right at b."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if I put the bracket on the right-hand side instead of a parenthesis, that means that we are including the point b. And continuous just means we don't have any gaps or jumps in the function over this closed interval. Now let's also assume that it's differentiable over the open interval between a and b. So now we're saying, well, it's okay if it's not differentiable right at a, or if it's not differentiable right at b. And differentiable just means that there's a defined derivative, that you can actually take the derivative at those points. So it's differentiable over the open interval between a and b. So those are the constraints we're gonna put on ourselves for the mean value theorem."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we're saying, well, it's okay if it's not differentiable right at a, or if it's not differentiable right at b. And differentiable just means that there's a defined derivative, that you can actually take the derivative at those points. So it's differentiable over the open interval between a and b. So those are the constraints we're gonna put on ourselves for the mean value theorem. And so let's just try to visualize this thing. So this is my function. That's the y-axis."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So those are the constraints we're gonna put on ourselves for the mean value theorem. And so let's just try to visualize this thing. So this is my function. That's the y-axis. And this right over here is the x-axis. And I'm gonna, it's the x-axis, and let me draw my interval. So that's a, and then this is b right over here."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the y-axis. And this right over here is the x-axis. And I'm gonna, it's the x-axis, and let me draw my interval. So that's a, and then this is b right over here. And so let's say our function looks something like this. Let's say it looks something, draw an arbitrary function right over here. Let's say my function looks something like that."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's a, and then this is b right over here. And so let's say our function looks something like this. Let's say it looks something, draw an arbitrary function right over here. Let's say my function looks something like that. So this point right over here, the x-value is a, and the y-value is f of a. F of a, this point right over here, the x-value is b, and the y-value, of course, is f of b. F of b. F of b. So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point, the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change. Now what does that mean visually?"}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say my function looks something like that. So this point right over here, the x-value is a, and the y-value is f of a. F of a, this point right over here, the x-value is b, and the y-value, of course, is f of b. F of b. F of b. So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point, the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change. Now what does that mean visually? So let's calculate the average change. The average change between point a and point b, well that's going to be the slope of the secant line. That's going to be the slope of the secant line."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what does that mean visually? So let's calculate the average change. The average change between point a and point b, well that's going to be the slope of the secant line. That's going to be the slope of the secant line. So this is the secant line, so think about its slope. All the mean value theorem tells us is that at some point in this interval, the slope of the tangent line is going to be the same as the slope of the secant line. And we can see just visually, it looks like right over here the slope of the tangent line looks like the same as the slope of the secant line."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's going to be the slope of the secant line. So this is the secant line, so think about its slope. All the mean value theorem tells us is that at some point in this interval, the slope of the tangent line is going to be the same as the slope of the secant line. And we can see just visually, it looks like right over here the slope of the tangent line looks like the same as the slope of the secant line. It also looks like the case right over here, the slope of the tangent line is equal to the slope of the secant line. And it makes intuitive sense. At some point, your instantaneous slope is going to be the same as the average slope."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can see just visually, it looks like right over here the slope of the tangent line looks like the same as the slope of the secant line. It also looks like the case right over here, the slope of the tangent line is equal to the slope of the secant line. And it makes intuitive sense. At some point, your instantaneous slope is going to be the same as the average slope. Now how would we write that mathematically? Well let's calculate, let's calculate, well let's calculate the average slope over this interval. Well the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y, our change in y right over here, over our change in x, over our change in x."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "At some point, your instantaneous slope is going to be the same as the average slope. Now how would we write that mathematically? Well let's calculate, let's calculate, well let's calculate the average slope over this interval. Well the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y, our change in y right over here, over our change in x, over our change in x. Well what is our change in y? Our change in y is f of b minus f of a, minus f of a, and that's going to be over, that is going to be over our change in x. Over b minus, b minus a."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y, our change in y right over here, over our change in x, over our change in x. Well what is our change in y? Our change in y is f of b minus f of a, minus f of a, and that's going to be over, that is going to be over our change in x. Over b minus, b minus a. Let me do that in that right color. So let's just remind ourselves what's going on here. So this right over here, this is the graph of y is equal to f of x."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Over b minus, b minus a. Let me do that in that right color. So let's just remind ourselves what's going on here. So this right over here, this is the graph of y is equal to f of x. We're saying that the slope of the secant line, or our average rate of change over the interval from a to b, is our change in, is our change in y, our change in y, that the Greek letter delta is just shorthand for change in y over our change in x. Over our change in x, which of course is equal to this. And the mean value theorem tells us that there exists, so if we know these two things about the function, then there exists, there exists some, some x value in between a and b."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here, this is the graph of y is equal to f of x. We're saying that the slope of the secant line, or our average rate of change over the interval from a to b, is our change in, is our change in y, our change in y, that the Greek letter delta is just shorthand for change in y over our change in x. Over our change in x, which of course is equal to this. And the mean value theorem tells us that there exists, so if we know these two things about the function, then there exists, there exists some, some x value in between a and b. So in the open interval between a and b, there exists some c, there exists some c, and we could say it's a member of the open interval between a and b, between a and b. Or we could say some c such that a is less than, such that a is less than c, which is less than b. So some c in this interval."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the mean value theorem tells us that there exists, so if we know these two things about the function, then there exists, there exists some, some x value in between a and b. So in the open interval between a and b, there exists some c, there exists some c, and we could say it's a member of the open interval between a and b, between a and b. Or we could say some c such that a is less than, such that a is less than c, which is less than b. So some c in this interval. So some c, some c in between it, where the instantaneous rate of change at, at that x value is the same as the average rate of change. So there exists some c in this open interval where the average rate of change is equal to the instantaneous rate of change at that point. That's all it's saying."}, {"video_title": "Mean value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So some c in this interval. So some c, some c in between it, where the instantaneous rate of change at, at that x value is the same as the average rate of change. So there exists some c in this open interval where the average rate of change is equal to the instantaneous rate of change at that point. That's all it's saying. And as we saw in this diagram right over here, this could be our c, or this could be our c as well. So nothing, nothing really, it looks, you know, you would say f is continuous over a, b, differentiable over, over, f is continuous over the closed interval, differentiable over the open interval, and you see all of this notation, you're like, what is that telling us? All it's saying is, at some point in the interval, the instantaneous rate of change is going to be the same as the average rate of change over the whole interval."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I wanna do in this video is use some of our calculus tools to see if we can come up with the same or maybe even a better result. So to do that, I'm gonna have to figure out the critical points of our volume as a function of x, and to do that, I need to take the derivative of the volume, so let me do that, and before I even do that, it'll simplify things, so I don't have to use some product rule in some way and then have to simplify that. Let me just multiply this expression out. So let's rewrite volume as a function of x is equal to, and I'll write it all in yellow, so it's going to be x times, I'll multiply these two binomials first, so 20 times 30 is 600. Then I have 20 times negative two x, which is negative 40x. Then I have negative two x times 30, which is negative 60x, and then I have negative two x times negative two x, which is positive four x squared. So this part over here simplifies, and I can change the order to four x squared minus 100x plus 600."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's rewrite volume as a function of x is equal to, and I'll write it all in yellow, so it's going to be x times, I'll multiply these two binomials first, so 20 times 30 is 600. Then I have 20 times negative two x, which is negative 40x. Then I have negative two x times 30, which is negative 60x, and then I have negative two x times negative two x, which is positive four x squared. So this part over here simplifies, and I can change the order to four x squared minus 100x plus 600. I just switched the order in which I'm writing them. So that's that, and so I can rewrite the volume of x as being equal to x times all of this business, which is, let me make sure I have enough space, let me do it a little higher, which is equal to four x to the third power minus 100x squared plus 600x. Now it'll be pretty straightforward to take the derivative."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this part over here simplifies, and I can change the order to four x squared minus 100x plus 600. I just switched the order in which I'm writing them. So that's that, and so I can rewrite the volume of x as being equal to x times all of this business, which is, let me make sure I have enough space, let me do it a little higher, which is equal to four x to the third power minus 100x squared plus 600x. Now it'll be pretty straightforward to take the derivative. So let's say that v prime of x, v prime of x is going to be equal to, I just have to use the power rule multiple times, so four times three is 12x to the three minus one power, 12x squared, minus 200 times x to the first power, which is just x, plus 600. And so now we just have to figure out when this is equal to zero. So we have to figure out when 12x squared minus 200x plus 600 is equal to zero."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now it'll be pretty straightforward to take the derivative. So let's say that v prime of x, v prime of x is going to be equal to, I just have to use the power rule multiple times, so four times three is 12x to the three minus one power, 12x squared, minus 200 times x to the first power, which is just x, plus 600. And so now we just have to figure out when this is equal to zero. So we have to figure out when 12x squared minus 200x plus 600 is equal to zero. What x values gets my derivative to be equal to zero? When is my slope equal to zero? I could also look for critical points where the derivative is undefined, but this derivative is defined, especially throughout my domain of x that I care about, between zero and 10."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have to figure out when 12x squared minus 200x plus 600 is equal to zero. What x values gets my derivative to be equal to zero? When is my slope equal to zero? I could also look for critical points where the derivative is undefined, but this derivative is defined, especially throughout my domain of x that I care about, between zero and 10. So I could try to factor this or try to simplify this a little bit, but I'm just gonna cut to the chase and try to use the quadratic formula here. So this is the x's that satisfy this. It's going to be, x is going to be equal to, so negative b, so it's 200, negative negative 200 is positive 200, plus or minus the square root of b squared, which is negative 200 squared, so I could just write that as 200 squared."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could also look for critical points where the derivative is undefined, but this derivative is defined, especially throughout my domain of x that I care about, between zero and 10. So I could try to factor this or try to simplify this a little bit, but I'm just gonna cut to the chase and try to use the quadratic formula here. So this is the x's that satisfy this. It's going to be, x is going to be equal to, so negative b, so it's 200, negative negative 200 is positive 200, plus or minus the square root of b squared, which is negative 200 squared, so I could just write that as 200 squared. Doesn't matter if it's negative 200 squared or 200 squared, I'm gonna get the same value. So, let me give myself some more space. So negative 200 squared, well that's going to be four with four zeros."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be, x is going to be equal to, so negative b, so it's 200, negative negative 200 is positive 200, plus or minus the square root of b squared, which is negative 200 squared, so I could just write that as 200 squared. Doesn't matter if it's negative 200 squared or 200 squared, I'm gonna get the same value. So, let me give myself some more space. So negative 200 squared, well that's going to be four with four zeros. One, two, three, four. So that's gonna be 40,000 minus four a c. So minus four times 12, 12 times 600, I still didn't give myself enough space, times 600, all of that over two times a. So all of that over 24."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative 200 squared, well that's going to be four with four zeros. One, two, three, four. So that's gonna be 40,000 minus four a c. So minus four times 12, 12 times 600, I still didn't give myself enough space, times 600, all of that over two times a. So all of that over 24. And I'll take out the calculator again to try to calculate this. So let me get out of graphing mode. All right, so first I'll try when I add the radical."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So all of that over 24. And I'll take out the calculator again to try to calculate this. So let me get out of graphing mode. All right, so first I'll try when I add the radical. So I'm going to get 200, 200 plus the square root of 40,000. I could have just written that, I could have just written that as 200 squared, but that's fine, 40,000 minus four times 12, times 600, times 600. And I get 305, which I then need to divide by 24."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so first I'll try when I add the radical. So I'm going to get 200, 200 plus the square root of 40,000. I could have just written that, I could have just written that as 200 squared, but that's fine, 40,000 minus four times 12, times 600, times 600. And I get 305, which I then need to divide by 24. Which I'll divide by 24. And I get 12.74. So one of my possible x's."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I get 305, which I then need to divide by 24. Which I'll divide by 24. And I get 12.74. So one of my possible x's. So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one of my possible x's. So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back. And so now let me do 200. I probably could have done this slightly more efficiently, but this is fine. Minus the square root of 40,000, one, two, three."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me get my calculator back. And so now let me do 200. I probably could have done this slightly more efficiently, but this is fine. Minus the square root of 40,000, one, two, three. Minus four times 12, times 600. And I get, that's just the numerator. And then I'm gonna divide that by 24."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Minus the square root of 40,000, one, two, three. Minus four times 12, times 600. And I get, that's just the numerator. And then I'm gonna divide that by 24. Divide that by 24. And I get 3.92. Did I do that right?"}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I'm gonna divide that by 24. Divide that by 24. And I get 3.92. Did I do that right? 200 minus 40,000 minus four times 12 times 600. All of that divided by 24. My previous answer divided by 24."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Did I do that right? 200 minus 40,000 minus four times 12 times 600. All of that divided by 24. My previous answer divided by 24. Gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use?"}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "My previous answer divided by 24. Gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? Well, 12, x equals 12.74 is outside of our, outside of our valid values for x. If x was equal to 12.74, we would cut past, we would completely cut past, the x's would start to overlap with each other. So x cannot be 12.74."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now which of these can I use? Well, 12, x equals 12.74 is outside of our, outside of our valid values for x. If x was equal to 12.74, we would cut past, we would completely cut past, the x's would start to overlap with each other. So x cannot be 12.74. So we get a critical point at x is equal to 3.92. And you could look at the graph, and you could say, oh look, that looks like a maximum value. But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92?"}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So x cannot be 12.74. So we get a critical point at x is equal to 3.92. And you could look at the graph, and you could say, oh look, that looks like a maximum value. But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92? Well, in order to do the second derivative test, you have to figure out what the second derivative is, so let's do that. V prime prime of x is going to be equal to, is going to be equal to 24x, 24 times x to the first, minus 200. Minus, minus 200."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92? Well, in order to do the second derivative test, you have to figure out what the second derivative is, so let's do that. V prime prime of x is going to be equal to, is going to be equal to 24x, 24 times x to the first, minus 200. Minus, minus 200. And you can just look at inspection that this number right over here is less than four, so this thing right over here is going to be less than 100. You subtract, you subtract 200, so we can write the second derivative at 3.92 is going to be less than zero. You can figure out what the exact value is if you like."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Minus, minus 200. And you can just look at inspection that this number right over here is less than four, so this thing right over here is going to be less than 100. You subtract, you subtract 200, so we can write the second derivative at 3.92 is going to be less than zero. You can figure out what the exact value is if you like. So because this is less than zero, we are concave downwards, concave downwards. Another way of saying it is that the slope is, the slope is decreasing the entire time, concave downwards. When the slope is decreasing the entire time, our shape looks like that."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can figure out what the exact value is if you like. So because this is less than zero, we are concave downwards, concave downwards. Another way of saying it is that the slope is, the slope is decreasing the entire time, concave downwards. When the slope is decreasing the entire time, our shape looks like that. The slope could start off high, lower, lower, gets to zero, even lower, lower, lower. And we even saw that on the graph right over here. And since it's concave downwards, that implies that our critical point that sits where, during where the interval is concave downward, that critical point is a local maximum, is a maximum."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "When the slope is decreasing the entire time, our shape looks like that. The slope could start off high, lower, lower, gets to zero, even lower, lower, lower. And we even saw that on the graph right over here. And since it's concave downwards, that implies that our critical point that sits where, during where the interval is concave downward, that critical point is a local maximum, is a maximum. So this is the x value at which our function attains our maximum. Now, what is that maximum value? Well, we could type that back in into our original expression for volume to figure what that is."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And since it's concave downwards, that implies that our critical point that sits where, during where the interval is concave downward, that critical point is a local maximum, is a maximum. So this is the x value at which our function attains our maximum. Now, what is that maximum value? Well, we could type that back in into our original expression for volume to figure what that is. So let's figure out what the volume, when we get to 3.92, is equal to. What is our maximum volume? So get the calculator back out."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we could type that back in into our original expression for volume to figure what that is. So let's figure out what the volume, when we get to 3.92, is equal to. What is our maximum volume? So get the calculator back out. And I could use, it's obviously roughly 3.92, I could use this exact value. Actually, let's, well, I'll just use 3.92 to get a rough sense of what our maximum value is, our maximum volume. So it'll be 3.92, 3.92."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So get the calculator back out. And I could use, it's obviously roughly 3.92, I could use this exact value. Actually, let's, well, I'll just use 3.92 to get a rough sense of what our maximum value is, our maximum volume. So it'll be 3.92, 3.92. I'll just use this expression for the volume as a function of x. 3.92 times 20 minus two times 3.92 times 30, 30 minus two times 3.92 gives us, and we deserve a drum roll now, gives us 1,056.3. So 1,056.3, which is a higher volume than we got when we just inspected it graphically."}, {"video_title": "Optimization box volume (Part 2)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it'll be 3.92, 3.92. I'll just use this expression for the volume as a function of x. 3.92 times 20 minus two times 3.92 times 30, 30 minus two times 3.92 gives us, and we deserve a drum roll now, gives us 1,056.3. So 1,056.3, which is a higher volume than we got when we just inspected it graphically. We probably could have gotten a little bit more precise if we zoomed in some, and then we would have gotten a little bit better of an answer. But there you have it. Analytically, we were able to actually get an even better answer than we were able to do, at least on that first pass graphically."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "For x is greater than or equal to zero, f of x is cosine of pi x. And we want to evaluate the definite integral from negative one to one of f of x dx. And you might immediately say, well, which of these versions of f of x am I going to take the antiderivative from? Because from negative one to zero, I would think about x plus one, but then from zero to one, I would think about cosine pi x. And if you were thinking that, you're thinking in the right direction. And the way that we can make this a little bit more straightforward is to actually split up this definite integral. This is going to be equal to the definite integral from negative one to zero of f of x dx plus the integral from zero to one of f of x dx."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because from negative one to zero, I would think about x plus one, but then from zero to one, I would think about cosine pi x. And if you were thinking that, you're thinking in the right direction. And the way that we can make this a little bit more straightforward is to actually split up this definite integral. This is going to be equal to the definite integral from negative one to zero of f of x dx plus the integral from zero to one of f of x dx. Now, why was it useful for me to split it up this way? And in particular, to split it up, split the interval from negative one to one, split it into two intervals from negative one to zero and zero to one? Well, I did that because zero, x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be equal to the definite integral from negative one to zero of f of x dx plus the integral from zero to one of f of x dx. Now, why was it useful for me to split it up this way? And in particular, to split it up, split the interval from negative one to one, split it into two intervals from negative one to zero and zero to one? Well, I did that because zero, x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x. So if you look at the interval from negative one to zero, f of x is x plus one. So f of x here is x plus one. And then when you go from zero to one, f of x is cosine pi x."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I did that because zero, x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x. So if you look at the interval from negative one to zero, f of x is x plus one. So f of x here is x plus one. And then when you go from zero to one, f of x is cosine pi x. So cosine of pi x. And so now we just have to evaluate each of these separately and add them together. So let's take the definite integral from negative one to zero of x plus one dx."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then when you go from zero to one, f of x is cosine pi x. So cosine of pi x. And so now we just have to evaluate each of these separately and add them together. So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two. I'm just incrementing the exponent and then dividing by that value. And then plus x, and you could view it as I'm doing the same thing."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two. I'm just incrementing the exponent and then dividing by that value. And then plus x, and you could view it as I'm doing the same thing. If this is x to the zero, it'll be x to the first, x to the first over one, which is just x. And I'm gonna evaluate that at zero and subtract from that it evaluated at one, sorry, it evaluated at negative one. And so this is going to be equal to, if I evaluate it at zero, let me do this in another color."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then plus x, and you could view it as I'm doing the same thing. If this is x to the zero, it'll be x to the first, x to the first over one, which is just x. And I'm gonna evaluate that at zero and subtract from that it evaluated at one, sorry, it evaluated at negative one. And so this is going to be equal to, if I evaluate it at zero, let me do this in another color. If I evaluate it at zero, it's going to be zero squared over two, which is, well, I'll just write it zero squared over two plus zero. Well, all of that's just gonna be equal to zero, minus it evaluated at, it evaluated at negative one. So minus negative one squared, negative one squared over two plus negative one."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to, if I evaluate it at zero, let me do this in another color. If I evaluate it at zero, it's going to be zero squared over two, which is, well, I'll just write it zero squared over two plus zero. Well, all of that's just gonna be equal to zero, minus it evaluated at, it evaluated at negative one. So minus negative one squared, negative one squared over two plus negative one. So negative one squared is just one. So it's 1 1 2 plus negative one, 1 1 2 plus negative one is, or 1 1 2 minus one is negative 1 1 2. So all of that is negative 1 1 2."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So minus negative one squared, negative one squared over two plus negative one. So negative one squared is just one. So it's 1 1 2 plus negative one, 1 1 2 plus negative one is, or 1 1 2 minus one is negative 1 1 2. So all of that is negative 1 1 2. But then we're subtracting negative 1 1 2. Zero minus negative 1 1 2 is going to be equal to positive 1 1 2. So this is going to be equal to positive 1 1 2."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So all of that is negative 1 1 2. But then we're subtracting negative 1 1 2. Zero minus negative 1 1 2 is going to be equal to positive 1 1 2. So this is going to be equal to positive 1 1 2. So this first part right over here is positive 1 1 2. And now let's evaluate the integral from zero to one of cosine pi, I don't need that first parenthesis, of cosine of pi x dx. What is this equal to?"}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to positive 1 1 2. So this first part right over here is positive 1 1 2. And now let's evaluate the integral from zero to one of cosine pi, I don't need that first parenthesis, of cosine of pi x dx. What is this equal to? Now, if we were just trying to find the antiderivative of cosine of x, it's pretty straightforward. We know that the derivative with respect to x of sine of x is equal to cosine of x. Cosine of x. But that's not what we have here."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is this equal to? Now, if we were just trying to find the antiderivative of cosine of x, it's pretty straightforward. We know that the derivative with respect to x of sine of x is equal to cosine of x. Cosine of x. But that's not what we have here. We have cosine of pi x. So there is a technique here, you could call it u substitution. You could say u is equal to pi x."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that's not what we have here. We have cosine of pi x. So there is a technique here, you could call it u substitution. You could say u is equal to pi x. If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x somehow. So the derivative with respect to x of sine of pi x would be what? Well, we would use the chain rule."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say u is equal to pi x. If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x somehow. So the derivative with respect to x of sine of pi x would be what? Well, we would use the chain rule. It would be the derivative of the outside function with respect to the inside, or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi. Or you could say the derivative of sine pi x is pi cosine of pi x. Now, we almost have that here, except we just need a pi."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we would use the chain rule. It would be the derivative of the outside function with respect to the inside, or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi. Or you could say the derivative of sine pi x is pi cosine of pi x. Now, we almost have that here, except we just need a pi. So what if we were to throw a pi right over here, but so we don't change the value, we also multiply by one over pi? So if you divide and multiply by the same number, you're not changing its value. One over pi times pi is just equal to one."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, we almost have that here, except we just need a pi. So what if we were to throw a pi right over here, but so we don't change the value, we also multiply by one over pi? So if you divide and multiply by the same number, you're not changing its value. One over pi times pi is just equal to one. But this is useful. This is useful because we now know that pi cosine pi x is the derivative of sine pi x. So this is all going to be equal to, this is equal to one."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "One over pi times pi is just equal to one. But this is useful. This is useful because we now know that pi cosine pi x is the derivative of sine pi x. So this is all going to be equal to, this is equal to one. Let me take that one over pi. So this is equal to one over pi times, now we're going to evaluate, so the antiderivative here we just said is sine, sine of pi x, and we're going to evaluate that at one and at zero. So this is going to be equal to one over pi, one over pi, not pi, my hand is not listening to my mouth, one over pi times sine of pi, sine of pi minus sine of, minus sine of pi times zero, which is just zero."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is all going to be equal to, this is equal to one. Let me take that one over pi. So this is equal to one over pi times, now we're going to evaluate, so the antiderivative here we just said is sine, sine of pi x, and we're going to evaluate that at one and at zero. So this is going to be equal to one over pi, one over pi, not pi, my hand is not listening to my mouth, one over pi times sine of pi, sine of pi minus sine of, minus sine of pi times zero, which is just zero. Well sine of pi, that's zero. Sine of zero is zero. So you're going to have one over pi times zero minus zero."}, {"video_title": "Definite integral of piecewise function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to one over pi, one over pi, not pi, my hand is not listening to my mouth, one over pi times sine of pi, sine of pi minus sine of, minus sine of pi times zero, which is just zero. Well sine of pi, that's zero. Sine of zero is zero. So you're going to have one over pi times zero minus zero. So this whole thing is just all going to be equal to zero. So this first part was 1 1 2. This second part right over here is equal to zero."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Where we left off in the last video, we'd actually set up our definite integral to figure out the volume of this figure. So now we just have to evaluate it. And really the hardest part is going to be simplifying that and that right over there. So let's get to it. So what is this thing squared? Well, we're going to have to do a little bit of polynomial multiplication here. So I'm going to go into that same color I had."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's get to it. So what is this thing squared? Well, we're going to have to do a little bit of polynomial multiplication here. So I'm going to go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus. Actually, let me write that in the order of the terms or the degree of the terms. It's negative x squared plus 2x plus 4."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm going to go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus. Actually, let me write that in the order of the terms or the degree of the terms. It's negative x squared plus 2x plus 4. I just switched the order of these things. Times negative x squared plus 2x plus 4. So we're just going to multiply these two things."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's negative x squared plus 2x plus 4. I just switched the order of these things. Times negative x squared plus 2x plus 4. So we're just going to multiply these two things. I shouldn't even write a multiplication symbol. It looks too much like an x. So 4 times 4 is 16."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're just going to multiply these two things. I shouldn't even write a multiplication symbol. It looks too much like an x. So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all of this."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all of this. So negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squared times negative x squared is positive x to the fourth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we just have to multiply negative x times all of this. So negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth. And now we just have to add up all of these terms. And we get x to the fourth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth. And now we just have to add up all of these terms. And we get x to the fourth. Add these two, minus 4x to the third. And then this cancels with this, but we still have that. So minus 4x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we get x to the fourth. Add these two, minus 4x to the third. And then this cancels with this, but we still have that. So minus 4x squared. You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So minus 4x squared. You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out. And now if we want to x minus 4, or 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x. We could actually do it this way as well."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's this expanded out. And now if we want to x minus 4, or 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x. We could actually do it this way as well. But that's just going to be 4 times 4, which is 16. Plus 4 times negative x, which is negative 4x. Negative x times 4, another negative 4x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could actually do it this way as well. But that's just going to be 4 times 4, which is 16. Plus 4 times negative x, which is negative 4x. Negative x times 4, another negative 4x. And then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But we're going to subtract this."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative x times 4, another negative 4x. And then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business. So let's just do it right over here, since we already have it set up. So we have this minus this."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business. So let's just do it right over here, since we already have it set up. So we have this minus this. So we have this minus, we're going to subtract this. Or we could add the negative of it. So we'll put negative x squared plus 8x minus 16."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have this minus this. So we have this minus, we're going to subtract this. Or we could add the negative of it. So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this. And we get, and I'll do this in a new color, let's see, we get x to the fourth minus 4x to the third power minus 5x squared plus 24x. And then these cancel out."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this. And we get, and I'll do this in a new color, let's see, we get x to the fourth minus 4x to the third power minus 5x squared plus 24x. And then these cancel out. So that's what we are left with. And so that's the inside of our integral. So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then these cancel out. So that's what we are left with. And so that's the inside of our integral. So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yep, from 0 to 3 of this dx. And then we had a pi out front here. We had a pi out front here."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yep, from 0 to 3 of this dx. And then we had a pi out front here. We had a pi out front here. So I'll just take that out of the integral. So times pi. Well, now we just have to take the antiderivative."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "We had a pi out front here. So I'll just take that out of the integral. So times pi. Well, now we just have to take the antiderivative. And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going to be minus x to the fourth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, now we just have to take the antiderivative. And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going to be minus x to the fourth. You can verify that. Derivative is 4x. And then you decrement the exponent 4x to the third."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be minus x to the fourth. You can verify that. Derivative is 4x. And then you decrement the exponent 4x to the third. So that works out. And then the antiderivative of this is negative 5 thirds x to the third. Just incremented the exponent and divided by that."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you decrement the exponent 4x to the third. So that works out. And then the antiderivative of this is negative 5 thirds x to the third. Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my opening and closing parentheses."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my opening and closing parentheses. We're going to evaluate that. Actually, let me do it as brackets. We're going to evaluate that from 0 to 3."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "I like to match colors for my opening and closing parentheses. We're going to evaluate that. Actually, let me do it as brackets. We're going to evaluate that from 0 to 3. So this is going to be equal to pi times, let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to evaluate that from 0 to 3. So this is going to be equal to pi times, let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see. 3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to, this times 3 is 243."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see. 3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to, this times 3 is 243. So it's going to be 243. It's going to be some hairy math. 243 over 5 minus, well, 3 to the fourth is 81."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "3 to the fifth is going to be equal to, this times 3 is 243. So it's going to be 243. It's going to be some hairy math. 243 over 5 minus, well, 3 to the fourth is 81. Let's see. We're going to have 3 to the third. So it's going to be minus 5 over 3 times 3 to the third power."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "243 over 5 minus, well, 3 to the fourth is 81. Let's see. We're going to have 3 to the third. So it's going to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. So let's just simplify that. So this is going to be equal to 45. Did I do that right?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "9 times 5 is 45. So let's just simplify that. So this is going to be equal to 45. Did I do that right? Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Did I do that right? Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9. 9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, 3 squared is 9. 9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0. But lucky for us, that's pretty simple. This evaluated at 0, 0, 0, 0, 0."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0. But lucky for us, that's pretty simple. This evaluated at 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies things a good bit. So now we just have to do some hairy fraction arithmetic. So let's do it."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "This evaluated at 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies things a good bit. So now we just have to do some hairy fraction arithmetic. So let's do it. So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see what we have."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do it. So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see what we have. We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see what we have. We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18. Did I do that right? So this is negative 126."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18. Did I do that right? So this is negative 126. And then negative 126, yes, it will be negative 18. So now we just have to write negative 18 over 5. Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is negative 126. And then negative 126, yes, it will be negative 18. So now we just have to write negative 18 over 5. Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5. So this whole thing has simplified to equal to pi times 243 over 5 minus 90 over 5, which is equal to pi. I deserve a drum roll now."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5. So this whole thing has simplified to equal to pi times 243 over 5 minus 90 over 5, which is equal to pi. I deserve a drum roll now. This was some hairy mathematics. So 243 minus 90 is going to be 153 over 5. Or we can write this as 153 pi over 5."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's write down a differential equation. The derivative of y with respect to x is equal to four y over x. And what we'll see in this video is the solution to a differential equation isn't a value or a set of values. It's a function or a set of functions. But before we go about actually trying to solve this or figure out all of the solutions, let's test whether certain equations, certain functions are solutions to this differential equation. So for example, if I have y is equal to four x, is this a solution to this differential equation? Pause the video and see if you can figure it out."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's a function or a set of functions. But before we go about actually trying to solve this or figure out all of the solutions, let's test whether certain equations, certain functions are solutions to this differential equation. So for example, if I have y is equal to four x, is this a solution to this differential equation? Pause the video and see if you can figure it out. Well, to see if this is a solution, what we have to do is figure out the derivative of y with respect to x, and see is that truly equal to four times y over x. And I'm gonna try to express everything in terms of x to see if I really have an equality there. So first, let's figure out the derivative of y with respect to x."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and see if you can figure it out. Well, to see if this is a solution, what we have to do is figure out the derivative of y with respect to x, and see is that truly equal to four times y over x. And I'm gonna try to express everything in terms of x to see if I really have an equality there. So first, let's figure out the derivative of y with respect to x. Well, that's just going to be equal to four. We've seen that many times before. And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put everything in terms of x."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So first, let's figure out the derivative of y with respect to x. Well, that's just going to be equal to four. We've seen that many times before. And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put everything in terms of x. So y is equal to four x. So instead of four y, I could write four times four x. All of that over x."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put everything in terms of x. So y is equal to four x. So instead of four y, I could write four times four x. All of that over x. Is this true? Well, that x cancels with that, and I'm gonna get four is equal to 16, which it clearly is not. And so this is not a solution."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that over x. Is this true? Well, that x cancels with that, and I'm gonna get four is equal to 16, which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to x to the fourth power?"}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well, we're going to do the same thing. What's the derivative of y with respect to x?"}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well, we're going to do the same thing. What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y. Instead of writing a y, I'm gonna write it all in terms of x."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y. Instead of writing a y, I'm gonna write it all in terms of x. So is that equal to four times x to the fourth, because x to the fourth is the same thing as y, divided by x. And so let's see, x to the fourth divided by x, that is going to be x to the third. And so you will indeed get four x to the third is equal to four x to the third."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of writing a y, I'm gonna write it all in terms of x. So is that equal to four times x to the fourth, because x to the fourth is the same thing as y, divided by x. And so let's see, x to the fourth divided by x, that is going to be x to the third. And so you will indeed get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It's not necessarily the only solution, but it is a solution to that differential equation."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you will indeed get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equal to two x."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this differential equation? Pause the video again and see if you can figure it out. Well, to figure that out, you have to say, well, what is f prime of x?"}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this differential equation? Pause the video again and see if you can figure it out. Well, to figure that out, you have to say, well, what is f prime of x? F prime of x is just going to be equal to two. And then test the equality. Is two, is f prime of x equal to f of x, which is two x, minus x, minus x."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to figure that out, you have to say, well, what is f prime of x? F prime of x is just going to be equal to two. And then test the equality. Is two, is f prime of x equal to f of x, which is two x, minus x, minus x. And so let's see, we are going to get two is equal to x. So you might be tempted to say, oh, hey, I just solved for x or something like that. But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is two, is f prime of x equal to f of x, which is two x, minus x, minus x. And so let's see, we are going to get two is equal to x. So you might be tempted to say, oh, hey, I just solved for x or something like that. But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put an incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let's look at another one."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put an incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let's look at another one. Let's say that we have f of x is equal to x plus one. Pause the video and see, is this a solution to our differential equation? Well, same drill."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's look at another one. Let's say that we have f of x is equal to x plus one. Pause the video and see, is this a solution to our differential equation? Well, same drill. f prime of x is going to be equal to one. And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, same drill. f prime of x is going to be equal to one. And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these. Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these. Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do it in a red color. Let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation. Pause the video again and see if you can figure it out."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do it in a red color. Let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation. Pause the video again and see if you can figure it out. All right, well, let's figure out the derivative here. f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing, and so, and then plus one, and the derivative of this with respect to x is just zero. And then let's substitute this into our original differential equation."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video again and see if you can figure it out. All right, well, let's figure out the derivative here. f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing, and so, and then plus one, and the derivative of this with respect to x is just zero. And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x plus one minus x, minus x? And if that x cancels out with that x, it is indeed."}, {"video_title": "Verifying solutions to differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x plus one minus x, minus x? And if that x cancels out with that x, it is indeed. They are indeed equal. So this is also a solution. So this, this is a solution."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cross sections of the solid perpendicular to the x-axis are rectangles whose height is x. Express the volume of the solid with a definite integral. So pause this video and see if you can have a go at that. All right, now what's interesting about this is they've just given us the equations for the graphs, but we haven't visualized them yet. And we need to visualize them, or at least I like to visualize them, so I can think about this region that they're talking about. So maybe a first thing to do is think about, well, where do these two lines intersect? So when do we have the same y value?"}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now what's interesting about this is they've just given us the equations for the graphs, but we haven't visualized them yet. And we need to visualize them, or at least I like to visualize them, so I can think about this region that they're talking about. So maybe a first thing to do is think about, well, where do these two lines intersect? So when do we have the same y value? Or another way to think about it is when does this thing equal four? So if we set them equal to each other, we have negative x squared plus six x minus one is equal to four. This will give us the x values where these two lines intersect."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when do we have the same y value? Or another way to think about it is when does this thing equal four? So if we set them equal to each other, we have negative x squared plus six x minus one is equal to four. This will give us the x values where these two lines intersect. And so we will get, if we wanna solve for x, we can subtract four from both sides. Negative x squared plus six x minus five is equal to zero. We can multiply both sides by negative one."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This will give us the x values where these two lines intersect. And so we will get, if we wanna solve for x, we can subtract four from both sides. Negative x squared plus six x minus five is equal to zero. We can multiply both sides by negative one. We will get x squared minus six x plus five is equal to zero and then this is pretty straightforward to factor. One times five is five, or actually I say negative one times negative five is five, and negative one plus negative five is negative six. So it's going to be x minus one times x minus five is equal to zero."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can multiply both sides by negative one. We will get x squared minus six x plus five is equal to zero and then this is pretty straightforward to factor. One times five is five, or actually I say negative one times negative five is five, and negative one plus negative five is negative six. So it's going to be x minus one times x minus five is equal to zero. And so these intersect when x is equal to one or x is equal to five. Since we have a negative out front of the second degree term right over here, we know it's going to be a downward opening parabola. And we know that we intersect y equals four when x is equal to one and x equals five."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be x minus one times x minus five is equal to zero. And so these intersect when x is equal to one or x is equal to five. Since we have a negative out front of the second degree term right over here, we know it's going to be a downward opening parabola. And we know that we intersect y equals four when x is equal to one and x equals five. And so the vertex must be right in between them. So the vertex is going to be at x equals three. So let's actually visualize this a little bit."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know that we intersect y equals four when x is equal to one and x equals five. And so the vertex must be right in between them. So the vertex is going to be at x equals three. So let's actually visualize this a little bit. So it's going to look something like this. I'll draw it with some perspective because we're gonna have to think about a three-dimensional shape. So that's our y-axis."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's actually visualize this a little bit. So it's going to look something like this. I'll draw it with some perspective because we're gonna have to think about a three-dimensional shape. So that's our y-axis. This is our x-axis. And let me draw some y values. So one, two, three, four, five, six, seven, eight."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's our y-axis. This is our x-axis. And let me draw some y values. So one, two, three, four, five, six, seven, eight. This is probably sufficient. Now we have y is equal to four, which is going to look something like this. So that is y is equal to four."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one, two, three, four, five, six, seven, eight. This is probably sufficient. Now we have y is equal to four, which is going to look something like this. So that is y is equal to four. And then we have y is equal to negative x squared plus six x minus one, which we know intersects y equals four at x equals one or x equals five. So let's see, one, two, three, four, five. So x equals one."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is y is equal to four. And then we have y is equal to negative x squared plus six x minus one, which we know intersects y equals four at x equals one or x equals five. So let's see, one, two, three, four, five. So x equals one. So we have that point right over there. One comma four, and then we have five comma four. And then we know the vertex is when x is equal to three."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So x equals one. So we have that point right over there. One comma four, and then we have five comma four. And then we know the vertex is when x is equal to three. So it might look something like this. We could substitute three back in here. So let's see, y will equal to negative nine, three squared, plus 18 minus one."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we know the vertex is when x is equal to three. So it might look something like this. We could substitute three back in here. So let's see, y will equal to negative nine, three squared, plus 18 minus one. And so what is that going to be? That's going to be y is going to be equal to eight. So we have the point three comma eight."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, y will equal to negative nine, three squared, plus 18 minus one. And so what is that going to be? That's going to be y is going to be equal to eight. So we have the point three comma eight. So this is five, six, seven, eight. Yep, right about there. And so we are dealing with a situation, we are dealing with a situation that looks something like this."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have the point three comma eight. So this is five, six, seven, eight. Yep, right about there. And so we are dealing with a situation, we are dealing with a situation that looks something like this. This is the region in question. So that's going to be the base of our solid. And they say cross sections of the solid perpendicular to the x-axis."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we are dealing with a situation, we are dealing with a situation that looks something like this. This is the region in question. So that's going to be the base of our solid. And they say cross sections of the solid perpendicular to the x-axis. So let me draw one of those cross sections. So this is a cross section perpendicular to the x-axis. Our rectangles whose height is x."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they say cross sections of the solid perpendicular to the x-axis. So let me draw one of those cross sections. So this is a cross section perpendicular to the x-axis. Our rectangles whose height is x. So this is going to have height x right over here. Height x. Now what is this, the width, I guess we could say, of this rectangle?"}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our rectangles whose height is x. So this is going to have height x right over here. Height x. Now what is this, the width, I guess we could say, of this rectangle? Well, it's going to be the difference between these two functions. It's going to be this upper function minus this lower function. So it's going to be, that right over there is going to be negative x squared plus six x minus one."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is this, the width, I guess we could say, of this rectangle? Well, it's going to be the difference between these two functions. It's going to be this upper function minus this lower function. So it's going to be, that right over there is going to be negative x squared plus six x minus one. And then minus four, minus the lower function. So that could be simplified as negative x squared plus six x minus five. And so if we wanna figure out the volume of this little section right over here, we'd multiply x times this."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be, that right over there is going to be negative x squared plus six x minus one. And then minus four, minus the lower function. So that could be simplified as negative x squared plus six x minus five. And so if we wanna figure out the volume of this little section right over here, we'd multiply x times this. And then we would multiply that times an infinitesimally small depth, a dx. And then we can just integrate from x equals one to x equals five. So let's do that."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we wanna figure out the volume of this little section right over here, we'd multiply x times this. And then we would multiply that times an infinitesimally small depth, a dx. And then we can just integrate from x equals one to x equals five. So let's do that. The volume of just this little slice over here is going to be the base, which is negative x squared plus six x minus five times the height, times x, times the depth, times dx. And then what we wanna do is we wanna sum up all of these. And so you could imagine right over here, you would have, or like right over here, you would have a cross section that looks like this."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. The volume of just this little slice over here is going to be the base, which is negative x squared plus six x minus five times the height, times x, times the depth, times dx. And then what we wanna do is we wanna sum up all of these. And so you could imagine right over here, you would have, or like right over here, you would have a cross section that looks like this. X is now much larger. The height is x. So now it looks something like this."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you could imagine right over here, you would have, or like right over here, you would have a cross section that looks like this. X is now much larger. The height is x. So now it looks something like this. So I'm just drawing two cross sections just so you get the idea. So these are the, this is any one cross section for a given x. But now we wanna integrate."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now it looks something like this. So I'm just drawing two cross sections just so you get the idea. So these are the, this is any one cross section for a given x. But now we wanna integrate. Our x is going from x equals one to x equals five. X equals one to x equals five. And there you have it."}, {"video_title": "Volume with cross sections squares and rectangles (no graph)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now we wanna integrate. Our x is going from x equals one to x equals five. X equals one to x equals five. And there you have it. We have expressed the volume of that solid as a definite integral. And it's worth noting that this definite integral, if you distribute this x, if you multiply it by all of these terms, it's very solvable. You don't need a, or it's very solvable without a calculator."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's try to see if we can visualize that. So this is my y-axis. That's my y-axis. This right over here. I'm going to make it my t-axis. We'll use x a little bit later. So I'll call this my t-axis."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here. I'm going to make it my t-axis. We'll use x a little bit later. So I'll call this my t-axis. And then let's say that this right over here is the graph of y is equal to f of t. y is equal to f of t. And we're saying it's continuous on the interval from a to b. So this is t is equal to a. This is t is equal to b."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll call this my t-axis. And then let's say that this right over here is the graph of y is equal to f of t. y is equal to f of t. And we're saying it's continuous on the interval from a to b. So this is t is equal to a. This is t is equal to b. So we're saying that it is continuous. It is continuous over this whole interval. Now, for fun, let's define a function capital F of x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is t is equal to b. So we're saying that it is continuous. It is continuous over this whole interval. Now, for fun, let's define a function capital F of x. And I will do it in blue. Let's define capital F of x as equal to the definite integral from a, from a as a lower bound, to x of f of t, of f of t, dt, where x is in this interval. Where a is less than or equal to x is less than or equal to b."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, for fun, let's define a function capital F of x. And I will do it in blue. Let's define capital F of x as equal to the definite integral from a, from a as a lower bound, to x of f of t, of f of t, dt, where x is in this interval. Where a is less than or equal to x is less than or equal to b. Or that's just another way of saying that x is in this interval right over here. Now, when you see this, you might say, oh, you know, the definite integral, this has to do with differentiation and antiderivatives and all that. But we don't know that yet."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Where a is less than or equal to x is less than or equal to b. Or that's just another way of saying that x is in this interval right over here. Now, when you see this, you might say, oh, you know, the definite integral, this has to do with differentiation and antiderivatives and all that. But we don't know that yet. All we know right now is that this is the area under the curve F between a and x. So between a and let's say this right over here, this right over here is x. So f of x is just this area right over here."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we don't know that yet. All we know right now is that this is the area under the curve F between a and x. So between a and let's say this right over here, this right over here is x. So f of x is just this area right over here. That's all we know about it. We don't know it has anything to do with antiderivatives just yet. That's what we're going to try to prove in this video."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of x is just this area right over here. That's all we know about it. We don't know it has anything to do with antiderivatives just yet. That's what we're going to try to prove in this video. So just for fun, let's take the derivative of f. And we're going to do it just using the definition of derivatives and see what we get when we take the derivative using the definition of derivatives. So we would get the derivative f prime of x. Well, this definition of derivatives, it's a limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's what we're going to try to prove in this video. So just for fun, let's take the derivative of f. And we're going to do it just using the definition of derivatives and see what we get when we take the derivative using the definition of derivatives. So we would get the derivative f prime of x. Well, this definition of derivatives, it's a limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x. This is just the definition of the derivative. Now, what is this equal to? Well, let me rewrite it using these integrals right up here."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this definition of derivatives, it's a limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x. This is just the definition of the derivative. Now, what is this equal to? Well, let me rewrite it using these integrals right up here. This is going to be equal to the limit as delta x approaches 0 of what's f of x plus delta x? Well, put x in right over here. You're going to get the definite integral from a to x plus delta x of f of t dt."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let me rewrite it using these integrals right up here. This is going to be equal to the limit as delta x approaches 0 of what's f of x plus delta x? Well, put x in right over here. You're going to get the definite integral from a to x plus delta x of f of t dt. And then from that, you are going to subtract this business, f of x, which we've already written as the definite integral from a to x of f of t dt. And then all of that is over delta x. Now, what does this represent?"}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're going to get the definite integral from a to x plus delta x of f of t dt. And then from that, you are going to subtract this business, f of x, which we've already written as the definite integral from a to x of f of t dt. And then all of that is over delta x. Now, what does this represent? Remember, we don't know anything about definite integrals or somehow dealing with something with an antiderivative and all that. We just know this is another way of saying the area under the curve f between a and x plus delta x. So it's the area under the curve f between a and x plus delta x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, what does this represent? Remember, we don't know anything about definite integrals or somehow dealing with something with an antiderivative and all that. We just know this is another way of saying the area under the curve f between a and x plus delta x. So it's the area under the curve f between a and x plus delta x. So it's this entire area right over here. So that's this part. We already know what this blue stuff is."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's the area under the curve f between a and x plus delta x. So it's this entire area right over here. So that's this part. We already know what this blue stuff is. This blue stuff, that same shade of blue, so this blue stuff right over here, this is equal to all of this business. We've already shaded this in. It's equal to all of this business right over here."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already know what this blue stuff is. This blue stuff, that same shade of blue, so this blue stuff right over here, this is equal to all of this business. We've already shaded this in. It's equal to all of this business right over here. So if you were to take all of this green area, which is from a to x plus delta x, and subtract out this blue area, which is exactly what we're doing in the numerator, what are you left with? Well, you're going to be left with what color have I not used yet? Maybe I will use this pink color."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's equal to all of this business right over here. So if you were to take all of this green area, which is from a to x plus delta x, and subtract out this blue area, which is exactly what we're doing in the numerator, what are you left with? Well, you're going to be left with what color have I not used yet? Maybe I will use this pink color. Well, no, I already used that. I'll use this purple color. You're going to be left with this area right over here."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe I will use this pink color. Well, no, I already used that. I'll use this purple color. You're going to be left with this area right over here. So what's another way of writing that? Well, another way of writing this area right over here is the definite integral between x and x plus delta x of f of t dt. So we can rewrite this entire expression, the derivative of capital F of x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're going to be left with this area right over here. So what's another way of writing that? Well, another way of writing this area right over here is the definite integral between x and x plus delta x of f of t dt. So we can rewrite this entire expression, the derivative of capital F of x. This is capital F prime of x. We can rewrite it now as being equal to the limit as delta x approaches 0. This I can write as 1 over delta x times the numerator."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can rewrite this entire expression, the derivative of capital F of x. This is capital F prime of x. We can rewrite it now as being equal to the limit as delta x approaches 0. This I can write as 1 over delta x times the numerator. We already figured out the numerator. The green area minus the blue area is just the purple area. And another way of denoting that area is this expression right over here."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "This I can write as 1 over delta x times the numerator. We already figured out the numerator. The green area minus the blue area is just the purple area. And another way of denoting that area is this expression right over here. So 1 over delta x times the definite integral from x to x plus delta x of f of t dt. Now, this expression is interesting. This might look familiar from the mean value theorem of definite integrals."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And another way of denoting that area is this expression right over here. So 1 over delta x times the definite integral from x to x plus delta x of f of t dt. Now, this expression is interesting. This might look familiar from the mean value theorem of definite integrals. The mean value theorem of definite integrals tells us, the mean value theorem of definite integrals tells us, there exists a c in the interval. So I could see where, I'll write it this way, where a is less than or equal to c, which is less than, or actually, let me make it clear, the interval that we now care about is between x and x plus delta x. Where x is less than or equal to c, which is less than or equal to x plus delta x, such that the function evaluated at c, so let me draw this c. So there's a c someplace over here."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "This might look familiar from the mean value theorem of definite integrals. The mean value theorem of definite integrals tells us, the mean value theorem of definite integrals tells us, there exists a c in the interval. So I could see where, I'll write it this way, where a is less than or equal to c, which is less than, or actually, let me make it clear, the interval that we now care about is between x and x plus delta x. Where x is less than or equal to c, which is less than or equal to x plus delta x, such that the function evaluated at c, so let me draw this c. So there's a c someplace over here. So if I were to take the function evaluated at the c, that's f of c right over here. So if I were to take the function evaluated at the c, which would essentially be the height of this line, and I multiply it times the base, this interval, if I multiply it times the interval, and this interval is just delta x, x plus delta x minus x is just delta x. So if we just multiply the height times the base, that this is going to be equal to the area under the curve, which is the definite integral from x to x plus delta x, x to x plus delta x of f of t dt."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Where x is less than or equal to c, which is less than or equal to x plus delta x, such that the function evaluated at c, so let me draw this c. So there's a c someplace over here. So if I were to take the function evaluated at the c, that's f of c right over here. So if I were to take the function evaluated at the c, which would essentially be the height of this line, and I multiply it times the base, this interval, if I multiply it times the interval, and this interval is just delta x, x plus delta x minus x is just delta x. So if we just multiply the height times the base, that this is going to be equal to the area under the curve, which is the definite integral from x to x plus delta x, x to x plus delta x of f of t dt. This is what the mean value theorem of integrals tells us. If f is a continuous function, there exists a c in this interval between our two endpoints, where the function evaluated at the c is essentially, you can view it as the mean height. And if you take that mean value of the function and you multiply it times the base, you're going to get the area of the curve."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we just multiply the height times the base, that this is going to be equal to the area under the curve, which is the definite integral from x to x plus delta x, x to x plus delta x of f of t dt. This is what the mean value theorem of integrals tells us. If f is a continuous function, there exists a c in this interval between our two endpoints, where the function evaluated at the c is essentially, you can view it as the mean height. And if you take that mean value of the function and you multiply it times the base, you're going to get the area of the curve. Or another way of rewriting this, you could say that f of c, there exists a c in that interval where f of c is equal to 1 over delta x. I'm just dividing both sides by delta x, times the definite integral from x to x plus delta x of f of t dt. And this is often viewed as the mean value of the function over the interval. Why is that?"}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you take that mean value of the function and you multiply it times the base, you're going to get the area of the curve. Or another way of rewriting this, you could say that f of c, there exists a c in that interval where f of c is equal to 1 over delta x. I'm just dividing both sides by delta x, times the definite integral from x to x plus delta x of f of t dt. And this is often viewed as the mean value of the function over the interval. Why is that? Well, this part right over here gives you the area, and then you divide the area by the base, and you get the mean height. Or another way you could say it is, if you were to take the height right over here, multiply it times the base, you get a rectangle that has the exact same area as the area under the curve. Well, this is useful, because this is exactly what we got as the derivative of f prime of x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Why is that? Well, this part right over here gives you the area, and then you divide the area by the base, and you get the mean height. Or another way you could say it is, if you were to take the height right over here, multiply it times the base, you get a rectangle that has the exact same area as the area under the curve. Well, this is useful, because this is exactly what we got as the derivative of f prime of x. So there must exist a c such that f of c is equal to this stuff, or we could say that the limit, and let me rewrite all of this down a new color. So there exists a c in the interval x to x plus delta x, where f prime of x, which we know is equal to this, we can now say is now equal to the limit as delta x approaches 0. And instead of writing this, we know that there's some c that's equal to all of this business, of f of c. Now we're in the home stretch."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is useful, because this is exactly what we got as the derivative of f prime of x. So there must exist a c such that f of c is equal to this stuff, or we could say that the limit, and let me rewrite all of this down a new color. So there exists a c in the interval x to x plus delta x, where f prime of x, which we know is equal to this, we can now say is now equal to the limit as delta x approaches 0. And instead of writing this, we know that there's some c that's equal to all of this business, of f of c. Now we're in the home stretch. We just have to figure out what the limit as delta x approaches 0 of f of c is. And the main realization is this part right over here. We know that c is always sandwiched in between x and x plus delta x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And instead of writing this, we know that there's some c that's equal to all of this business, of f of c. Now we're in the home stretch. We just have to figure out what the limit as delta x approaches 0 of f of c is. And the main realization is this part right over here. We know that c is always sandwiched in between x and x plus delta x. And intuitively, you could tell that, look, as delta x approaches 0, as this green line right over here moves more and more to the left, as it approaches this blue line, the c has to be in between. And so the c is going to approach x. So we know intuitively that c approaches x as delta x approaches 0."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that c is always sandwiched in between x and x plus delta x. And intuitively, you could tell that, look, as delta x approaches 0, as this green line right over here moves more and more to the left, as it approaches this blue line, the c has to be in between. And so the c is going to approach x. So we know intuitively that c approaches x as delta x approaches 0. Or another way of saying it is that f of c is going to approach f of x as delta x approaches 0. And so intuitively, we could say that this is going to be equal to f of x. Now you might say, OK, that's intuitively, but we're kind of working on a little bit of a proof here, Sal."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know intuitively that c approaches x as delta x approaches 0. Or another way of saying it is that f of c is going to approach f of x as delta x approaches 0. And so intuitively, we could say that this is going to be equal to f of x. Now you might say, OK, that's intuitively, but we're kind of working on a little bit of a proof here, Sal. Let me know for sure that x is going to approach c. Don't just do this little thing where you drew this diagram and it makes sense that c is going to have to get closer and closer to x. And if you want that, you could just resort to the squeeze theorem. And to resort to the squeeze theorem, you just have to view c as a function of delta x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now you might say, OK, that's intuitively, but we're kind of working on a little bit of a proof here, Sal. Let me know for sure that x is going to approach c. Don't just do this little thing where you drew this diagram and it makes sense that c is going to have to get closer and closer to x. And if you want that, you could just resort to the squeeze theorem. And to resort to the squeeze theorem, you just have to view c as a function of delta x. And it really is. Depending on your delta x, c is going to be further to the left or to the right, possibly. And so I can just rewrite this expression as x is less than or equal to c as a function of delta x, which is less than or equal to x plus delta x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to resort to the squeeze theorem, you just have to view c as a function of delta x. And it really is. Depending on your delta x, c is going to be further to the left or to the right, possibly. And so I can just rewrite this expression as x is less than or equal to c as a function of delta x, which is less than or equal to x plus delta x. So now you see that c is always sandwiched between x and x plus delta x. Well what's the limit of x as delta x approaches 0? Well x isn't dependent on delta x in any way."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so I can just rewrite this expression as x is less than or equal to c as a function of delta x, which is less than or equal to x plus delta x. So now you see that c is always sandwiched between x and x plus delta x. Well what's the limit of x as delta x approaches 0? Well x isn't dependent on delta x in any way. So this is just going to be equal to x. What's the limit of x plus delta x as delta x approaches 0? Well as delta x approaches 0, this is just going to be equal to x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well x isn't dependent on delta x in any way. So this is just going to be equal to x. What's the limit of x plus delta x as delta x approaches 0? Well as delta x approaches 0, this is just going to be equal to x. So if this approaches x as delta x approaches 0, and it's less than this function, and if this approaches x as delta x approaches 0, and it's always greater than this, then we know from the squeeze theorem or the sandwich theorem that the limit as delta x approaches 0 of c as a function of delta x is going to be equal to x as well. It has to approach the same thing that that and that is. It's sandwiched in between."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well as delta x approaches 0, this is just going to be equal to x. So if this approaches x as delta x approaches 0, and it's less than this function, and if this approaches x as delta x approaches 0, and it's always greater than this, then we know from the squeeze theorem or the sandwich theorem that the limit as delta x approaches 0 of c as a function of delta x is going to be equal to x as well. It has to approach the same thing that that and that is. It's sandwiched in between. And so that's a slight, we resort to the sandwich theorem. It's a little bit more rigorous to get to this exact result. As delta x approaches 0, c approaches x."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's sandwiched in between. And so that's a slight, we resort to the sandwich theorem. It's a little bit more rigorous to get to this exact result. As delta x approaches 0, c approaches x. If c is approaching x, then f of c is going to approach f of x. And then we essentially have our proof. f is a continuous function."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "As delta x approaches 0, c approaches x. If c is approaching x, then f of c is going to approach f of x. And then we essentially have our proof. f is a continuous function. We defined f in this way, capital F in this way. And we were able to use just the definition of the derivative to figure out that the derivative of capital F of x is equal to f of x. And once again, why is this a big deal?"}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "f is a continuous function. We defined f in this way, capital F in this way. And we were able to use just the definition of the derivative to figure out that the derivative of capital F of x is equal to f of x. And once again, why is this a big deal? Well, it tells you that if you have any continuous function f, and that's what we assumed. We assumed that f is continuous over the interval. There exists some function."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, why is this a big deal? Well, it tells you that if you have any continuous function f, and that's what we assumed. We assumed that f is continuous over the interval. There exists some function. There exists a function. You can just define the function this way as the area under the curve between some end point or the beginning of the interval and some x. If you define a function in that way, the derivative of this function is going to be equal to your continuous function."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "There exists some function. There exists a function. You can just define the function this way as the area under the curve between some end point or the beginning of the interval and some x. If you define a function in that way, the derivative of this function is going to be equal to your continuous function. Or another way of saying it is that you always have an antiderivative, that any continuous function has an antiderivative. And so it's a couple of cool things. Any continuous function has an antiderivative."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you define a function in that way, the derivative of this function is going to be equal to your continuous function. Or another way of saying it is that you always have an antiderivative, that any continuous function has an antiderivative. And so it's a couple of cool things. Any continuous function has an antiderivative. It's going to be that capital F of x. And this is why it's called the fundamental theorem of calculus. It ties together these two ideas."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Any continuous function has an antiderivative. It's going to be that capital F of x. And this is why it's called the fundamental theorem of calculus. It ties together these two ideas. And you have differential calculus. You have the idea of a derivative. And then in integral calculus, you have the idea of an integral."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "It ties together these two ideas. And you have differential calculus. You have the idea of a derivative. And then in integral calculus, you have the idea of an integral. Before this proof, all we viewed an integral as is the area under the curve. It was just literally a notation to say the area under the curve. But now we've been able to make a connection, that there's a connection between the integral and the derivative, or a connection between the integral and the antiderivative in particular."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then in integral calculus, you have the idea of an integral. Before this proof, all we viewed an integral as is the area under the curve. It was just literally a notation to say the area under the curve. But now we've been able to make a connection, that there's a connection between the integral and the derivative, or a connection between the integral and the antiderivative in particular. So it connects all of calculus together in a very, very, very powerful. And we're so used to it now. And now we can say almost a somewhat obvious way."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now we've been able to make a connection, that there's a connection between the integral and the derivative, or a connection between the integral and the antiderivative in particular. So it connects all of calculus together in a very, very, very powerful. And we're so used to it now. And now we can say almost a somewhat obvious way. But it wasn't obvious. Remember, we always think of integrals as somehow doing an antiderivative. But it wasn't clear."}, {"video_title": "Proof of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we can say almost a somewhat obvious way. But it wasn't obvious. Remember, we always think of integrals as somehow doing an antiderivative. But it wasn't clear. If you just viewed an integral as only an area, you would have to go through this process and say, wow, no. It's connected. It's connected to the process of taking a derivative."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that the derivative with respect to x of cosine of x is equal to negative sine of x. And so what we want to do in this video is find the derivatives of the other basic trig functions. So in particular, we know, let's figure out what the derivative with respect to x, let's first do tangent of x. Tangent of x. Well, this is the same thing as trying to find the derivative with respect to x of, well, tangent of x is just sine of x. Sine of x over cosine of x. And since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this or to figure out what this is going to be. The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x, times the bottom function, which is cosine of x, so times cosine of x, minus the top function, which is sine of x, the top function, which is sine of x, sine of x, times the derivative of the bottom function, so the derivative of cosine of x is negative sine of x."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is the same thing as trying to find the derivative with respect to x of, well, tangent of x is just sine of x. Sine of x over cosine of x. And since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this or to figure out what this is going to be. The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x, times the bottom function, which is cosine of x, so times cosine of x, minus the top function, which is sine of x, the top function, which is sine of x, sine of x, times the derivative of the bottom function, so the derivative of cosine of x is negative sine of x. So I could put the sine of x there, but where the negative can just cancel that out, and it's going to be over, over the bottom function squared. So cosine squared of x. Now what is this?"}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x, times the bottom function, which is cosine of x, so times cosine of x, minus the top function, which is sine of x, the top function, which is sine of x, sine of x, times the derivative of the bottom function, so the derivative of cosine of x is negative sine of x. So I could put the sine of x there, but where the negative can just cancel that out, and it's going to be over, over the bottom function squared. So cosine squared of x. Now what is this? Well, what we have here, this is just cosine squared of x. This is just sine squared of x. And we know from the Pythagorean identity, and this really just comes out of the unit circle definition, that cosine squared of x plus sine squared of x, well that's going to be equal to one for any x."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is this? Well, what we have here, this is just cosine squared of x. This is just sine squared of x. And we know from the Pythagorean identity, and this really just comes out of the unit circle definition, that cosine squared of x plus sine squared of x, well that's going to be equal to one for any x. So all of this is equal to one. And so we end up with one over cosine squared x, which is the same thing as secant of x squared. One over cosine of x is secant, so this is just secant of x squared."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know from the Pythagorean identity, and this really just comes out of the unit circle definition, that cosine squared of x plus sine squared of x, well that's going to be equal to one for any x. So all of this is equal to one. And so we end up with one over cosine squared x, which is the same thing as secant of x squared. One over cosine of x is secant, so this is just secant of x squared. So that was pretty straightforward. Now let's just do the inverse of the, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent. So that was fun, so let's do that."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "One over cosine of x is secant, so this is just secant of x squared. So that was pretty straightforward. Now let's just do the inverse of the, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent. So that was fun, so let's do that. D dx of cotangent, not cosine, of cotangent of x, well, same idea. That's the derivative with respect to x. And this time, let me make some sufficiently large brackets."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that was fun, so let's do that. D dx of cotangent, not cosine, of cotangent of x, well, same idea. That's the derivative with respect to x. And this time, let me make some sufficiently large brackets. So now this is cosine of x over sine of x. Over sine of x. But once again, we can use the quotient rule here."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this time, let me make some sufficiently large brackets. So now this is cosine of x over sine of x. Over sine of x. But once again, we can use the quotient rule here. So this is going to be the derivative of the top function, which is negative, we're doing that magenta color, that is negative sine of x times the bottom function. So times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the derivative of the bottom function, which is just going to be another cosine of x, and then all of that over the bottom function squared. So sine of x squared."}, {"video_title": "Derivatives of tan(x) and cot(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But once again, we can use the quotient rule here. So this is going to be the derivative of the top function, which is negative, we're doing that magenta color, that is negative sine of x times the bottom function. So times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the derivative of the bottom function, which is just going to be another cosine of x, and then all of that over the bottom function squared. So sine of x squared. Now what does this simplify to? Up here, let's see, this is sine squared of x, although we have a negative there, minus cosine squared of x, but we could factor out the negative, and this would be negative sine squared of x plus cosine squared of x. Well, this is just one by the Pythagorean identity, and so this is negative one over sine squared x."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "The result that I hope to show you or give you an intuition for in this video is something that we will use in the proof of the chain rule, or in a proof of the chain rule. Actually, we may do more than one proof of the chain rule. But the result we're gonna look at is if we have some function u, which is a function of x, and we know that it is continuous, it is continuous at x equals c. So if we know this, then that's going to imply that the change in u goes to zero as our change in x in this region around c goes to zero. This is what I want to get an intuition for, that if u is continuous at c, then as our change in x around c gets smaller, or it gets smaller and smaller and smaller as it approaches zero, then our change in u approaches zero as well. So let's just, to think about this, or even kind of prove it to ourselves a little bit more rigorously, let's think about what it means to be continuous at x equals c. Well, the definition of continuity is, so this literally is the same thing as saying that the limit as x approaches c of u of x is equal to u of c. That the limit that our function approaches as x approaches c is equal to the value of the function at c. We don't have a point discontinuity or a jump discontinuity. If we had a jump discontinuity, then the limit wouldn't exist, and we've seen that in previous videos. Now I'm just going to manipulate this algebraically, so it essentially gives us this conclusion right over here."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is what I want to get an intuition for, that if u is continuous at c, then as our change in x around c gets smaller, or it gets smaller and smaller and smaller as it approaches zero, then our change in u approaches zero as well. So let's just, to think about this, or even kind of prove it to ourselves a little bit more rigorously, let's think about what it means to be continuous at x equals c. Well, the definition of continuity is, so this literally is the same thing as saying that the limit as x approaches c of u of x is equal to u of c. That the limit that our function approaches as x approaches c is equal to the value of the function at c. We don't have a point discontinuity or a jump discontinuity. If we had a jump discontinuity, then the limit wouldn't exist, and we've seen that in previous videos. Now I'm just going to manipulate this algebraically, so it essentially gives us this conclusion right over here. So this we can rewrite. It's important to realize that u of c, this is just going to be some value. It looks like I've kind of, maybe this is a function of x or something, but no, this is just going to be some value."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now I'm just going to manipulate this algebraically, so it essentially gives us this conclusion right over here. So this we can rewrite. It's important to realize that u of c, this is just going to be some value. It looks like I've kind of, maybe this is a function of x or something, but no, this is just going to be some value. I've inputted c here, and I've evaluated the function of that, and so this is going to be some number. It could be five or seven or pi or negative one, but it's just going to be some value, some constant. So I can treat it like a constant."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like I've kind of, maybe this is a function of x or something, but no, this is just going to be some value. I've inputted c here, and I've evaluated the function of that, and so this is going to be some number. It could be five or seven or pi or negative one, but it's just going to be some value, some constant. So I can treat it like a constant. So this is going to be the same thing as saying the limit as x approaches c of u of x minus u of c is equal to zero. And actually, in the video where we prove that differentiability implies continuity, we started with this, and we proved that right over there. We showed that these two are equivalent things."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I can treat it like a constant. So this is going to be the same thing as saying the limit as x approaches c of u of x minus u of c is equal to zero. And actually, in the video where we prove that differentiability implies continuity, we started with this, and we proved that right over there. We showed that these two are equivalent things. But hopefully you can even think about the intuition. This is just, if the limit as u of x approaches, the limit of u of x as x approaches c is equal to this, then when you evaluate this limit, the limit as x approaches c, well, this thing is going to approach u of c, because we saw it right up here. u of c minus u of c is indeed going to be equal to zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "We showed that these two are equivalent things. But hopefully you can even think about the intuition. This is just, if the limit as u of x approaches, the limit of u of x as x approaches c is equal to this, then when you evaluate this limit, the limit as x approaches c, well, this thing is going to approach u of c, because we saw it right up here. u of c minus u of c is indeed going to be equal to zero. So hopefully you don't feel like it's too much of a stretch, and you can just subtract u of c from both sides and apply properties of limits, and you can get this result as well. But this is interesting, because this essentially can take us to this, that the idea that as our change in x gets smaller and smaller and smaller, as it approaches zero, then our change in our function is also going to approach zero. Now let's just look and take a, let's just graph this or visualize this to get a sense of that."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "u of c minus u of c is indeed going to be equal to zero. So hopefully you don't feel like it's too much of a stretch, and you can just subtract u of c from both sides and apply properties of limits, and you can get this result as well. But this is interesting, because this essentially can take us to this, that the idea that as our change in x gets smaller and smaller and smaller, as it approaches zero, then our change in our function is also going to approach zero. Now let's just look and take a, let's just graph this or visualize this to get a sense of that. So this is our x-axis. Whoops, that's our x-axis. Let's call that our u-axis maybe."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's just look and take a, let's just graph this or visualize this to get a sense of that. So this is our x-axis. Whoops, that's our x-axis. Let's call that our u-axis maybe. And I did u intentionally, because that's the variable we'll use in our proof of the chain rule video. And let's say this right over here is our function. Let's say this right over here, that is c. This right over here is u of c. u of c. And then let's just take some arbitrary x over here."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's call that our u-axis maybe. And I did u intentionally, because that's the variable we'll use in our proof of the chain rule video. And let's say this right over here is our function. Let's say this right over here, that is c. This right over here is u of c. u of c. And then let's just take some arbitrary x over here. So some arbitrary x, and then this right over here, this right over here is u of x. is u of x. So if we define, if we define our change, if we define, let me do this. If we define our change in, our change in u is equal to u of x minus u of c, which makes sense, because this is our change in u."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say this right over here, that is c. This right over here is u of c. u of c. And then let's just take some arbitrary x over here. So some arbitrary x, and then this right over here, this right over here is u of x. is u of x. So if we define, if we define our change, if we define, let me do this. If we define our change in, our change in u is equal to u of x minus u of c, which makes sense, because this is our change in u. So let's say this is going to be u of x minus u of c. And if we define our change in x is equal to x minus c, which it is in this case, it is x minus c. It is x minus c. Then we can rewrite this limit right over here. Instead of saying the limit as x approaches c, we could write the limit as delta x approaches zero, because if x approaches c, then delta x is going to approach zero. So we could write this, the limit as delta x approaches zero of delta u, of delta u, is going to be equal to zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we define our change in, our change in u is equal to u of x minus u of c, which makes sense, because this is our change in u. So let's say this is going to be u of x minus u of c. And if we define our change in x is equal to x minus c, which it is in this case, it is x minus c. It is x minus c. Then we can rewrite this limit right over here. Instead of saying the limit as x approaches c, we could write the limit as delta x approaches zero, because if x approaches c, then delta x is going to approach zero. So we could write this, the limit as delta x approaches zero of delta u, of delta u, is going to be equal to zero. This, we define this as our change in u, and it is our change in u. So this is equal to zero. So another way of thinking about this is as delta x approaches zero, our change in u, our change in the function, is going to approach zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could write this, the limit as delta x approaches zero of delta u, of delta u, is going to be equal to zero. This, we define this as our change in u, and it is our change in u. So this is equal to zero. So another way of thinking about this is as delta x approaches zero, our change in u, our change in the function, is going to approach zero. So as delta x approaches zero, as delta x approaches zero, delta u approaches zero. And that's what we wrote over here. Delta u approaches zero as delta x approaches zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0   AP Calculus AB   Khan Academy.mp3", "Sentence": "So another way of thinking about this is as delta x approaches zero, our change in u, our change in the function, is going to approach zero. So as delta x approaches zero, as delta x approaches zero, delta u approaches zero. And that's what we wrote over here. Delta u approaches zero as delta x approaches zero. And in a lot of ways, this is hopefully common sense. We're dealing with a continuous function. As you get smaller and smaller, and you could just think of it this way, as you get smaller and smaller changes in x's, as our change in x gets smaller, and smaller, and smaller, and smaller, because it's continuous, and you wouldn't be able to say this for a discontinuous function, but because it's continuous, or you wouldn't be able to say this for some discontinuous functions."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What I would like to explore in this video is to see if we could figure out what the derivative of y is with respect to x. If y is equal to the inverse sine, the inverse sine of x. And like always, I encourage you to pause this video and try to figure this out on your own. And I will give you two hints. First hint is, well, we don't know what the derivative of sine inverse of x is, but we do know what the derivative of the sine of something is. And so maybe if you rearrange this and use some implicit differentiation, maybe you can figure out what dy dx is. Remember, this right over here, this right over here is our goal."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And I will give you two hints. First hint is, well, we don't know what the derivative of sine inverse of x is, but we do know what the derivative of the sine of something is. And so maybe if you rearrange this and use some implicit differentiation, maybe you can figure out what dy dx is. Remember, this right over here, this right over here is our goal. We essentially want to figure out the derivative of this with respect to x. So I'm assuming you've had a go at it, so let's work through this together. So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Remember, this right over here, this right over here is our goal. We essentially want to figure out the derivative of this with respect to x. So I'm assuming you've had a go at it, so let's work through this together. So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x. Sine of y is equal to x. So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We could take the derivative of both sides with respect to x."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x. Sine of y is equal to x. So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We could take the derivative of both sides with respect to x. So derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. Well, what's the derivative of the left-hand side with respect to x going to be? And here we just apply the chain rule."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We could take the derivative of both sides with respect to x. So derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. Well, what's the derivative of the left-hand side with respect to x going to be? And here we just apply the chain rule. It's going to be the derivative of sine of y with respect to y, which is going to be cosine of y times the derivative of y with respect to x. So times dy dx. Times dy dx."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And here we just apply the chain rule. It's going to be the derivative of sine of y with respect to y, which is going to be cosine of y times the derivative of y with respect to x. So times dy dx. Times dy dx. And the right-hand side, what's the derivative of x with respect to x? Well, that's obviously just going to be equal to 1. And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Times dy dx. And the right-hand side, what's the derivative of x with respect to x? Well, that's obviously just going to be equal to 1. And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y. Now, this still isn't that satisfying because I have the derivative in terms of y. So let's see if we can re-express it in terms of x. So how could we do that?"}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y. Now, this still isn't that satisfying because I have the derivative in terms of y. So let's see if we can re-express it in terms of x. So how could we do that? Well, we already know that x is equal to sine of y. Let me rewrite it. We already know that x is equal to sine of y."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So how could we do that? Well, we already know that x is equal to sine of y. Let me rewrite it. We already know that x is equal to sine of y. So if we could rewrite this bottom expression in terms, instead of cosine of y, if we could use our trigonometric identities to rewrite it in terms of sine of y, then we'll be in good shape because x is equal to sine of y. Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We already know that x is equal to sine of y. So if we could rewrite this bottom expression in terms, instead of cosine of y, if we could use our trigonometric identities to rewrite it in terms of sine of y, then we'll be in good shape because x is equal to sine of y. Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1. Or if we want to solve for cosine of y, subtract sine squared of y from both sides, we know that cosine squared of y is equal to 1 minus sine squared of y, or that cosine of y, just take the principal root of both sides, is equal to the principal root of 1 minus sine squared of y. So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. Now why is this useful?"}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1. Or if we want to solve for cosine of y, subtract sine squared of y from both sides, we know that cosine squared of y is equal to 1 minus sine squared of y, or that cosine of y, just take the principal root of both sides, is equal to the principal root of 1 minus sine squared of y. So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. Now why is this useful? Well, sine of y is just x. So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. I could write it as sine y squared."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Now why is this useful? Well, sine of y is just x. So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. I could write it as sine y squared. We know that this thing right over here is x. So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. So 1 minus x squared."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I could write it as sine y squared. We know that this thing right over here is x. So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. So 1 minus x squared. So there you have it. The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. So let me just make that very clear."}, {"video_title": "Derivative of inverse sine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So 1 minus x squared. So there you have it. The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. So let me just make that very clear. If you were to take the derivative with respect to x of both sides of this, you would get dy dx is equal to this on the right-hand side. Or we could say the derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. Now you could always reprove this if your memory starts to fail you, and actually that is the best way to really internalize this."}, {"video_title": "Worked example range of solution curve from slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "If the initial condition is zero comma six, what is the range of the solution curve, y is equal to f of x, for x is greater than or equal to zero? So we have a slope field here for a differential equation, and we're saying, okay, if we have a solution where the initial condition is zero comma six, so zero comma six is part of that solution. So, let's see, zero comma six, so this is part of the solution, and we want to know the range of the solution curve. So the solution curve, you can eyeball a little bit by looking at the slope field. So as x, remember, x is gonna be greater than or equal to zero, so it's going to include this point right over here, and as x increases, you can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing at a slower and slower rate, and it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger and larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y values that this is going to take on, y is going to be greater than four."}, {"video_title": "Worked example range of solution curve from slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the solution curve, you can eyeball a little bit by looking at the slope field. So as x, remember, x is gonna be greater than or equal to zero, so it's going to include this point right over here, and as x increases, you can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing at a slower and slower rate, and it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger and larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y values that this is going to take on, y is going to be greater than four. It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on. Another way I could have written this is four is less than y, is less than or equal to six."}, {"video_title": "Worked example range of solution curve from slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the range, the y values that this is going to take on, y is going to be greater than four. It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on. Another way I could have written this is four is less than y, is less than or equal to six. Either way, this is a way of describing the range, the y values that the solution will take on for x being greater than or equal to zero. If they said for all x's, well, then you might have been able to go back this way and keep going, but they're saying the range of the solution curve for x is greater than or equal to zero so we won't consider those values of x less than zero. So there you go."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So let's just say you wanted to find a sum of some terms. And these terms have a pattern. So let's say you want to find a sum of the first 10 numbers. You could say 1 plus 2 plus 3 plus, and you go all the way to plus 9 plus 10. And I clearly could have even written this whole thing out. But you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers. So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "You could say 1 plus 2 plus 3 plus, and you go all the way to plus 9 plus 10. And I clearly could have even written this whole thing out. But you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers. So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100. So mathematicians said, well, let's find some notation instead of having to do this dot, dot, dot thing, which you will see sometimes done, so that we can more cleanly express these types of sums. And that's where sigma notation comes from. So this sum up here right over here, this first one, it could be represented as sigma."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100. So mathematicians said, well, let's find some notation instead of having to do this dot, dot, dot thing, which you will see sometimes done, so that we can more cleanly express these types of sums. And that's where sigma notation comes from. So this sum up here right over here, this first one, it could be represented as sigma. This is a capital sigma. It's a Greek letter right over here. And what you do is you define an index."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So this sum up here right over here, this first one, it could be represented as sigma. This is a capital sigma. It's a Greek letter right over here. And what you do is you define an index. And you could start your index at some value. So let's say your index starts at 1. I'll just use i for index."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And what you do is you define an index. And you could start your index at some value. So let's say your index starts at 1. I'll just use i for index. So let's say that i starts at 1, and I'm going to go to 10. So i starts at 1, and it goes to 10. And I'm going to sum up the i's."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "I'll just use i for index. So let's say that i starts at 1, and I'm going to go to 10. So i starts at 1, and it goes to 10. And I'm going to sum up the i's. So how does this translate into this right over here? Well, what you do is you start wherever the index is. The index is at 1."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And I'm going to sum up the i's. So how does this translate into this right over here? Well, what you do is you start wherever the index is. The index is at 1. Set i equal to 1. Write the 1 down. And then you increment the index."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "The index is at 1. Set i equal to 1. Write the 1 down. And then you increment the index. And so i will then be equal to 2. i is 2. Put the 2 down. And you're summing each of these terms as you go."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And then you increment the index. And so i will then be equal to 2. i is 2. Put the 2 down. And you're summing each of these terms as you go. And you go all the way until i is equal to 10. All the way until i is equal to 10. So given what I just told you, I encourage you to pause this video and write the sigma notation for this sum right over here."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "And you're summing each of these terms as you go. And you go all the way until i is equal to 10. All the way until i is equal to 10. So given what I just told you, I encourage you to pause this video and write the sigma notation for this sum right over here. Assuming you've given a go at it, well, this would be the sum. The first term, well, it might be easy to just say, we'll start at i equals 1 again. i equals 1."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So given what I just told you, I encourage you to pause this video and write the sigma notation for this sum right over here. Assuming you've given a go at it, well, this would be the sum. The first term, well, it might be easy to just say, we'll start at i equals 1 again. i equals 1. But now we're not going to stop until i equals 100. And we're going to sum up all of the i's. Let's do another example."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "i equals 1. But now we're not going to stop until i equals 100. And we're going to sum up all of the i's. Let's do another example. Let's imagine the sum. Let's imagine the sum from i equals 0 to 50 of, I don't know, let me say pi i squared. What would this sum look like?"}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Let's do another example. Let's imagine the sum. Let's imagine the sum from i equals 0 to 50 of, I don't know, let me say pi i squared. What would this sum look like? And once again, I encourage you to pause the video and write it out. I kind of expand out the sum. Well, let's just go step by step."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "What would this sum look like? And once again, I encourage you to pause the video and write it out. I kind of expand out the sum. Well, let's just go step by step. When i equals 0, this will be pi times 0 squared. And that's clearly 0, but I'll write it out. Pi times 0 squared."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Well, let's just go step by step. When i equals 0, this will be pi times 0 squared. And that's clearly 0, but I'll write it out. Pi times 0 squared. Then we increase our i. And we make sure that we haven't hit this, that our i isn't already this top boundary right over here or this top value. So now we said i equals 1."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Pi times 0 squared. Then we increase our i. And we make sure that we haven't hit this, that our i isn't already this top boundary right over here or this top value. So now we said i equals 1. Pi times 1 squared. So plus pi times 1 squared. Well, is 1 our top value right over here where we stop?"}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So now we said i equals 1. Pi times 1 squared. So plus pi times 1 squared. Well, is 1 our top value right over here where we stop? No. So we keep going. So then we go i equals 2."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "Well, is 1 our top value right over here where we stop? No. So we keep going. So then we go i equals 2. Pi times 2 squared. So plus pi times 2 squared. I think you see the pattern here."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "So then we go i equals 2. Pi times 2 squared. So plus pi times 2 squared. I think you see the pattern here. And we're just going to keep going all the way until at some point we're going to keep incrementing our i. i is going to be 49. So it's going to be pi times 49 squared. And then finally, we increment i. i equals becomes 50."}, {"video_title": "Sigma notation for sums   Sequences, series and induction   Precalculus   Khan Academy.mp3", "Sentence": "I think you see the pattern here. And we're just going to keep going all the way until at some point we're going to keep incrementing our i. i is going to be 49. So it's going to be pi times 49 squared. And then finally, we increment i. i equals becomes 50. And we're going to have plus pi times 50 squared. And then we say, OK, our i is finally equal to this top boundary. And now we can stop."}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "We're given a differential equation right over here, cosine of y plus two, this whole thing, times the derivative of y with respect to x, is equal to two x, and we're given that for a particular solution, when x is equal to one, y of one is equal to zero. And we're asked, what is x when y is equal to pi? So the first thing I like to look at when I see a differential equation is, is it separable? Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So just like that, I've been able to, all I did is I multiplied both sides of this times dx."}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get?"}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared."}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero."}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero zero, two times zero zero, all of that's just going to be zero. So we get zero is equal to one plus c, or c is equal to negative one."}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero zero, two times zero zero, all of that's just going to be zero. So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one."}, {"video_title": "Worked example separable equation with an implicit solution   Khan Academy.mp3", "Sentence": "So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one. And now what is x when y is equal to pi? So sine of pi plus two times pi is equal to x squared minus one. See sine of pi is equal to zero, and so we get, see we can add one to both sides, and we get two pi plus one is equal to x squared, or we could say that x is equal to the plus or minus square root of two pi plus one."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that y is equal to five minus three x over x squared plus three x. And we want to figure out what is the derivative of y with respect to x. Now, it might immediately jump out at you that look, look, y is being defined as a rational expression here, as the quotient of two different expressions. Or you could even view this as two different functions. You could view this one up here as u of x. So you could say this is the same thing. This is the same thing as u of x over, you could view the one in the denominator as v of x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could even view this as two different functions. You could view this one up here as u of x. So you could say this is the same thing. This is the same thing as u of x over, you could view the one in the denominator as v of x. So that one right there is v of x. And so if you're taking the derivative of something that can be expressed in this way as the quotient of two different functions, well then you could use the quotient rule. And I'll give you my little aside, like I always do, the quotient rule, if you ever forget it, it can be derived from the product rule."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the same thing as u of x over, you could view the one in the denominator as v of x. So that one right there is v of x. And so if you're taking the derivative of something that can be expressed in this way as the quotient of two different functions, well then you could use the quotient rule. And I'll give you my little aside, like I always do, the quotient rule, if you ever forget it, it can be derived from the product rule. And we have videos there, because the product rule's a little bit easier to remember. But what I can do is just say, look, dy dx, if y is just u of x over v of x, I'm just gonna restate the quotient rule. This is going to be, this is going to be the derivative of the function in the numerator."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll give you my little aside, like I always do, the quotient rule, if you ever forget it, it can be derived from the product rule. And we have videos there, because the product rule's a little bit easier to remember. But what I can do is just say, look, dy dx, if y is just u of x over v of x, I'm just gonna restate the quotient rule. This is going to be, this is going to be the derivative of the function in the numerator. So d dx of u of x times the function in the denominator, times v of x, minus, minus, I'll do the, minus the function in the numerator, u of x, times the derivative of the function in the denominator, times d dx, v of x, and we're almost there, and then over, over, the function in the denominator squared. The function in the denominator squared. So this might look messy, but all we have to do now is think about, well, what is the derivative of u of x?"}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be, this is going to be the derivative of the function in the numerator. So d dx of u of x times the function in the denominator, times v of x, minus, minus, I'll do the, minus the function in the numerator, u of x, times the derivative of the function in the denominator, times d dx, v of x, and we're almost there, and then over, over, the function in the denominator squared. The function in the denominator squared. So this might look messy, but all we have to do now is think about, well, what is the derivative of u of x? What is the derivative of v of x? And we should just be able to substitute those things back into this expression we just wrote down. So let's do that."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this might look messy, but all we have to do now is think about, well, what is the derivative of u of x? What is the derivative of v of x? And we should just be able to substitute those things back into this expression we just wrote down. So let's do that. So the derivative with respect to x of u of x, of u of x, is equal to, let's see, five minus three x, the derivative of five is zero. The derivative of negative three x, well, that's just gonna be negative three. That's just negative three."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So the derivative with respect to x of u of x, of u of x, is equal to, let's see, five minus three x, the derivative of five is zero. The derivative of negative three x, well, that's just gonna be negative three. That's just negative three. If any of that looks completely unfamiliar to you, I encourage you to review the derivative properties and maybe the power rule. And now let's think about what is the derivative with respect to x, derivative with respect to x, of v of x, of v of x. Well, derivative of x squared, we just bring that exponent out front."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's just negative three. If any of that looks completely unfamiliar to you, I encourage you to review the derivative properties and maybe the power rule. And now let's think about what is the derivative with respect to x, derivative with respect to x, of v of x, of v of x. Well, derivative of x squared, we just bring that exponent out front. It's gonna be two times x to the two minus one, or two x to the first power, or just two x. And then the derivative of three x is just three. So two x plus three."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, derivative of x squared, we just bring that exponent out front. It's gonna be two times x to the two minus one, or two x to the first power, or just two x. And then the derivative of three x is just three. So two x plus three. And now we know everything we need to substitute back in here. The derivative of u with respect to x, this right over here is just negative three. V of x, this we know is x squared plus three x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So two x plus three. And now we know everything we need to substitute back in here. The derivative of u with respect to x, this right over here is just negative three. V of x, this we know is x squared plus three x. We know that this right over here is v of x. And then u of x we know is five minus three x. Five minus three x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "V of x, this we know is x squared plus three x. We know that this right over here is v of x. And then u of x we know is five minus three x. Five minus three x. The derivative of v with respect to x we know is two x plus three. Two x plus three. And then finally, v of x we know is x squared plus three x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Five minus three x. The derivative of v with respect to x we know is two x plus three. Two x plus three. And then finally, v of x we know is x squared plus three x. So this is x squared plus three x. And so what do we get? Well, we are going to get, and it's gonna look a little bit hairy, it's going to be equal to negative, I'll focus, so first we have, first we have this business up here."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, v of x we know is x squared plus three x. So this is x squared plus three x. And so what do we get? Well, we are going to get, and it's gonna look a little bit hairy, it's going to be equal to negative, I'll focus, so first we have, first we have this business up here. So it's negative three times x squared plus three x. So I'm just gonna distribute the negative three. So it's negative three x squared minus nine x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we are going to get, and it's gonna look a little bit hairy, it's going to be equal to negative, I'll focus, so first we have, first we have this business up here. So it's negative three times x squared plus three x. So I'm just gonna distribute the negative three. So it's negative three x squared minus nine x. And then from that we are going to subtract the product of these two expressions. And so let's see, what is that going to be? Well, we have a five times two x, which is 10 x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's negative three x squared minus nine x. And then from that we are going to subtract the product of these two expressions. And so let's see, what is that going to be? Well, we have a five times two x, which is 10 x. A five times three, which is 15. We have a negative three x times two x. So that is going to be negative six x squared."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we have a five times two x, which is 10 x. A five times three, which is 15. We have a negative three x times two x. So that is going to be negative six x squared. Minus six x squared. And then a negative three x times three. So negative nine x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is going to be negative six x squared. Minus six x squared. And then a negative three x times three. So negative nine x. And let's see, we can simplify that a little bit. 10 x minus nine x, well that's just going to leave us with x. So if we, 10 x minus nine x is just going to be x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative nine x. And let's see, we can simplify that a little bit. 10 x minus nine x, well that's just going to leave us with x. So if we, 10 x minus nine x is just going to be x. And then in our denominator, we're almost there. In our denominator, we could just write that as x plus three x, x squared plus three x squared. Or if we want, we could expand it out."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we, 10 x minus nine x is just going to be x. And then in our denominator, we're almost there. In our denominator, we could just write that as x plus three x, x squared plus three x squared. Or if we want, we could expand it out. I'll just leave it like that. x squared plus three x squared. And so if we want to, let's just simplify, or attempt to simplify this a little bit."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or if we want, we could expand it out. I'll just leave it like that. x squared plus three x squared. And so if we want to, let's just simplify, or attempt to simplify this a little bit. It's going to be negative three x squared minus nine x. And then, let's see, you're going to have a negative minus x, minus x. And then minus 15."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we want to, let's just simplify, or attempt to simplify this a little bit. It's going to be negative three x squared minus nine x. And then, let's see, you're going to have a negative minus x, minus x. And then minus 15. And then minus negative six x squared. So plus six x squared. All of that, all of that, over x squared plus three x squared."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then minus 15. And then minus negative six x squared. So plus six x squared. All of that, all of that, over x squared plus three x squared. Or x squared plus three x squared. I should say it that way. And let's see, this numerator I can simplify a little bit."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that, all of that, over x squared plus three x squared. Or x squared plus three x squared. I should say it that way. And let's see, this numerator I can simplify a little bit. Negative three x squared plus six x squared, that's going to be positive three x squared. And then we have, I'll do it in orange. We have negative nine x minus an x, well that's going to be minus 10 x."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, this numerator I can simplify a little bit. Negative three x squared plus six x squared, that's going to be positive three x squared. And then we have, I'll do it in orange. We have negative nine x minus an x, well that's going to be minus 10 x. Minus 10 x. And then we have minus 15. So minus 15."}, {"video_title": "Differentiating rational functions   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have negative nine x minus an x, well that's going to be minus 10 x. Minus 10 x. And then we have minus 15. So minus 15. And so there you have it. We finally have finished. This is all going to be equal to, this is all going to be equal to three x squared minus 10 x minus 15 over x squared plus three x squared."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we already know if we somehow tried to apply that anti-power rule, that inverse power rule over here, we would get something that's not defined. We would get x to the 0 over 0. It doesn't make any sense. And you might have been saying, OK, well, I know what to do in this case. When we first learned about derivatives, we know that the derivative, let me do this in yellow, the derivative with respect to x of the natural log of x is equal to 1 over x. So why can't we just say that the antiderivative of this right over here is equal to the natural log of x plus c? And this isn't necessarily wrong."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you might have been saying, OK, well, I know what to do in this case. When we first learned about derivatives, we know that the derivative, let me do this in yellow, the derivative with respect to x of the natural log of x is equal to 1 over x. So why can't we just say that the antiderivative of this right over here is equal to the natural log of x plus c? And this isn't necessarily wrong. The problem here is that it's not broad enough. And when I say it's not broad enough, is that the domain over here for our original function, that we're taking the antiderivative of, is all real numbers except for x equals 0. So over here, x cannot be equal to 0."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this isn't necessarily wrong. The problem here is that it's not broad enough. And when I say it's not broad enough, is that the domain over here for our original function, that we're taking the antiderivative of, is all real numbers except for x equals 0. So over here, x cannot be equal to 0. While the domain over here is only positive numbers. So over here, x, for this expression, x has to be greater than 0. So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here, x cannot be equal to 0. While the domain over here is only positive numbers. So over here, x, for this expression, x has to be greater than 0. So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of. So it would be nice if we could find an antiderivative that is defined everywhere that our original function is. So pretty much everywhere except for x equaling 0. So how can we rearrange this a little bit so that it could be defined for negative values as well?"}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of. So it would be nice if we could find an antiderivative that is defined everywhere that our original function is. So pretty much everywhere except for x equaling 0. So how can we rearrange this a little bit so that it could be defined for negative values as well? Well, one possibility is to think about the natural log of the absolute value of x. So I'll put a little question mark here because we don't really know what the derivative of this thing is going to be. And I'm not going to rigorously prove it here, but I will give you kind of the conceptual understanding."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how can we rearrange this a little bit so that it could be defined for negative values as well? Well, one possibility is to think about the natural log of the absolute value of x. So I'll put a little question mark here because we don't really know what the derivative of this thing is going to be. And I'm not going to rigorously prove it here, but I will give you kind of the conceptual understanding. So to understand it, let's plot the natural log of x. And I had done this ahead of time. So that right over there is roughly what the graph of the natural log of x looks like."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm not going to rigorously prove it here, but I will give you kind of the conceptual understanding. So to understand it, let's plot the natural log of x. And I had done this ahead of time. So that right over there is roughly what the graph of the natural log of x looks like. So what would the natural log of the absolute value of x is going to look like? Well, for positive x's, it's going to look just like this. For positive x's, you take the absolute value of it."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that right over there is roughly what the graph of the natural log of x looks like. So what would the natural log of the absolute value of x is going to look like? Well, for positive x's, it's going to look just like this. For positive x's, you take the absolute value of it. It's just the same thing as taking that original value. So it's going to look just like that for positive x's. But now this is also going to be defined for negative x's."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "For positive x's, you take the absolute value of it. It's just the same thing as taking that original value. So it's going to look just like that for positive x's. But now this is also going to be defined for negative x's. If you're taking the absolute value of negative 1, that evaluates to just 1. So it's just the natural log of 1. So you're going to be right there."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now this is also going to be defined for negative x's. If you're taking the absolute value of negative 1, that evaluates to just 1. So it's just the natural log of 1. So you're going to be right there. As you get closer and closer and closer to 0 from the negative side, you're just going to take the absolute value. So it's essentially going to be exactly this curve for natural log of x. But the left side of the natural log of the absolute value of x is going to be its mirror image if you were to reflect around the y-axis."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you're going to be right there. As you get closer and closer and closer to 0 from the negative side, you're just going to take the absolute value. So it's essentially going to be exactly this curve for natural log of x. But the left side of the natural log of the absolute value of x is going to be its mirror image if you were to reflect around the y-axis. It's going to look something like this. So what's nice about this function is you see it's defined everywhere except for, I'm trying to draw it as symmetrically as possible, it's defined everywhere except for x equals 0. So if you combine this pink part and this part on the right, if you combine both of these, you get y is equal to the natural log of the absolute value of x."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the left side of the natural log of the absolute value of x is going to be its mirror image if you were to reflect around the y-axis. It's going to look something like this. So what's nice about this function is you see it's defined everywhere except for, I'm trying to draw it as symmetrically as possible, it's defined everywhere except for x equals 0. So if you combine this pink part and this part on the right, if you combine both of these, you get y is equal to the natural log of the absolute value of x. Now let's think about its derivative. We already know what the derivative of the natural log of x is. And for positive values of x, let me write this down."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you combine this pink part and this part on the right, if you combine both of these, you get y is equal to the natural log of the absolute value of x. Now let's think about its derivative. We already know what the derivative of the natural log of x is. And for positive values of x, let me write this down. For x is greater than 0, we get the natural log of the absolute value of x is equal to the natural log of x. And we would also know, since these two are equal for x is greater than 0, for x is greater than 0, the derivative of the natural log of the absolute value of x is going to be equal to the derivative of the natural log of x, which is equal to 1 over x for x greater than 0. So let's plot that."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "And for positive values of x, let me write this down. For x is greater than 0, we get the natural log of the absolute value of x is equal to the natural log of x. And we would also know, since these two are equal for x is greater than 0, for x is greater than 0, the derivative of the natural log of the absolute value of x is going to be equal to the derivative of the natural log of x, which is equal to 1 over x for x greater than 0. So let's plot that. I'll do that in green. It's equal to 1 over x. So 1 over x, we've seen it before."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's plot that. I'll do that in green. It's equal to 1 over x. So 1 over x, we've seen it before. It looks something like this. So let me, my best attempt to draw it. Both vertical and horizontal asymptotes."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 1 over x, we've seen it before. It looks something like this. So let me, my best attempt to draw it. Both vertical and horizontal asymptotes. So it looks something like this. So this right over here is 1 over x for x is greater than 0. So this is 1 over x when x is greater than 0."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "Both vertical and horizontal asymptotes. So it looks something like this. So this right over here is 1 over x for x is greater than 0. So this is 1 over x when x is greater than 0. So all it's saying here, and you can see it pretty clearly, is the slope right over here, the slope of the tangent line is 1. And so you see, when you look at the derivative, the slope right over here, the derivative should be equal to 1 here. When you get close to 0, you have a very, very steep positive slope here."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is 1 over x when x is greater than 0. So all it's saying here, and you can see it pretty clearly, is the slope right over here, the slope of the tangent line is 1. And so you see, when you look at the derivative, the slope right over here, the derivative should be equal to 1 here. When you get close to 0, you have a very, very steep positive slope here. And so you see, you have a very high value for its derivative. And then as you move away from 0, it's still steep, but it becomes less and less and less steep all the way until you get to 1. And then it keeps getting less and less and less steep, but it never quite gets to an absolutely flat slope."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you get close to 0, you have a very, very steep positive slope here. And so you see, you have a very high value for its derivative. And then as you move away from 0, it's still steep, but it becomes less and less and less steep all the way until you get to 1. And then it keeps getting less and less and less steep, but it never quite gets to an absolutely flat slope. And that's what you see its derivative doing. Now what is the natural log of absolute value of x doing right over here? When we are out here, our slope is very close to 0."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it keeps getting less and less and less steep, but it never quite gets to an absolutely flat slope. And that's what you see its derivative doing. Now what is the natural log of absolute value of x doing right over here? When we are out here, our slope is very close to 0. It's symmetric. The slope here is essentially the negative of the slope here. I could do it maybe clearer showing it right over here."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "When we are out here, our slope is very close to 0. It's symmetric. The slope here is essentially the negative of the slope here. I could do it maybe clearer showing it right over here. Whatever the slope is right over here, it's the exact negative of whatever the slope is at a symmetric point on the other side. So if on the other side, the slope is right over here, over here, it's going to be the negative of that. So it's going to be right over there."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could do it maybe clearer showing it right over here. Whatever the slope is right over here, it's the exact negative of whatever the slope is at a symmetric point on the other side. So if on the other side, the slope is right over here, over here, it's going to be the negative of that. So it's going to be right over there. And then the slope just gets more and more and more negative. Right over here, the slope is a positive 1. Over here, it's going to be a negative 1."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be right over there. And then the slope just gets more and more and more negative. Right over here, the slope is a positive 1. Over here, it's going to be a negative 1. So right over here, our slope is a negative 1. And then as we get closer and closer to 0, it's just going to get more and more and more negative. So the derivative of the natural log of the absolute value of x for x is less than 0 looks something like this."}, {"video_title": "Indefinite integral of 1 x   AP Calculus AB   Khan Academy.mp3", "Sentence": "Over here, it's going to be a negative 1. So right over here, our slope is a negative 1. And then as we get closer and closer to 0, it's just going to get more and more and more negative. So the derivative of the natural log of the absolute value of x for x is less than 0 looks something like this. And you see, and once again, it's not a ultra rigorous proof, but what you see is that the derivative of the natural log of the absolute value of x is equal to 1 over x for all x's not equaling 0. So what you're seeing, or hopefully you can visualize, that the derivative, let me write it this way, the derivative of the natural log of the absolute value of x is indeed equal to 1 over x for all x does not equal 0. So this is a much more satisfying antiderivative for 1 over x."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So we have the differential equation, the derivative of y with respect to x is equal to y over six times four minus y. And what we have plotted right over here is the slope field, or a slope field, for this differential equation. And we can verify that this indeed is a slope field for this differential equation. Let's draw a little table here. So let's just verify a few points. So let's say x, y, and dy, dx. So let's say we start with, I don't know, let's start with this point right over here, one comma one."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "Let's draw a little table here. So let's just verify a few points. So let's say x, y, and dy, dx. So let's say we start with, I don't know, let's start with this point right over here, one comma one. When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5. And we see indeed on this slope field, they depicted the slope there. So if a solution goes to that point, right at that point, its slope would be 1 1.5."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So let's say we start with, I don't know, let's start with this point right over here, one comma one. When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5. And we see indeed on this slope field, they depicted the slope there. So if a solution goes to that point, right at that point, its slope would be 1 1.5. And as you see, it's actually only dependent on the y value. It doesn't matter what x is. As long as y is one, dy, dx is going to be 1 1.5."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So if a solution goes to that point, right at that point, its slope would be 1 1.5. And as you see, it's actually only dependent on the y value. It doesn't matter what x is. As long as y is one, dy, dx is going to be 1 1.5. And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5. And as long as y is one, all of these sampled points right over here all have a slope of 1 1.5. So just looking at that, that makes us feel that this slope field is consistent with this differential equation."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "As long as y is one, dy, dx is going to be 1 1.5. And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5. And as long as y is one, all of these sampled points right over here all have a slope of 1 1.5. So just looking at that, that makes us feel that this slope field is consistent with this differential equation. But let's try a few other points just to feel a little bit better about it. And then we will use the slope field to actually visualize some solutions. So let's do an interesting point."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So just looking at that, that makes us feel that this slope field is consistent with this differential equation. But let's try a few other points just to feel a little bit better about it. And then we will use the slope field to actually visualize some solutions. So let's do an interesting point. Let's say we have this point. Actually, no, that's at a half point. Let's say we have this, let's see, I want to do, let's say we do this point right over here."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So let's do an interesting point. Let's say we have this point. Actually, no, that's at a half point. Let's say we have this, let's see, I want to do, let's say we do this point right over here. So that's x is equal to one and y is equal to six. And we see the way the differential equation is defined, it doesn't matter what our x is. It's really dependent on the y that's going to drive the slope."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "Let's say we have this, let's see, I want to do, let's say we do this point right over here. So that's x is equal to one and y is equal to six. And we see the way the differential equation is defined, it doesn't matter what our x is. It's really dependent on the y that's going to drive the slope. But we have six over six, which is one, times four minus six, which is negative two. So it's negative two. So we should have a slope of negative two."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "It's really dependent on the y that's going to drive the slope. But we have six over six, which is one, times four minus six, which is negative two. So it's negative two. So we should have a slope of negative two. And it looks like that's what they depicted. So as long as y is six, we should have a slope of negative two. Have a slope of negative two."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So we should have a slope of negative two. And it looks like that's what they depicted. So as long as y is six, we should have a slope of negative two. Have a slope of negative two. And you see that in the slope field. So hopefully you feel pretty good that this is the slope field for this differential equation. If you don't, I encourage you to keep, keep verifying these points here."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "Have a slope of negative two. And you see that in the slope field. So hopefully you feel pretty good that this is the slope field for this differential equation. If you don't, I encourage you to keep, keep verifying these points here. But now let's actually use this slope field. Let's actually use this to visualize solutions to this differential equation based on points that the solution might go through. So let's say that we have a solution that goes through this point right over here."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "If you don't, I encourage you to keep, keep verifying these points here. But now let's actually use this slope field. Let's actually use this to visualize solutions to this differential equation based on points that the solution might go through. So let's say that we have a solution that goes through this point right over here. So what is that solution likely to look like? And once again, this is going to be a rough approximation. Well, right at that point, it's going to have a slope, just as the slope field shows."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So let's say that we have a solution that goes through this point right over here. So what is that solution likely to look like? And once again, this is going to be a rough approximation. Well, right at that point, it's going to have a slope, just as the slope field shows. And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that. So this, if I keep going up at this point, when y is equal to two, I should be parallel to all of these, these segments on the slope field at y is equal to two. And then it looks like the slope starts to decrease as we approach y is equal to four."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "Well, right at that point, it's going to have a slope, just as the slope field shows. And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that. So this, if I keep going up at this point, when y is equal to two, I should be parallel to all of these, these segments on the slope field at y is equal to two. And then it looks like the slope starts to decrease as we approach y is equal to four. And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero. And of course, we see that because if when y equals zero, this whole thing is zero, so our derivative's going to be zero. So a reasonable solution might look something like this."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "And then it looks like the slope starts to decrease as we approach y is equal to four. And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero. And of course, we see that because if when y equals zero, this whole thing is zero, so our derivative's going to be zero. So a reasonable solution might look something like this. So this gives us a clue. Well, look, if a solution goes through this point, this right over here might, might be what it looks like. But what if it goes through, I don't know, what if it goes through this point right over here?"}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So a reasonable solution might look something like this. So this gives us a clue. Well, look, if a solution goes through this point, this right over here might, might be what it looks like. But what if it goes through, I don't know, what if it goes through this point right over here? Well then, it might look like, it might look like this by the same exact logic. So it might look like this. So just like that, we're starting to get a sense."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "But what if it goes through, I don't know, what if it goes through this point right over here? Well then, it might look like, it might look like this by the same exact logic. So it might look like this. So just like that, we're starting to get a sense. We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation. But what's interesting about this slope field is it looks like there's some, you know, there's some interesting stuff, you know, if we have, if our solution includes points between, where the y value's between zero and four, it looks like we're going to have solutions like this, but what if we had y values that were larger than that or that were less than that or exactly, exactly zero or four? So for example, what if we had a solution that went through this point right over here?"}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So just like that, we're starting to get a sense. We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation. But what's interesting about this slope field is it looks like there's some, you know, there's some interesting stuff, you know, if we have, if our solution includes points between, where the y value's between zero and four, it looks like we're going to have solutions like this, but what if we had y values that were larger than that or that were less than that or exactly, exactly zero or four? So for example, what if we had a solution that went through this point right over here? Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change. And as long as our y value doesn't change, our y value's going to stay at four, so our slope is going to stay zero. So we actually already found, this is actually a solution to the differential equation."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So for example, what if we had a solution that went through this point right over here? Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change. And as long as our y value doesn't change, our y value's going to stay at four, so our slope is going to stay zero. So we actually already found, this is actually a solution to the differential equation. It's y is equal to four is a solution to this differential equation. So y is equal to four. y is equal to four."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So we actually already found, this is actually a solution to the differential equation. It's y is equal to four is a solution to this differential equation. So y is equal to four. y is equal to four. And you can verify that that is a solution. When y is equal to four, this right-hand side is going to be zero, and the derivative is zero for y is equal to four. So that is a solution to the differential equation."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "y is equal to four. And you can verify that that is a solution. When y is equal to four, this right-hand side is going to be zero, and the derivative is zero for y is equal to four. So that is a solution to the differential equation. And the same thing for y is equal to zero. That is also a solution to the differential equation. Now, what if we included points, what if we included this point up here?"}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "So that is a solution to the differential equation. And the same thing for y is equal to zero. That is also a solution to the differential equation. Now, what if we included points, what if we included this point up here? And actually, let me do it in a different color so that you could see it. Let's say our solution included that point. Well, then it might look something, it might look something like this."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "Now, what if we included points, what if we included this point up here? And actually, let me do it in a different color so that you could see it. Let's say our solution included that point. Well, then it might look something, it might look something like this. And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses. So a solution that includes the point zero five might look something like this. And once again, it's just another clue."}, {"video_title": "Slope field to visualize solutions   First order differential equations   Khan Academy.mp3", "Sentence": "Well, then it might look something, it might look something like this. And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses. So a solution that includes the point zero five might look something like this. And once again, it's just another clue. A solution that includes the point zero negative one and a half might look something like, might look something like this. So anyway, hopefully this gives you a better appreciation for why slope fields are interesting. If you have a differential equation that just involves the first derivative and some x's and y's, this one only involves the first derivative and y's, we can plot a slope field like this, not too much trouble."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The function g is defined over the real numbers. This table gives select values of g. What is a reasonable estimate for the limit as x approaches five of g of x? So pause this video, look at this table. It gives us the x values as we approach five from values less than five, and as we approach five from values greater than five, it even tells us what g of x is at x equals five. And so given that, what is a reasonable estimate for this limit? All right, now let's work through this together. So let's think about what g of x seems to be approaching as x approaches five from values less than five."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It gives us the x values as we approach five from values less than five, and as we approach five from values greater than five, it even tells us what g of x is at x equals five. And so given that, what is a reasonable estimate for this limit? All right, now let's work through this together. So let's think about what g of x seems to be approaching as x approaches five from values less than five. Let's see, at four, it's at 3.37, 4.9, it's a little higher, it's at 3.5, 4.99, it's at 3.66, 4.999, so very close to five. We're only a thousandth away. We're at 3.68."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what g of x seems to be approaching as x approaches five from values less than five. Let's see, at four, it's at 3.37, 4.9, it's a little higher, it's at 3.5, 4.99, it's at 3.66, 4.999, so very close to five. We're only a thousandth away. We're at 3.68. But then at five, all of a sudden, it looks like we're kind of jumping to 6.37. And once again, I'm making an inference here. These are just sample points of this function."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're at 3.68. But then at five, all of a sudden, it looks like we're kind of jumping to 6.37. And once again, I'm making an inference here. These are just sample points of this function. We don't know exactly what the function is. But then if we approach five from values greater than five, at six, we're at 3.97, at 5.1, we're at 3.84, 5.01, 3.7, 5.001, we are at 3.68. So a thousandth below five and a thousandth above five, we're at 3.68, but then at five, all of a sudden, we're at 6.37."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "These are just sample points of this function. We don't know exactly what the function is. But then if we approach five from values greater than five, at six, we're at 3.97, at 5.1, we're at 3.84, 5.01, 3.7, 5.001, we are at 3.68. So a thousandth below five and a thousandth above five, we're at 3.68, but then at five, all of a sudden, we're at 6.37. So my most reasonable estimate would be, well, it looks like we are approaching 3.68 when we are approaching from values less than five, and we're approaching 3.68 from values, as we approach five from values greater than five. It doesn't matter that the value of five is 6.37. The limit would be 3.68, or a reasonable estimate for the limit would be 3.68."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So a thousandth below five and a thousandth above five, we're at 3.68, but then at five, all of a sudden, we're at 6.37. So my most reasonable estimate would be, well, it looks like we are approaching 3.68 when we are approaching from values less than five, and we're approaching 3.68 from values, as we approach five from values greater than five. It doesn't matter that the value of five is 6.37. The limit would be 3.68, or a reasonable estimate for the limit would be 3.68. And this is probably the most tempting distractor here, because if you were to just substitute five, what is g of five? It tells us 6.37. But the limit does not have to be what the actual function equals at that point."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The limit would be 3.68, or a reasonable estimate for the limit would be 3.68. And this is probably the most tempting distractor here, because if you were to just substitute five, what is g of five? It tells us 6.37. But the limit does not have to be what the actual function equals at that point. Let me draw what this might look like. And so an example of this. So if this is five right over here, at the point five, the value of my function is 6.37."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the limit does not have to be what the actual function equals at that point. Let me draw what this might look like. And so an example of this. So if this is five right over here, at the point five, the value of my function is 6.37. So let's say that this right over here is 6.37. So that's the value of my function right over there. So 6.37, but as we approach five, so that's four, actually let me spread out a little bit."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if this is five right over here, at the point five, the value of my function is 6.37. So let's say that this right over here is 6.37. So that's the value of my function right over there. So 6.37, but as we approach five, so that's four, actually let me spread out a little bit. This obviously is not drawing to scale. But as we approach five, so if that's 6.37, then at four, 3.37 is about here, and it looks like it's approaching 3.68. So 3.68, actually let me draw that."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 6.37, but as we approach five, so that's four, actually let me spread out a little bit. This obviously is not drawing to scale. But as we approach five, so if that's 6.37, then at four, 3.37 is about here, and it looks like it's approaching 3.68. So 3.68, actually let me draw that. So 3.68 is gonna be roughly that. 3.68 is gonna be roughly that. So the graph, the graph might look something like this."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 3.68, actually let me draw that. So 3.68 is gonna be roughly that. 3.68 is gonna be roughly that. So the graph, the graph might look something like this. We could infer it looks like it's doing something like this, where it's approaching 3.68 from values less than five and values greater than five. But right at five, our value is 6.37. I don't know for sure if this is what the graph looks like."}, {"video_title": "Estimating limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the graph, the graph might look something like this. We could infer it looks like it's doing something like this, where it's approaching 3.68 from values less than five and values greater than five. But right at five, our value is 6.37. I don't know for sure if this is what the graph looks like. Once again, we're just getting some sample points. But this would be a reasonable inference. And so you can see our limit, we are approaching 3.68, even though the value of the function is something different."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The particle's position x of t is not explicitly given. The velocity of the particle is given by v of t is equal to all of this business right over here. The acceleration of the particle is given by a of t is equal to all of this business over here. They actually didn't have to give us that because the acceleration is just the derivative of the velocity. They don't tell us the position function, but they tell us where we start off. x of zero is equal to two. Fair enough."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "They actually didn't have to give us that because the acceleration is just the derivative of the velocity. They don't tell us the position function, but they tell us where we start off. x of zero is equal to two. Fair enough. Now let's do part a. Is the speed of the particle increasing or decreasing at time t equals 5.5? Give a reason for your answer."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Fair enough. Now let's do part a. Is the speed of the particle increasing or decreasing at time t equals 5.5? Give a reason for your answer. It looks like they did something a little sneaky here because they gave us a velocity function and then they ask about a speed. You might say, wait, aren't those the same thing? I would say, no, they aren't quite the same thing."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Give a reason for your answer. It looks like they did something a little sneaky here because they gave us a velocity function and then they ask about a speed. You might say, wait, aren't those the same thing? I would say, no, they aren't quite the same thing. Velocity is a magnitude and a direction. It's a vector quantity. Speed is just a magnitude."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I would say, no, they aren't quite the same thing. Velocity is a magnitude and a direction. It's a vector quantity. Speed is just a magnitude. It is a scalar quantity. To see the difference, you could have a velocity, and this isn't maybe particular to this problem because they don't give us the units, but you could have a velocity of negative 5 meters per second. Maybe if we're talking about on the x-axis, this would mean we're moving leftward at 5 meters per second on the x-axis."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Speed is just a magnitude. It is a scalar quantity. To see the difference, you could have a velocity, and this isn't maybe particular to this problem because they don't give us the units, but you could have a velocity of negative 5 meters per second. Maybe if we're talking about on the x-axis, this would mean we're moving leftward at 5 meters per second on the x-axis. The magnitude is 5 meters per second. This is the magnitude. The direction is specified by the negative number."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe if we're talking about on the x-axis, this would mean we're moving leftward at 5 meters per second on the x-axis. The magnitude is 5 meters per second. This is the magnitude. The direction is specified by the negative number. That is the direction. Your velocity could be negative 5 meters per second, but your speed would just be 5 meters per second. Your speed is 5 meters per second whether you're going to the left or to the right."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The direction is specified by the negative number. That is the direction. Your velocity could be negative 5 meters per second, but your speed would just be 5 meters per second. Your speed is 5 meters per second whether you're going to the left or to the right. Your velocity, you actually care whether you're going to the left or the right. Let's just keep that in mind while we try to solve this problem. The best way to figure out whether our rate of change is increasing or decreasing is to look at the acceleration because acceleration is really just the rate of change of velocity."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Your speed is 5 meters per second whether you're going to the left or to the right. Your velocity, you actually care whether you're going to the left or the right. Let's just keep that in mind while we try to solve this problem. The best way to figure out whether our rate of change is increasing or decreasing is to look at the acceleration because acceleration is really just the rate of change of velocity. Then we can think a little bit about this velocity versus speed question. What is the acceleration at time 5.5? Get the calculator out."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The best way to figure out whether our rate of change is increasing or decreasing is to look at the acceleration because acceleration is really just the rate of change of velocity. Then we can think a little bit about this velocity versus speed question. What is the acceleration at time 5.5? Get the calculator out. We can use calculators for this part of the AP exam. I assume they intend us to because this isn't something that's easy to calculate by hand. The acceleration at time 5.5, we just have to say t is 5.5 and evaluate this function."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Get the calculator out. We can use calculators for this part of the AP exam. I assume they intend us to because this isn't something that's easy to calculate by hand. The acceleration at time 5.5, we just have to say t is 5.5 and evaluate this function. 1 half, I'll just write.5 times e to the t over 4. t is 5.5. 5.5 divided by 4 and then times cosine of 5.5 divided by 4. Gives us.38."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The acceleration at time 5.5, we just have to say t is 5.5 and evaluate this function. 1 half, I'll just write.5 times e to the t over 4. t is 5.5. 5.5 divided by 4 and then times cosine of 5.5 divided by 4. Gives us.38. Did I do that right? We have.5 times e to the 5.5 divided by 4 times cosine. Sorry, I made a mistake."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Gives us.38. Did I do that right? We have.5 times e to the 5.5 divided by 4 times cosine. Sorry, I made a mistake. That looked a little strange. It's not cosine of 5.5 divided by 4. It's cosine of e to the 5.5 divided by 4."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sorry, I made a mistake. That looked a little strange. It's not cosine of 5.5 divided by 4. It's cosine of e to the 5.5 divided by 4. Let's look at that. That's one parenthesis I closed. Now that is the second parenthesis that I've closed."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's cosine of e to the 5.5 divided by 4. Let's look at that. That's one parenthesis I closed. Now that is the second parenthesis that I've closed. I get roughly negative 1.36. This is approximately equal to negative 1.36 if I round it. We don't care about so much as the actual value."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now that is the second parenthesis that I've closed. I get roughly negative 1.36. This is approximately equal to negative 1.36 if I round it. We don't care about so much as the actual value. What we really care about is its sign. The acceleration at time 5.5 is negative, which tells us that the velocity is decreasing. Now, you might be tempted to say we're done, but remember they're not asking us is the velocity of the particle increasing or decreasing."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't care about so much as the actual value. What we really care about is its sign. The acceleration at time 5.5 is negative, which tells us that the velocity is decreasing. Now, you might be tempted to say we're done, but remember they're not asking us is the velocity of the particle increasing or decreasing. They're asking us is the speed of the particle increasing or decreasing. If you're saying, hey, how did I know that? Just remember acceleration is just the rate of change of velocity."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, you might be tempted to say we're done, but remember they're not asking us is the velocity of the particle increasing or decreasing. They're asking us is the speed of the particle increasing or decreasing. If you're saying, hey, how did I know that? Just remember acceleration is just the rate of change of velocity. If acceleration is negative, that means the rate of change of velocity is negative. It's just going down. Anyway, how do we address this speed issue?"}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Just remember acceleration is just the rate of change of velocity. If acceleration is negative, that means the rate of change of velocity is negative. It's just going down. Anyway, how do we address this speed issue? How do we think about it? There's two scenarios. If our velocity is positive at time 5.5, so if we have a positive velocity, so let's say our velocity is 5 meters per second, although they don't give us units here, so I won't use units."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Anyway, how do we address this speed issue? How do we think about it? There's two scenarios. If our velocity is positive at time 5.5, so if we have a positive velocity, so let's say our velocity is 5 meters per second, although they don't give us units here, so I won't use units. Let's say our velocity is 5, and then it's negative, so at some point our velocity is going to be something smaller. That means that the speed would also be decreasing. If we have a positive velocity, then the fact that acceleration is negative means that both velocity and speed would be decreasing."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If our velocity is positive at time 5.5, so if we have a positive velocity, so let's say our velocity is 5 meters per second, although they don't give us units here, so I won't use units. Let's say our velocity is 5, and then it's negative, so at some point our velocity is going to be something smaller. That means that the speed would also be decreasing. If we have a positive velocity, then the fact that acceleration is negative means that both velocity and speed would be decreasing. On the other hand, if we had a negative velocity, if we have a negative velocity at time t equals 5.5, then the fact that it's decreasing means that we're getting even more negative. If we're getting even more negative, then that means the speed is increasing. The magnitude is increasing in the leftward direction."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we have a positive velocity, then the fact that acceleration is negative means that both velocity and speed would be decreasing. On the other hand, if we had a negative velocity, if we have a negative velocity at time t equals 5.5, then the fact that it's decreasing means that we're getting even more negative. If we're getting even more negative, then that means the speed is increasing. The magnitude is increasing in the leftward direction. What we really need to do, beyond just evaluating the acceleration at time 5.5, we also have to evaluate the velocity to see if it's going in the left or the rightward direction. Let's evaluate the velocity. The velocity at 5.5, and we'll just get our calculator out again, velocity at 5.5, this is our velocity function, is going to be equal to 2 times the sine of, let me write it this way just because I want to make sure I get my parentheses right, 2 times the sine, let me write it this way, 2 sine of e to the 5.5, that's our time, divided by 4, so I did that part right over here, I'm going to close the 2 as well, and then plus 1."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The magnitude is increasing in the leftward direction. What we really need to do, beyond just evaluating the acceleration at time 5.5, we also have to evaluate the velocity to see if it's going in the left or the rightward direction. Let's evaluate the velocity. The velocity at 5.5, and we'll just get our calculator out again, velocity at 5.5, this is our velocity function, is going to be equal to 2 times the sine of, let me write it this way just because I want to make sure I get my parentheses right, 2 times the sine, let me write it this way, 2 sine of e to the 5.5, that's our time, divided by 4, so I did that part right over here, I'm going to close the 2 as well, and then plus 1. This is our velocity. Our velocity at time 5.5 is negative, so negative 0.45, roughly. This is negative, velocity is negative."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The velocity at 5.5, and we'll just get our calculator out again, velocity at 5.5, this is our velocity function, is going to be equal to 2 times the sine of, let me write it this way just because I want to make sure I get my parentheses right, 2 times the sine, let me write it this way, 2 sine of e to the 5.5, that's our time, divided by 4, so I did that part right over here, I'm going to close the 2 as well, and then plus 1. This is our velocity. Our velocity at time 5.5 is negative, so negative 0.45, roughly. This is negative, velocity is negative. We have the scenario where the velocity is negative, which means we're going in the left direction. The fact that the velocity is also decreasing means that over time, at least at this point in time, as we go forward in time, it will become even more negative, and it will become even more negative if we wait a little bit longer. That means that the magnitude of the velocity is increasing, it's just going in the leftward direction."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is negative, velocity is negative. We have the scenario where the velocity is negative, which means we're going in the left direction. The fact that the velocity is also decreasing means that over time, at least at this point in time, as we go forward in time, it will become even more negative, and it will become even more negative if we wait a little bit longer. That means that the magnitude of the velocity is increasing, it's just going in the leftward direction. The magnitude of the velocity is increasing, although it's going in the leftward direction, that means that the speed is increasing. The velocity, this is one of those interesting scenarios, the velocity is decreasing, but the speed, which is what they're asking us in the question, speed is increasing. If you wanted to do it really quick with all of this explanation I gave you, you would say, hey look, what's acceleration?"}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That means that the magnitude of the velocity is increasing, it's just going in the leftward direction. The magnitude of the velocity is increasing, although it's going in the leftward direction, that means that the speed is increasing. The velocity, this is one of those interesting scenarios, the velocity is decreasing, but the speed, which is what they're asking us in the question, speed is increasing. If you wanted to do it really quick with all of this explanation I gave you, you would say, hey look, what's acceleration? Is it positive or negative? You would evaluate it, say hey, it's negative. You know velocity is decreasing."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you wanted to do it really quick with all of this explanation I gave you, you would say, hey look, what's acceleration? Is it positive or negative? You would evaluate it, say hey, it's negative. You know velocity is decreasing. Then you would say, hey, what is velocity? Is it positive or negative? You evaluate it, you say it's negative."}, {"video_title": "2011 Calculus AB free response #1a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "You know velocity is decreasing. Then you would say, hey, what is velocity? Is it positive or negative? You evaluate it, you say it's negative. You have a negative value that is decreasing, so it's becoming more negative. That means its magnitude is increasing, or speed is increasing. ."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we look at her solution and then they ask us, is Olga's work correct? If not, what's her mistake? So pause this video and see if you can figure this out. All right, let's just follow her work. So here she's trying to take the first derivative. So you would apply the chain rule. It would be four times x minus two to the third power times the derivative of x minus two, which is just one."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's just follow her work. So here she's trying to take the first derivative. So you would apply the chain rule. It would be four times x minus two to the third power times the derivative of x minus two, which is just one. So this checks out. Then you take the derivative of this. It would be three times four, which would be 12 times x minus two to the second power times the derivative of x minus two, which is just one, which is exactly what she has here, 12 times x minus two to the second power."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would be four times x minus two to the third power times the derivative of x minus two, which is just one. So this checks out. Then you take the derivative of this. It would be three times four, which would be 12 times x minus two to the second power times the derivative of x minus two, which is just one, which is exactly what she has here, 12 times x minus two to the second power. That checks out. So step one's looking good for Olga. Step two, the solution of the second derivative equaling zero is x equals two."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would be three times four, which would be 12 times x minus two to the second power times the derivative of x minus two, which is just one, which is exactly what she has here, 12 times x minus two to the second power. That checks out. So step one's looking good for Olga. Step two, the solution of the second derivative equaling zero is x equals two. That looks right. The second derivative is 12 times x minus two squared and we wanna make that equal to zero. This is only going to be true when x is equal to two."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "Step two, the solution of the second derivative equaling zero is x equals two. That looks right. The second derivative is 12 times x minus two squared and we wanna make that equal to zero. This is only going to be true when x is equal to two. So step two is looking good. So step three, Olga says f has an inflection point at x equals two. So she's basing this just on the fact that the second derivative is zero when x is equal to two."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is only going to be true when x is equal to two. So step two is looking good. So step three, Olga says f has an inflection point at x equals two. So she's basing this just on the fact that the second derivative is zero when x is equal to two. She's basing this just on the fact that f prime prime of two is equal to zero. Now I have a problem with this because the fact that your second derivative is zero at x equals two, that makes two a nice candidate to check out, but you can't immediately say that we have an inflection point there. Remember, an inflection point is where we go from being concave upwards to concave downwards or concave downwards to concave upwards."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "So she's basing this just on the fact that the second derivative is zero when x is equal to two. She's basing this just on the fact that f prime prime of two is equal to zero. Now I have a problem with this because the fact that your second derivative is zero at x equals two, that makes two a nice candidate to check out, but you can't immediately say that we have an inflection point there. Remember, an inflection point is where we go from being concave upwards to concave downwards or concave downwards to concave upwards. And speaking in the language of the second derivative, it means that the second derivative changes signs as we go from below x equals two to above x equals two. But we have to test that because it's not necessarily always the case. So let's actually test it."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, an inflection point is where we go from being concave upwards to concave downwards or concave downwards to concave upwards. And speaking in the language of the second derivative, it means that the second derivative changes signs as we go from below x equals two to above x equals two. But we have to test that because it's not necessarily always the case. So let's actually test it. Let's think about some intervals. Intervals. So let's think about the interval when we go from negative infinity to two."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's actually test it. Let's think about some intervals. Intervals. So let's think about the interval when we go from negative infinity to two. And let's think about the interval where we go from two to positive infinity. If you want, you could have some test values. You could think about the sign, sign of our second derivative."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about the interval when we go from negative infinity to two. And let's think about the interval where we go from two to positive infinity. If you want, you could have some test values. You could think about the sign, sign of our second derivative. And then based on that, you could think about concavity. Concavity of f. So let's think about what's happening. So you could take a test value, let's say one is in this interval, and let's say three is in this interval."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could think about the sign, sign of our second derivative. And then based on that, you could think about concavity. Concavity of f. So let's think about what's happening. So you could take a test value, let's say one is in this interval, and let's say three is in this interval. And you could say one minus two squared is going to be, let's see, that's negative one squared, which is one. And then you're just going to, this is just going to be 12. So this is going to be positive."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you could take a test value, let's say one is in this interval, and let's say three is in this interval. And you could say one minus two squared is going to be, let's see, that's negative one squared, which is one. And then you're just going to, this is just going to be 12. So this is going to be positive. And if you tried three, three minus two squared is one times 12. Well, that's also going to be positive. And so you're going to be concave upwards, at least at these test values, it looks like on either side of two, that the sign of the second derivative is positive on either side of two."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be positive. And if you tried three, three minus two squared is one times 12. Well, that's also going to be positive. And so you're going to be concave upwards, at least at these test values, it looks like on either side of two, that the sign of the second derivative is positive on either side of two. And you might say, well, maybe I just need to find closer values. But if you inspect the second derivative here, you can see that this is never going to be negative. In fact, for any value other than x equals two, this value right over here, since we're, even if x minus two is negative, you're squaring it, which will make this entire thing positive, and then multiplying it times a positive value."}, {"video_title": "Mistakes when finding inflection points not checking candidates   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you're going to be concave upwards, at least at these test values, it looks like on either side of two, that the sign of the second derivative is positive on either side of two. And you might say, well, maybe I just need to find closer values. But if you inspect the second derivative here, you can see that this is never going to be negative. In fact, for any value other than x equals two, this value right over here, since we're, even if x minus two is negative, you're squaring it, which will make this entire thing positive, and then multiplying it times a positive value. So for any value other than x equals two, the sign of our second derivative is positive, which means that we're going to be concave upwards. And so we actually don't have an inflection point at x equals two, because we are not switching signs as we go from values less than x equals two to values greater than x equals two. Our second derivative is not switching signs."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're now going to tackle part B of the potato problem. And it says use the second derivative of eight with respect to time to determine whether your answer in part A is an underestimate or an overestimate of the internal temperature of the potato at time t is equal to three. So in part A, we found the equation of the tangent line at time equals zero, and we used that to estimate what our internal temperature would be at time is equal to three. So how is the second derivative going to help us think whether that was an overestimate or an underestimate? Well, the second derivative can help us know about concavity. It'll let us know, well, is our slope increasing over this interval, or is our slope decreasing? And then we can use that to estimate whether we overestimated or not."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how is the second derivative going to help us think whether that was an overestimate or an underestimate? Well, the second derivative can help us know about concavity. It'll let us know, well, is our slope increasing over this interval, or is our slope decreasing? And then we can use that to estimate whether we overestimated or not. So first, let's just find the second derivative. So we have the first derivative written up here. Let me just rewrite it, and I'll distribute the negative 1 4th, because it'll be a little bit more straightforward then."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can use that to estimate whether we overestimated or not. So first, let's just find the second derivative. So we have the first derivative written up here. Let me just rewrite it, and I'll distribute the negative 1 4th, because it'll be a little bit more straightforward then. So if I write the derivative of h with respect to time is equal to negative 1 4th times our internal temperature, which itself is a function of time, and then negative 1 4th times negative 27, that would be plus 27 over four. Let me scroll down a little bit. So now, let me leave that graph up there, because I think that might be useful."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just rewrite it, and I'll distribute the negative 1 4th, because it'll be a little bit more straightforward then. So if I write the derivative of h with respect to time is equal to negative 1 4th times our internal temperature, which itself is a function of time, and then negative 1 4th times negative 27, that would be plus 27 over four. Let me scroll down a little bit. So now, let me leave that graph up there, because I think that might be useful. What is the second derivative going to be with respect to time? So I'll write it right over here. D, the second derivative of h with respect to time is going to be equal to, well, the derivative of this first term with respect to time, it's going to be the derivative of this with respect to h times the derivative of h with respect to time."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now, let me leave that graph up there, because I think that might be useful. What is the second derivative going to be with respect to time? So I'll write it right over here. D, the second derivative of h with respect to time is going to be equal to, well, the derivative of this first term with respect to time, it's going to be the derivative of this with respect to h times the derivative of h with respect to time. So this is equal to negative 1 4th, that's the derivative of this term with respect to h, and then we want to multiply that times the derivative of h with respect to time. This comes just straight out of the chain rule. And then the derivative of a constant, how does that change with respect to time?"}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "D, the second derivative of h with respect to time is going to be equal to, well, the derivative of this first term with respect to time, it's going to be the derivative of this with respect to h times the derivative of h with respect to time. So this is equal to negative 1 4th, that's the derivative of this term with respect to h, and then we want to multiply that times the derivative of h with respect to time. This comes just straight out of the chain rule. And then the derivative of a constant, how does that change with respect to time? It's not gonna change, that is just going to be zero. So just like that, we were able to find the second derivative of h with respect to time. And now, what does this tell us?"}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the derivative of a constant, how does that change with respect to time? It's not gonna change, that is just going to be zero. So just like that, we were able to find the second derivative of h with respect to time. And now, what does this tell us? Well, we talked about in the previous video that over the interval that we care about, well, actually, I can show you from this graph. Over the interval that we care about, for t greater than zero, it says that our internal temperature is always going to be greater than 27. And so when you look at this expression here, or when you look at this expression here for dh dt, we talked in the previous video how this is always going to be negative here, because h is always going to be greater than 27, so that part's going to be positive, but then we're gonna multiply it by a negative 1 4th."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now, what does this tell us? Well, we talked about in the previous video that over the interval that we care about, well, actually, I can show you from this graph. Over the interval that we care about, for t greater than zero, it says that our internal temperature is always going to be greater than 27. And so when you look at this expression here, or when you look at this expression here for dh dt, we talked in the previous video how this is always going to be negative here, because h is always going to be greater than 27, so that part's going to be positive, but then we're gonna multiply it by a negative 1 4th. So our slope, dh dt, our derivative of our temperature with respect to time is always going to be negative. So we could write that this or this, this is going to be negative, let me write it this way, since h is greater than 27, for t is greater than zero, we know that dh dt is negative, is negative. So we could say that this right over here, since dh dt is less than zero, for t is greater than zero, the second derivative of h with respect to t is going to be greater than zero, greater than zero, for t is greater than zero."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so when you look at this expression here, or when you look at this expression here for dh dt, we talked in the previous video how this is always going to be negative here, because h is always going to be greater than 27, so that part's going to be positive, but then we're gonna multiply it by a negative 1 4th. So our slope, dh dt, our derivative of our temperature with respect to time is always going to be negative. So we could write that this or this, this is going to be negative, let me write it this way, since h is greater than 27, for t is greater than zero, we know that dh dt is negative, is negative. So we could say that this right over here, since dh dt is less than zero, for t is greater than zero, the second derivative of h with respect to t is going to be greater than zero, greater than zero, for t is greater than zero. And so what does that mean? It means that we are, if your second derivative is positive, that means you're concave upward, concave upward, which means slope is increasing, slope increasing, slope increasing, or you could say slope becoming less negative, slope becoming less negative. Now what does that mean?"}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say that this right over here, since dh dt is less than zero, for t is greater than zero, the second derivative of h with respect to t is going to be greater than zero, greater than zero, for t is greater than zero. And so what does that mean? It means that we are, if your second derivative is positive, that means you're concave upward, concave upward, which means slope is increasing, slope increasing, slope increasing, or you could say slope becoming less negative, slope becoming less negative. Now what does that mean? And you could see it intuitively, if the slope is becoming less and less negative, then that means when we approximated what our temperature is at t equals three, we used a really negative slope, when in reality our slope is getting less and less and less negative. So what we would have done is we would have over-decreased our temperature from t equals zero to t equals three. So that would mean that we would have underestimated it."}, {"video_title": "2017 AP Calculus AB BC 4b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what does that mean? And you could see it intuitively, if the slope is becoming less and less negative, then that means when we approximated what our temperature is at t equals three, we used a really negative slope, when in reality our slope is getting less and less and less negative. So what we would have done is we would have over-decreased our temperature from t equals zero to t equals three. So that would mean that we would have underestimated it. So let me write that down. And I'm running out of a little bit of space, but let me write it right over here. So that implies, so this implies that underestimate, underestimate, underestimate in part A."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we've already seen that the derivative with respect to x of e to the x is equal to e to the x, which is a pretty amazing thing. One of the many things that makes e somewhat special, that when you have an exponential with your base right over here as e, the derivative of it, the slope at any point, is equal to the value of that actual function. But now let's see, let's start exploring when we have other bases. Can we somehow figure out what is the derivative, what is the derivative with respect to x when we have a to the x, where a could be any number? Is there some way to figure this out? And maybe using our knowledge that the derivative of e to the x is e to the x. Well, can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base?"}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can we somehow figure out what is the derivative, what is the derivative with respect to x when we have a to the x, where a could be any number? Is there some way to figure this out? And maybe using our knowledge that the derivative of e to the x is e to the x. Well, can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base? Well, you could view a, you could view a as being equal to e, let me write it this way, well, I'll write it, a as being equal to e to the natural log of a. Now, I want you to, if this isn't obvious to you, I really want you to think about it. What is the natural log of a?"}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base? Well, you could view a, you could view a as being equal to e, let me write it this way, well, I'll write it, a as being equal to e to the natural log of a. Now, I want you to, if this isn't obvious to you, I really want you to think about it. What is the natural log of a? The natural log, the natural log of a is the power you need to raise e to to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e to to get to a, well, then you're just going to get to a. So really think about this."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the natural log of a? The natural log, the natural log of a is the power you need to raise e to to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e to to get to a, well, then you're just going to get to a. So really think about this. Don't just accept this as a leap of faith. It should make sense to you, and it just comes out of really what a logarithm is. And so we can replace a with this whole expression here."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So really think about this. Don't just accept this as a leap of faith. It should make sense to you, and it just comes out of really what a logarithm is. And so we can replace a with this whole expression here. We can, if a is the same thing as e to the natural log of a, well, then this is going to be, then this is going to be equal to the derivative with respect to x of e, e to the natural log, I keep writing la, to the natural log of a, and then we're gonna raise that to the xth power. We're gonna raise that to the xth power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color coding it."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can replace a with this whole expression here. We can, if a is the same thing as e to the natural log of a, well, then this is going to be, then this is going to be equal to the derivative with respect to x of e, e to the natural log, I keep writing la, to the natural log of a, and then we're gonna raise that to the xth power. We're gonna raise that to the xth power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a, times x power."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a, times x power. Times x power. And now we can use the chain rule to evaluate this derivative. So what we will do is we will first take the derivative of the outside function, so e to the natural log of a times x, with respect to the inside function, with respect to natural log of a times x."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's going to be the same thing as e to the natural log of a, natural log of a, times x power. Times x power. And now we can use the chain rule to evaluate this derivative. So what we will do is we will first take the derivative of the outside function, so e to the natural log of a times x, with respect to the inside function, with respect to natural log of a times x. And so this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well, natural log of a might not immediately jump out to you but that's just going to be a number."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we will do is we will first take the derivative of the outside function, so e to the natural log of a times x, with respect to the inside function, with respect to natural log of a times x. And so this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well, natural log of a might not immediately jump out to you but that's just going to be a number. So that's just going to be, so times, if it was the derivative of three x, it would just be three. If it's the derivative of natural log of a times x, it's just going to be natural log, natural log of a. And so this is going to give us the natural log of a times e to the natural log of a, and I'm going to write it like this, natural log of a to the x power."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, natural log of a might not immediately jump out to you but that's just going to be a number. So that's just going to be, so times, if it was the derivative of three x, it would just be three. If it's the derivative of natural log of a times x, it's just going to be natural log, natural log of a. And so this is going to give us the natural log of a times e to the natural log of a, and I'm going to write it like this, natural log of a to the x power. Well, we've already seen this, let me, this right over here is just a. So it all simplifies, it all simplifies to the natural log of a times, times a to the x, which is a pretty neat result. So if you're taking the derivative of e to the x, it's just going to be e to the x."}, {"video_title": "Exponential functions differentiation intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to give us the natural log of a times e to the natural log of a, and I'm going to write it like this, natural log of a to the x power. Well, we've already seen this, let me, this right over here is just a. So it all simplifies, it all simplifies to the natural log of a times, times a to the x, which is a pretty neat result. So if you're taking the derivative of e to the x, it's just going to be e to the x. If you're taking the derivative of the natural log, if you're taking the derivative of a to the x, it's just going to be the natural log of a times a to the x. And so we can now use this result to actually take the derivatives of, of these types of expressions with bases other than e. So if I want to find the derivative with respect to x of eight times three to the x power, well, what's that going to be? Well, that's just going to be eight times, and then the derivative of this right over here is going to be, based on what we just saw, it's going to be the natural log of our base, natural log of three times three to the x, times three to the x."}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let g be the function g of x is equal to one over x. Let capital F be a function defined as the product of those other two functions. What is capital F prime of negative one? Well, we can just apply the product rule here. Let me just rewrite, let me just essentially state the product rule. Capital F prime of x is going to be equal to, since capital F of x is the product of these two functions, we apply the product rule, this is going to be f prime of x times g of x plus f of x times g prime of x. If we want to evaluate this at negative one, capital F prime at negative one is equal to f prime of negative one times g of negative one plus function f evaluated at negative one times the derivative of g evaluated at negative one."}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can just apply the product rule here. Let me just rewrite, let me just essentially state the product rule. Capital F prime of x is going to be equal to, since capital F of x is the product of these two functions, we apply the product rule, this is going to be f prime of x times g of x plus f of x times g prime of x. If we want to evaluate this at negative one, capital F prime at negative one is equal to f prime of negative one times g of negative one plus function f evaluated at negative one times the derivative of g evaluated at negative one. Now let's see if we can figure these things out. Do they tell us this anywhere? Can we figure this out, f prime of negative one?"}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we want to evaluate this at negative one, capital F prime at negative one is equal to f prime of negative one times g of negative one plus function f evaluated at negative one times the derivative of g evaluated at negative one. Now let's see if we can figure these things out. Do they tell us this anywhere? Can we figure this out, f prime of negative one? Well, they tell us right over here, f prime of negative one is equal to five. This is equal to five. Now, let's actually stick with f. What is f of negative one?"}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can we figure this out, f prime of negative one? Well, they tell us right over here, f prime of negative one is equal to five. This is equal to five. Now, let's actually stick with f. What is f of negative one? Well, they tell that to us right over here. f of negative one is equal to three. So f of negative one is equal to three."}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, let's actually stick with f. What is f of negative one? Well, they tell that to us right over here. f of negative one is equal to three. So f of negative one is equal to three. Now, g of negative one and g prime of negative one, they don't give it to us explicitly here, but we can figure it out. We know that, well, if g of x is equal to this, g of negative one is equal to one over negative one, which is equal to negative one. So this is equal to negative one."}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of negative one is equal to three. Now, g of negative one and g prime of negative one, they don't give it to us explicitly here, but we can figure it out. We know that, well, if g of x is equal to this, g of negative one is equal to one over negative one, which is equal to negative one. So this is equal to negative one. And then last but not least, if we want to find g prime of negative one, we just have to take the derivative of this. So g prime of x, actually, let me just rewrite g of x. g of x, one over x is just the same thing as x to the negative one. So we're gonna use a power rule to figure out g prime of x is equal to, bring that exponent out front, negative one times x to the, and then decrement the exponent negative two power."}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to negative one. And then last but not least, if we want to find g prime of negative one, we just have to take the derivative of this. So g prime of x, actually, let me just rewrite g of x. g of x, one over x is just the same thing as x to the negative one. So we're gonna use a power rule to figure out g prime of x is equal to, bring that exponent out front, negative one times x to the, and then decrement the exponent negative two power. So g prime of negative one, of negative one is equal to negative one times negative one to the negative two power. And that's just the same thing as negative one over negative one squared. This is one, so this is just all going to evaluate to negative one."}, {"video_title": "Worked example Product rule with mixed implicit & explicit   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna use a power rule to figure out g prime of x is equal to, bring that exponent out front, negative one times x to the, and then decrement the exponent negative two power. So g prime of negative one, of negative one is equal to negative one times negative one to the negative two power. And that's just the same thing as negative one over negative one squared. This is one, so this is just all going to evaluate to negative one. So this is negative one. And so we have five times negative one, which is negative five, plus three times negative one, which is negative three, which is equal to negative eight. So f prime of negative one is equal to negative eight."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Pause the video and try to figure that out. All right, now let's do this together. So we know that a function is decreasing when its derivative is negative. Or another way to say it, it's going to be decreasing when f prime of x is less than zero. So what is f prime of x? Well, we could use the derivative rules and derivative properties we know. We apply the power rule to x to the sixth."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Or another way to say it, it's going to be decreasing when f prime of x is less than zero. So what is f prime of x? Well, we could use the derivative rules and derivative properties we know. We apply the power rule to x to the sixth. We bring the six out front or multiply the one coefficient here times six to get six x to the fifth, decrement that exponent, minus, bring the five times the three, minus 15 x to the, we decrement the five, so x to the fourth. And we need to figure out over what intervals is this going to be less than zero? And now let's see how we can simplify this a little bit."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "We apply the power rule to x to the sixth. We bring the six out front or multiply the one coefficient here times six to get six x to the fifth, decrement that exponent, minus, bring the five times the three, minus 15 x to the, we decrement the five, so x to the fourth. And we need to figure out over what intervals is this going to be less than zero? And now let's see how we can simplify this a little bit. Both of these terms are divisible by x to the fourth and they're both divisible by three. So let's factor out a three x to the fourth times, you factor out a three x to the fourth here, you're left with a two x. And then over here you have minus five has to be less than zero."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "And now let's see how we can simplify this a little bit. Both of these terms are divisible by x to the fourth and they're both divisible by three. So let's factor out a three x to the fourth times, you factor out a three x to the fourth here, you're left with a two x. And then over here you have minus five has to be less than zero. Any interval where this is true, we are going to be decreasing. Now, how do we get this to be less than zero? Well, if I take the product of two things and it's less than zero, that means that they have to have different signs, either one's positive and the other is negative or one's negative and the other's positive."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "And then over here you have minus five has to be less than zero. Any interval where this is true, we are going to be decreasing. Now, how do we get this to be less than zero? Well, if I take the product of two things and it's less than zero, that means that they have to have different signs, either one's positive and the other is negative or one's negative and the other's positive. So we have two situations. So we could say either, either three x to the fourth is greater than zero and, and two x minus five is less than zero. So that's one situation."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Well, if I take the product of two things and it's less than zero, that means that they have to have different signs, either one's positive and the other is negative or one's negative and the other's positive. So we have two situations. So we could say either, either three x to the fourth is greater than zero and, and two x minus five is less than zero. So that's one situation. I'll do this in a different color. Or, and I'll do this one in a different color, three x to the fourth is less than zero and two x minus five is greater than zero. Actually, let me stay on the second case first."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "So that's one situation. I'll do this in a different color. Or, and I'll do this one in a different color, three x to the fourth is less than zero and two x minus five is greater than zero. Actually, let me stay on the second case first. Are there any situations where three x to the fourth can be less than zero? You take any number, you take it to the fourth power, even if it's a negative, it's going to become a positive. So you can't get a negative expression right over here."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Actually, let me stay on the second case first. Are there any situations where three x to the fourth can be less than zero? You take any number, you take it to the fourth power, even if it's a negative, it's going to become a positive. So you can't get a negative expression right over here. So actually, the second condition is impossible to obtain. You can't get any situation for any x where three x to the fourth is less than zero. So we can rule this one out."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "So you can't get a negative expression right over here. So actually, the second condition is impossible to obtain. You can't get any situation for any x where three x to the fourth is less than zero. So we can rule this one out. And so this is our best hope. So under what conditions is three x to the fourth greater than zero? Well, if you divide both sides by three, you get x to the fourth is greater than zero."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "So we can rule this one out. And so this is our best hope. So under what conditions is three x to the fourth greater than zero? Well, if you divide both sides by three, you get x to the fourth is greater than zero. And if you think about it, this is gonna be true for any x value that is not equal to zero. Even if you have a negative value there, if you have a negative one, you take it to the fourth power, it becomes a positive one. Only zero will be equal to zero when you take it to the fourth power."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Well, if you divide both sides by three, you get x to the fourth is greater than zero. And if you think about it, this is gonna be true for any x value that is not equal to zero. Even if you have a negative value there, if you have a negative one, you take it to the fourth power, it becomes a positive one. Only zero will be equal to zero when you take it to the fourth power. So one way we could say this is going to be true for any non-zero x, or we could just say x does not equal zero. And this is a little bit more straightforward. We add five to both sides, we get two x is less than five."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Only zero will be equal to zero when you take it to the fourth power. So one way we could say this is going to be true for any non-zero x, or we could just say x does not equal zero. And this is a little bit more straightforward. We add five to both sides, we get two x is less than five. Divide both sides by two, you get x is less than 5 1\u20442. So it might be tempting to say, all right, the intervals that matter are all the x's less than 5 1\u20442, but x cannot be equal to zero. Now, is that the entire interval where our function is decreasing?"}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "We add five to both sides, we get two x is less than five. Divide both sides by two, you get x is less than 5 1\u20442. So it might be tempting to say, all right, the intervals that matter are all the x's less than 5 1\u20442, but x cannot be equal to zero. Now, is that the entire interval where our function is decreasing? Well, let's think about what happens at zero itself. We're decreasing over the interval from negative infinity all the way up to zero, and we're also decreasing from zero to 5 1\u20442. And so if we're decreasing right to the left of zero, we're decreasing right to the right of zero, we're actually going to be decreasing at zero at, we're also going to be decreasing at zero as well."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "Now, is that the entire interval where our function is decreasing? Well, let's think about what happens at zero itself. We're decreasing over the interval from negative infinity all the way up to zero, and we're also decreasing from zero to 5 1\u20442. And so if we're decreasing right to the left of zero, we're decreasing right to the right of zero, we're actually going to be decreasing at zero at, we're also going to be decreasing at zero as well. So there's something interesting here. Even though the derivative at x equals zero is going to be equal to zero, we are still decreasing there. And so the interval that we care about, the interval over which we're decreasing is just x is less than 5 1\u20442."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "And so if we're decreasing right to the left of zero, we're decreasing right to the right of zero, we're actually going to be decreasing at zero at, we're also going to be decreasing at zero as well. So there's something interesting here. Even though the derivative at x equals zero is going to be equal to zero, we are still decreasing there. And so the interval that we care about, the interval over which we're decreasing is just x is less than 5 1\u20442. And we can see that by graphing the function. I graphed it on Desmos. And you can see here that the function is decreasing from negative infinity, it's decreasing at a slower and slower rate, we get to zero, still decreasing to the left of zero, and then it continues to decrease to the right of zero."}, {"video_title": "Finding decreasing interval given the function   Calculus   Khan Academy.mp3", "Sentence": "And so the interval that we care about, the interval over which we're decreasing is just x is less than 5 1\u20442. And we can see that by graphing the function. I graphed it on Desmos. And you can see here that the function is decreasing from negative infinity, it's decreasing at a slower and slower rate, we get to zero, still decreasing to the left of zero, and then it continues to decrease to the right of zero. So any x value to the left of zero, the value of the function is going to be larger than f of zero. And x to the right of zero, the value of the function is going to be less than the function at zero. So it's actually decreasing through zero, even though the slope of the tangent line at zero is zero, even though it's non-negative."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I want to do in this video is start with the abstract, I guess you could call it formula for the chain rule, and then learn to apply it in the concrete setting. So let's start off with some function, some expression, that can be expressed as the product of two functions, or the composition of two functions. So it can be expressed as f of g of x. So it's a function that can be expressed as a composition or expression that can be expressed as a composition of two functions. Let me get that same color. I want the colors to be accurate. And my goal is to take the derivative of this business, the derivative with respect to x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's a function that can be expressed as a composition or expression that can be expressed as a composition of two functions. Let me get that same color. I want the colors to be accurate. And my goal is to take the derivative of this business, the derivative with respect to x. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. And we can write that as f prime of not x, but f prime of g of x, of the inner function, f prime of g of x times the derivative of the inner function with respect to x. Now this might seem all very abstract and mathy."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And my goal is to take the derivative of this business, the derivative with respect to x. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. And we can write that as f prime of not x, but f prime of g of x, of the inner function, f prime of g of x times the derivative of the inner function with respect to x. Now this might seem all very abstract and mathy. How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this might seem all very abstract and mathy. How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so that this really is the composition of f of x and g of x? Well, we could define f of x as being equal to the square root of x. And if we defined g of x as being equal to 3x squared minus x, then what is f of g of x?"}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so that this really is the composition of f of x and g of x? Well, we could define f of x as being equal to the square root of x. And if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to, and I'm going to try to keep all the colors accurate. Hopefully it'll help with the understanding. f of g of x is equal to, where everywhere you see the x, you replace with the g of x, the principal root of g of x, which is equal to the principal root of, we've defined g of x right over here, 3x squared minus x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to, and I'm going to try to keep all the colors accurate. Hopefully it'll help with the understanding. f of g of x is equal to, where everywhere you see the x, you replace with the g of x, the principal root of g of x, which is equal to the principal root of, we've defined g of x right over here, 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "f of g of x is equal to, where everywhere you see the x, you replace with the g of x, the principal root of g of x, which is equal to the principal root of, we've defined g of x right over here, 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to, this is the same thing as x to the 1 half power, so we can just apply the power rule."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to, this is the same thing as x to the 1 half power, so we can just apply the power rule. So it's going to be 1 half times x to the, and then we just take 1 away from the exponent, 1 half minus 1 is negative 1 half. And so what is f prime of g of x? f prime of g of x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "f prime of x is equal to, this is the same thing as x to the 1 half power, so we can just apply the power rule. So it's going to be 1 half times x to the, and then we just take 1 away from the exponent, 1 half minus 1 is negative 1 half. And so what is f prime of g of x? f prime of g of x. Well, where everywhere in the derivative we saw an x, we can replace it with a g of x. It's going to be 1 half times, instead of an x to the negative 1 half, we can write a g of x. We can write a g of x to the 1 half."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "f prime of g of x. Well, where everywhere in the derivative we saw an x, we can replace it with a g of x. It's going to be 1 half times, instead of an x to the negative 1 half, we can write a g of x. We can write a g of x to the 1 half. And this is just going to be equal to, let me write it right over here, it's going to be equal to 1 half times all of this business to the negative 1 half power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can write a g of x to the 1 half. And this is just going to be equal to, let me write it right over here, it's going to be equal to 1 half times all of this business to the negative 1 half power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here, I will, let me square it off in green. What we're trying to solve right over here, f prime of g of x, we have just figured out, is exactly this thing right over here. It's the derivative of f of the outer function with respect to the inner function."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "f prime of g of x is equal to this. So this part right over here, I will, let me square it off in green. What we're trying to solve right over here, f prime of g of x, we have just figured out, is exactly this thing right over here. It's the derivative of f of the outer function with respect to the inner function. So let me write it. It is equal to 1 half times g of x to the negative 1 half times 3x squared minus x. This is exactly this, based on how we've defined f of x and how we've defined g of x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's the derivative of f of the outer function with respect to the inner function. So let me write it. It is equal to 1 half times g of x to the negative 1 half times 3x squared minus x. This is exactly this, based on how we've defined f of x and how we've defined g of x. Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1 half power. So the derivative of that whole thing with respect to your something is going to be 1 half times that something to the negative 1 half power. That's essentially what we're saying."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is exactly this, based on how we've defined f of x and how we've defined g of x. Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1 half power. So the derivative of that whole thing with respect to your something is going to be 1 half times that something to the negative 1 half power. That's essentially what we're saying. But now we have to take the derivative of our something with respect to x. The derivative of our something with respect to x. And that's more straightforward."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's essentially what we're saying. But now we have to take the derivative of our something with respect to x. The derivative of our something with respect to x. And that's more straightforward. g prime of x, we just use the power rule for each of these terms, is equal to 6x to the first, or just 6x, minus 1. So this part right over here is just going to be 6x minus 1. Just to be clear."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's more straightforward. g prime of x, we just use the power rule for each of these terms, is equal to 6x to the first, or just 6x, minus 1. So this part right over here is just going to be 6x minus 1. Just to be clear. This right over here is this right over here. And we're multiplying. And we're done."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Just to be clear. This right over here is this right over here. And we're multiplying. And we're done. We have just applied the power rule. So just to review, it's the derivative of the outer function with respect to the inner. So instead of having 1 half x to the negative 1 half, it's 1 half g of x to the negative 1 half times the derivative of the inner function with respect to x times the derivative of g with respect to x, which is right over there."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we'll assume that the top of the ladder kind of glides along the side of the wall. It stays kind of in contact with the wall and moves straight down. And we see right over here, the arrow is moving straight down. And our question is, how fast is it moving straight down at that moment? So let's think about this a little bit. What do we know and what do we not know? So if we call the distance between the base of the wall and the base of the ladder, let's call that x."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And our question is, how fast is it moving straight down at that moment? So let's think about this a little bit. What do we know and what do we not know? So if we call the distance between the base of the wall and the base of the ladder, let's call that x. We know right now x is equal to 8 feet. We also know the rate at which x is changing with respect to time. The rate at which x is changing with respect to time is 4 feet per second."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we call the distance between the base of the wall and the base of the ladder, let's call that x. We know right now x is equal to 8 feet. We also know the rate at which x is changing with respect to time. The rate at which x is changing with respect to time is 4 feet per second. So we could call this dx dt. Now let's call the distance between the top of the ladder and the base of the ladder h. Let's call that h. So what we're really trying to figure out is what dh dt is, given that we know all of this other information. So let's see if we can come up with a relationship between x and h, and then take the derivative with respect to time, maybe using the chain rule, and see if we can solve for dh dt knowing all of this information."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The rate at which x is changing with respect to time is 4 feet per second. So we could call this dx dt. Now let's call the distance between the top of the ladder and the base of the ladder h. Let's call that h. So what we're really trying to figure out is what dh dt is, given that we know all of this other information. So let's see if we can come up with a relationship between x and h, and then take the derivative with respect to time, maybe using the chain rule, and see if we can solve for dh dt knowing all of this information. Well, we know the relationship between x and h at any time because of the Pythagorean theorem. We can assume this is a right angle. So we know that x squared plus h squared is going to be equal to the length of the ladder squared, is going to be equal to 100."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can come up with a relationship between x and h, and then take the derivative with respect to time, maybe using the chain rule, and see if we can solve for dh dt knowing all of this information. Well, we know the relationship between x and h at any time because of the Pythagorean theorem. We can assume this is a right angle. So we know that x squared plus h squared is going to be equal to the length of the ladder squared, is going to be equal to 100. And what we care about is the rate at which these things change with respect to time. So let's take the derivative with respect to time of both sides of this. So we're doing a little bit of implicit differentiation."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that x squared plus h squared is going to be equal to the length of the ladder squared, is going to be equal to 100. And what we care about is the rate at which these things change with respect to time. So let's take the derivative with respect to time of both sides of this. So we're doing a little bit of implicit differentiation. So what's the derivative with respect to time of x squared? Well, the derivative of x squared with respect to x is 2x. And we're going to have to multiply that times the derivative of x with respect to t, dx dt."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're doing a little bit of implicit differentiation. So what's the derivative with respect to time of x squared? Well, the derivative of x squared with respect to x is 2x. And we're going to have to multiply that times the derivative of x with respect to t, dx dt. Just to be clear, this is a chain rule. This is the derivative of x squared with respect to x, which is 2x, times dx dt to get the derivative of x squared with respect to time, just the chain rule. Now similarly, what's the derivative of h squared with respect to time?"}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to have to multiply that times the derivative of x with respect to t, dx dt. Just to be clear, this is a chain rule. This is the derivative of x squared with respect to x, which is 2x, times dx dt to get the derivative of x squared with respect to time, just the chain rule. Now similarly, what's the derivative of h squared with respect to time? Well, that's just going to be 2h. The derivative of h squared with respect to h is 2h times the derivative of h with respect to time. Once again, this right over here is the derivative of h squared with respect to h times the derivative of h with respect to time, which gives us the derivative of h squared with respect to time."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now similarly, what's the derivative of h squared with respect to time? Well, that's just going to be 2h. The derivative of h squared with respect to h is 2h times the derivative of h with respect to time. Once again, this right over here is the derivative of h squared with respect to h times the derivative of h with respect to time, which gives us the derivative of h squared with respect to time. And what do we get on the right-hand side of our equation? Well, the length of our ladder isn't changing. This 100 isn't going to change with respect to time."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, this right over here is the derivative of h squared with respect to h times the derivative of h with respect to time, which gives us the derivative of h squared with respect to time. And what do we get on the right-hand side of our equation? Well, the length of our ladder isn't changing. This 100 isn't going to change with respect to time. The derivative of a constant is just equal 0. So now we have it, a relationship between the rate of change of h with respect to time, the rate of change of x with respect to time, and then at a given point in time when the length of x is x and h is h. But do we know what h is when x is equal to 8 feet? Well, we can figure it out."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This 100 isn't going to change with respect to time. The derivative of a constant is just equal 0. So now we have it, a relationship between the rate of change of h with respect to time, the rate of change of x with respect to time, and then at a given point in time when the length of x is x and h is h. But do we know what h is when x is equal to 8 feet? Well, we can figure it out. When x is equal to 8 feet, we can use the Pythagorean theorem again. We get 8 feet squared plus h squared is going to be equal to 100. So 8 squared is 64."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can figure it out. When x is equal to 8 feet, we can use the Pythagorean theorem again. We get 8 feet squared plus h squared is going to be equal to 100. So 8 squared is 64. Subtract it from both sides. You get 8 squared is equal to 36. Take the positive square root."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 8 squared is 64. Subtract it from both sides. You get 8 squared is equal to 36. Take the positive square root. A negative square root doesn't make sense because then the ladder would be below the ground. It would be somehow underground. So we get h is equal to 6."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Take the positive square root. A negative square root doesn't make sense because then the ladder would be below the ground. It would be somehow underground. So we get h is equal to 6. So this is something that was essentially given by the problem. So now we know. We can look at this original thing right over here."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get h is equal to 6. So this is something that was essentially given by the problem. So now we know. We can look at this original thing right over here. We know what x is. That was given. Right now, x is 8 feet."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can look at this original thing right over here. We know what x is. That was given. Right now, x is 8 feet. We know the rate of change of x with respect to time. It's 4 feet per second. We know what h is right now."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right now, x is 8 feet. We know the rate of change of x with respect to time. It's 4 feet per second. We know what h is right now. It is 6. So then we can solve for the rate of h with respect to time. So let's do that."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what h is right now. It is 6. So then we can solve for the rate of h with respect to time. So let's do that. So we get 2 times 8 feet times 4 feet per second plus 2h. So plus 2h is going to be plus 2 times our height right now is 6 times the rate at which our height is changing with respect to t is equal to 0. And so we get 2 times 8 times 4 is 64 plus 12 dh dt is equal to 0."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So we get 2 times 8 feet times 4 feet per second plus 2h. So plus 2h is going to be plus 2 times our height right now is 6 times the rate at which our height is changing with respect to t is equal to 0. And so we get 2 times 8 times 4 is 64 plus 12 dh dt is equal to 0. We can subtract 64 from both sides. We get 12 times the derivative of h with respect to time is equal to negative 64. And then we just have to divide both sides by 12."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we get 2 times 8 times 4 is 64 plus 12 dh dt is equal to 0. We can subtract 64 from both sides. We get 12 times the derivative of h with respect to time is equal to negative 64. And then we just have to divide both sides by 12. And so now we get a little bit of a drum roll. The derivative, the rate of change of h with respect to time is equal to negative 64 divided by 12 is equal to negative 64 over 12, which is the same thing as negative 16 over 3. Yep, that's right, which is equal to negative 5 and 1 third feet per second."}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we just have to divide both sides by 12. And so now we get a little bit of a drum roll. The derivative, the rate of change of h with respect to time is equal to negative 64 divided by 12 is equal to negative 64 over 12, which is the same thing as negative 16 over 3. Yep, that's right, which is equal to negative 5 and 1 third feet per second. So we're done. But let's just do a reality check. Does that make sense that we got a negative value right over here?"}, {"video_title": "Related rates Falling ladder   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yep, that's right, which is equal to negative 5 and 1 third feet per second. So we're done. But let's just do a reality check. Does that make sense that we got a negative value right over here? Well, our height is decreasing. So it completely makes sense that its rate of change is indeed negative. And we're done."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is u substitution a possibility here? Well for u substitution we want to look for an expression and its derivative. Well what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? Well du is going to be the derivative of the natural log of x with respect to x, which is just one over x dx. This is an equivalent statement to saying that du dx is equal to one over x. So do we see a one over x dx anywhere in this original expression?"}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what would du be equal to in that scenario? Well du is going to be the derivative of the natural log of x with respect to x, which is just one over x dx. This is an equivalent statement to saying that du dx is equal to one over x. So do we see a one over x dx anywhere in this original expression? Well it's kind of hiding, it's not so obvious, but this x in the denominator is essentially a one over x and then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi, I should do that in a different color since I've already used, let me take the pi and just stick it out front."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So do we see a one over x dx anywhere in this original expression? Well it's kind of hiding, it's not so obvious, but this x in the denominator is essentially a one over x and then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi, I should do that in a different color since I've already used, let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral and so this becomes the integral of, and let me write the one over natural log of x first, one over the natural log of x times one over x dx. Now it becomes a little bit clearer. These are completely equivalent statements."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first thing I'm going to do is I'm going to take the pi, I should do that in a different color since I've already used, let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral and so this becomes the integral of, and let me write the one over natural log of x first, one over the natural log of x times one over x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that yes, u substitution will work over here. If we set our u equal to natural log of x, then our du is one over x dx, one over x dx. Our du is one over x dx."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "These are completely equivalent statements. But this makes it clear that yes, u substitution will work over here. If we set our u equal to natural log of x, then our du is one over x dx, one over x dx. Our du is one over x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of one over u, natural log of x is u, we set that equal to natural log of x, times du, times du. Now this becomes pretty straightforward."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our du is one over x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of one over u, natural log of x is u, we set that equal to natural log of x, times du, times du. Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log, the natural log of the absolute value of u, so that we can handle even negative values of u, the natural log of the absolute value of u plus c, just in case we had a constant factor out here, plus c. And we're almost done."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log, the natural log of the absolute value of u, so that we can handle even negative values of u, the natural log of the absolute value of u plus c, just in case we had a constant factor out here, plus c. And we're almost done. We just have to unsubstitute for the u. U is equal to natural log of x. So we end up with this kind of neat looking expression. This entire indefinite integral we have simplified, we have evaluated it."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be equal to pi times the natural log, the natural log of the absolute value of u, so that we can handle even negative values of u, the natural log of the absolute value of u plus c, just in case we had a constant factor out here, plus c. And we're almost done. We just have to unsubstitute for the u. U is equal to natural log of x. So we end up with this kind of neat looking expression. This entire indefinite integral we have simplified, we have evaluated it. And it is now equal to pi times the natural log, the natural log of the absolute value of u, but u is just the natural log of x. The natural log of x, and then we have this plus c right over here. And we could have assumed that from the get-go, this original expression was only defined for positive values of x, because you have to take the natural log here and it wasn't an absolute value."}, {"video_title": "_-substitution logarithmic function   AP Calculus AB   Khan Academy.mp3", "Sentence": "This entire indefinite integral we have simplified, we have evaluated it. And it is now equal to pi times the natural log, the natural log of the absolute value of u, but u is just the natural log of x. The natural log of x, and then we have this plus c right over here. And we could have assumed that from the get-go, this original expression was only defined for positive values of x, because you have to take the natural log here and it wasn't an absolute value. But now, so we can leave this as just a natural log of x. But this also works for the situations now, because we're taking the absolute value of that, where the natural log of x might have been a negative number. For example, if it was a natural log of.5, or who knows, whatever it might be."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is start to think about how we describe position in one dimension as a function of time. So we could say our position, and we're gonna think about position on the x-axis as a function of time, and we could define it by some expression. Let's say in this situation, it is going to be our time to the third power minus three times our time squared plus five, and this is going to apply for our time being non-negative because the idea of negative time, at least for now, is a bit strange. So let's think about what this right over here is describing, and to help us do that, we could set up a little bit of a table to understand that depending on what time we are, let's say that time is in seconds, what is going to be our position along our x-axis? So at time equals zero, x of zero is just going to be five. At time one, you're gonna have one minus three plus five, so that is going to be, see, one minus three is negative two plus five is going to be, we're gonna be at position three, and then at time two, we are going to be at eight minus 12 plus five, so we're going to be at position one, and then at time t equals three, it's gonna be 27 minus 27 plus five. We're gonna be back at five, and so this can at least help us understand what's going on for the first three seconds."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what this right over here is describing, and to help us do that, we could set up a little bit of a table to understand that depending on what time we are, let's say that time is in seconds, what is going to be our position along our x-axis? So at time equals zero, x of zero is just going to be five. At time one, you're gonna have one minus three plus five, so that is going to be, see, one minus three is negative two plus five is going to be, we're gonna be at position three, and then at time two, we are going to be at eight minus 12 plus five, so we're going to be at position one, and then at time t equals three, it's gonna be 27 minus 27 plus five. We're gonna be back at five, and so this can at least help us understand what's going on for the first three seconds. So let me draw our positive x-axis. So, say it looks something like that, and this is x equals zero, this is our x-axis, x equals one, two, three, four, and five. And now let's play out how this particle that's being described is moving along the x-axis."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna be back at five, and so this can at least help us understand what's going on for the first three seconds. So let me draw our positive x-axis. So, say it looks something like that, and this is x equals zero, this is our x-axis, x equals one, two, three, four, and five. And now let's play out how this particle that's being described is moving along the x-axis. So we're gonna start here, and we're gonna go one, two, three. Let's do it again. We're going to go one, two, three."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's play out how this particle that's being described is moving along the x-axis. So we're gonna start here, and we're gonna go one, two, three. Let's do it again. We're going to go one, two, three. The way I just moved my mouse, if we assume that I got the time roughly right, is how that particle would move. And we can graph this as well. So, for example, it would look like this."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to go one, two, three. The way I just moved my mouse, if we assume that I got the time roughly right, is how that particle would move. And we can graph this as well. So, for example, it would look like this. We are starting at time t equals zero. Our position, this is our vertical axis, our y-axis, but we're just saying y is going to be equal to our position along the x-axis. So that's a little bit counterintuitive, because we're talking about our position in the left-right dimension, and here you're seeing it start off in the vertical dimension."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, for example, it would look like this. We are starting at time t equals zero. Our position, this is our vertical axis, our y-axis, but we're just saying y is going to be equal to our position along the x-axis. So that's a little bit counterintuitive, because we're talking about our position in the left-right dimension, and here you're seeing it start off in the vertical dimension. But you see the same thing. At time t equals one, our position has gone down to three, then it goes down further. At time equals two, our position is down to one."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's a little bit counterintuitive, because we're talking about our position in the left-right dimension, and here you're seeing it start off in the vertical dimension. But you see the same thing. At time t equals one, our position has gone down to three, then it goes down further. At time equals two, our position is down to one. And then we switch direction, and then over the next, if we say that time is in seconds, over the next second, we get back to five. Now, an interesting thing to think about in the context of calculus is, well, what is our velocity at any point in time? And velocity, as you might remember, is the derivative of position."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "At time equals two, our position is down to one. And then we switch direction, and then over the next, if we say that time is in seconds, over the next second, we get back to five. Now, an interesting thing to think about in the context of calculus is, well, what is our velocity at any point in time? And velocity, as you might remember, is the derivative of position. So let me write that down. So we're gonna be thinking about velocity as a function of time. And you could view velocity as the first derivative of position with respect to time, which is just going to be equal to, we're gonna apply the power rule and some derivative properties multiple times."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And velocity, as you might remember, is the derivative of position. So let me write that down. So we're gonna be thinking about velocity as a function of time. And you could view velocity as the first derivative of position with respect to time, which is just going to be equal to, we're gonna apply the power rule and some derivative properties multiple times. If this is unfamiliar to you, I encourage you to review it. But this is going to be three t squared minus six t, and then plus zero, and we're going to restrict the domain as well, for t is greater than or equal to zero. And then we can plot that."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could view velocity as the first derivative of position with respect to time, which is just going to be equal to, we're gonna apply the power rule and some derivative properties multiple times. If this is unfamiliar to you, I encourage you to review it. But this is going to be three t squared minus six t, and then plus zero, and we're going to restrict the domain as well, for t is greater than or equal to zero. And then we can plot that. It would look like that. Now, let's see if this curve makes intuitive sense. We mentioned that one second, two second, three seconds."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can plot that. It would look like that. Now, let's see if this curve makes intuitive sense. We mentioned that one second, two second, three seconds. So we're starting moving to the left. And the convention is, if you're moving to the left, you have negative velocity. And if you're moving to the right, you have positive velocity."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We mentioned that one second, two second, three seconds. So we're starting moving to the left. And the convention is, if you're moving to the left, you have negative velocity. And if you're moving to the right, you have positive velocity. And you can see here, our velocity immediately gets more and more negative until we get to one second. And then it stays negative, but it's getting less and less negative until we get to two seconds. And at two seconds, our velocity becomes positive."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you're moving to the right, you have positive velocity. And you can see here, our velocity immediately gets more and more negative until we get to one second. And then it stays negative, but it's getting less and less negative until we get to two seconds. And at two seconds, our velocity becomes positive. And that makes sense, because at two seconds was when our velocity switched directions to the rightward direction. So our velocity is getting more and more negative, less and less negative, and then we switch directions, and we go just like that. And we see it right over here."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And at two seconds, our velocity becomes positive. And that makes sense, because at two seconds was when our velocity switched directions to the rightward direction. So our velocity is getting more and more negative, less and less negative, and then we switch directions, and we go just like that. And we see it right over here. Now, one thing to keep in mind when we're thinking about velocity as a function of time is that velocity and speed are two different things. Speed, speed, let me write it over here. Speed is equal to, if you're thinking about it in one dimension, you could think about it as the absolute value of your velocity as a function of time, or your magnitude of velocity as a function of time."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we see it right over here. Now, one thing to keep in mind when we're thinking about velocity as a function of time is that velocity and speed are two different things. Speed, speed, let me write it over here. Speed is equal to, if you're thinking about it in one dimension, you could think about it as the absolute value of your velocity as a function of time, or your magnitude of velocity as a function of time. So in the beginning, even though your velocity is becoming more and more negative, your speed is actually increasing. Your speed is increasing to the left, then your speed is decreasing, you slow down, and then your speed is increasing as we go to the right. And we'll do some worked examples that work through that a little bit more."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Speed is equal to, if you're thinking about it in one dimension, you could think about it as the absolute value of your velocity as a function of time, or your magnitude of velocity as a function of time. So in the beginning, even though your velocity is becoming more and more negative, your speed is actually increasing. Your speed is increasing to the left, then your speed is decreasing, you slow down, and then your speed is increasing as we go to the right. And we'll do some worked examples that work through that a little bit more. Now, the last concept we will touch on in this video is the idea of acceleration. And acceleration, you could view as the rate of change of velocity with respect to time. So acceleration as a function of time is just going to be the first derivative of velocity with respect to time, which is equal to the second derivative of position with respect to time."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we'll do some worked examples that work through that a little bit more. Now, the last concept we will touch on in this video is the idea of acceleration. And acceleration, you could view as the rate of change of velocity with respect to time. So acceleration as a function of time is just going to be the first derivative of velocity with respect to time, which is equal to the second derivative of position with respect to time. It's just going to be the derivative of this expression. So once again, using the power rule here, that's going to be six t, and then using the power rule here minus six, and once again, we will restrict the domain. And we can graph that as well."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So acceleration as a function of time is just going to be the first derivative of velocity with respect to time, which is equal to the second derivative of position with respect to time. It's just going to be the derivative of this expression. So once again, using the power rule here, that's going to be six t, and then using the power rule here minus six, and once again, we will restrict the domain. And we can graph that as well. And we would see right over here, this is y is equal to acceleration as a function of time. And you can see at time equals zero, our acceleration is quite negative, it is negative six. And then it becomes less and less and less negative."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can graph that as well. And we would see right over here, this is y is equal to acceleration as a function of time. And you can see at time equals zero, our acceleration is quite negative, it is negative six. And then it becomes less and less and less negative. And then our acceleration actually becomes positive at t equals one. Now, does that make sense? Well, we're going one, two, three."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it becomes less and less and less negative. And then our acceleration actually becomes positive at t equals one. Now, does that make sense? Well, we're going one, two, three. You might say, wait, we didn't switch directions until we get to the second second. But remember, after we get to the first second, our velocity in the negative direction becomes less negative, which means that our acceleration is positive. If that's a little confusing, pause the video and really think through that."}, {"video_title": "Introduction to one-dimensional motion with calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're going one, two, three. You might say, wait, we didn't switch directions until we get to the second second. But remember, after we get to the first second, our velocity in the negative direction becomes less negative, which means that our acceleration is positive. If that's a little confusing, pause the video and really think through that. So our acceleration is negative, then positive, and then positive continues. And so this is just to give you an intuition. In the next few videos, we'll do several worked examples that help us dive deeper into this idea of studying motion and position, into this idea of studying motion in one dimension."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So are there x values where g takes on an absolute maximum value or an absolute minimum value? Sometimes you might call them a global maximum or a global minimum. So the first thing I like to think about is, well, what's the domain for which g is actually defined? And we know that in the natural log of x, the input into natural log, it has to be greater than zero. So the domain is all real numbers greater than zero. So x has to be greater than zero. Anything lateral log of zero is not defined."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know that in the natural log of x, the input into natural log, it has to be greater than zero. So the domain is all real numbers greater than zero. So x has to be greater than zero. Anything lateral log of zero is not defined. There is no power that you could take e to to get to zero. And natural log of negative numbers is not defined. So that is the domain."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Anything lateral log of zero is not defined. There is no power that you could take e to to get to zero. And natural log of negative numbers is not defined. So that is the domain. The domain is all real numbers such that all real numbers x is greater than zero. So our absolute extrema have to be within that domain. So to find these, let's see if we can find some local extrema and see if any of them are good candidates for absolute extrema."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is the domain. The domain is all real numbers such that all real numbers x is greater than zero. So our absolute extrema have to be within that domain. So to find these, let's see if we can find some local extrema and see if any of them are good candidates for absolute extrema. And we can find our local extrema by looking at critical points or critical values. So let's take the derivative of g. So g prime, I'm just gonna use a new color just for kicks. Alright, so g prime of x is equal to, we can use the product rule here."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So to find these, let's see if we can find some local extrema and see if any of them are good candidates for absolute extrema. And we can find our local extrema by looking at critical points or critical values. So let's take the derivative of g. So g prime, I'm just gonna use a new color just for kicks. Alright, so g prime of x is equal to, we can use the product rule here. So derivative of x squared, which is two x, times the natural log of x, plus x squared times the derivative of natural log of x, so that is one over x. And I can just rewrite that, x squared times one over x. And we're gonna assume that x is positive."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, so g prime of x is equal to, we can use the product rule here. So derivative of x squared, which is two x, times the natural log of x, plus x squared times the derivative of natural log of x, so that is one over x. And I can just rewrite that, x squared times one over x. And we're gonna assume that x is positive. So that is going to be, that is going to be just, and actually I didn't even have to make that assumption for what I'm about to do, x squared divided by x is just going to be x. Alright, and so that is g prime. So now let's think about the critical points. Critical points are where the derivative, they're points in the domain, so they're gonna have to satisfy x is greater than zero, such that g prime is either undefined or it is equal to zero."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're gonna assume that x is positive. So that is going to be, that is going to be just, and actually I didn't even have to make that assumption for what I'm about to do, x squared divided by x is just going to be x. Alright, and so that is g prime. So now let's think about the critical points. Critical points are where the derivative, they're points in the domain, so they're gonna have to satisfy x is greater than zero, such that g prime is either undefined or it is equal to zero. So let's first think about when g prime is equal to zero. So let's set it equal to zero. Two x natural log of x plus x is equal to zero."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Critical points are where the derivative, they're points in the domain, so they're gonna have to satisfy x is greater than zero, such that g prime is either undefined or it is equal to zero. So let's first think about when g prime is equal to zero. So let's set it equal to zero. Two x natural log of x plus x is equal to zero. Well we can subtract, we could subtract x from both sides of that. And so we get two x natural log of x is equal to negative x. See, if we divide both sides by x, and we can do that, we know x isn't gonna be zero, our domain is x is greater than zero."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two x natural log of x plus x is equal to zero. Well we can subtract, we could subtract x from both sides of that. And so we get two x natural log of x is equal to negative x. See, if we divide both sides by x, and we can do that, we know x isn't gonna be zero, our domain is x is greater than zero. So this is going to be, actually let's divide both sides by two x. So then we get the natural log of x is equal to negative 1 1 2. Negative x divided by two x is negative 1 1 2."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "See, if we divide both sides by x, and we can do that, we know x isn't gonna be zero, our domain is x is greater than zero. So this is going to be, actually let's divide both sides by two x. So then we get the natural log of x is equal to negative 1 1 2. Negative x divided by two x is negative 1 1 2. Or we could say that x is equal to e to the negative 1 1 2 is equal to x. Remember, natural log is just log base e. So e to the negative 1 1 2, which we could also write like that, e to the negative 1 1 2, or one over the square root of e. So that's a point at which g, at which our derivative, I should say, is equal to zero. It is a critical point or critical value for our original function g. So, and that's the only place where g prime is equal to zero."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative x divided by two x is negative 1 1 2. Or we could say that x is equal to e to the negative 1 1 2 is equal to x. Remember, natural log is just log base e. So e to the negative 1 1 2, which we could also write like that, e to the negative 1 1 2, or one over the square root of e. So that's a point at which g, at which our derivative, I should say, is equal to zero. It is a critical point or critical value for our original function g. So, and that's the only place where g prime is equal to zero. Are there any other points where g prime is undefined? And there have to be points within the domain. So let's see, what would make this undefined?"}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is a critical point or critical value for our original function g. So, and that's the only place where g prime is equal to zero. Are there any other points where g prime is undefined? And there have to be points within the domain. So let's see, what would make this undefined? Two x to the x, that you can evaluate for any x. Natural log of x, once again, is only going to be defined for x greater than zero. But that's, we've already restricted ourselves to that domain."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, what would make this undefined? Two x to the x, that you can evaluate for any x. Natural log of x, once again, is only going to be defined for x greater than zero. But that's, we've already restricted ourselves to that domain. So within the domain, any point in the domain, our derivative is actually going to be defined. So given that, let's see what's happening on either side of this critical point. On either side of this critical point."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that's, we've already restricted ourselves to that domain. So within the domain, any point in the domain, our derivative is actually going to be defined. So given that, let's see what's happening on either side of this critical point. On either side of this critical point. And I could draw a little number line here to really help us visualize this. So, if this is negative one, this is zero, this is, let's see, one e to the, this is gonna be like one over, oh boy, this is, this is going to be a little bit less than one. So let's see, let me plot one here, and then two here."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "On either side of this critical point. And I could draw a little number line here to really help us visualize this. So, if this is negative one, this is zero, this is, let's see, one e to the, this is gonna be like one over, oh boy, this is, this is going to be a little bit less than one. So let's see, let me plot one here, and then two here. And so we have a critical point at one over the square root of e. Let me put it right over there. One over the square root of e. And we know that we're only defined from, for all x is greater than zero. So let's think about the interval between zero and this critical point, right over here."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, let me plot one here, and then two here. And so we have a critical point at one over the square root of e. Let me put it right over there. One over the square root of e. And we know that we're only defined from, for all x is greater than zero. So let's think about the interval between zero and this critical point, right over here. So the open interval from zero to one over the square root of e. Let's think about whether g prime is positive or negative there. And then let's think about it for, for x greater than one over the square root of e. So that's the interval from one over the square root of e to infinity. So over that yellow interval, well let's just try out a value that is in there."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about the interval between zero and this critical point, right over here. So the open interval from zero to one over the square root of e. Let's think about whether g prime is positive or negative there. And then let's think about it for, for x greater than one over the square root of e. So that's the interval from one over the square root of e to infinity. So over that yellow interval, well let's just try out a value that is in there. So let's just try g prime of, I don't know, let's try g prime of 0.1. Z prime of 0.1 is definitely going to be in this interval. And so it's going to be equal to two point, two times 0.1 is equal to, is equal to 0.2 times the natural log of 0.1 plus 0.1."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over that yellow interval, well let's just try out a value that is in there. So let's just try g prime of, I don't know, let's try g prime of 0.1. Z prime of 0.1 is definitely going to be in this interval. And so it's going to be equal to two point, two times 0.1 is equal to, is equal to 0.2 times the natural log of 0.1 plus 0.1. And let's see, this right over here, this is going to be a negative value. In fact, it's going to be quite, it's definitely going to be greater than negative one. Because e to the negative one only gets you to, let's see, e to the negative one is one over e. So that's one over 2.7."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so it's going to be equal to two point, two times 0.1 is equal to, is equal to 0.2 times the natural log of 0.1 plus 0.1. And let's see, this right over here, this is going to be a negative value. In fact, it's going to be quite, it's definitely going to be greater than negative one. Because e to the negative one only gets you to, let's see, e to the negative one is one over e. So that's one over 2.7. So one over 2.7 is going to be, so this is going to be around, around 0.3 or 0.4. So in order to get 0.1, so this will be around 0.3 to 0.4. So in order to get to 0.1, you have to be even more negative."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because e to the negative one only gets you to, let's see, e to the negative one is one over e. So that's one over 2.7. So one over 2.7 is going to be, so this is going to be around, around 0.3 or 0.4. So in order to get 0.1, so this will be around 0.3 to 0.4. So in order to get to 0.1, you have to be even more negative. So this is going to be, this is going to be, I guess you could say, less than negative one. So if this is less than negative one, I'm multiplying it times 0.2, I'm going to get a negative value that is less than negative 0.2. And if I'm adding 0.1 to it, well, I'm still going to get a negative value."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in order to get to 0.1, you have to be even more negative. So this is going to be, this is going to be, I guess you could say, less than negative one. So if this is less than negative one, I'm multiplying it times 0.2, I'm going to get a negative value that is less than negative 0.2. And if I'm adding 0.1 to it, well, I'm still going to get a negative value. So over this yellow interval, g prime of x is less than zero. And it would be, I should have gotten a calculator, or I could have gotten a calculator out. I could have just evaluated a lot easier."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if I'm adding 0.1 to it, well, I'm still going to get a negative value. So over this yellow interval, g prime of x is less than zero. And it would be, I should have gotten a calculator, or I could have gotten a calculator out. I could have just evaluated a lot easier. So g prime of x is less than zero in this interval. And let's see in this blue interval what's going on. And this will be easier."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could have just evaluated a lot easier. So g prime of x is less than zero in this interval. And let's see in this blue interval what's going on. And this will be easier. We could just try out the value one. So g prime of one is equal to two times the natural log of one plus one. Natural log of one is just zero."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this will be easier. We could just try out the value one. So g prime of one is equal to two times the natural log of one plus one. Natural log of one is just zero. So all of this just simplifies to one. So over this blue interval, I sampled a point there, g prime of x is greater than zero. So it looks like our function is decreasing from zero to one over square root of e, and then we increase after that."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Natural log of one is just zero. So all of this just simplifies to one. So over this blue interval, I sampled a point there, g prime of x is greater than zero. So it looks like our function is decreasing from zero to one over square root of e, and then we increase after that. And we increase for all x's that are greater than one over the square root of e. And so our function is going to hit, if we're decreasing into that and then increasing after that, we're hitting a global minimum point or an absolute minimum point at x equals one over the square root of e. So let me write this down. We hit a absolute minimum absolute minimum, minimum at x equals one over the square root of e. And there is no absolute maximum. As we get above one over the square root of e, we are just going to, think about what's going to be happening here."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks like our function is decreasing from zero to one over square root of e, and then we increase after that. And we increase for all x's that are greater than one over the square root of e. And so our function is going to hit, if we're decreasing into that and then increasing after that, we're hitting a global minimum point or an absolute minimum point at x equals one over the square root of e. So let me write this down. We hit a absolute minimum absolute minimum, minimum at x equals one over the square root of e. And there is no absolute maximum. As we get above one over the square root of e, we are just going to, think about what's going to be happening here. We're just going to, one, we know that our function just keeps on increasing and increasing and increasing forever. And you can look at even this. X squared is just gonna get unbounded towards infinity."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "As we get above one over the square root of e, we are just going to, think about what's going to be happening here. We're just going to, one, we know that our function just keeps on increasing and increasing and increasing forever. And you can look at even this. X squared is just gonna get unbounded towards infinity. And natural log of x is gonna grow slower than x squared, but it's still going to go unbounded towards infinity. So there's no global or no absolute maximum, no absolute maximum point. And now let's look at the graph of this to feel good about what we just did analytically without looking at it graphically."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "X squared is just gonna get unbounded towards infinity. And natural log of x is gonna grow slower than x squared, but it's still going to go unbounded towards infinity. So there's no global or no absolute maximum, no absolute maximum point. And now let's look at the graph of this to feel good about what we just did analytically without looking at it graphically. And I looked at it ahead of time. So let me copy and paste it. And so this is the graph of our function."}, {"video_title": "Absolute minima & maxima (entire domain)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's look at the graph of this to feel good about what we just did analytically without looking at it graphically. And I looked at it ahead of time. So let me copy and paste it. And so this is the graph of our function. So as we can see, when this point right over here, this is when, this is one over the square root of e. It's not obvious from looking at it that it's that point. X equals one over the square root of e. And we can see that it is indeed an absolute minimum point here. And there is no absolute maximum point."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And we see why that is. If x is equal to positive or negative 2, then x squared is going to be equal to positive 4. And 4 minus 4 is 0. And then we're going to have a 0 in the denominator. And that's not defined. We don't know what happens when you divide. Or we've never defined what happens when you divide by 0."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we're going to have a 0 in the denominator. And that's not defined. We don't know what happens when you divide. Or we've never defined what happens when you divide by 0. So they say what value should be assigned to f of negative 2 to make f of x continuous at that point? So to think about that, let's try to actually simplify f of x. So f of x, I'll just rewrite it, is equal to."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Or we've never defined what happens when you divide by 0. So they say what value should be assigned to f of negative 2 to make f of x continuous at that point? So to think about that, let's try to actually simplify f of x. So f of x, I'll just rewrite it, is equal to. Actually, let me just start simplifying right from the get-go. So in the numerator, I can factor out a 6 out of every one of those terms. So it's 6 times x squared plus 3x plus 2 over."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So f of x, I'll just rewrite it, is equal to. Actually, let me just start simplifying right from the get-go. So in the numerator, I can factor out a 6 out of every one of those terms. So it's 6 times x squared plus 3x plus 2 over. And in the denominator, this is a difference of squares. This is x plus 2 times x minus 2. And then we can factor this expression up here."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's 6 times x squared plus 3x plus 2 over. And in the denominator, this is a difference of squares. This is x plus 2 times x minus 2. And then we can factor this expression up here. So this is going to be equal to 6 times. Let me do it a different color. So we think of two numbers that if I take their product, I get 2."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we can factor this expression up here. So this is going to be equal to 6 times. Let me do it a different color. So we think of two numbers that if I take their product, I get 2. If I take their sum, I get 3. The most obvious one is 2 and 1. So this is 6 times x plus 2 times x plus 1."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So we think of two numbers that if I take their product, I get 2. If I take their sum, I get 3. The most obvious one is 2 and 1. So this is 6 times x plus 2 times x plus 1. When you take the product there, you'll get x squared plus 3x plus 2, and then all of that over x plus 2 times x minus 2. Now, if we know that x does not equal negative 2, then we can divide both the numerator and the denominator by x plus 2. The reason why I'm making that constraint is that if x were to be equal to negative 2, then x plus 2 is going to be equal to 0."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is 6 times x plus 2 times x plus 1. When you take the product there, you'll get x squared plus 3x plus 2, and then all of that over x plus 2 times x minus 2. Now, if we know that x does not equal negative 2, then we can divide both the numerator and the denominator by x plus 2. The reason why I'm making that constraint is that if x were to be equal to negative 2, then x plus 2 is going to be equal to 0. And you won't be able to do that. You can't. We don't know what it means to divide something by 0."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "The reason why I'm making that constraint is that if x were to be equal to negative 2, then x plus 2 is going to be equal to 0. And you won't be able to do that. You can't. We don't know what it means to divide something by 0. So we could say that this is going to be equal to. So we can divide the numerator and the denominator by x plus 2. But we have to assume that x is not equal to negative 2."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "We don't know what it means to divide something by 0. So we could say that this is going to be equal to. So we can divide the numerator and the denominator by x plus 2. But we have to assume that x is not equal to negative 2. So this is equal to 6 times. So we're going to divide by x plus 2 in the numerator, x plus 2 in the denominator. So it's going to be 6 times x plus 1 over x minus 2 over x minus 2."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "But we have to assume that x is not equal to negative 2. So this is equal to 6 times. So we're going to divide by x plus 2 in the numerator, x plus 2 in the denominator. So it's going to be 6 times x plus 1 over x minus 2 over x minus 2. And we have to put the constraint here because now we've changed it. Now, this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function, we have to constrain it."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be 6 times x plus 1 over x minus 2 over x minus 2. And we have to put the constraint here because now we've changed it. Now, this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function, we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "But in order to be equivalent to the original function, we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0. So you could say for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what can we assign f of negative 2 to make the function continuous at the point? Well, the function is completely equivalent to this expression except that the function is not defined at x equals negative 2."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "This one also isn't defined at positive 2 because you're dividing by 0. So you could say for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what can we assign f of negative 2 to make the function continuous at the point? Well, the function is completely equivalent to this expression except that the function is not defined at x equals negative 2. So that's why we had to put that constraint here if we wanted this to be the same thing as our original function. But if we wanted to re-engineer the function so it is continuous at that point, well, then we just have to set f of x equal to whatever this expression would have been when x is equal to negative 2. So let's think about that."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, the function is completely equivalent to this expression except that the function is not defined at x equals negative 2. So that's why we had to put that constraint here if we wanted this to be the same thing as our original function. But if we wanted to re-engineer the function so it is continuous at that point, well, then we just have to set f of x equal to whatever this expression would have been when x is equal to negative 2. So let's think about that. So 6 times negative 2 plus 1 over negative 2 minus 2 is equal to, this is 6 times negative 1, so it's negative 6 over negative 4, which is equal to 3 halves. So if we redefine f of x, if we say f of x is equal to 6x squared plus 18x plus 12 over x squared minus 4 for x not equal positive or negative 2, and it's equal to 3 halves for x equals negative 2, now this function is going to be the exact same thing as this right over here. This f of x, this new one, this new definition, this extended definition of our original one, is now equivalent to this expression."}, {"video_title": "Defining a function at a point to make it continuous   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's think about that. So 6 times negative 2 plus 1 over negative 2 minus 2 is equal to, this is 6 times negative 1, so it's negative 6 over negative 4, which is equal to 3 halves. So if we redefine f of x, if we say f of x is equal to 6x squared plus 18x plus 12 over x squared minus 4 for x not equal positive or negative 2, and it's equal to 3 halves for x equals negative 2, now this function is going to be the exact same thing as this right over here. This f of x, this new one, this new definition, this extended definition of our original one, is now equivalent to this expression. It is equal to 6 times x plus 1 over x minus 2. But just to answer their question, what value should be assigned to f of negative 2 to make f of x continuous at that point? Well, f of x should be, or f of negative 2 should be 3 halves."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here it says Nate tried to find the derivative of x squared plus five x times sine of x. Here is his work. Is Nate's work correct? If not, what's his mistake? So pause the video and see if you can answer this. Is Nate's work correct? And if not, what's his mistake?"}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If not, what's his mistake? So pause the video and see if you can answer this. Is Nate's work correct? And if not, what's his mistake? All right, so I'm assuming you've had a go at it. So let's work through this step by step. So over here he's just trying to apply the derivative operator to the expression, which is exactly what he needed to do."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if not, what's his mistake? All right, so I'm assuming you've had a go at it. So let's work through this step by step. So over here he's just trying to apply the derivative operator to the expression, which is exactly what he needed to do. He's trying to find the derivative of this thing. And he says, okay, this is a product of two expressions. And then he says, okay, well, this is gonna be the same thing as the derivative, or this is the same thing as the product of the derivatives."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here he's just trying to apply the derivative operator to the expression, which is exactly what he needed to do. He's trying to find the derivative of this thing. And he says, okay, this is a product of two expressions. And then he says, okay, well, this is gonna be the same thing as the derivative, or this is the same thing as the product of the derivatives. Now this is a problem. You are probably familiar, if I take the derivative of the sum of two things, so the derivative with respect to x of f of x plus g of x, that indeed is equal to the derivative of the first, f prime of x, plus the derivative of the second. But that is not true if we are dealing with the product of functions."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then he says, okay, well, this is gonna be the same thing as the derivative, or this is the same thing as the product of the derivatives. Now this is a problem. You are probably familiar, if I take the derivative of the sum of two things, so the derivative with respect to x of f of x plus g of x, that indeed is equal to the derivative of the first, f prime of x, plus the derivative of the second. But that is not true if we are dealing with the product of functions. The derivative with respect to x of f of x, g of x, is not necessarily, maybe there's some very special circumstances, but in general, it's not going to be just the product of the derivative. It's not going to be just f prime of x, g prime of x. Here we would want to apply the product rule."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that is not true if we are dealing with the product of functions. The derivative with respect to x of f of x, g of x, is not necessarily, maybe there's some very special circumstances, but in general, it's not going to be just the product of the derivative. It's not going to be just f prime of x, g prime of x. Here we would want to apply the product rule. This is going to be equal to, this is going to be equal to the derivative of the first function times the second function, plus the derivative, well let me write it this way, plus the first function, not taking its derivative, times the derivative of the second function. So he should have applied the product rule here. And so let's do that, just to see what his answer should have been."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here we would want to apply the product rule. This is going to be equal to, this is going to be equal to the derivative of the first function times the second function, plus the derivative, well let me write it this way, plus the first function, not taking its derivative, times the derivative of the second function. So he should have applied the product rule here. And so let's do that, just to see what his answer should have been. So what he should have done here, I'll get my correcting red pen out here, say no, that's not what he should have done. He says let's take the derivative of this first thing. So actually let me do it, color code it."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's do that, just to see what his answer should have been. So what he should have done here, I'll get my correcting red pen out here, say no, that's not what he should have done. He says let's take the derivative of this first thing. So actually let me do it, color code it. So the derivative of this is two x plus five. So it should have been two x plus five, times the second thing, so times sine of x, times, let me do it in another color, times sine of x. And then to that, he would add the first thing, which is x squared plus five x, times the derivative of the second thing."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So actually let me do it, color code it. So the derivative of this is two x plus five. So it should have been two x plus five, times the second thing, so times sine of x, times, let me do it in another color, times sine of x. And then to that, he would add the first thing, which is x squared plus five x, times the derivative of the second thing. So the derivative of sine of x is cosine of x. So this is what we should have been seeing at this step right over here. He shouldn't have just taken the product of the derivatives, he should have applied the product rule."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then to that, he would add the first thing, which is x squared plus five x, times the derivative of the second thing. So the derivative of sine of x is cosine of x. So this is what we should have been seeing at this step right over here. He shouldn't have just taken the product of the derivatives, he should have applied the product rule. So his work is not correct, and his mistake is that he didn't apply the product rule. He just assumed that the derivative of the products is the same thing as the product of the derivatives. Let's do more examples."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "He shouldn't have just taken the product of the derivatives, he should have applied the product rule. So his work is not correct, and his mistake is that he didn't apply the product rule. He just assumed that the derivative of the products is the same thing as the product of the derivatives. Let's do more examples. Okay, so let's see. It says, Katie tried to find the derivative of two x squared minus four, all of that to the third power. Here is her work."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do more examples. Okay, so let's see. It says, Katie tried to find the derivative of two x squared minus four, all of that to the third power. Here is her work. Is Katie's work correct? If not, what is her mistake? So once again, pause the video, see if you can figure it out."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here is her work. Is Katie's work correct? If not, what is her mistake? So once again, pause the video, see if you can figure it out. All right, now let's inspect Katie's work. So she's taking the derivative of this. And let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, pause the video, see if you can figure it out. All right, now let's inspect Katie's work. So she's taking the derivative of this. And let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression. And that is close to applying the chain rule properly, but it's not applying the chain rule properly. So her work is not correct, and her mistake is she's not correctly applying the chain rule. Just as a review, the chain rule says, look, if we're trying to take the derivative with respect to x of f of g of x, f of g of x, f of g of x, that this is going to be equal to, this is going to be equal to the derivative of the whole thing with respect to g of x."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression. And that is close to applying the chain rule properly, but it's not applying the chain rule properly. So her work is not correct, and her mistake is she's not correctly applying the chain rule. Just as a review, the chain rule says, look, if we're trying to take the derivative with respect to x of f of g of x, f of g of x, f of g of x, that this is going to be equal to, this is going to be equal to the derivative of the whole thing with respect to g of x. So I could write that as f prime of g of x, f prime of g of x, times the derivative of the inner function with respect to x, times g prime of x. So over here, we could view our f function as a thing that takes its input and takes it to the third power. And so this right over here is f prime of g of x."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Just as a review, the chain rule says, look, if we're trying to take the derivative with respect to x of f of g of x, f of g of x, f of g of x, that this is going to be equal to, this is going to be equal to the derivative of the whole thing with respect to g of x. So I could write that as f prime of g of x, f prime of g of x, times the derivative of the inner function with respect to x, times g prime of x. So over here, we could view our f function as a thing that takes its input and takes it to the third power. And so this right over here is f prime of g of x. So this thing is the f prime of g of x, but she forgot to multiply it by the derivative of the inner function with respect to x. So she forgot to multiply this times the derivative of two x squared minus four with respect to x, which is going to be, let's see, the derivative of two x squared, power rule, two times two is four, so it's gonna be four x to the first. And then the derivative of negative four is just zero, so it's just gonna be times four x."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this right over here is f prime of g of x. So this thing is the f prime of g of x, but she forgot to multiply it by the derivative of the inner function with respect to x. So she forgot to multiply this times the derivative of two x squared minus four with respect to x, which is going to be, let's see, the derivative of two x squared, power rule, two times two is four, so it's gonna be four x to the first. And then the derivative of negative four is just zero, so it's just gonna be times four x. So that's what she needed to do in order for it to be correct. So she had to have this times four x here, times four x. So not correct, she didn't correctly apply the chain rule."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the derivative of negative four is just zero, so it's just gonna be times four x. So that's what she needed to do in order for it to be correct. So she had to have this times four x here, times four x. So not correct, she didn't correctly apply the chain rule. So let's do another one of these. So here it says Nejoman tried to find the derivative of sine of seven x squared plus four x. Here is his work."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So not correct, she didn't correctly apply the chain rule. So let's do another one of these. So here it says Nejoman tried to find the derivative of sine of seven x squared plus four x. Here is his work. Is Nejoman's work correct? If not, what is his mistake? Pause the video, see if you can figure it out."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here is his work. Is Nejoman's work correct? If not, what is his mistake? Pause the video, see if you can figure it out. All right, so it's the derivative of sine of this expression. So you'd wanna use the chain rule. In fact, using the chain rule, you wanna find the derivative of the outside function with respect to the inside."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video, see if you can figure it out. All right, so it's the derivative of sine of this expression. So you'd wanna use the chain rule. In fact, using the chain rule, you wanna find the derivative of the outside function with respect to the inside. So the derivative of sine of something with respect to that something is gonna be cosine of that something. So that's right, that's right. And then you wanna multiply that times the derivative of the inside with respect to x."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "In fact, using the chain rule, you wanna find the derivative of the outside function with respect to the inside. So the derivative of sine of something with respect to that something is gonna be cosine of that something. So that's right, that's right. And then you wanna multiply that times the derivative of the inside with respect to x. So the derivative of seven x squared is 14 x. Derivative of four x is four. So this is actually, that step looks good."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you wanna multiply that times the derivative of the inside with respect to x. So the derivative of seven x squared is 14 x. Derivative of four x is four. So this is actually, that step looks good. But then Nejoman does something strange over here. This is the cosine of seven x squared plus four x, and then that whole thing times 14 x plus four. But they get confused where, just looking at these parentheses, and this tends to happen sometimes."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is actually, that step looks good. But then Nejoman does something strange over here. This is the cosine of seven x squared plus four x, and then that whole thing times 14 x plus four. But they get confused where, just looking at these parentheses, and this tends to happen sometimes. This is actually one of these key errors that the folks at the college board, the AP folks, told us about. Is that when dealing with these transcendental functions, cosine, sine, tangent, natural log, that are written like this, and people see the parentheses and then see another parentheses, their brain just says, oh, let me multiply these two expressions in parentheses. But that's not right, because if we were to add parentheses, this is what this is implying."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But they get confused where, just looking at these parentheses, and this tends to happen sometimes. This is actually one of these key errors that the folks at the college board, the AP folks, told us about. Is that when dealing with these transcendental functions, cosine, sine, tangent, natural log, that are written like this, and people see the parentheses and then see another parentheses, their brain just says, oh, let me multiply these two expressions in parentheses. But that's not right, because if we were to add parentheses, this is what this is implying. So you can't just take the 14 x plus four and multiply it by this, and assuming you're taking the cosine of the whole thing. So this is where Nejoman makes the mistake. The work is not correct, and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that's not right, because if we were to add parentheses, this is what this is implying. So you can't just take the 14 x plus four and multiply it by this, and assuming you're taking the cosine of the whole thing. So this is where Nejoman makes the mistake. The work is not correct, and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing. Let's do one more of these. I find these strangely fun. All right, this one is involved."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The work is not correct, and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing. Let's do one more of these. I find these strangely fun. All right, this one is involved. Tom tried to find the derivative of the square root of x over x to the fourth. Here is his work. Is Tom's work correct?"}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, this one is involved. Tom tried to find the derivative of the square root of x over x to the fourth. Here is his work. Is Tom's work correct? If not, what's his mistake? Pause the video and see if you can figure that out. So it looks like he's trying to apply the quotient rule."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is Tom's work correct? If not, what's his mistake? Pause the video and see if you can figure that out. So it looks like he's trying to apply the quotient rule. So applying the quotient rule, you would, in the numerator, you would take the derivative of the first expression times the second expression, and then minus the first expression times the derivative of the second expression, all of that over the, or I should say, the derivative of the numerator expression over the, or times the denominator expression, minus the numerator expression times the derivative of the denominator expression, all of that over the denominator expression squared. So this looks correct, actually. It's a correct application of the quotient rule."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks like he's trying to apply the quotient rule. So applying the quotient rule, you would, in the numerator, you would take the derivative of the first expression times the second expression, and then minus the first expression times the derivative of the second expression, all of that over the, or I should say, the derivative of the numerator expression over the, or times the denominator expression, minus the numerator expression times the derivative of the denominator expression, all of that over the denominator expression squared. So this looks correct, actually. It's a correct application of the quotient rule. It looks like Tom is correctly simplifying. So the derivative of x to the 1 1\u20442 is 1 1\u20442 x to the negative 1 1\u20442, so that looks right. Derivative of x to the fourth is four x to the third, so that looks right."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's a correct application of the quotient rule. It looks like Tom is correctly simplifying. So the derivative of x to the 1 1\u20442 is 1 1\u20442 x to the negative 1 1\u20442, so that looks right. Derivative of x to the fourth is four x to the third, so that looks right. All of this looks algebraically right. And let's see, when you simplify this, so let's see, x to the negative 1 1\u20442 times x to the fourth is indeed x to, well, that's going to be x to the, oh, this correlates, so this simplifies to that, which looks correct, and that simplifies to that, which looks correct. We're just using exponent properties there."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative of x to the fourth is four x to the third, so that looks right. All of this looks algebraically right. And let's see, when you simplify this, so let's see, x to the negative 1 1\u20442 times x to the fourth is indeed x to, well, that's going to be x to the, oh, this correlates, so this simplifies to that, which looks correct, and that simplifies to that, which looks correct. We're just using exponent properties there. And then divided everything by, let's see, oh, then everything is in terms of x to 3.5, so we're going to have negative 3.5 x to 3.5, and then you use exponent properties. So actually, it looks like he did everything correctly. This is the right answer."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're just using exponent properties there. And then divided everything by, let's see, oh, then everything is in terms of x to 3.5, so we're going to have negative 3.5 x to 3.5, and then you use exponent properties. So actually, it looks like he did everything correctly. This is the right answer. Now, so his work is correct. He did not make any mistakes, but I do have a bone to pick, so to speak, with Tom, because he didn't have to apply the quotient rule here. He did all of this hairy calculus and algebra, but there could have been a very simple simplification he could have made up here, and this is a key thing to realize."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the right answer. Now, so his work is correct. He did not make any mistakes, but I do have a bone to pick, so to speak, with Tom, because he didn't have to apply the quotient rule here. He did all of this hairy calculus and algebra, but there could have been a very simple simplification he could have made up here, and this is a key thing to realize. He could have said, hey, you know what? This is the same thing as the derivative with respect to x of x to the 1.5, that's what the square root of x is, times x to the negative fourth power. That's what one over x to the fourth is."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "He did all of this hairy calculus and algebra, but there could have been a very simple simplification he could have made up here, and this is a key thing to realize. He could have said, hey, you know what? This is the same thing as the derivative with respect to x of x to the 1.5, that's what the square root of x is, times x to the negative fourth power. That's what one over x to the fourth is. And so let me color code it. So that is the same thing as that, and that is the same thing as that. And you wouldn't even have to use the product rule here."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's what one over x to the fourth is. And so let me color code it. So that is the same thing as that, and that is the same thing as that. And you wouldn't even have to use the product rule here. You could simplify this even further. This is the same thing as the derivative with respect to x of, just we have the same base. We can add the exponents."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you wouldn't even have to use the product rule here. You could simplify this even further. This is the same thing as the derivative with respect to x of, just we have the same base. We can add the exponents. We're taking the product, so it's gonna be x to the negative 3.5. And so you could just use the power rule. So this is going to be equal to, bring the negative 3.5 out front, negative 3.5 x to the, and then we just decrement this by one."}, {"video_title": "Differentiating functions Find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can add the exponents. We're taking the product, so it's gonna be x to the negative 3.5. And so you could just use the power rule. So this is going to be equal to, bring the negative 3.5 out front, negative 3.5 x to the, and then we just decrement this by one. We subtract one from that, negative 4.5 power. So as you can see, he could have gotten this answer much, much, much, much, much, much quicker, but he didn't make any mistakes. There's a little bit of a judgment error just immediately going forth with the quotient rule, which gets quite hairy quite fast."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And all it tells us is that if we have a function that can be expressed as the product of two functions, so let's say it can be expressed as f of x times g of x, and we want to take the derivative of this function, we want to take the derivative of it, that it's going to be equal to the derivative of one of these functions, f prime of x, let's say the derivative of the first one, times the second function, plus the first function, not taking its derivative, times the derivative of the second function. So here we have two terms. In each term, we took the derivative of one of the functions and not the other. And we multiply the derivative of the first function times the second function plus just the first function times the derivative of the second function. Now let's see if we can actually apply this to actually find the derivative of something. So let's say we are dealing with x squared times cosine of x. Or let's say, well, yeah, sure."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we multiply the derivative of the first function times the second function plus just the first function times the derivative of the second function. Now let's see if we can actually apply this to actually find the derivative of something. So let's say we are dealing with x squared times cosine of x. Or let's say, well, yeah, sure. Let's do x squared times sine of x. We could have done it either way. And we are curious about taking the derivative of this."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or let's say, well, yeah, sure. Let's do x squared times sine of x. We could have done it either way. And we are curious about taking the derivative of this. We are curious about what its derivative is. Well, we might immediately recognize that this can be expressed as a product of two functions. We could set f of x is equal to x squared."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we are curious about taking the derivative of this. We are curious about what its derivative is. Well, we might immediately recognize that this can be expressed as a product of two functions. We could set f of x is equal to x squared. So that is f of x right over there. And we could set g of x to be equal to sine of x. And there we have it."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could set f of x is equal to x squared. So that is f of x right over there. And we could set g of x to be equal to sine of x. And there we have it. We have our f of x times g of x. And we could think about what these individual derivatives are. The derivative of f of x is just going to be equal to 2x by the power rule."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there we have it. We have our f of x times g of x. And we could think about what these individual derivatives are. The derivative of f of x is just going to be equal to 2x by the power rule. And the derivative of g of x is just the derivative of sine of x. And we covered this when we just talked about common derivatives. Derivative of sine of x is cosine of x."}, {"video_title": "Product rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of f of x is just going to be equal to 2x by the power rule. And the derivative of g of x is just the derivative of sine of x. And we covered this when we just talked about common derivatives. Derivative of sine of x is cosine of x. And so now we're ready to apply the product rule. This is going to be equal to f prime of x times g of x. So f prime of x, the derivative of f, is 2x times g of x, which is sine of x, plus just our function f, which is x squared, times the derivative of g, times cosine of x."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to evaluate the definite integral from negative one to negative two of 16 minus x to the third over x to the third dx. Now at first this might seem daunting. I have this rational expression. I have x's in the numerators and x's in the denominators, but we just have to remember, we just have to do some algebraic manipulation and this is going to seem a lot more tractable. This is the same thing as the definite integral from negative one to negative two of 16 over x to the third minus x to the third over x to the third. Minus x to the third over x to the third dx. And now what is that going to be equal to?"}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I have x's in the numerators and x's in the denominators, but we just have to remember, we just have to do some algebraic manipulation and this is going to seem a lot more tractable. This is the same thing as the definite integral from negative one to negative two of 16 over x to the third minus x to the third over x to the third. Minus x to the third over x to the third dx. And now what is that going to be equal to? That is going to be equal to the definite integral from negative one to negative two of, I could write this first term right over here. Let me do this in a different color. I could write this as 16x to the negative three."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now what is that going to be equal to? That is going to be equal to the definite integral from negative one to negative two of, I could write this first term right over here. Let me do this in a different color. I could write this as 16x to the negative three. x to the negative three. And this second one, we have minus x to the third over x to the third. Well, x to the third is just over, x to the third over x to the third is just going to be equal to one."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could write this as 16x to the negative three. x to the negative three. And this second one, we have minus x to the third over x to the third. Well, x to the third is just over, x to the third over x to the third is just going to be equal to one. So this is going to be minus one dx. So dx. And so what is this going to be equal to?"}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, x to the third is just over, x to the third over x to the third is just going to be equal to one. So this is going to be minus one dx. So dx. And so what is this going to be equal to? Well, let's take the antiderivative of each of these parts and then we're going to have to evaluate them at the different bounds. So let's see. The antiderivative of 16x to the negative three."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what is this going to be equal to? Well, let's take the antiderivative of each of these parts and then we're going to have to evaluate them at the different bounds. So let's see. The antiderivative of 16x to the negative three. We're just going to do the power rule for derivatives in reverse. You could view this as the power rule of integration or power rule of taking the antiderivative where what you do is you're going to increase our exponent by one. So you're going to go from negative three to negative two."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of 16x to the negative three. We're just going to do the power rule for derivatives in reverse. You could view this as the power rule of integration or power rule of taking the antiderivative where what you do is you're going to increase our exponent by one. So you're going to go from negative three to negative two. And then you're going to divide by that amount, by negative two. So it's going to be 16 divided by negative two times x to the negative two. All I did is I increased the exponent and I divided by that amount."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you're going to go from negative three to negative two. And then you're going to divide by that amount, by negative two. So it's going to be 16 divided by negative two times x to the negative two. All I did is I increased the exponent and I divided by that amount. So that's the antiderivative here. And 16 divided by negative two, that is just negative eight. So we have negative eight x to the negative two."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "All I did is I increased the exponent and I divided by that amount. So that's the antiderivative here. And 16 divided by negative two, that is just negative eight. So we have negative eight x to the negative two. And then the antiderivative of negative one, well, that's just negative x. Negative, negative x. Negative x."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have negative eight x to the negative two. And then the antiderivative of negative one, well, that's just negative x. Negative, negative x. Negative x. And actually you could, you might just know that. And hey, if I take the derivative of negative x, I get negative one. Or if you viewed this as negative x to the zero power, because that's what one is, well, it's the same thing."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative x. And actually you could, you might just know that. And hey, if I take the derivative of negative x, I get negative one. Or if you viewed this as negative x to the zero power, because that's what one is, well, it's the same thing. You increase the exponent by one to get x to the first power. And then you divide by one. And so, I mean, you could view it as that right over there."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or if you viewed this as negative x to the zero power, because that's what one is, well, it's the same thing. You increase the exponent by one to get x to the first power. And then you divide by one. And so, I mean, you could view it as that right over there. But either way, you get to negative or minus x. And so now we want to evaluate that. We're going to evaluate that at the bounds and take the difference."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so, I mean, you could view it as that right over there. But either way, you get to negative or minus x. And so now we want to evaluate that. We're going to evaluate that at the bounds and take the difference. So we're going to evaluate that at negative two and then subtract from that this evaluated at negative one. And let me do those in two different colors just so we can see what's going on. So we're gonna evaluate it at negative two and we're going to evaluate it at negative one."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to evaluate that at the bounds and take the difference. So we're going to evaluate that at negative two and then subtract from that this evaluated at negative one. And let me do those in two different colors just so we can see what's going on. So we're gonna evaluate it at negative two and we're going to evaluate it at negative one. So let's first evaluate it at negative two. So this is going to be equal to, this is going to be equal to, when you evaluate it at negative two, it's going to be negative eight, negative eight times x to the negative two, so negative two to the negative two power, minus negative two, and from that we're going to subtract it, evaluate it at negative one. So it's going to be negative eight times negative one to the negative two power, minus negative one."}, {"video_title": "Definite integral of rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna evaluate it at negative two and we're going to evaluate it at negative one. So let's first evaluate it at negative two. So this is going to be equal to, this is going to be equal to, when you evaluate it at negative two, it's going to be negative eight, negative eight times x to the negative two, so negative two to the negative two power, minus negative two, and from that we're going to subtract it, evaluate it at negative one. So it's going to be negative eight times negative one to the negative two power, minus negative one. Alright, so what is this going to be? So negative two to the negative two, so negative two to the negative two is equal to one over negative two squared, which is equal to 1 4th. So this is equal to positive 1 4th, but then negative eight times positive 1 4th is going to be equal to negative two, and then we have negative two minus negative two, so that's negative two plus two, and so everything I've just done in this purplish color, that is just going to be zero, and then if we look at what's going on in the orange, when we evaluate at negative one, let's see, negative one to the negative two power, well that's one over negative one squared, well this is all just going to be one, and so we're going to have negative eight plus one, which is equal to negative seven."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "And the way we did it, we said, OK, well, let's find the slope between that point and then another point that's not too far away from that point. And we got the slope of the secant line. And it looks all fancy, but this is just the y value of the point that's not too far away. And this is just the y value of the point in question. So this is just your change in y. And then you divide that by your change in x. So in the example we did, h was the difference between our two x values."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "And this is just the y value of the point in question. So this is just your change in y. And then you divide that by your change in x. So in the example we did, h was the difference between our two x values. This distance was h. And that gave us the point, the slope of that line. We said, hey, what if we take the limit as this point right here gets closer and closer to this point? If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So in the example we did, h was the difference between our two x values. This distance was h. And that gave us the point, the slope of that line. We said, hey, what if we take the limit as this point right here gets closer and closer to this point? If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line. And we define that as the derivative of our function. We said that's equal to f prime of x. So let's see if we can apply this in this video to maybe make things a little bit more concrete in your head."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line. And we define that as the derivative of our function. We said that's equal to f prime of x. So let's see if we can apply this in this video to maybe make things a little bit more concrete in your head. So let me do one. First I'll do a particular case where I want to find the slope at exactly some point. So let's say, let me draw my axes again."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's see if we can apply this in this video to maybe make things a little bit more concrete in your head. So let me do one. First I'll do a particular case where I want to find the slope at exactly some point. So let's say, let me draw my axes again. Let's draw some axes right there. And let's say I have the curve. This is the curve."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say, let me draw my axes again. Let's draw some axes right there. And let's say I have the curve. This is the curve. y is equal to x squared. So this is my y-axis. This is my x-axis."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the curve. y is equal to x squared. So this is my y-axis. This is my x-axis. And I want to know the slope at the point x is equal to 3. When I say the slope, you can imagine a tangent line here. Do it in a light yellow."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This is my x-axis. And I want to know the slope at the point x is equal to 3. When I say the slope, you can imagine a tangent line here. Do it in a light yellow. You can imagine a tangent line that goes just like that. And it would just barely graze the curve at that point. But what is the slope of that tangent line?"}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Do it in a light yellow. You can imagine a tangent line that goes just like that. And it would just barely graze the curve at that point. But what is the slope of that tangent line? What is the slope of that tangent line? Which is the same as the slope of the curve right at that point. So to do it, I'm going to actually do this exact technique that we did before."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "But what is the slope of that tangent line? What is the slope of that tangent line? Which is the same as the slope of the curve right at that point. So to do it, I'm going to actually do this exact technique that we did before. And then we'll generalize it so you don't have to do it every time for a particular number. So let's take some other point here. Let's call this 3 plus delta x. I'm changing the notation because in some books you'll see an h, some books you'll see a delta x."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So to do it, I'm going to actually do this exact technique that we did before. And then we'll generalize it so you don't have to do it every time for a particular number. So let's take some other point here. Let's call this 3 plus delta x. I'm changing the notation because in some books you'll see an h, some books you'll see a delta x. Doesn't hurt to be exposed to both of them. So this is 3 plus delta x. So first of all, what is this point right here?"}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's call this 3 plus delta x. I'm changing the notation because in some books you'll see an h, some books you'll see a delta x. Doesn't hurt to be exposed to both of them. So this is 3 plus delta x. So first of all, what is this point right here? This is a curve y is equal to x squared. So f of x is 3 squared. This is the point 9."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So first of all, what is this point right here? This is a curve y is equal to x squared. So f of x is 3 squared. This is the point 9. This is the point 3, 9 right here. Let me draw that out. Well, I'll draw it out in a second."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the point 9. This is the point 3, 9 right here. Let me draw that out. Well, I'll draw it out in a second. And what is this point right here? So if we were to go all the way up here, what is that point? Well, here our x is now 3 plus delta x."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, I'll draw it out in a second. And what is this point right here? So if we were to go all the way up here, what is that point? Well, here our x is now 3 plus delta x. It's the same thing as this one right here. It's x naught plus h. I could have called this 3 plus h just as easily. So it's 3 plus delta x up there."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, here our x is now 3 plus delta x. It's the same thing as this one right here. It's x naught plus h. I could have called this 3 plus h just as easily. So it's 3 plus delta x up there. So what's the y value going to be? Well, whatever x value is, it's on the curve, it's going to be that squared. So it's going to be, this is going to be the point 3 plus delta x squared."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's 3 plus delta x up there. So what's the y value going to be? Well, whatever x value is, it's on the curve, it's going to be that squared. So it's going to be, this is going to be the point 3 plus delta x squared. So let's figure out the slope of this secant line. Let me zoom in a little bit because that might help. So if I zoom in on just this part of the curve, it might look like that."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be, this is going to be the point 3 plus delta x squared. So let's figure out the slope of this secant line. Let me zoom in a little bit because that might help. So if I zoom in on just this part of the curve, it might look like that. And then I have one point here, and then I have the other point is up here. Let me see if I can draw that. That's the secant line."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So if I zoom in on just this part of the curve, it might look like that. And then I have one point here, and then I have the other point is up here. Let me see if I can draw that. That's the secant line. Just like that. This was the point over here. This point is the point 3, 9."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "That's the secant line. Just like that. This was the point over here. This point is the point 3, 9. And then this point up here is the point 3 plus delta x. So just some larger number than 3. And then it's going to be that number squared."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This point is the point 3, 9. And then this point up here is the point 3 plus delta x. So just some larger number than 3. And then it's going to be that number squared. So it's going to be 3 plus delta x squared. What is that? That's going to be 9."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "And then it's going to be that number squared. So it's going to be 3 plus delta x squared. What is that? That's going to be 9. I'm just FOILing this out, or you do the distributive property twice. If a plus b squared is a squared plus 2ab plus b squared, so it's going to be 9 plus 2 times the product of these things. So plus 6 delta x, and then plus delta x squared."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "That's going to be 9. I'm just FOILing this out, or you do the distributive property twice. If a plus b squared is a squared plus 2ab plus b squared, so it's going to be 9 plus 2 times the product of these things. So plus 6 delta x, and then plus delta x squared. Plus delta x plus delta x squared. That's the coordinate of the second line. This looks complicated, but I just took this x value and I squared it, because it's on the line y is equal to x squared."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So plus 6 delta x, and then plus delta x squared. Plus delta x plus delta x squared. That's the coordinate of the second line. This looks complicated, but I just took this x value and I squared it, because it's on the line y is equal to x squared. So the slope of this line, the slope of the secant line is going to be the change in y divided by the change in x. So the change in y is just going to be this guy's y value, which is 9 plus 6 delta x plus delta x squared. That's this guy's y value."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This looks complicated, but I just took this x value and I squared it, because it's on the line y is equal to x squared. So the slope of this line, the slope of the secant line is going to be the change in y divided by the change in x. So the change in y is just going to be this guy's y value, which is 9 plus 6 delta x plus delta x squared. That's this guy's y value. Minus this guy's y value. I'll do it in green. So minus 9."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "That's this guy's y value. Minus this guy's y value. I'll do it in green. So minus 9. That's your change in y. And you want to divide that by your change in x. Well, what is your change in x?"}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So minus 9. That's your change in y. And you want to divide that by your change in x. Well, what is your change in x? This is actually going to be pretty convenient. This larger x value, we started with this point on the top. So we have to start with this point on the bottom."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, what is your change in x? This is actually going to be pretty convenient. This larger x value, we started with this point on the top. So we have to start with this point on the bottom. So it's going to be 3 plus delta x. And then what's this x value? Well, it's minus 3."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So we have to start with this point on the bottom. So it's going to be 3 plus delta x. And then what's this x value? Well, it's minus 3. That's his x value. So what does this simplify to? The numerator, this 9 and that 9 cancel out."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, it's minus 3. That's his x value. So what does this simplify to? The numerator, this 9 and that 9 cancel out. We get a 9 minus 9. And the denominator, what happens? This 3 and minus 3 cancel out."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "The numerator, this 9 and that 9 cancel out. We get a 9 minus 9. And the denominator, what happens? This 3 and minus 3 cancel out. So the change in x actually end up becoming this delta x, which makes sense, because this delta x is essentially how much more this guy is than that guy. So that should be the change in x, delta x. So the slope of my secant line has simplified to 6 times my change in x plus my change in x squared, all of that over my change in x."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This 3 and minus 3 cancel out. So the change in x actually end up becoming this delta x, which makes sense, because this delta x is essentially how much more this guy is than that guy. So that should be the change in x, delta x. So the slope of my secant line has simplified to 6 times my change in x plus my change in x squared, all of that over my change in x. Now we can simplify this even more. Let's divide the numerator and the denominator by our change in x. And I will switch colors just to ease the monotony."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So the slope of my secant line has simplified to 6 times my change in x plus my change in x squared, all of that over my change in x. Now we can simplify this even more. Let's divide the numerator and the denominator by our change in x. And I will switch colors just to ease the monotony. So my slope of my tangent of my secant line, the one that goes through both of these, is going to be equal if you divide the numerator and denominator. This becomes 6. I'm just dividing numerator and denominator by delta x."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "And I will switch colors just to ease the monotony. So my slope of my tangent of my secant line, the one that goes through both of these, is going to be equal if you divide the numerator and denominator. This becomes 6. I'm just dividing numerator and denominator by delta x. Plus 6 plus delta x. So that is the slope of the secant line. So slope is equal to 6 plus delta x."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "I'm just dividing numerator and denominator by delta x. Plus 6 plus delta x. So that is the slope of the secant line. So slope is equal to 6 plus delta x. That's this one right here. That's this reddish line that I've drawn right there. So if this number right here, if the delta x was 1, if these were the points 3 and 4, then my slope would be 6 plus 1."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So slope is equal to 6 plus delta x. That's this one right here. That's this reddish line that I've drawn right there. So if this number right here, if the delta x was 1, if these were the points 3 and 4, then my slope would be 6 plus 1. Because I'm picking a point 4 where the delta x here would have to be 1. So the slope would be 7. So we have a general formula for no matter what my delta x is, I can find the slope between 3 and 3 plus delta x, between those two points."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So if this number right here, if the delta x was 1, if these were the points 3 and 4, then my slope would be 6 plus 1. Because I'm picking a point 4 where the delta x here would have to be 1. So the slope would be 7. So we have a general formula for no matter what my delta x is, I can find the slope between 3 and 3 plus delta x, between those two points. Now, we wanted to find the slope at exactly that point right there. So let's see what happens when delta x gets smaller and smaller. This is what delta x is right now."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So we have a general formula for no matter what my delta x is, I can find the slope between 3 and 3 plus delta x, between those two points. Now, we wanted to find the slope at exactly that point right there. So let's see what happens when delta x gets smaller and smaller. This is what delta x is right now. It's this distance. But if delta x got a little bit smaller, then it would start, the secant line would look like that. Got even smaller, the secant line would look like that."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "This is what delta x is right now. It's this distance. But if delta x got a little bit smaller, then it would start, the secant line would look like that. Got even smaller, the secant line would look like that. It gets even smaller, then we're getting pretty close to the slope of the tangent line. The tangent line is this thing right here that I want to find the slope of. So let's find the limit as our delta x approaches 0."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "Got even smaller, the secant line would look like that. It gets even smaller, then we're getting pretty close to the slope of the tangent line. The tangent line is this thing right here that I want to find the slope of. So let's find the limit as our delta x approaches 0. So the limit as delta x approaches 0 of our slope of the secant line of 6 plus delta x is equal to what? This is pretty straightforward. You could just set this equal to 0 and it's equal to 6."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's find the limit as our delta x approaches 0. So the limit as delta x approaches 0 of our slope of the secant line of 6 plus delta x is equal to what? This is pretty straightforward. You could just set this equal to 0 and it's equal to 6. So the slope of our tangent line at the point x is equal to 3 right there is equal to 6. And another way we could write this, if we wrote that f of x is equal to x squared, we now know that the derivative, or the slope, of the tangent line of this function at the point 3. I just only evaluated it at the point 3 right there."}, {"video_title": "Calculating slope of tangent line using derivative definition   Differential Calculus   Khan Academy.mp3", "Sentence": "You could just set this equal to 0 and it's equal to 6. So the slope of our tangent line at the point x is equal to 3 right there is equal to 6. And another way we could write this, if we wrote that f of x is equal to x squared, we now know that the derivative, or the slope, of the tangent line of this function at the point 3. I just only evaluated it at the point 3 right there. That that is equal to 6. I haven't yet come up with a general formula for the slope of this line at any point. And I'm going to do that in the next video."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write that down. So dy dx is equal to y over three y squared minus x. Three y squared minus x. And now what I want to do is essentially take the derivative of both sides again. Now there's a bunch of ways that we could try to tackle it, but the way it's written right now, if I take the derivative of the right-hand side, I'm gonna have to apply the quotient rule, which really just comes from the product rule, but it gets pretty hairy. So let's see if I can simplify my task a little bit. I'm going to have to do implicit differentiation one way or the other."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now what I want to do is essentially take the derivative of both sides again. Now there's a bunch of ways that we could try to tackle it, but the way it's written right now, if I take the derivative of the right-hand side, I'm gonna have to apply the quotient rule, which really just comes from the product rule, but it gets pretty hairy. So let's see if I can simplify my task a little bit. I'm going to have to do implicit differentiation one way or the other. So what if I multiplied both sides of this times three y squared minus x? Then I get three y squared minus x times dy dx is equal to y. And so now taking the derivative of both sides of this is going to be a little bit more straightforward."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm going to have to do implicit differentiation one way or the other. So what if I multiplied both sides of this times three y squared minus x? Then I get three y squared minus x times dy dx is equal to y. And so now taking the derivative of both sides of this is going to be a little bit more straightforward. In fact, if I want, I could, well actually, let me just do it like this. And so let's apply our derivative operator to both sides. Let's give ourselves some space."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now taking the derivative of both sides of this is going to be a little bit more straightforward. In fact, if I want, I could, well actually, let me just do it like this. And so let's apply our derivative operator to both sides. Let's give ourselves some space. So I'm gonna apply my derivative operator to both sides. I'm gonna take the derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. And so here I'll apply the product rule."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's give ourselves some space. So I'm gonna apply my derivative operator to both sides. I'm gonna take the derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. And so here I'll apply the product rule. First I'll take the derivative of this expression and then multiply that times dy dx and then I'll take the derivative of this expression and multiply it times this. So the derivative of three y squared minus x with respect to x, well that is going to be, so if I take the derivative of three y squared with respect to y, it's going to be six y. And then I have to take the derivative of y with respect to x, so times dy dx."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so here I'll apply the product rule. First I'll take the derivative of this expression and then multiply that times dy dx and then I'll take the derivative of this expression and multiply it times this. So the derivative of three y squared minus x with respect to x, well that is going to be, so if I take the derivative of three y squared with respect to y, it's going to be six y. And then I have to take the derivative of y with respect to x, so times dy dx. As we say, if this is completely unfamiliar to you, I encourage you to watch the several videos on Khan Academy on implicit differentiation, which is really just an extension of the chain rule. I took the derivative of three y squared with respect to y and took the derivative of y with respect to x. Now if I take the derivative of negative x with respect to x, that's going to be negative one."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I have to take the derivative of y with respect to x, so times dy dx. As we say, if this is completely unfamiliar to you, I encourage you to watch the several videos on Khan Academy on implicit differentiation, which is really just an extension of the chain rule. I took the derivative of three y squared with respect to y and took the derivative of y with respect to x. Now if I take the derivative of negative x with respect to x, that's going to be negative one. Fair enough. And so I took the derivative of this, I'm gonna multiply it times that, so times dy dx. And then to that, I'm going to add the derivative of this with respect to x."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now if I take the derivative of negative x with respect to x, that's going to be negative one. Fair enough. And so I took the derivative of this, I'm gonna multiply it times that, so times dy dx. And then to that, I'm going to add the derivative of this with respect to x. Well that's just going to be the second derivative of y with respect to x times this. And all I did is apply the product rule. Three y squared minus x."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then to that, I'm going to add the derivative of this with respect to x. Well that's just going to be the second derivative of y with respect to x times this. And all I did is apply the product rule. Three y squared minus x. Once again, I'll say it the third time. Derivative of this times this plus the derivative of this times that. Alright, that's going to be equal to, on the right hand side, derivative of y with respect to x is just dy dx."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Three y squared minus x. Once again, I'll say it the third time. Derivative of this times this plus the derivative of this times that. Alright, that's going to be equal to, on the right hand side, derivative of y with respect to x is just dy dx. Now if I want to, I could solve for the second derivative of y with respect to x, but what could be even better than that is if I substitute everything else with numbers, because then it's just going to be a nice, easy numerical equation to solve. So we know that we want to figure out our second derivative when y is equal to one. So y is equal to one, so that's going to be equal to one."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, that's going to be equal to, on the right hand side, derivative of y with respect to x is just dy dx. Now if I want to, I could solve for the second derivative of y with respect to x, but what could be even better than that is if I substitute everything else with numbers, because then it's just going to be a nice, easy numerical equation to solve. So we know that we want to figure out our second derivative when y is equal to one. So y is equal to one, so that's going to be equal to one. And this is going to be equal to one. X is equal to negative one. So let me underline that."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So y is equal to one, so that's going to be equal to one. And this is going to be equal to one. X is equal to negative one. So let me underline that. X is equal to negative one. So x is negative one. Are there any other x's here?"}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me underline that. X is equal to negative one. So x is negative one. Are there any other x's here? And what's dy dx? Well, dy dx, dy dx is going to be equal to one over three times one, which is three minus, minus negative one. So this is equal to 1 4th."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Are there any other x's here? And what's dy dx? Well, dy dx, dy dx is going to be equal to one over three times one, which is three minus, minus negative one. So this is equal to 1 4th. In fact, that's I think where we evaluated in the beginning. Yep, at the point negative one comma one. At the point negative one comma one."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to 1 4th. In fact, that's I think where we evaluated in the beginning. Yep, at the point negative one comma one. At the point negative one comma one. So that's dy dx is 1 4th. So this is 1 4th. This is 1 4th."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "At the point negative one comma one. So that's dy dx is 1 4th. So this is 1 4th. This is 1 4th. And this is 1 4th. And now we can solve for the second, can solve for the second derivative. Just going to make sure I don't make any careless mistakes."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 1 4th. And this is 1 4th. And now we can solve for the second, can solve for the second derivative. Just going to make sure I don't make any careless mistakes. So it's six times one times 1 4th is going to be 1.5, or six times one times 1 4th is 6 4ths, minus one is going to be, so 6 4ths minus 4 4ths is 2 4ths. So this is going to be 1 half times 1 4th, times 1 4th, that's this over here, plus the second derivative of y with respect to x. And here I have one minus negative one."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Just going to make sure I don't make any careless mistakes. So it's six times one times 1 4th is going to be 1.5, or six times one times 1 4th is 6 4ths, minus one is going to be, so 6 4ths minus 4 4ths is 2 4ths. So this is going to be 1 half times 1 4th, times 1 4th, that's this over here, plus the second derivative of y with respect to x. And here I have one minus negative one. So that's one plus one, so times two. So maybe I'll write it this way. Plus two times the second derivative of y with respect to x is equal to 1 4th."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And here I have one minus negative one. So that's one plus one, so times two. So maybe I'll write it this way. Plus two times the second derivative of y with respect to x is equal to 1 4th. And so let's see, I have 1 8th plus two times the second derivative is equal to 1 4th. And let's see, I could subtract 1 8th from both sides. So subtract 1 8th, subtract 1 8th."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus two times the second derivative of y with respect to x is equal to 1 4th. And so let's see, I have 1 8th plus two times the second derivative is equal to 1 4th. And let's see, I could subtract 1 8th from both sides. So subtract 1 8th, subtract 1 8th. These cancel, I get two times the second derivative of y with respect to x is equal to 1 4th is the same thing as 2 8ths. So 2 8ths minus 1 8th is 1 8th. And then I can divide both sides by two, and we get a little bit of a drum roll here."}, {"video_title": "2015 AP Calculus AB 6c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So subtract 1 8th, subtract 1 8th. These cancel, I get two times the second derivative of y with respect to x is equal to 1 4th is the same thing as 2 8ths. So 2 8ths minus 1 8th is 1 8th. And then I can divide both sides by two, and we get a little bit of a drum roll here. You divide both sides by two, or you could say multiply both sides times 1 half. And you are going to get, you are going to get the second derivative of y with respect to x when x is negative one and y is equal to one is one over 16. And we're done."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the particle's velocity v of t at t is equal to two? So pause this video, see if you can figure that out. Well, the key thing to realize is that your velocity is a function of time, is the derivative of position, and so this is going to be equal to, we just take the derivative with respect to t up here. So derivative of t to the third with respect to t is three t squared. If that's unfamiliar, I encourage you to review the power rule. The derivative of negative four t squared with respect to t is negative eight t, and derivative of three t with respect to t is plus three. Derivative of a constant doesn't change with respect to time, so that's just zero."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So derivative of t to the third with respect to t is three t squared. If that's unfamiliar, I encourage you to review the power rule. The derivative of negative four t squared with respect to t is negative eight t, and derivative of three t with respect to t is plus three. Derivative of a constant doesn't change with respect to time, so that's just zero. And so here we have velocity as a function of time, and so if we wanna know our velocity at time t equals two, we just substitute two wherever we see the t's, so it's gonna be three times four, three times two squared, so it's 12 minus eight times two, minus 16, plus three, which is equal to negative one. And you might say, well, negative one by itself doesn't sound like a velocity. Well, if they gave us units, if they told us that x was in meters and that t was in seconds, well, then x would be, well, I already said, would be in meters, and velocity would be negative one meters per second."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative of a constant doesn't change with respect to time, so that's just zero. And so here we have velocity as a function of time, and so if we wanna know our velocity at time t equals two, we just substitute two wherever we see the t's, so it's gonna be three times four, three times two squared, so it's 12 minus eight times two, minus 16, plus three, which is equal to negative one. And you might say, well, negative one by itself doesn't sound like a velocity. Well, if they gave us units, if they told us that x was in meters and that t was in seconds, well, then x would be, well, I already said, would be in meters, and velocity would be negative one meters per second. You might also be saying, well, what does the negative mean? Well, that means that we are moving to the left. Remember, we're moving along the x-axis, so if our velocity is negative, that means that x is decreasing or we are moving to the left."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if they gave us units, if they told us that x was in meters and that t was in seconds, well, then x would be, well, I already said, would be in meters, and velocity would be negative one meters per second. You might also be saying, well, what does the negative mean? Well, that means that we are moving to the left. Remember, we're moving along the x-axis, so if our velocity is negative, that means that x is decreasing or we are moving to the left. What is the particle's acceleration, a of t, at t equals three? So pause this video again and see if you can do that. Well, here, the realization is that acceleration is a function of time."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, we're moving along the x-axis, so if our velocity is negative, that means that x is decreasing or we are moving to the left. What is the particle's acceleration, a of t, at t equals three? So pause this video again and see if you can do that. Well, here, the realization is that acceleration is a function of time. It's just the derivative of velocity, which is the second derivative of our position, which is just going to be equal to the derivative of this right over here. And so I'm just going to get derivative of three t squared with respect to t is six t, derivative of negative eight t with respect to t is minus eight and derivative of constants is zero. So it's just going to be six t minus eight."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, here, the realization is that acceleration is a function of time. It's just the derivative of velocity, which is the second derivative of our position, which is just going to be equal to the derivative of this right over here. And so I'm just going to get derivative of three t squared with respect to t is six t, derivative of negative eight t with respect to t is minus eight and derivative of constants is zero. So it's just going to be six t minus eight. So our acceleration at time t equals three is going to be six times three, which is 18 minus eight. So minus eight, which is going to be equal to positive 10. All right, now they ask us, what is the direction of the particle's motion at t equals two?"}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's just going to be six t minus eight. So our acceleration at time t equals three is going to be six times three, which is 18 minus eight. So minus eight, which is going to be equal to positive 10. All right, now they ask us, what is the direction of the particle's motion at t equals two? Well, I already talked about this, but pause this video and see if you can answer that yourself. Well, we've already looked at the sign right over here. The fact that we have a negative sign on our velocity means we are moving towards the left."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now they ask us, what is the direction of the particle's motion at t equals two? Well, I already talked about this, but pause this video and see if you can answer that yourself. Well, we've already looked at the sign right over here. The fact that we have a negative sign on our velocity means we are moving towards the left. So I'll fill that in right over there. At t equals three, is the particle's speed increasing, decreasing, or neither? So pause this video and try to answer that."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The fact that we have a negative sign on our velocity means we are moving towards the left. So I'll fill that in right over there. At t equals three, is the particle's speed increasing, decreasing, or neither? So pause this video and try to answer that. All right, now we have to be very careful here. If it says is the particle's velocity increasing, decreasing, or neither, then we would just have to look at the acceleration. We see that the acceleration is positive, and so we know that the velocity is increasing."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video and try to answer that. All right, now we have to be very careful here. If it says is the particle's velocity increasing, decreasing, or neither, then we would just have to look at the acceleration. We see that the acceleration is positive, and so we know that the velocity is increasing. But here they're not saying velocity, they're saying speed. And just as a reminder, speed is the magnitude of velocity. So for example, at time t equals two, our velocity is negative one."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We see that the acceleration is positive, and so we know that the velocity is increasing. But here they're not saying velocity, they're saying speed. And just as a reminder, speed is the magnitude of velocity. So for example, at time t equals two, our velocity is negative one. If the units were meters in second, it would be negative one meters per second. But our speed would just be one meter per second. Speed, you're not talking about the direction, so you would not have that sign there."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, at time t equals two, our velocity is negative one. If the units were meters in second, it would be negative one meters per second. But our speed would just be one meter per second. Speed, you're not talking about the direction, so you would not have that sign there. And so in order to figure out if the speed is increasing or decreasing or neither, if the acceleration is positive and the velocity is positive, that means the magnitude of your velocity is increasing. So that means your speed is increasing. If your velocity is negative and your acceleration is also negative, that also means that your speed is increasing."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Speed, you're not talking about the direction, so you would not have that sign there. And so in order to figure out if the speed is increasing or decreasing or neither, if the acceleration is positive and the velocity is positive, that means the magnitude of your velocity is increasing. So that means your speed is increasing. If your velocity is negative and your acceleration is also negative, that also means that your speed is increasing. But if your velocity and acceleration have different signs, well that means that your speed is decreasing. The magnitude of your velocity would be coming less. So let's look at our velocity at time t equals three."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If your velocity is negative and your acceleration is also negative, that also means that your speed is increasing. But if your velocity and acceleration have different signs, well that means that your speed is decreasing. The magnitude of your velocity would be coming less. So let's look at our velocity at time t equals three. Our velocity at time three, we just go back right over here. It's gonna be three times nine, which is 27, three times three squared, minus 24 plus three, plus three. So this is going to be equal to six."}, {"video_title": "Worked example Motion problems with derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at our velocity at time t equals three. Our velocity at time three, we just go back right over here. It's gonna be three times nine, which is 27, three times three squared, minus 24 plus three, plus three. So this is going to be equal to six. So our velocity and acceleration are both, you could say, in the same direction. They are both positive. And so our velocity is only going to become more positive, or the magnitude of our velocity is only going to increase."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is get some practice taking derivatives with the power rule. So let's say we need to take the derivative with respect to x of one over x. What is that going to be equal to? Pause this video and try to figure it out. So at first you might say, how does the power rule apply here? The power rule, just to remind ourselves, it tells us that if we're taking the derivative of x to the n with respect to x, so if we're taking the derivative of that, that that's going to be equal to, we take the exponent, bring it out front, and we've proven it in other videos, but this is going to be n times x to the, and then we decrement the exponent, so n minus one. But this does not look like that, and the key is to appreciate that one over x is the same thing as x to the negative one."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause this video and try to figure it out. So at first you might say, how does the power rule apply here? The power rule, just to remind ourselves, it tells us that if we're taking the derivative of x to the n with respect to x, so if we're taking the derivative of that, that that's going to be equal to, we take the exponent, bring it out front, and we've proven it in other videos, but this is going to be n times x to the, and then we decrement the exponent, so n minus one. But this does not look like that, and the key is to appreciate that one over x is the same thing as x to the negative one. So this is going to be the derivative with respect to x of x to the negative one. And now this looks a lot more like what you might be used to where this is going to be equal to, you take our exponent, bring it out front, so that's negative one, times x to the negative one minus one, negative one minus one, and so this is going to be equal to negative x to the negative two, and we're done. Let's do another example."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this does not look like that, and the key is to appreciate that one over x is the same thing as x to the negative one. So this is going to be the derivative with respect to x of x to the negative one. And now this looks a lot more like what you might be used to where this is going to be equal to, you take our exponent, bring it out front, so that's negative one, times x to the negative one minus one, negative one minus one, and so this is going to be equal to negative x to the negative two, and we're done. Let's do another example. Let's say that we're told that f of x is equal to the cube root of x, and we wanna figure out what f prime of x is equal to. Pause the video and see if you can figure it out again. Well, once again, you might say, hey, how do I take the derivative of something like this, especially if my goal, or if I'm thinking that maybe the power rule might be useful?"}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do another example. Let's say that we're told that f of x is equal to the cube root of x, and we wanna figure out what f prime of x is equal to. Pause the video and see if you can figure it out again. Well, once again, you might say, hey, how do I take the derivative of something like this, especially if my goal, or if I'm thinking that maybe the power rule might be useful? And the idea is to rewrite this as an exponent. If you could rewrite the cube root as x to the 1 3rd power. And so the derivative, you take the 1 3rd, bring it out front, so it's 1 3rd, x to the 1 3rd minus one power."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, once again, you might say, hey, how do I take the derivative of something like this, especially if my goal, or if I'm thinking that maybe the power rule might be useful? And the idea is to rewrite this as an exponent. If you could rewrite the cube root as x to the 1 3rd power. And so the derivative, you take the 1 3rd, bring it out front, so it's 1 3rd, x to the 1 3rd minus one power. And so this is going to be 1 3rd times x to the 1 3rd minus one is negative 2 3rds, negative 2 3rd power, and we are done. And hopefully through these examples, you're seeing that the power rule is incredibly powerful. You can tackle a far broader range of derivatives than you might have initially thought."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the derivative, you take the 1 3rd, bring it out front, so it's 1 3rd, x to the 1 3rd minus one power. And so this is going to be 1 3rd times x to the 1 3rd minus one is negative 2 3rds, negative 2 3rd power, and we are done. And hopefully through these examples, you're seeing that the power rule is incredibly powerful. You can tackle a far broader range of derivatives than you might have initially thought. Let's do another example, and I'll make this one really nice and hairy. Let's say we wanna figure out the derivative with respect to x of the cube root of x squared. What is this going to be?"}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can tackle a far broader range of derivatives than you might have initially thought. Let's do another example, and I'll make this one really nice and hairy. Let's say we wanna figure out the derivative with respect to x of the cube root of x squared. What is this going to be? And actually, let's just not figure out what the derivative is. Let's figure out the derivative at x equals eight. Pause this video again and see if you can figure that out."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is this going to be? And actually, let's just not figure out what the derivative is. Let's figure out the derivative at x equals eight. Pause this video again and see if you can figure that out. Well, what we're gonna do is first just figure out what this is, and then we're going to evaluate it at x equals eight. And the key thing to appreciate is this is the same thing, and we're just gonna do what we did up here, as the derivative with respect to x. Instead of saying the cube root of x squared, we could say this is x squared to the 1 3rd power, which is the same thing as the derivative with respect to x of, well, x squared, if I raise something to an exponent and then raise that to an exponent, I could just take the product of the exponents."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause this video again and see if you can figure that out. Well, what we're gonna do is first just figure out what this is, and then we're going to evaluate it at x equals eight. And the key thing to appreciate is this is the same thing, and we're just gonna do what we did up here, as the derivative with respect to x. Instead of saying the cube root of x squared, we could say this is x squared to the 1 3rd power, which is the same thing as the derivative with respect to x of, well, x squared, if I raise something to an exponent and then raise that to an exponent, I could just take the product of the exponents. And so this is just going to be x to the two times 1 3rd power, or to the 2 3rds power. And now this is just going to be equal to, I'll do it right over here, bring the 2 3rds out front, 2 3rds times x to the, what's 2 3rds minus one? Well, that's 2 3rds minus 3 3rd, or it would be negative 1 3rd power."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of saying the cube root of x squared, we could say this is x squared to the 1 3rd power, which is the same thing as the derivative with respect to x of, well, x squared, if I raise something to an exponent and then raise that to an exponent, I could just take the product of the exponents. And so this is just going to be x to the two times 1 3rd power, or to the 2 3rds power. And now this is just going to be equal to, I'll do it right over here, bring the 2 3rds out front, 2 3rds times x to the, what's 2 3rds minus one? Well, that's 2 3rds minus 3 3rd, or it would be negative 1 3rd power. And we wanna know what happens at x equals eight, so let's just evaluate that. That's going to be 2 3rds times x is equal to eight to the negative 1 3rd power. Well, what's eight to the 1 3rd power?"}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's 2 3rds minus 3 3rd, or it would be negative 1 3rd power. And we wanna know what happens at x equals eight, so let's just evaluate that. That's going to be 2 3rds times x is equal to eight to the negative 1 3rd power. Well, what's eight to the 1 3rd power? Eight to the 1 3rd power is going to be equal to two, and so eight to the negative 1 3rd power is 1 1.5. Actually, let me just do that step by step. So this is going to be equal to 2 3rds times, we could do it this way, one over eight to the 1 3rd power."}, {"video_title": "Power rule (with rewriting the expression)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's eight to the 1 3rd power? Eight to the 1 3rd power is going to be equal to two, and so eight to the negative 1 3rd power is 1 1.5. Actually, let me just do that step by step. So this is going to be equal to 2 3rds times, we could do it this way, one over eight to the 1 3rd power. And so this is just one over two, 2 3rds times 1 1.5. Well, that's just going to be equal to 1 3rd. And we're done."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now, if we want to take the antiderivative of the stuff on the inside, we're just going to do, essentially the power rule, you could use this as the power rule of integrals, or it's the reverse of the power rule for derivatives, where we increase this exponent by one, and then we divide by that increased exponent. So this is going to be equal to 12 times x to the 1 3rd plus one, let me do it in another color just so we can keep track of it. X to the 1 3rd plus one, and then we're going to divide by 1 3rd plus one. And so what's 1 3rd plus one? Well, that's 4 3rds, 1 3rd plus 3 3rds, that's 4 3rds. So I could write it this way, I could write this x to the 4 3rds divided by 4 3rds. And this is going to be, and I'm going to evaluate this at the bounds."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what's 1 3rd plus one? Well, that's 4 3rds, 1 3rd plus 3 3rds, that's 4 3rds. So I could write it this way, I could write this x to the 4 3rds divided by 4 3rds. And this is going to be, and I'm going to evaluate this at the bounds. So I'm going to evaluate this at, and I'll do this in different colors, I'm going to evaluate it at eight, and I'm going to evaluate it at negative one, and I'm going to subtract it, evaluate it at negative one from this expression evaluated at eight. And so what is this going to be equal to? Well, actually, let me simplify a little bit more."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is going to be, and I'm going to evaluate this at the bounds. So I'm going to evaluate this at, and I'll do this in different colors, I'm going to evaluate it at eight, and I'm going to evaluate it at negative one, and I'm going to subtract it, evaluate it at negative one from this expression evaluated at eight. And so what is this going to be equal to? Well, actually, let me simplify a little bit more. What is 12 divided by 4 3rds? So 12, I'll do it right, well, I'll do it right over here. 12 over 4 3rds is equal to 12 times three over four, which we could use 12 over one times three over four."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, actually, let me simplify a little bit more. What is 12 divided by 4 3rds? So 12, I'll do it right, well, I'll do it right over here. 12 over 4 3rds is equal to 12 times three over four, which we could use 12 over one times three over four. 12 divided by four is three, so this is going to be equal to nine. 3 4ths of 12 is nine. So this, we could rewrite this, we could write this as nine x to the 4 3rd power."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "12 over 4 3rds is equal to 12 times three over four, which we could use 12 over one times three over four. 12 divided by four is three, so this is going to be equal to nine. 3 4ths of 12 is nine. So this, we could rewrite this, we could write this as nine x to the 4 3rd power. So if we evaluate it at eight, this is going to be nine times eight to the 4 3rds power, and from that, we're going to subtract, it evaluated at negative one. So this is going to be nine times negative one to the 4 3rd power. So what is eight to the 4 3rd power?"}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this, we could rewrite this, we could write this as nine x to the 4 3rd power. So if we evaluate it at eight, this is going to be nine times eight to the 4 3rds power, and from that, we're going to subtract, it evaluated at negative one. So this is going to be nine times negative one to the 4 3rd power. So what is eight to the 4 3rd power? I'll do it over here. So eight to the 4 3rds is equal to eight to the 1 3rd to the 4th power. These are just exponent properties here."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is eight to the 4 3rd power? I'll do it over here. So eight to the 4 3rds is equal to eight to the 1 3rd to the 4th power. These are just exponent properties here. Eight to the 1 3rd, the cube root of eight, or eight to the 1 3rd power, that's two, because two to the 3rd power is eight. And two to the 4th power, well, two to the 4th power is equal to 16. So eight to the 4 3rds is 16."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "These are just exponent properties here. Eight to the 1 3rd, the cube root of eight, or eight to the 1 3rd power, that's two, because two to the 3rd power is eight. And two to the 4th power, well, two to the 4th power is equal to 16. So eight to the 4 3rds is 16. And what's negative one to the 4 3rds? Well, same idea. Negative one to the 4 3rds is equal to negative one, there's several ways you could do it."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So eight to the 4 3rds is 16. And what's negative one to the 4 3rds? Well, same idea. Negative one to the 4 3rds is equal to negative one, there's several ways you could do it. You could say negative one to the 4th, and then the cube root of that, or the cube root of negative one, and then raise that to the 4th power, either way. So let's do it the first way. Negative one to the 4th, and then take the cube root of that."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative one to the 4 3rds is equal to negative one, there's several ways you could do it. You could say negative one to the 4th, and then the cube root of that, or the cube root of negative one, and then raise that to the 4th power, either way. So let's do it the first way. Negative one to the 4th, and then take the cube root of that. Well, negative one to the 4th is just one, and then one to the 1 3rd power, well, that's just going to be equal to one. So what we have here in blue, that's just equal to one. So we have nine times 16 minus nine times one."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative one to the 4th, and then take the cube root of that. Well, negative one to the 4th is just one, and then one to the 1 3rd power, well, that's just going to be equal to one. So what we have here in blue, that's just equal to one. So we have nine times 16 minus nine times one. Well, that's just going to be nine times 15. We have 16 nines, and then we're gonna take away a nine. So that's gonna be nine times 15."}, {"video_title": "Definite integral of radical function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have nine times 16 minus nine times one. Well, that's just going to be nine times 15. We have 16 nines, and then we're gonna take away a nine. So that's gonna be nine times 15. So what is that? That is going to be equal to, nine times 15 is 90 plus 45, which is equal to 135. 135, and we're done."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's right over there. Use the data in the table to approximate the rate at which the temperature of the tea is changing at time t equals 3.5. Show the computations that lead to your answer. So they give us a bunch of points. Let's just graph this just so we can visualize what this data is telling us a little bit, and I suspect that this might be useful. So this right over here is our temperature axis. This is our temperature at any moment in degrees Celsius, and then this is our time axis, and they gave us the time 0, 2, 5, 9, and 10, or they gave us the temperature at those times."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they give us a bunch of points. Let's just graph this just so we can visualize what this data is telling us a little bit, and I suspect that this might be useful. So this right over here is our temperature axis. This is our temperature at any moment in degrees Celsius, and then this is our time axis, and they gave us the time 0, 2, 5, 9, and 10, or they gave us the temperature at those times. So this is 0, 2, 5 is a little bit further, 5, 9, and then 10. And at time 0, the temperature is 66 degrees Celsius, 66, so it's right over there. At time 2, it is 60 degrees Celsius."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our temperature at any moment in degrees Celsius, and then this is our time axis, and they gave us the time 0, 2, 5, 9, and 10, or they gave us the temperature at those times. So this is 0, 2, 5 is a little bit further, 5, 9, and then 10. And at time 0, the temperature is 66 degrees Celsius, 66, so it's right over there. At time 2, it is 60 degrees Celsius. At time 5, it is 52 degrees Celsius. So time 5, maybe it's right over here. This is 52 degrees Celsius."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "At time 2, it is 60 degrees Celsius. At time 5, it is 52 degrees Celsius. So time 5, maybe it's right over here. This is 52 degrees Celsius. At time 9, or after 9 minutes, I should say, it's at 44 degrees Celsius. So this is 44 degrees Celsius, and after 10 minutes, it's at 43 degrees Celsius. So this is really just a graph showing how it cools off over those 10 minutes."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 52 degrees Celsius. At time 9, or after 9 minutes, I should say, it's at 44 degrees Celsius. So this is 44 degrees Celsius, and after 10 minutes, it's at 43 degrees Celsius. So this is really just a graph showing how it cools off over those 10 minutes. That's what the data is telling us. It's kind of a curve like this, and we've sampled it at these points. So going back to part A, use the data in the table to approximate the rate at which the temperature of the T is changing at time T equals 3.5."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is really just a graph showing how it cools off over those 10 minutes. That's what the data is telling us. It's kind of a curve like this, and we've sampled it at these points. So going back to part A, use the data in the table to approximate the rate at which the temperature of the T is changing at time T equals 3.5. So the rate of change is really just the slope of this curve at time T is equal to 3.5. So really, we just want to find the slope at that point right over there. We want to find the slope."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So going back to part A, use the data in the table to approximate the rate at which the temperature of the T is changing at time T equals 3.5. So the rate of change is really just the slope of this curve at time T is equal to 3.5. So really, we just want to find the slope at that point right over there. We want to find the slope. And we don't know the actual function, so the best way to approximate the slope at that point is really just find the slope between minute 5 and minute 2. So the rate of change, we could say that the rate is going to be approximately our change in temperature over those 3 minutes. So h of 5 minus h of 2, and this is obviously going to be in degrees Celsius, over the number of minutes that changed."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We want to find the slope. And we don't know the actual function, so the best way to approximate the slope at that point is really just find the slope between minute 5 and minute 2. So the rate of change, we could say that the rate is going to be approximately our change in temperature over those 3 minutes. So h of 5 minus h of 2, and this is obviously going to be in degrees Celsius, over the number of minutes that changed. So we end at 5 minutes and we started at 2 minutes. And so the rate is going to be approximately at 5 minutes, our temperature is 52 degrees. At 2 minutes, our temperature is 60 degrees."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So h of 5 minus h of 2, and this is obviously going to be in degrees Celsius, over the number of minutes that changed. So we end at 5 minutes and we started at 2 minutes. And so the rate is going to be approximately at 5 minutes, our temperature is 52 degrees. At 2 minutes, our temperature is 60 degrees. And then this is over a change in 3 minutes. 5 minus 2 is 3. So this gets us, let me scroll to the right a little bit, this gives us negative 8 degrees Celsius over 3 minutes."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "At 2 minutes, our temperature is 60 degrees. And then this is over a change in 3 minutes. 5 minus 2 is 3. So this gets us, let me scroll to the right a little bit, this gives us negative 8 degrees Celsius over 3 minutes. Or the rate at 3 and a half minutes is going to be approximately negative 8 thirds degrees Celsius per minute. So that is part A. And then part B."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this gets us, let me scroll to the right a little bit, this gives us negative 8 degrees Celsius over 3 minutes. Or the rate at 3 and a half minutes is going to be approximately negative 8 thirds degrees Celsius per minute. So that is part A. And then part B. Using correct units, explain the meaning of 1 over 10 times the definite integral from 0 to 10 of h of t dt in the context of this problem. Use a trapezoidal sum with the 4 subintervals indicated by the table to estimate this thing. So the integral from 0 to 10 of h of t is really the area of this entire, under this curve right over here."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then part B. Using correct units, explain the meaning of 1 over 10 times the definite integral from 0 to 10 of h of t dt in the context of this problem. Use a trapezoidal sum with the 4 subintervals indicated by the table to estimate this thing. So the integral from 0 to 10 of h of t is really the area of this entire, under this curve right over here. That's the integral, that's this part right over here. And then we're going to divide that by 10. So what this is really giving us, this expression right over here is the average temperature."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the integral from 0 to 10 of h of t is really the area of this entire, under this curve right over here. That's the integral, that's this part right over here. And then we're going to divide that by 10. So what this is really giving us, this expression right over here is the average temperature. This is the average temperature over that time period. The integral of the temperature function divided by the total amount of time that has elapsed. And then using correct units, well the average temperature is once again going to be in degrees Celsius."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what this is really giving us, this expression right over here is the average temperature. This is the average temperature over that time period. The integral of the temperature function divided by the total amount of time that has elapsed. And then using correct units, well the average temperature is once again going to be in degrees Celsius. And it makes sense because this right over here is in degrees Celsius. You're multiplying it by time right over here. But then you could divide by, you're dividing I'm assuming by time right over here."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then using correct units, well the average temperature is once again going to be in degrees Celsius. And it makes sense because this right over here is in degrees Celsius. You're multiplying it by time right over here. But then you could divide by, you're dividing I'm assuming by time right over here. So this is times minutes divided by minutes. So then you just get degrees Celsius again. So this expression right over here is just the average temperature over those 10 minutes."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then you could divide by, you're dividing I'm assuming by time right over here. So this is times minutes divided by minutes. So then you just get degrees Celsius again. So this expression right over here is just the average temperature over those 10 minutes. Then they say use a trapezoidal sum with the 4 subintervals indicated by the table to estimate this. So a trapezoidal sum is, we don't know the exact function here so we won't be able to analytically evaluate this definite integral. But what we can do is divide this area into 4 sections."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this expression right over here is just the average temperature over those 10 minutes. Then they say use a trapezoidal sum with the 4 subintervals indicated by the table to estimate this. So a trapezoidal sum is, we don't know the exact function here so we won't be able to analytically evaluate this definite integral. But what we can do is divide this area into 4 sections. They tell us to use this, 4 subintervals. And we'll essentially divide it into 4 trapezoids. So this is one trapezoid right over here."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what we can do is divide this area into 4 sections. They tell us to use this, 4 subintervals. And we'll essentially divide it into 4 trapezoids. So this is one trapezoid right over here. So this is one trapezoid right over there. That's my first trapezoid from 0 to 2 on the base. And you see on the left end of that trapezoid its height is 66."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is one trapezoid right over here. So this is one trapezoid right over there. That's my first trapezoid from 0 to 2 on the base. And you see on the left end of that trapezoid its height is 66. The right end is 60. And then the next trapezoid will go from 2 to 5. Let me do that in a color that contrasts a little bit better."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you see on the left end of that trapezoid its height is 66. The right end is 60. And then the next trapezoid will go from 2 to 5. Let me do that in a color that contrasts a little bit better. The next trapezoid will go from 2 to 5. And then the third trapezoid goes from 5 to 9. And then the fourth trapezoid goes from 9 to 10."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do that in a color that contrasts a little bit better. The next trapezoid will go from 2 to 5. And then the third trapezoid goes from 5 to 9. And then the fourth trapezoid goes from 9 to 10. So if I want to approximate the definite integral part right here, before we divide by 10, I just need to find the area of these 4 trapezoids. So let's do that. The area of this first trapezoid is going to be the base, which is 2, times the average height."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the fourth trapezoid goes from 9 to 10. So if I want to approximate the definite integral part right here, before we divide by 10, I just need to find the area of these 4 trapezoids. So let's do that. The area of this first trapezoid is going to be the base, which is 2, times the average height. The height on the left side of this trapezoid is 66. The height on the right side of this trapezoid is 60. The average height right over here is going to be 63."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The area of this first trapezoid is going to be the base, which is 2, times the average height. The height on the left side of this trapezoid is 66. The height on the right side of this trapezoid is 60. The average height right over here is going to be 63. Just the average of 60 and 66. So this area right over here is 126. That's the area of the green part right over there."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The average height right over here is going to be 63. Just the average of 60 and 66. So this area right over here is 126. That's the area of the green part right over there. By the same logic, the area of this orange part, I want to do that in the orange color, the base right over here is 3. And then the average height, the height over here is 60. The height over here is 52."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the area of the green part right over there. By the same logic, the area of this orange part, I want to do that in the orange color, the base right over here is 3. And then the average height, the height over here is 60. The height over here is 52. They're 8 apart, so the average is going to be 4 away from each of them. The average is going to be 56. So the area here is going to be 3 times 56, which is 150 plus 18, 150 plus 18 is 168."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The height over here is 52. They're 8 apart, so the average is going to be 4 away from each of them. The average is going to be 56. So the area here is going to be 3 times 56, which is 150 plus 18, 150 plus 18 is 168. That's the area of this orange trapezoid. And then the area of this blue trapezoid, the base right over here is 4. And then the average height, the height here is 52, the height here is 44."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area here is going to be 3 times 56, which is 150 plus 18, 150 plus 18 is 168. That's the area of this orange trapezoid. And then the area of this blue trapezoid, the base right over here is 4. And then the average height, the height here is 52, the height here is 44. So the average of 52 and 44, they are 8 apart, so then it's just going to be 4 from each of these, so it's 48. So the average height here is 48. And so 4 times 48 is 160 plus 32, so it's 192."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the average height, the height here is 52, the height here is 44. So the average of 52 and 44, they are 8 apart, so then it's just going to be 4 from each of these, so it's 48. So the average height here is 48. And so 4 times 48 is 160 plus 32, so it's 192. And then finally, this last trapezoid, its base is only 1. Its base right over here is only 1. And then its height is 43 at the left side, 43 at the right side."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so 4 times 48 is 160 plus 32, so it's 192. And then finally, this last trapezoid, its base is only 1. Its base right over here is only 1. And then its height is 43 at the left side, 43 at the right side. So its average height is 43.5 times 1. So that's just going to be 43.5. And so if we add up the areas of all these trapezoids, we have a pretty good approximation for the definite integral."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then its height is 43 at the left side, 43 at the right side. So its average height is 43.5 times 1. So that's just going to be 43.5. And so if we add up the areas of all these trapezoids, we have a pretty good approximation for the definite integral. I'll just use the calculator for this part right over here. So we have 126 plus 168 plus 192 plus 43.5 gives us 529.5. So this is equal to 529.5."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we add up the areas of all these trapezoids, we have a pretty good approximation for the definite integral. I'll just use the calculator for this part right over here. So we have 126 plus 168 plus 192 plus 43.5 gives us 529.5. So this is equal to 529.5. And so that's our approximation for the definite integral part, but then we also have to divide it by 10, or we have to multiply it times 1 tenth. So we've evaluated, so this part right over here, we got 529.5 as our approximation. It's not going to be exact, but using the trapezoidal sum, the sum of the areas of these trapezoids, and now we have to multiply it by 1 tenth."}, {"video_title": "2011 Calculus AB free response #2 (a & b)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to 529.5. And so that's our approximation for the definite integral part, but then we also have to divide it by 10, or we have to multiply it times 1 tenth. So we've evaluated, so this part right over here, we got 529.5 as our approximation. It's not going to be exact, but using the trapezoidal sum, the sum of the areas of these trapezoids, and now we have to multiply it by 1 tenth. So let's do that. Or we can just divide by 10, which actually we don't need a calculator for that, 52.95. So this whole thing evaluates to 52.95."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "Think of your derivative properties, especially the derivative of logarithmic functions, especially logarithmic functions, in this case, with base 10. If someone just writes log without the base, you can just assume that that is a 10 right over there. So pause the video and see if you can work through it. All right, so the key here is to remember that if I have f of x, let me do it over here, I'll do it over here, f of x, and I want to find f prime of, let's say f prime of some number, let's say a, this is going to be equal to the limit as x, or sorry, as h approaches zero is one way of thinking about it, as h approaches zero of f of a plus h minus f of a, all of that over h. So this looks pretty close to that limit definition, except we have these fives here. But lucky for us, we can factor out those fives, and we could factor them out out front here, but if you just have a scalar times the expression, we know from our limit properties that we can actually take those out of the limit themselves. So let's do that. Let's take both of these fives and factor them out."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "All right, so the key here is to remember that if I have f of x, let me do it over here, I'll do it over here, f of x, and I want to find f prime of, let's say f prime of some number, let's say a, this is going to be equal to the limit as x, or sorry, as h approaches zero is one way of thinking about it, as h approaches zero of f of a plus h minus f of a, all of that over h. So this looks pretty close to that limit definition, except we have these fives here. But lucky for us, we can factor out those fives, and we could factor them out out front here, but if you just have a scalar times the expression, we know from our limit properties that we can actually take those out of the limit themselves. So let's do that. Let's take both of these fives and factor them out. And so this whole thing is going to simplify to five times the limit as h approaches zero of log of two plus h minus, minus log of two, all of that over h. Minus log of two, all of that over h. Now, you might recognize what we have in yellow here. Let's think about it. What this is, if we had f of x is equal to log of x, and we wanted to know what f prime of, well actually, let's say f prime of two is, well, this would be the limit as h approaches zero of log of two plus h, two plus h, minus log of two, minus log of two, all of that over h. So this is really just a, what we see here, this by definition, this right over here is f prime of two."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "Let's take both of these fives and factor them out. And so this whole thing is going to simplify to five times the limit as h approaches zero of log of two plus h minus, minus log of two, all of that over h. Minus log of two, all of that over h. Now, you might recognize what we have in yellow here. Let's think about it. What this is, if we had f of x is equal to log of x, and we wanted to know what f prime of, well actually, let's say f prime of two is, well, this would be the limit as h approaches zero of log of two plus h, two plus h, minus log of two, minus log of two, all of that over h. So this is really just a, what we see here, this by definition, this right over here is f prime of two. If f of x is log of x, this is f prime of two. F prime of two. So can we figure that out?"}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "What this is, if we had f of x is equal to log of x, and we wanted to know what f prime of, well actually, let's say f prime of two is, well, this would be the limit as h approaches zero of log of two plus h, two plus h, minus log of two, minus log of two, all of that over h. So this is really just a, what we see here, this by definition, this right over here is f prime of two. If f of x is log of x, this is f prime of two. F prime of two. So can we figure that out? If f of x is log of x, what is f prime of x? F prime of x, we don't need to use the limit definition. In fact, the limit definition is quite hard to evaluate this limit, but we know how to take the derivative of logarithmic functions."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "So can we figure that out? If f of x is log of x, what is f prime of x? F prime of x, we don't need to use the limit definition. In fact, the limit definition is quite hard to evaluate this limit, but we know how to take the derivative of logarithmic functions. So f prime of x is going to be equal to one over the natural log of our base, our base here, we already talked about that, that is 10. So one over natural log of 10 times, times x. If this was a natural log, well then this would be one over natural log of e times x, natural log of e is just one, so that's where you get the one over x, but if you have any other base, you put the natural log of that base right over here in the denominator."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "In fact, the limit definition is quite hard to evaluate this limit, but we know how to take the derivative of logarithmic functions. So f prime of x is going to be equal to one over the natural log of our base, our base here, we already talked about that, that is 10. So one over natural log of 10 times, times x. If this was a natural log, well then this would be one over natural log of e times x, natural log of e is just one, so that's where you get the one over x, but if you have any other base, you put the natural log of that base right over here in the denominator. So what is f prime of two? F prime of two is one over the natural log of 10 times two. So this whole thing has simplified, this whole thing is equal to five times this business."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "If this was a natural log, well then this would be one over natural log of e times x, natural log of e is just one, so that's where you get the one over x, but if you have any other base, you put the natural log of that base right over here in the denominator. So what is f prime of two? F prime of two is one over the natural log of 10 times two. So this whole thing has simplified, this whole thing is equal to five times this business. So I could actually just write it as, it's equal to five over, five over the natural log of 10, natural log of 10 times two. I could have written it as two natural log of 10s. The key here for this type of exercise, you might be like, oh let me see if I can evaluate this limit, but you're like, well, this looks a lot like the derivative of a logarithmic function, especially the derivative when x is equal to two, if we could just factor these fives out."}, {"video_title": "Derivatives expressed as limits   Advanced derivatives   AP Calculus BC   Khan Academy.mp3", "Sentence": "So this whole thing has simplified, this whole thing is equal to five times this business. So I could actually just write it as, it's equal to five over, five over the natural log of 10, natural log of 10 times two. I could have written it as two natural log of 10s. The key here for this type of exercise, you might be like, oh let me see if I can evaluate this limit, but you're like, well, this looks a lot like the derivative of a logarithmic function, especially the derivative when x is equal to two, if we could just factor these fives out. So you factor out the five, you say, hey, this is the derivative of log of x when x is equal to two. And so we know how to take the derivative of log of x. If you don't know, we have videos where we prove this, where we take the derivatives of logarithms with bases other than e, and you just use that to actually, and you find the derivative, then you evaluate it at two, and then you're done."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "In order to have the rigorous proof of these properties, we need a rigorous definition of what a limit is. And we're not doing that in this tutorial. We'll do that in the tutorial on the epsilon delta definition of limits. But most of these should be fairly intuitive. And they're very helpful for simplifying limit problems in the future. So let's say we know that the limit of some function f of x as x approaches c is equal to capital L. And let's say that we also know that the limit of some other function, let's say g of x as x approaches c, is equal to capital M. Now, given that, what would be the limit of f of x plus g of x as x approaches c? Well, and you could look at this visually."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But most of these should be fairly intuitive. And they're very helpful for simplifying limit problems in the future. So let's say we know that the limit of some function f of x as x approaches c is equal to capital L. And let's say that we also know that the limit of some other function, let's say g of x as x approaches c, is equal to capital M. Now, given that, what would be the limit of f of x plus g of x as x approaches c? Well, and you could look at this visually. If you look at the graphs of two arbitrary functions, you would essentially just add those two functions. It'll be pretty clear that this is going to be equal to, and once again, I'm not doing a rigorous proof. I'm just really giving you the properties here."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, and you could look at this visually. If you look at the graphs of two arbitrary functions, you would essentially just add those two functions. It'll be pretty clear that this is going to be equal to, and once again, I'm not doing a rigorous proof. I'm just really giving you the properties here. This is going to be the limit of f of x as x approaches c plus the limit of g of x as x approaches c, which is equal to, well, this right over here is, let me do that in that same color, this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule or the sum property of limits. And we could come up with a very similar one with differences."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm just really giving you the properties here. This is going to be the limit of f of x as x approaches c plus the limit of g of x as x approaches c, which is equal to, well, this right over here is, let me do that in that same color, this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule or the sum property of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x is just going to be L minus M. It's just the limit of f of x as x approaches c minus the limit of g of x as x approaches c. So it's just going to be L minus M. It's often called the difference rule or the difference property of limits. And these, once again, are very, very, hopefully, reasonably intuitive. Now what happens if you take the product of the functions?"}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x is just going to be L minus M. It's just the limit of f of x as x approaches c minus the limit of g of x as x approaches c. So it's just going to be L minus M. It's often called the difference rule or the difference property of limits. And these, once again, are very, very, hopefully, reasonably intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well, lucky for us, this is going to be equal to the limit of f of x as x approaches c times the limit of g of x as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to L times M. Same thing if instead of having a function here, we had a constant. So if we just had the limit of k times f of x as x approaches c, where k is just some constant, this is going to be the same thing as k times the limit of f of x as x approaches c. And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well, lucky for us, this is going to be equal to the limit of f of x as x approaches c times the limit of g of x as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to L times M. Same thing if instead of having a function here, we had a constant. So if we just had the limit of k times f of x as x approaches c, where k is just some constant, this is going to be the same thing as k times the limit of f of x as x approaches c. And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x, this is the exact same thing as the limit of f of x as x approaches c divided by the limit of g of x as x approaches c, which is going to be equal to L over M. And finally, this is sometimes called the quotient property."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we just had the limit of k times f of x as x approaches c, where k is just some constant, this is going to be the same thing as k times the limit of f of x as x approaches c. And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x, this is the exact same thing as the limit of f of x as x approaches c divided by the limit of g of x as x approaches c, which is going to be equal to L over M. And finally, this is sometimes called the quotient property. Finally, we'll look at the exponent property. So if I have the limit of f of x to some power, and actually let me even write it as a fractional power, to the r over s power, where both r and s are integers, then the limit of f of x to the r over s power as x approaches c is going to be the exact same thing as the limit of f of x as x approaches c raised to the r over s power. Once again, when r and s are both integers, and s is not equal to 0, otherwise this exponent would not make much sense."}, {"video_title": "Limit properties   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we have the limit as x approaches c of f of x divided by g of x, this is the exact same thing as the limit of f of x as x approaches c divided by the limit of g of x as x approaches c, which is going to be equal to L over M. And finally, this is sometimes called the quotient property. Finally, we'll look at the exponent property. So if I have the limit of f of x to some power, and actually let me even write it as a fractional power, to the r over s power, where both r and s are integers, then the limit of f of x to the r over s power as x approaches c is going to be the exact same thing as the limit of f of x as x approaches c raised to the r over s power. Once again, when r and s are both integers, and s is not equal to 0, otherwise this exponent would not make much sense. And this is the same thing as L to the r over s power. This is equal to L to the r over s power. So using these, we can actually find the limit of many, many, many things."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is think about limits involving trigonometric functions. So let's just start with a fairly straightforward one. Let's find the limit as x approaches pi of sine of x. Pause the video and see if you can figure this out. Well, with both sine of x and cosine of x, they're defined for all real numbers, so their domain is all real numbers. You can put any real number in here for x and it will give you an output. It is defined."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and see if you can figure this out. Well, with both sine of x and cosine of x, they're defined for all real numbers, so their domain is all real numbers. You can put any real number in here for x and it will give you an output. It is defined. And they are also continuous over their entire domain. In fact, all of the trigonometric functions are continuous over their entire domain. And so for sine of x, because it's continuous and it's defined at sine of pi, we would say that this is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is defined. And they are also continuous over their entire domain. In fact, all of the trigonometric functions are continuous over their entire domain. And so for sine of x, because it's continuous and it's defined at sine of pi, we would say that this is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero. And we could do a similar exercise with cosine of x. So if I were to say, what's the limit as x approaches, I'll just take an arbitrary angle, x approaches pi over four of cosine of x? Well, once again, cosine of x is defined for all real numbers."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so for sine of x, because it's continuous and it's defined at sine of pi, we would say that this is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero. And we could do a similar exercise with cosine of x. So if I were to say, what's the limit as x approaches, I'll just take an arbitrary angle, x approaches pi over four of cosine of x? Well, once again, cosine of x is defined for all real numbers. X can be any real number. It's also continuous. So for cosine of x, this limit is just going to be cosine of pi over four."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, once again, cosine of x is defined for all real numbers. X can be any real number. It's also continuous. So for cosine of x, this limit is just going to be cosine of pi over four. And that is going to be equal to square root of two over two. This is one of those useful angles to know the sine and cosine of. If you think in degrees, this is a 45-degree angle."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for cosine of x, this limit is just going to be cosine of pi over four. And that is going to be equal to square root of two over two. This is one of those useful angles to know the sine and cosine of. If you think in degrees, this is a 45-degree angle. And in general, if I'm dealing with a sine or a cosine, the limit as x approaches a of sine of x is equal to sine of a. Once again, this is going to be true for any a, any real number a. And I can make a similar statement about cosine of x."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you think in degrees, this is a 45-degree angle. And in general, if I'm dealing with a sine or a cosine, the limit as x approaches a of sine of x is equal to sine of a. Once again, this is going to be true for any a, any real number a. And I can make a similar statement about cosine of x. Limit as x approaches a of cosine of x is equal to cosine of a. Now, I've been saying it over and over. That's because both of their domains are all real numbers."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I can make a similar statement about cosine of x. Limit as x approaches a of cosine of x is equal to cosine of a. Now, I've been saying it over and over. That's because both of their domains are all real numbers. They are defined for all real numbers that you put in. And they're continuous on their entire domain. But now let's do slightly more involved trigonometric functions, or ones that aren't defined for all real numbers, that their domains are constrained just a little bit more."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's because both of their domains are all real numbers. They are defined for all real numbers that you put in. And they're continuous on their entire domain. But now let's do slightly more involved trigonometric functions, or ones that aren't defined for all real numbers, that their domains are constrained just a little bit more. So let's say if we were to take the limit as x approaches pi of tangent of x. What is this going to be equal to? Well, this is the same thing as the limit as x approaches pi."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now let's do slightly more involved trigonometric functions, or ones that aren't defined for all real numbers, that their domains are constrained just a little bit more. So let's say if we were to take the limit as x approaches pi of tangent of x. What is this going to be equal to? Well, this is the same thing as the limit as x approaches pi. Tangent of x is sine of x over cosine of x. And so both of these are defined for pi. And so we could just substitute pi in."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is the same thing as the limit as x approaches pi. Tangent of x is sine of x over cosine of x. And so both of these are defined for pi. And so we could just substitute pi in. And we just want to ensure that we don't get a zero in the denominator, because that would make it undefined. So we get sine of pi over cosine of pi, which is equal to zero over negative one, which is completely fine. If it was negative one over zero, we'd be in trouble."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we could just substitute pi in. And we just want to ensure that we don't get a zero in the denominator, because that would make it undefined. So we get sine of pi over cosine of pi, which is equal to zero over negative one, which is completely fine. If it was negative one over zero, we'd be in trouble. But this is just going to be equal to zero. So that works out. But if I were to ask you, what is the limit as x approaches pi over two of tangent of x?"}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If it was negative one over zero, we'd be in trouble. But this is just going to be equal to zero. So that works out. But if I were to ask you, what is the limit as x approaches pi over two of tangent of x? Pause the video and try to work that out. Well, think about it. This is the limit as x approaches pi over two of sine of x over cosine of x."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if I were to ask you, what is the limit as x approaches pi over two of tangent of x? Pause the video and try to work that out. Well, think about it. This is the limit as x approaches pi over two of sine of x over cosine of x. Now sine of pi over two is one, but cosine of pi over two is zero. So if you were to just substitute it in, this would give you one over zero. And one way to think about it is, pi over two is not in the domain of tangent of x."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the limit as x approaches pi over two of sine of x over cosine of x. Now sine of pi over two is one, but cosine of pi over two is zero. So if you were to just substitute it in, this would give you one over zero. And one way to think about it is, pi over two is not in the domain of tangent of x. And so this limit actually turns out it doesn't exist. In general, if we're dealing with sine cosine tangent or cosecant secant or cotangent, if we're taking a limit to a point that is in their domain, then the value of the limit is going to be the same thing as the value of the function at that point. If you're taking a limit to a point that's not in their domain, there's a good chance that we're not going to have a limit."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And one way to think about it is, pi over two is not in the domain of tangent of x. And so this limit actually turns out it doesn't exist. In general, if we're dealing with sine cosine tangent or cosecant secant or cotangent, if we're taking a limit to a point that is in their domain, then the value of the limit is going to be the same thing as the value of the function at that point. If you're taking a limit to a point that's not in their domain, there's a good chance that we're not going to have a limit. So here, there is no limit. And a way to deduce that is that pi over two is not in tangent of x's domain. If you were to graph tan of x, you would see a vertical asymptote at pi over two."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you're taking a limit to a point that's not in their domain, there's a good chance that we're not going to have a limit. So here, there is no limit. And a way to deduce that is that pi over two is not in tangent of x's domain. If you were to graph tan of x, you would see a vertical asymptote at pi over two. Let's do one more of these. So let's say the limit as x approaches pi of cotangent of x. Pause the video and see if you can figure out what that's going to be."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you were to graph tan of x, you would see a vertical asymptote at pi over two. Let's do one more of these. So let's say the limit as x approaches pi of cotangent of x. Pause the video and see if you can figure out what that's going to be. Well, one way to think about it, cotangent of x, it's one over tangent of x, it's cosine of x over sine of x. It's sine of x. This is the limit as x approaches pi of this."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and see if you can figure out what that's going to be. Well, one way to think about it, cotangent of x, it's one over tangent of x, it's cosine of x over sine of x. It's sine of x. This is the limit as x approaches pi of this. And is pi in the domain of cotangent of x? Well, no, if you were to just substitute pi in, you're gonna get negative one over zero. And so that is not in the domain of cotangent of x."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the limit as x approaches pi of this. And is pi in the domain of cotangent of x? Well, no, if you were to just substitute pi in, you're gonna get negative one over zero. And so that is not in the domain of cotangent of x. If you were to plot it, you would see a vertical asymptote right over there. And so we have no limit. We have no limit."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that is not in the domain of cotangent of x. If you were to plot it, you would see a vertical asymptote right over there. And so we have no limit. We have no limit. So once again, this is not in the domain of that. And so good chance that we have no limit. When the thing we're taking the limit to is in the domain of the trigonometric function, we're going to have a defined limit."}, {"video_title": "Limits of trigonometric functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have no limit. So once again, this is not in the domain of that. And so good chance that we have no limit. When the thing we're taking the limit to is in the domain of the trigonometric function, we're going to have a defined limit. And sine and cosine in particular are defined for all real numbers. And they're continuous over all real numbers. So you take the limit to anything for them, it's going to be defined, and it's going to be the value of the function at that point."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, so they give us the derivative in terms of x. So maybe we could take the antiderivative of the derivative to find our original function. So let's do that. So we could say that f of x, f of x is going to be equal to the antiderivative, or we could say the indefinite integral of f prime of x, which is equal to 24 over x to the third. I could write it over like this, 24 over x to the third. But to help me process it a little bit more, I'm gonna write it, I'm gonna write this as 24 x to the negative three, because then it'll become a little harder how to take that antiderivative d dx. And so what is the antiderivative of 24 x to the negative three?"}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say that f of x, f of x is going to be equal to the antiderivative, or we could say the indefinite integral of f prime of x, which is equal to 24 over x to the third. I could write it over like this, 24 over x to the third. But to help me process it a little bit more, I'm gonna write it, I'm gonna write this as 24 x to the negative three, because then it'll become a little harder how to take that antiderivative d dx. And so what is the antiderivative of 24 x to the negative three? Well, we're just going to do the power rule in reverse. So what we're going to do is we're going to increase the exponent. So let me just rewrite it."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what is the antiderivative of 24 x to the negative three? Well, we're just going to do the power rule in reverse. So what we're going to do is we're going to increase the exponent. So let me just rewrite it. It's gonna be 24 x to the, we're going to increase the exponent by one. So it's gonna be x to the negative three plus one. And then we're going to divide by that increased exponent."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me just rewrite it. It's gonna be 24 x to the, we're going to increase the exponent by one. So it's gonna be x to the negative three plus one. And then we're going to divide by that increased exponent. So negative three plus one. And so that is going to be negative three plus one is x to the negative two, and then we divide by negative two. And if you're in doubt about what we just did, where we're kind of doing the power rule in reverse, now take the power rule, take the derivative of this using the power rule."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're going to divide by that increased exponent. So negative three plus one. And so that is going to be negative three plus one is x to the negative two, and then we divide by negative two. And if you're in doubt about what we just did, where we're kind of doing the power rule in reverse, now take the power rule, take the derivative of this using the power rule. Negative two times 24 over negative two is just gonna be 24, and then you decrement that exponent, you go to negative three. So are we done here? Is this f of x?"}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you're in doubt about what we just did, where we're kind of doing the power rule in reverse, now take the power rule, take the derivative of this using the power rule. Negative two times 24 over negative two is just gonna be 24, and then you decrement that exponent, you go to negative three. So are we done here? Is this f of x? Well, f of x might involve a constant. So let's put a constant out here, because notice, if you were to take the derivative of this thing here, the derivative of 24x to the negative two over negative two, we already established is 24x to the negative three. But then if you take the derivative of a constant, well, that just disappears, so you don't see it when you look at the derivative."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is this f of x? Well, f of x might involve a constant. So let's put a constant out here, because notice, if you were to take the derivative of this thing here, the derivative of 24x to the negative two over negative two, we already established is 24x to the negative three. But then if you take the derivative of a constant, well, that just disappears, so you don't see it when you look at the derivative. So we have to make sure that there might be a constant. And I have a feeling, based on the information that they've given us, we're going to make use of that constant. So let me rewrite f of x."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then if you take the derivative of a constant, well, that just disappears, so you don't see it when you look at the derivative. So we have to make sure that there might be a constant. And I have a feeling, based on the information that they've given us, we're going to make use of that constant. So let me rewrite f of x. So we know that f of x can be expressed as 24 divided by negative two. It's negative 12x to the negative two plus some constant. So how do we figure out that constant?"}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me rewrite f of x. So we know that f of x can be expressed as 24 divided by negative two. It's negative 12x to the negative two plus some constant. So how do we figure out that constant? Well, they have told us what f of two is. F of two is equal to 12. So let's write this down."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how do we figure out that constant? Well, they have told us what f of two is. F of two is equal to 12. So let's write this down. So when, so f, so f of two is equal to 12, which is equal to, well, we just have to put two in everywhere we see an x. That's going to be negative two times two to the negative two power plus c. And so 12 is equal to, what is this? Two to the negative two."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's write this down. So when, so f, so f of two is equal to 12, which is equal to, well, we just have to put two in everywhere we see an x. That's going to be negative two times two to the negative two power plus c. And so 12 is equal to, what is this? Two to the negative two. Two to the negative two is equal to one over two squared, which is equal to 1 4th. So this is negative 12 times 1 4th. Negative 12 times 1 4th is negative three."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two to the negative two. Two to the negative two is equal to one over two squared, which is equal to 1 4th. So this is negative 12 times 1 4th. Negative 12 times 1 4th is negative three. So it's negative three plus c. Now we can add three to both sides to solve for c. We get 15 is equal to, 15 is equal to c. So, or c is equal to 15. That is equal to 15. And so now we can write our f of x as we get f of x is equal to negative 12, and I could even write that as negative 12 over x squared if we like."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative 12 times 1 4th is negative three. So it's negative three plus c. Now we can add three to both sides to solve for c. We get 15 is equal to, 15 is equal to c. So, or c is equal to 15. That is equal to 15. And so now we can write our f of x as we get f of x is equal to negative 12, and I could even write that as negative 12 over x squared if we like. Negative 12 over x squared plus 15. And now using that, we can evaluate f of negative one. f of negative one."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now we can write our f of x as we get f of x is equal to negative 12, and I could even write that as negative 12 over x squared if we like. Negative 12 over x squared plus 15. And now using that, we can evaluate f of negative one. f of negative one. Wherever we see an x, we put a negative one there. So this is going to be negative one squared. So f of negative one is equal to 12 divided by, or negative 12 divided by negative one squared."}, {"video_title": "Finding specific antiderivatives rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "f of negative one. Wherever we see an x, we put a negative one there. So this is going to be negative one squared. So f of negative one is equal to 12 divided by, or negative 12 divided by negative one squared. Well, negative one squared is just one. So it's gonna be negative 12 plus 15, which is equal to three. And we're done."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "So my first question to you is, is this going to be a good case for u substitution? Well, when you look here, we, maybe the natural thing to set to be equal to u is seven x plus nine, but do I see its derivative anywhere over here? Well, let's see. If I set u to be equal to seven x plus nine, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to seven. Derivative of seven x is seven, derivative of nine is zero. So do we see a seven lying around anywhere over here?"}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I set u to be equal to seven x plus nine, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to seven. Derivative of seven x is seven, derivative of nine is zero. So do we see a seven lying around anywhere over here? Well, we don't. But what could we do in order to have a seven lying around but not change the value of the integral? Well, the neat thing, and we've seen this multiple times, is when you're evaluating integrals, scalars can go in and outside of the integral very easily."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "So do we see a seven lying around anywhere over here? Well, we don't. But what could we do in order to have a seven lying around but not change the value of the integral? Well, the neat thing, and we've seen this multiple times, is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of, let's say, some scalar a times f of x dx, dx, this is the same thing as a times the integral of f of x dx. That the integral of the scalar times the function is equal to the scalar times the integral of the function. So let me put this aside right over here."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the neat thing, and we've seen this multiple times, is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of, let's say, some scalar a times f of x dx, dx, this is the same thing as a times the integral of f of x dx. That the integral of the scalar times the function is equal to the scalar times the integral of the function. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a seven showing up? Well, we can multiply and divide by seven. So imagine doing this."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a seven showing up? Well, we can multiply and divide by seven. So imagine doing this. Let's rewrite our original integral. The original, so let me draw a little arrow here just to go around that aside. We could write our original integral as being equal to the integral of 1 7th times seven times the square root of seven x plus nine dx."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "So imagine doing this. Let's rewrite our original integral. The original, so let me draw a little arrow here just to go around that aside. We could write our original integral as being equal to the integral of 1 7th times seven times the square root of seven x plus nine dx. And if we want to, we could take the 1 7th outside of the integral. We don't have to, but we could rewrite this as 1 7th times the integral of seven times the square root of seven x plus nine dx. So now if we set u equal to seven x plus nine, do we have its derivative laying around?"}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could write our original integral as being equal to the integral of 1 7th times seven times the square root of seven x plus nine dx. And if we want to, we could take the 1 7th outside of the integral. We don't have to, but we could rewrite this as 1 7th times the integral of seven times the square root of seven x plus nine dx. So now if we set u equal to seven x plus nine, do we have its derivative laying around? Well, sure, the seven is right over here. We know that du, if we want to write in differential form, du is equal to seven times dx. So du is equal to seven times dx."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now if we set u equal to seven x plus nine, do we have its derivative laying around? Well, sure, the seven is right over here. We know that du, if we want to write in differential form, du is equal to seven times dx. So du is equal to seven times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the seven x plus nine. That is our u."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "So du is equal to seven times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the seven x plus nine. That is our u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1 7th times the integral of, and I'll just take the seven and put it in the back so we can just write the square root of u, square root of u, du, seven times dx is du. And we can rewrite this if we want as u to the 1 1\u20442 power."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is our u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1 7th times the integral of, and I'll just take the seven and put it in the back so we can just write the square root of u, square root of u, du, seven times dx is du. And we can rewrite this if we want as u to the 1 1\u20442 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1 7th times the integral of u to the 1 1\u20442 power du. And let me just make it clear."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can rewrite this if we want as u to the 1 1\u20442 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1 7th times the integral of u to the 1 1\u20442 power du. And let me just make it clear. This u I could have written in white if I wanted the same color, and this du is the same du right over here. So what is the antiderivative of u to the 1 1\u20442 power? Well, we increment u's power by one."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me just make it clear. This u I could have written in white if I wanted the same color, and this du is the same du right over here. So what is the antiderivative of u to the 1 1\u20442 power? Well, we increment u's power by one. So this is going to be equal to, and let me not forget this 1 7th out front. So it's going to be 1 7th times, 1 7th times, if we increment the power here, it's going to be u to the 3 1\u20442. 1 1\u20442 plus one is 1 1\u20442 or 3 1\u20442."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we increment u's power by one. So this is going to be equal to, and let me not forget this 1 7th out front. So it's going to be 1 7th times, 1 7th times, if we increment the power here, it's going to be u to the 3 1\u20442. 1 1\u20442 plus one is 1 1\u20442 or 3 1\u20442. So it's going to be u to the 3 1\u20442, u to the 3 1\u20442. And then we're gonna multiply this new thing times the reciprocal of 3 1\u20442, which is 2 1\u20443. And I encourage you to verify the derivative of 2 1\u20443 u to the 3 1\u20442 is indeed u to the 1 1\u20442."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "1 1\u20442 plus one is 1 1\u20442 or 3 1\u20442. So it's going to be u to the 3 1\u20442, u to the 3 1\u20442. And then we're gonna multiply this new thing times the reciprocal of 3 1\u20442, which is 2 1\u20443. And I encourage you to verify the derivative of 2 1\u20443 u to the 3 1\u20442 is indeed u to the 1 1\u20442. And so we have that, and since we're multiplying 1 7th times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1 7th."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to verify the derivative of 2 1\u20443 u to the 3 1\u20442 is indeed u to the 1 1\u20442. And so we have that, and since we're multiplying 1 7th times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1 7th. So it would get 1 7th times 2 1\u20443 is 2 over 21, 2 over 21 u to the 3 1\u20442. And 1 7th times some constant, well, that's just going to be some constant. And so I could write a constant like that."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we want, we can distribute the 1 7th. So it would get 1 7th times 2 1\u20443 is 2 over 21, 2 over 21 u to the 3 1\u20442. And 1 7th times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c one, and then I could call this c two, but it really is just some arbitrary constant. And we're done. Oh, actually, no, we aren't done."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so I could write a constant like that. I could call that c one, and then I could call this c two, but it really is just some arbitrary constant. And we're done. Oh, actually, no, we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2 over 21 times u to the 3 1\u20442."}, {"video_title": "_-substitution multiplying by a constant   AP Calculus AB   Khan Academy.mp3", "Sentence": "Oh, actually, no, we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2 over 21 times u to the 3 1\u20442. And we already know what u is equal to. u is equal to 7x plus nine. Let me put a new color in here just to ease the monotony."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I have the function g here, it's expressed as a fourth degree polynomial, and I want to think about the intervals over which g is either concave upwards or concave downwards. Now let's just remind ourselves what these things look like. So concave, concave upwards, is an interval, an interval where you're concave upwards, is an interval over which the slope is increasing, and it tends to look like an upward opening u like that, and you can see here that the slope over here is negative, and then as x increases it becomes less negative, it actually approaches zero, it becomes zero, then it crosses zero, it becomes slightly positive, more positive, even more positive, so you can see the slope is constantly increasing. And if you think about it in terms of derivatives, it means that your first derivative is increasing over that interval, and in order for your first derivative to be increasing over that interval, your second derivative, f prime prime of x, actually let me write it as g, because we're using g in this example, in order for your first derivative to be increasing, your, let me write this, so g, so concave upward means that your first derivative increasing, increasing, which means, which means that your second derivative is greater than zero. And concave downward is the opposite. Concave downward, downward, is an interval, or you're going to be concave downward over an interval when your slope is decreasing. So g prime of x is decreasing, or we can say that our second derivative, our second derivative is less than zero."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you think about it in terms of derivatives, it means that your first derivative is increasing over that interval, and in order for your first derivative to be increasing over that interval, your second derivative, f prime prime of x, actually let me write it as g, because we're using g in this example, in order for your first derivative to be increasing, your, let me write this, so g, so concave upward means that your first derivative increasing, increasing, which means, which means that your second derivative is greater than zero. And concave downward is the opposite. Concave downward, downward, is an interval, or you're going to be concave downward over an interval when your slope is decreasing. So g prime of x is decreasing, or we can say that our second derivative, our second derivative is less than zero. It's less than zero. And once again, I could draw it on this. So when x is lower, we have a, look, we have a, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, it's approaching zero, it becomes zero, then it becomes negative, and then even more negative, and then even more negative."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g prime of x is decreasing, or we can say that our second derivative, our second derivative is less than zero. It's less than zero. And once again, I could draw it on this. So when x is lower, we have a, look, we have a, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, it's approaching zero, it becomes zero, then it becomes negative, and then even more negative, and then even more negative. So as you see, our slope is constantly decreasing as x increases here. So in order to think about the intervals where g is either concave upward or concave downward, what we need to do is let's find the second derivative of g, and then let's think about the points at which the second, over the second, where the second derivative can go from being, from going, from being positive to negative or negative to positive, and those will be places where it's either undefined or where the second derivative is equal to zero. And then let's see what's happening in the interval between, and then we'll know."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x is lower, we have a, look, we have a, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, it's approaching zero, it becomes zero, then it becomes negative, and then even more negative, and then even more negative. So as you see, our slope is constantly decreasing as x increases here. So in order to think about the intervals where g is either concave upward or concave downward, what we need to do is let's find the second derivative of g, and then let's think about the points at which the second, over the second, where the second derivative can go from being, from going, from being positive to negative or negative to positive, and those will be places where it's either undefined or where the second derivative is equal to zero. And then let's see what's happening in the interval between, and then we'll know. What, over what intervals are we concave upward or concave downward? So let's do that. So let's first, let's take the first derivative, g prime of x, it's gonna apply the power rule a lot, four times negative one is negative four, x to the third power, plus, okay, so you're gonna have two times six is plus 12, x to the first power, you just write it as x, and then minus two, I could say minus two, x to the zeroth power, but that's just minus two, and then the derivative of negative three of a constant is just zero."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let's see what's happening in the interval between, and then we'll know. What, over what intervals are we concave upward or concave downward? So let's do that. So let's first, let's take the first derivative, g prime of x, it's gonna apply the power rule a lot, four times negative one is negative four, x to the third power, plus, okay, so you're gonna have two times six is plus 12, x to the first power, you just write it as x, and then minus two, I could say minus two, x to the zeroth power, but that's just minus two, and then the derivative of negative three of a constant is just zero. And now I can take the second derivative, g prime prime of x is going to be equal to three times negative four is negative 12x squared, decrement the exponent, plus 12. And so let's see, where could this be undefined? Well, the second derivative is just a quadratic expression here which would be defined for any x, so it's not gonna be undefined anywhere."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's first, let's take the first derivative, g prime of x, it's gonna apply the power rule a lot, four times negative one is negative four, x to the third power, plus, okay, so you're gonna have two times six is plus 12, x to the first power, you just write it as x, and then minus two, I could say minus two, x to the zeroth power, but that's just minus two, and then the derivative of negative three of a constant is just zero. And now I can take the second derivative, g prime prime of x is going to be equal to three times negative four is negative 12x squared, decrement the exponent, plus 12. And so let's see, where could this be undefined? Well, the second derivative is just a quadratic expression here which would be defined for any x, so it's not gonna be undefined anywhere. So interesting points where we could transition from going from a negative to a positive or a positive to a negative second derivative is where this thing could be equal to zero. So let's figure that out. So let's figure out where negative 12x plus 12 could be equal to zero."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the second derivative is just a quadratic expression here which would be defined for any x, so it's not gonna be undefined anywhere. So interesting points where we could transition from going from a negative to a positive or a positive to a negative second derivative is where this thing could be equal to zero. So let's figure that out. So let's figure out where negative 12x plus 12 could be equal to zero. See, we could subtract 12 from both sides, and we get negative 12x squared is equal to negative 12. Divide both sides by negative 12, you get x squared is equal to one, or x could be equal to the plus or minus, or x could be equal to the plus or minus square root of one, which is, of course, just one. So the second derivative at plus or minus one is equal to zero."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's figure out where negative 12x plus 12 could be equal to zero. See, we could subtract 12 from both sides, and we get negative 12x squared is equal to negative 12. Divide both sides by negative 12, you get x squared is equal to one, or x could be equal to the plus or minus, or x could be equal to the plus or minus square root of one, which is, of course, just one. So the second derivative at plus or minus one is equal to zero. So either between plus or minus one or on either side of them, we are going to be, we could be concave upward or concave downward. So let's think about this. And to think about this, I'm gonna make a number line."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the second derivative at plus or minus one is equal to zero. So either between plus or minus one or on either side of them, we are going to be, we could be concave upward or concave downward. So let's think about this. And to think about this, I'm gonna make a number line. Let me find a nice, soothing color here. All right, that's a nice, soothing color. And let's say, I should make the number line a little bit bigger."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to think about this, I'm gonna make a number line. Let me find a nice, soothing color here. All right, that's a nice, soothing color. And let's say, I should make the number line a little bit bigger. So there we go. Let's utilize the screen space. And so if this is zero, this is negative one, this is negative two, this is positive one, this is positive two, we know that at x equals negative one and x equals one, our second derivative is equal to zero."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's say, I should make the number line a little bit bigger. So there we go. Let's utilize the screen space. And so if this is zero, this is negative one, this is negative two, this is positive one, this is positive two, we know that at x equals negative one and x equals one, our second derivative is equal to zero. So let's think about what's happening in between those places to see if our second derivative is positive or negative. And from that, we'll be able to say where it's concave upward or concave downward. So on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if this is zero, this is negative one, this is negative two, this is positive one, this is positive two, we know that at x equals negative one and x equals one, our second derivative is equal to zero. So let's think about what's happening in between those places to see if our second derivative is positive or negative. And from that, we'll be able to say where it's concave upward or concave downward. So on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one. Well, let's just try a value in that interval to see whether our second derivative is positive or negative. And let's see, an easy value there could be negative two. It's in that interval."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one. Well, let's just try a value in that interval to see whether our second derivative is positive or negative. And let's see, an easy value there could be negative two. It's in that interval. So let's take g prime prime of negative two, which is equal to negative 12 times four, because negative two squared is positive four. So it's negative 48 plus 12, so it's equal to negative 36. The important thing to realize then is well, if over here it's negative, then over this whole interval, because it's not crossing through zero or it's not discontinuous at any of these points, that's why we picked this interval, that over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's in that interval. So let's take g prime prime of negative two, which is equal to negative 12 times four, because negative two squared is positive four. So it's negative 48 plus 12, so it's equal to negative 36. The important thing to realize then is well, if over here it's negative, then over this whole interval, because it's not crossing through zero or it's not discontinuous at any of these points, that's why we picked this interval, that over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards. So concave, concave downward, concave downward. Now let's go to the interval between negative one and one. So this is the open interval between negative one and one."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The important thing to realize then is well, if over here it's negative, then over this whole interval, because it's not crossing through zero or it's not discontinuous at any of these points, that's why we picked this interval, that over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards. So concave, concave downward, concave downward. Now let's go to the interval between negative one and one. So this is the open interval between negative one and one. And let's try a value there. Let's just try zero will be easy to compute. G prime prime of zero."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the open interval between negative one and one. And let's try a value there. Let's just try zero will be easy to compute. G prime prime of zero. Well, when x is zero, this is zero, so it's just going to be equal to 12. The important thing to realize is our second derivative here is greater than zero, so we are concave upward, concave upward on this interval between negative one and one. And then finally, let's look at the interval where x is greater than one."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "G prime prime of zero. Well, when x is zero, this is zero, so it's just going to be equal to 12. The important thing to realize is our second derivative here is greater than zero, so we are concave upward, concave upward on this interval between negative one and one. And then finally, let's look at the interval where x is greater than one. So this is the interval from one to infinity, if we want to view it that way. And let's just try the value. Let's try g prime prime of two, because that's in the interval."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, let's look at the interval where x is greater than one. So this is the interval from one to infinity, if we want to view it that way. And let's just try the value. Let's try g prime prime of two, because that's in the interval. And g prime prime of two is going to be the same thing as g prime prime of negative two, because whether you have a negative two or a positive two, you square, it becomes four. So you're going to have four times negative 12, which is negative 48, plus 12, which is negative 36. Which is negative 36."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's try g prime prime of two, because that's in the interval. And g prime prime of two is going to be the same thing as g prime prime of negative two, because whether you have a negative two or a positive two, you square, it becomes four. So you're going to have four times negative 12, which is negative 48, plus 12, which is negative 36. Which is negative 36. And so once again on this interval you are concave, concave downward. Now let's, I graphed this ahead of time. Let's see if what we just established is actually consistent with what the graph actually looks like."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Which is negative 36. And so once again on this interval you are concave, concave downward. Now let's, I graphed this ahead of time. Let's see if what we just established is actually consistent with what the graph actually looks like. We were able to come up with these insights about the concavity without graphing it. But now it's kind of satisfying to take a look at a graph. And actually let me see if I can match up the intervals."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if what we just established is actually consistent with what the graph actually looks like. We were able to come up with these insights about the concavity without graphing it. But now it's kind of satisfying to take a look at a graph. And actually let me see if I can match up the intervals. So actually this is pretty closely matched right over here. And so this is, actually let me make it a little bit smaller, all right. And so let me move my bounding box."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually let me see if I can match up the intervals. So actually this is pretty closely matched right over here. And so this is, actually let me make it a little bit smaller, all right. And so let me move my bounding box. So I'm saying that I'm concave downward between negative infinity, negative infinity, all the way until, all the way until negative one. All the way until this point right over here. So all the way until that point."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let me move my bounding box. So I'm saying that I'm concave downward between negative infinity, negative infinity, all the way until, all the way until negative one. All the way until this point right over here. So all the way until that point. And that looks right. It looks like the slope is constantly decreasing all the way until we get to x equals negative one. And then the slope starts increasing."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So all the way until that point. And that looks right. It looks like the slope is constantly decreasing all the way until we get to x equals negative one. And then the slope starts increasing. The slope starts increasing from there, from there all the way, and right at x we're transitioning. So I'm gonna leave a little, I won't color in that. And so here our slope is increasing."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the slope starts increasing. The slope starts increasing from there, from there all the way, and right at x we're transitioning. So I'm gonna leave a little, I won't color in that. And so here our slope is increasing. Do that same color. Our slope is increasing, increasing, increasing, increasing, increasing all the way until we get to x equals one. And then our slope starts decreasing again."}, {"video_title": "Analyzing concavity (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so here our slope is increasing. Do that same color. Our slope is increasing, increasing, increasing, increasing, increasing all the way until we get to x equals one. And then our slope starts decreasing again. And we get back into concave downwards. Whoops, I wanna do that in that orange color. We get back into concave downwards."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is an appropriate calculus-based justification for the fact that g is concave up on the interval, the open interval from five to 10? So concave up. So before I even think about what it means to be concave up, let's just make sure we understand this relationship between g and f. One way to understand it is if we took the derivative of both sides of this equation, we would get that g prime of x is equal to f of x. The derivative of this with respect to x would just be f of x. In fact, the whole reason why we introduce this variable t here is this thing right over here is actually a function of x, because x is this upper bound. And it would've been weird if we had x as an upper bound, or at least confusing, and we're also integrating with respect to x. So we just have to pick kind of another placeholder variable that doesn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of this with respect to x would just be f of x. In fact, the whole reason why we introduce this variable t here is this thing right over here is actually a function of x, because x is this upper bound. And it would've been weird if we had x as an upper bound, or at least confusing, and we're also integrating with respect to x. So we just have to pick kind of another placeholder variable that doesn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose. But this is still right over here. This is a function of x. But when you take the derivative of both sides, you realize that the function f, which is graphed here, and if this were the x-axis, then this would be f of x."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to pick kind of another placeholder variable that doesn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose. But this is still right over here. This is a function of x. But when you take the derivative of both sides, you realize that the function f, which is graphed here, and if this were the x-axis, then this would be f of x. If this is the t-axis, then this is y is equal to f of t. But generally, this is the graph of our function f, which you could also view as the graph of g prime. If this is x, this would be g prime of x. And so we're thinking about the interval from, the open interval from five to 10, and we have g's derivative graphed here, and we wanna know a calculus-based justification from this graph that lets us know that g is concave up."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "But when you take the derivative of both sides, you realize that the function f, which is graphed here, and if this were the x-axis, then this would be f of x. If this is the t-axis, then this is y is equal to f of t. But generally, this is the graph of our function f, which you could also view as the graph of g prime. If this is x, this would be g prime of x. And so we're thinking about the interval from, the open interval from five to 10, and we have g's derivative graphed here, and we wanna know a calculus-based justification from this graph that lets us know that g is concave up. So what does it mean to be concave up? Well, that means that your slope of tangent line, of tangent, tangent, slope of tangent is increasing, or another way of thinking about it, your derivative, derivative is increasing. Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we're thinking about the interval from, the open interval from five to 10, and we have g's derivative graphed here, and we wanna know a calculus-based justification from this graph that lets us know that g is concave up. So what does it mean to be concave up? Well, that means that your slope of tangent line, of tangent, tangent, slope of tangent is increasing, or another way of thinking about it, your derivative, derivative is increasing. Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval. And so we here, we have a graph of the derivative, and it is indeed increasing over that interval. So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval. The derivative is increasing on that interval, which means that the original function is concave up."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval. And so we here, we have a graph of the derivative, and it is indeed increasing over that interval. So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval. The derivative is increasing on that interval, which means that the original function is concave up. F is positive on that interval. That's not sufficient, that's not a sufficient calculus-based justification, because if your derivative is positive, that just means your original function is increasing, it doesn't tell you that your original function is concave up. F is concave up on the interval."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative is increasing on that interval, which means that the original function is concave up. F is positive on that interval. That's not sufficient, that's not a sufficient calculus-based justification, because if your derivative is positive, that just means your original function is increasing, it doesn't tell you that your original function is concave up. F is concave up on the interval. Well, just because your derivative is concave up doesn't mean that your original function is concave up. In fact, you could have a situation like this where you're concave up over that interval, but for much of that interval right over here, if this was our graph of f or g prime, we are decreasing, and if we're decreasing over much of that interval, then actually on this part, our original function would be concave down. The graph of g has a cup u shape on the interval."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "F is concave up on the interval. Well, just because your derivative is concave up doesn't mean that your original function is concave up. In fact, you could have a situation like this where you're concave up over that interval, but for much of that interval right over here, if this was our graph of f or g prime, we are decreasing, and if we're decreasing over much of that interval, then actually on this part, our original function would be concave down. The graph of g has a cup u shape on the interval. Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification. Let's do more of these. So this next one says, so we have the exact same setup, which actually all of these examples will have, g of x is equal to this thing here."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "The graph of g has a cup u shape on the interval. Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification. Let's do more of these. So this next one says, so we have the exact same setup, which actually all of these examples will have, g of x is equal to this thing here. What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight? So once again, they've graphed f here, which is the same thing as the derivative of g, and so if we have the graph of the derivative, how can we say, how do we know that we have a relative minimum at x equals eight? Well, if the fact that we cross, that we're at the x-axis, that x equals, that y is equal to zero, that the function, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero, but that alone does not tell us we have a relative minimum point."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this next one says, so we have the exact same setup, which actually all of these examples will have, g of x is equal to this thing here. What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight? So once again, they've graphed f here, which is the same thing as the derivative of g, and so if we have the graph of the derivative, how can we say, how do we know that we have a relative minimum at x equals eight? Well, if the fact that we cross, that we're at the x-axis, that x equals, that y is equal to zero, that the function, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero, but that alone does not tell us we have a relative minimum point. In order to have a relative minimum point, our derivative has to cross from being negative to positive. Why is that valuable? Because think about, if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing, goes from decreasing to increasing, and so you would have a relative minimum point."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if the fact that we cross, that we're at the x-axis, that x equals, that y is equal to zero, that the function, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero, but that alone does not tell us we have a relative minimum point. In order to have a relative minimum point, our derivative has to cross from being negative to positive. Why is that valuable? Because think about, if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing, goes from decreasing to increasing, and so you would have a relative minimum point. And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point. F is negative before x equals eight and positive after x equals eight. That's exactly what we just described."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because think about, if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing, goes from decreasing to increasing, and so you would have a relative minimum point. And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point. F is negative before x equals eight and positive after x equals eight. That's exactly what we just described. Let's see about these. F is concave up on the interval around x equals six. Well, x equals six is a little bit unrelated to that."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's exactly what we just described. Let's see about these. F is concave up on the interval around x equals six. Well, x equals six is a little bit unrelated to that. There's an interval in the graph of g around x equals eight where g of eight is the smallest value. Well, this would be a justification for a relative minimum, but it is not calculus-based. So let's get, I'll rule that one out as well."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, x equals six is a little bit unrelated to that. There's an interval in the graph of g around x equals eight where g of eight is the smallest value. Well, this would be a justification for a relative minimum, but it is not calculus-based. So let's get, I'll rule that one out as well. Let's do one more of these. So same setup, although we have a different f and g here, and we see it every time with the graph. What is an appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12?"}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's get, I'll rule that one out as well. Let's do one more of these. So same setup, although we have a different f and g here, and we see it every time with the graph. What is an appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12? So the positive on the closed interval from seven to 12. So this is interesting. Let's just remind ourselves."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is an appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12? So the positive on the closed interval from seven to 12. So this is interesting. Let's just remind ourselves. Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x. So if we think about what happens when x is equal to seven, x is equal to seven, when x is equal to seven, or another way of thinking about it, g of seven is going to be the integral from zero to seven of f of t dt. And so the integral from zero to seven, if this was a t-axis, and once again, t is just kind of a placeholder variable to help us keep this x up here, but we're really talking about this area right over here."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just remind ourselves. Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x. So if we think about what happens when x is equal to seven, x is equal to seven, when x is equal to seven, or another way of thinking about it, g of seven is going to be the integral from zero to seven of f of t dt. And so the integral from zero to seven, if this was a t-axis, and once again, t is just kind of a placeholder variable to help us keep this x up here, but we're really talking about this area right over here. And because from zero to seven, this function is above the x-axis, this is going to be a positive area. This is a positive area. And as we go from seven to 12, we're not adding any more area, but we're also not taking any away."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the integral from zero to seven, if this was a t-axis, and once again, t is just kind of a placeholder variable to help us keep this x up here, but we're really talking about this area right over here. And because from zero to seven, this function is above the x-axis, this is going to be a positive area. This is a positive area. And as we go from seven to 12, we're not adding any more area, but we're also not taking any away. So actually, g of seven all the way to g of 12 is going to be the same positive value because we're not adding any more value. When I say g of 12, g of 12 is going to be actually equal to g of seven because, once again, no added area right here, positive or negative. So let's see which of these choices match."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as we go from seven to 12, we're not adding any more area, but we're also not taking any away. So actually, g of seven all the way to g of 12 is going to be the same positive value because we're not adding any more value. When I say g of 12, g of 12 is going to be actually equal to g of seven because, once again, no added area right here, positive or negative. So let's see which of these choices match. For any x value in the interval from seven to 12, the value of f of x is zero. That is true, but that doesn't mean that we were positive. For example, before that interval, if our function did something like this, then we would have had negative area up to that point, and so these would be negative values."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see which of these choices match. For any x value in the interval from seven to 12, the value of f of x is zero. That is true, but that doesn't mean that we were positive. For example, before that interval, if our function did something like this, then we would have had negative area up to that point, and so these would be negative values. So I would rule that out. For any x value in the interval from seven to 12, the closed interval, the value of g of x is positive. For any x value in the interval from seven to 12, the value of g of x is positive."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "For example, before that interval, if our function did something like this, then we would have had negative area up to that point, and so these would be negative values. So I would rule that out. For any x value in the interval from seven to 12, the closed interval, the value of g of x is positive. For any x value in the interval from seven to 12, the value of g of x is positive. That is true, so I like this one. Let me see these other ones. F is positive over the closed interval from zero to seven, and it is non-negative over seven to 12."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "For any x value in the interval from seven to 12, the value of g of x is positive. That is true, so I like this one. Let me see these other ones. F is positive over the closed interval from zero to seven, and it is non-negative over seven to 12. I like this one as well. And actually, the reason why I would rule out this first one, this first one has nothing to do with the derivative, and so it's not a calculus-based justification, so I would rule that one out. This one is good."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "F is positive over the closed interval from zero to seven, and it is non-negative over seven to 12. I like this one as well. And actually, the reason why I would rule out this first one, this first one has nothing to do with the derivative, and so it's not a calculus-based justification, so I would rule that one out. This one is good. This is the exact rationale that I was talking about. F is positive from zero to seven, so it develops all this positive area, and it's non-negative over the interval, and so we are going to stay positive this entire time for g, which is the area under f and above the x-axis from zero to our whatever x we wanna pick. So I like this choice here."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084   AP Calculus AB   Khan Academy.mp3", "Sentence": "This one is good. This is the exact rationale that I was talking about. F is positive from zero to seven, so it develops all this positive area, and it's non-negative over the interval, and so we are going to stay positive this entire time for g, which is the area under f and above the x-axis from zero to our whatever x we wanna pick. So I like this choice here. F is neither concave up nor concave down over the interval from, the closed interval from seven to 12. No, that doesn't really help us in saying that g is positive over that interval. So there you go, choice C."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "That means that if I have two sets of numbers, let's say one set right over there, that's another set right over there, and if we view that first set as the domain of g, so if you start with some x right over here, g is going to map from that x to another value which we would call g of x, g of x. That's what the function g does. Now if h is the inverse of g, and frankly vice versa, then h could go from that point g of x back to x. So h would do this. H would get us back to our original value. So that's what the function h would do. And so we could view this point right over here, we could view it as x, so that is x, but we could also view it as h of g of x."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So h would do this. H would get us back to our original value. So that's what the function h would do. And so we could view this point right over here, we could view it as x, so that is x, but we could also view it as h of g of x. So we could also view this as h of g of x. And I did all of that so that we can really feel good about this idea. If someone tells you that g and h are inverse functions, that means that h of g of x is x."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we could view this point right over here, we could view it as x, so that is x, but we could also view it as h of g of x. So we could also view this as h of g of x. And I did all of that so that we can really feel good about this idea. If someone tells you that g and h are inverse functions, that means that h of g of x is x. So g of x is x, h of g of x, h of g of x is equal to x, or you could have gone the other way around. You could have started with a, well you could have done it multiple different ways, but also g of h of x. I could have just swapped these letters here, the letters h and g are somewhat arbitrary. So you could have also said that g of h of x is equal to x."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "If someone tells you that g and h are inverse functions, that means that h of g of x is x. So g of x is x, h of g of x, h of g of x is equal to x, or you could have gone the other way around. You could have started with a, well you could have done it multiple different ways, but also g of h of x. I could have just swapped these letters here, the letters h and g are somewhat arbitrary. So you could have also said that g of h of x is equal to x. So g of h of x is equal to x. And then they give us some information. The following table lists a few values of g, h, and g prime."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you could have also said that g of h of x is equal to x. So g of h of x is equal to x. And then they give us some information. The following table lists a few values of g, h, and g prime. Alright, so they want us to evaluate h prime of three. They don't even give us h prime of three, how do we figure it out? They gave us g prime and h and g. How do we figure this out?"}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "The following table lists a few values of g, h, and g prime. Alright, so they want us to evaluate h prime of three. They don't even give us h prime of three, how do we figure it out? They gave us g prime and h and g. How do we figure this out? Well here we're going to actually derive something based on the chain rule. And this isn't the type of problem that you'll see a lot of, but it is interesting. So we're going to work through it, and there's a chance that you might see it in your calculus class."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "They gave us g prime and h and g. How do we figure this out? Well here we're going to actually derive something based on the chain rule. And this isn't the type of problem that you'll see a lot of, but it is interesting. So we're going to work through it, and there's a chance that you might see it in your calculus class. So let's start with either one of these expressions up here. So let's start with the expression, well let's start with the, let's do this one over here. So if we have g of h of x is equal to x, h of x is equal to x, so we put that h of x back there, which is by definition true if g and h's are inverses."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to work through it, and there's a chance that you might see it in your calculus class. So let's start with either one of these expressions up here. So let's start with the expression, well let's start with the, let's do this one over here. So if we have g of h of x is equal to x, h of x is equal to x, so we put that h of x back there, which is by definition true if g and h's are inverses. Well now let's take the derivative of both sides of this. So let's take the derivative with respect to x of both sides, derivative with respect to x. And on the left hand side, well we just apply the chain rule."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we have g of h of x is equal to x, h of x is equal to x, so we put that h of x back there, which is by definition true if g and h's are inverses. Well now let's take the derivative of both sides of this. So let's take the derivative with respect to x of both sides, derivative with respect to x. And on the left hand side, well we just apply the chain rule. So this would be g prime of h of x, g prime of h of x, times h prime of x, that's just the chain rule right over there, and that would be equal to, what's the derivative with respect to x of x? Well that's just going to be equal to one. So now it's interesting."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And on the left hand side, well we just apply the chain rule. So this would be g prime of h of x, g prime of h of x, times h prime of x, that's just the chain rule right over there, and that would be equal to, what's the derivative with respect to x of x? Well that's just going to be equal to one. So now it's interesting. We need to figure out what h prime of three is. So we can figure out what h of three is, and then we can use that to figure out what g prime of whatever h, g prime of h of three is. And so we should be able to figure out h prime of x."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now it's interesting. We need to figure out what h prime of three is. So we can figure out what h of three is, and then we can use that to figure out what g prime of whatever h, g prime of h of three is. And so we should be able to figure out h prime of x. Or we could just rewrite it this way. We could rewrite that h prime of x is equal to, is equal to one over g prime of h of x. Now in some circles, they might encourage you to memorize this, and maybe for the sake of doing this exercise on Khan Academy, you might want to memorize it."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we should be able to figure out h prime of x. Or we could just rewrite it this way. We could rewrite that h prime of x is equal to, is equal to one over g prime of h of x. Now in some circles, they might encourage you to memorize this, and maybe for the sake of doing this exercise on Khan Academy, you might want to memorize it. But I'll tell you, 20 years after I took calculus, or almost 25 years after I took calculus, this is not something that I retain in my long term memory, but I did retain that you can derive this from just what the definition of inverse functions actually are. But we can use this now if we want to figure out what h prime of three is. H prime of three is going to be equal to one over, one over, one over g prime of h of three, which I'm guessing that they have given us."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now in some circles, they might encourage you to memorize this, and maybe for the sake of doing this exercise on Khan Academy, you might want to memorize it. But I'll tell you, 20 years after I took calculus, or almost 25 years after I took calculus, this is not something that I retain in my long term memory, but I did retain that you can derive this from just what the definition of inverse functions actually are. But we can use this now if we want to figure out what h prime of three is. H prime of three is going to be equal to one over, one over, one over g prime of h of three, which I'm guessing that they have given us. So h of three, when x is three, h is four. So that is h of three there. So h of three is four."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "H prime of three is going to be equal to one over, one over, one over g prime of h of three, which I'm guessing that they have given us. So h of three, when x is three, h is four. So that is h of three there. So h of three is four. So now we just have to figure out g prime of four. Well lucky for us, they have given us, when x is equal to four, g prime, g prime is equal to 1 1 2. So g prime of four is equal to 1 1 2."}, {"video_title": "Derivatives of inverse functions from table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So h of three is four. So now we just have to figure out g prime of four. Well lucky for us, they have given us, when x is equal to four, g prime, g prime is equal to 1 1 2. So g prime of four is equal to 1 1 2. So h prime of three is equal to one over 1 1 2. So one over 1 1 2. One divided by 1 1 2 is the same thing as one times two."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "At time t is equal to two, the town's population is 1,200 people. What is the town's population at t is equal to seven? Which expression can we use to solve the problem? So they don't want us to actually answer the question, they just want us to set up the expression using some symbols from calculus. So why don't you pause this video and try to think about it. So let's just remind us what they've given us. They've given us the rate function right over here."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they don't want us to actually answer the question, they just want us to set up the expression using some symbols from calculus. So why don't you pause this video and try to think about it. So let's just remind us what they've given us. They've given us the rate function right over here. And so if you want to find a change in population from one time to another time, what you could do is you could take the integral of the rate function from the starting time, t is equal to two years, to t is equal to seven years. So we're gonna take the integral of the rate function, and what this is going to tell us, this is going to tell us the change in population from time two to time seven. So this is, you could say, let me just write this, this is the change, I'll use delta for change, or let me just write it out."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "They've given us the rate function right over here. And so if you want to find a change in population from one time to another time, what you could do is you could take the integral of the rate function from the starting time, t is equal to two years, to t is equal to seven years. So we're gonna take the integral of the rate function, and what this is going to tell us, this is going to tell us the change in population from time two to time seven. So this is, you could say, let me just write this, this is the change, I'll use delta for change, or let me just write it out. Change in population, change in population, population, but they don't want us to, they're not asking us for the change in population. They want us to know what is the town's population at t is equal to seven. So what you would want is, you would want what your population is at t equals two, plus the change in population from two to seven, to get you your population at seven."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is, you could say, let me just write this, this is the change, I'll use delta for change, or let me just write it out. Change in population, change in population, population, but they don't want us to, they're not asking us for the change in population. They want us to know what is the town's population at t is equal to seven. So what you would want is, you would want what your population is at t equals two, plus the change in population from two to seven, to get you your population at seven. So they tell us the population at time t equals two, the town's population is 1,200 people. So if you want the population at t is equal to seven, it's going to be 1,200 plus how whatever the change in population, if you take the integral of the rate function, you are, and this is the rate of population, this integral is gonna give you the change in population from time t equals two to t equals seven. So we can see clearly that is choice D right over here."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what you would want is, you would want what your population is at t equals two, plus the change in population from two to seven, to get you your population at seven. So they tell us the population at time t equals two, the town's population is 1,200 people. So if you want the population at t is equal to seven, it's going to be 1,200 plus how whatever the change in population, if you take the integral of the rate function, you are, and this is the rate of population, this integral is gonna give you the change in population from time t equals two to t equals seven. So we can see clearly that is choice D right over here. These other choices, we can look at them really quick, choice B is just the change in population. So that's assuming that this is, and this is actually increasing. So this would tell us how much of the population increase from t equals two to t equals seven."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can see clearly that is choice D right over here. These other choices, we can look at them really quick, choice B is just the change in population. So that's assuming that this is, and this is actually increasing. So this would tell us how much of the population increase from t equals two to t equals seven. So that's not what we want. We want what the actual population is. This is how much the population increases from time zero to time seven."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this would tell us how much of the population increase from t equals two to t equals seven. So that's not what we want. We want what the actual population is. This is how much the population increases from time zero to time seven. Now you might say, well, wouldn't that be the town's population? Well, that would be the town's population if they had no people at time zero, but you can't assume that. Maybe the town got settled by 10 people or by 1,000 people or who knows whatever else."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is how much the population increases from time zero to time seven. Now you might say, well, wouldn't that be the town's population? Well, that would be the town's population if they had no people at time zero, but you can't assume that. Maybe the town got settled by 10 people or by 1,000 people or who knows whatever else. So right over there. And this is taking the derivative of the rate function, which is, it's actually a little hard to think about what is this. This is the rate of change of the rate at time seven minus the rate of change of the rate at time two."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe the town got settled by 10 people or by 1,000 people or who knows whatever else. So right over there. And this is taking the derivative of the rate function, which is, it's actually a little hard to think about what is this. This is the rate of change of the rate at time seven minus the rate of change of the rate at time two. So I would rule that one out as well. Let's do one more of these. So here we have, here we have the depth of water in a tank is changing at a rate of, r of t is equal to 0.3 t, centimeters per minute, where t is the time in minutes."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the rate of change of the rate at time seven minus the rate of change of the rate at time two. So I would rule that one out as well. Let's do one more of these. So here we have, here we have the depth of water in a tank is changing at a rate of, r of t is equal to 0.3 t, centimeters per minute, where t is the time in minutes. At time t equals zero, the depth of the water is 35 centimeters. What is the change in the water's depth during the fourth minute? So let's pause the video again and see if you can figure this out again and figure out what expression can we use to solve the problem."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here we have, here we have the depth of water in a tank is changing at a rate of, r of t is equal to 0.3 t, centimeters per minute, where t is the time in minutes. At time t equals zero, the depth of the water is 35 centimeters. What is the change in the water's depth during the fourth minute? So let's pause the video again and see if you can figure this out again and figure out what expression can we use to solve the problem. The problem being, what is the change in the water's depth during the fourth minute? Alright, so we've just talked about if you're trying to find the change in a value, you could take the integral of the rate function over the appropriate time. So we're talking about during the fourth minute."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's pause the video again and see if you can figure this out again and figure out what expression can we use to solve the problem. The problem being, what is the change in the water's depth during the fourth minute? Alright, so we've just talked about if you're trying to find the change in a value, you could take the integral of the rate function over the appropriate time. So we're talking about during the fourth minute. So we definitely wanna take the integral of the rate function. And we just have to think about the bounds and all of the choices here taking the integral of the rate function. So really the interesting part is during the fourth minute."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're talking about during the fourth minute. So we definitely wanna take the integral of the rate function. And we just have to think about the bounds and all of the choices here taking the integral of the rate function. So really the interesting part is during the fourth minute. So let me just draw a little number line here and we can think about what the fourth minute looks like. Or actually, let me just draw the whole thing. So let's say this is r of t right over there."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So really the interesting part is during the fourth minute. So let me just draw a little number line here and we can think about what the fourth minute looks like. Or actually, let me just draw the whole thing. So let's say this is r of t right over there. We could say y is equal to r of t. And this is t. And let's see, the first minute goes from zero to one, second minute goes from one to two, third minute goes from two to three, fourth minute goes from three to four. The rate function, this actually just looks like a straight up linear rate function, looks something like this. And so what is the fourth minute?"}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say this is r of t right over there. We could say y is equal to r of t. And this is t. And let's see, the first minute goes from zero to one, second minute goes from one to two, third minute goes from two to three, fourth minute goes from three to four. The rate function, this actually just looks like a straight up linear rate function, looks something like this. And so what is the fourth minute? Well the fourth minute is, the first minute is this one, second, third, fourth. The fourth minute is going from minute three to minute four. So what we wanna do is the expression that gives us this area right over here under the rate curve."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what is the fourth minute? Well the fourth minute is, the first minute is this one, second, third, fourth. The fourth minute is going from minute three to minute four. So what we wanna do is the expression that gives us this area right over here under the rate curve. Well this is, our lower bound's going to be three and our upper bound is going to be four. And so there you have it. It is this first choice."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we wanna do is the expression that gives us this area right over here under the rate curve. Well this is, our lower bound's going to be three and our upper bound is going to be four. And so there you have it. It is this first choice. You might have been tempted here if you got a little bit confused, hey maybe the fourth minute is after we've crossed that t is equal to four, but no, that would be the fifth minute. This would tell us, this would tell us our change over the first four minutes, not just during the fourth minute. And then this, well this is just gonna be zero."}, {"video_title": "Analyzing problems involving definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is this first choice. You might have been tempted here if you got a little bit confused, hey maybe the fourth minute is after we've crossed that t is equal to four, but no, that would be the fifth minute. This would tell us, this would tell us our change over the first four minutes, not just during the fourth minute. And then this, well this is just gonna be zero. If you're taking, what is the change in the value from three to three? Well it didn't change at all in that, because it's essentially at an instant. There's no time that passes from three to three."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "So right over here, we actually have the graph of our function h. This is a graph y is equal to h of x. And we don't have graphed the first derivative, but we do have graphed the second derivative right here in this orange color, h prime prime. So they're telling us, given that h prime of negative four is equal to zero. So that's saying that given that the first derivative at x equals negative four is equal to zero, and you can see that the slope of the tangent line when x is equal to negative four does indeed equal zero. So given that, what is a calculus-based, let me underline that, a calculus-based justification for the fact that h has a relative maximum at x equals negative four? So this first one says that the second derivative at x equals negative four is negative. Now what does that tell us?"}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's saying that given that the first derivative at x equals negative four is equal to zero, and you can see that the slope of the tangent line when x is equal to negative four does indeed equal zero. So given that, what is a calculus-based, let me underline that, a calculus-based justification for the fact that h has a relative maximum at x equals negative four? So this first one says that the second derivative at x equals negative four is negative. Now what does that tell us? If the second derivative is negative, that means that the first derivative is decreasing, which is another way of saying that we are dealing with a situation where, at least at x equals negative four, we are concave downwards, downwards, which means that the general shape of our curve is going to look something like this around x equals negative four. And if the slope at x equals negative four is zero, well, that tells us that, yes, we indeed are dealing with a relative maximum point. If the second derivative at that point was positive, then we would be concave upwards, and then if our derivative is zero there, we'd say, okay, that's a relative minimum point."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what does that tell us? If the second derivative is negative, that means that the first derivative is decreasing, which is another way of saying that we are dealing with a situation where, at least at x equals negative four, we are concave downwards, downwards, which means that the general shape of our curve is going to look something like this around x equals negative four. And if the slope at x equals negative four is zero, well, that tells us that, yes, we indeed are dealing with a relative maximum point. If the second derivative at that point was positive, then we would be concave upwards, and then if our derivative is zero there, we'd say, okay, that's a relative minimum point. But this is indeed true. The second derivative is negative at x equals negative four, which means we are concave downwards, which means that we are a upside-down u, and that point where the derivative is zero is indeed a relative maximum. So let me, so that is the answer, and we're done."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "If the second derivative at that point was positive, then we would be concave upwards, and then if our derivative is zero there, we'd say, okay, that's a relative minimum point. But this is indeed true. The second derivative is negative at x equals negative four, which means we are concave downwards, which means that we are a upside-down u, and that point where the derivative is zero is indeed a relative maximum. So let me, so that is the answer, and we're done. But let's just rule out the other ones. H increases before x equals negative four. That is indeed true."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me, so that is the answer, and we're done. But let's just rule out the other ones. H increases before x equals negative four. That is indeed true. Before x equals negative four, we are increasing. And H decreases after it. That is true."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is indeed true. Before x equals negative four, we are increasing. And H decreases after it. That is true. And that is one rationale for thinking that, hey, we must have a maximum point, assuming that our function is continuous, at x equals negative four. So this is true. It is a justification for relative maximum, but it is not calculus-based, and so that's why we can rule this one out."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is true. And that is one rationale for thinking that, hey, we must have a maximum point, assuming that our function is continuous, at x equals negative four. So this is true. It is a justification for relative maximum, but it is not calculus-based, and so that's why we can rule this one out. The second derivative has a relative minimum at x equals negative four. Well, it does indeed seem to be true. There's a relative minimum there, but that's not a justification for why this is, why H of negative four, or why we have a relative maximum at x equals negative four."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is a justification for relative maximum, but it is not calculus-based, and so that's why we can rule this one out. The second derivative has a relative minimum at x equals negative four. Well, it does indeed seem to be true. There's a relative minimum there, but that's not a justification for why this is, why H of negative four, or why we have a relative maximum at x equals negative four. For example, this, you could have a relative minimum in your second derivative, but your second derivative could still be positive there. So what if the second derivative was like that? That would still be a relative minimum, but if it was positive at that point, then you would be concave upwards, which would mean that at x equals negative four, your original function wouldn't have a maximum point."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "There's a relative minimum there, but that's not a justification for why this is, why H of negative four, or why we have a relative maximum at x equals negative four. For example, this, you could have a relative minimum in your second derivative, but your second derivative could still be positive there. So what if the second derivative was like that? That would still be a relative minimum, but if it was positive at that point, then you would be concave upwards, which would mean that at x equals negative four, your original function wouldn't have a maximum point. It would have a minimum point. And so just a relative minimum isn't enough. In order to know that you are dealing with a relative maximum, you would have to know that the second derivative is negative there."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "That would still be a relative minimum, but if it was positive at that point, then you would be concave upwards, which would mean that at x equals negative four, your original function wouldn't have a maximum point. It would have a minimum point. And so just a relative minimum isn't enough. In order to know that you are dealing with a relative maximum, you would have to know that the second derivative is negative there. Now this fourth choice, H prime prime is concave up. It does indeed look like the second derivative is concave up but that by itself does not justify that the original function is concave up. For example, well, I could use this example right here."}, {"video_title": "Justification using second derivative maximum point   AP Calculus AB   Khan Academy.mp3", "Sentence": "In order to know that you are dealing with a relative maximum, you would have to know that the second derivative is negative there. Now this fourth choice, H prime prime is concave up. It does indeed look like the second derivative is concave up but that by itself does not justify that the original function is concave up. For example, well, I could use this example right here. This is a potential second derivative that is concave upwards, but it is positive the entire time. And if your second derivative is positive the entire time, that means that your first derivative is increasing the entire time, which means that your original function is going to be concave upwards the entire time. And so if you're concave upwards the entire time, then you would not have a relative maximum at x equals negative four."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to figure out the points on that curve where the tangent line is vertical. So let's just remind ourselves what the slope of a tangent line is, or what it isn't, I guess. This may be a better way to think about it. If you have a horizontal line, so let me draw a horizontal line. If you have a horizontal line like that, well then your slope is zero, or you could say your rate of change of y with respect to x is equal to zero. But what about a vertical line? If you have a vertical line like that, what is your rate of change of y with respect to x?"}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you have a horizontal line, so let me draw a horizontal line. If you have a horizontal line like that, well then your slope is zero, or you could say your rate of change of y with respect to x is equal to zero. But what about a vertical line? If you have a vertical line like that, what is your rate of change of y with respect to x? Well, some people might say it's infinity, or you could say it's undefined. It's undefined in some way, because at some point, or one way to think about it, you're gonna try to divide by zero because you have a huge change in y over no change in x. Or another way to think about it that's a little bit more in line with that is you could say that your change in x with respect to change in y, notice I took the reciprocal, so now we're talking about a change in the derivative of x with respect to y is equal to zero."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you have a vertical line like that, what is your rate of change of y with respect to x? Well, some people might say it's infinity, or you could say it's undefined. It's undefined in some way, because at some point, or one way to think about it, you're gonna try to divide by zero because you have a huge change in y over no change in x. Or another way to think about it that's a little bit more in line with that is you could say that your change in x with respect to change in y, notice I took the reciprocal, so now we're talking about a change in the derivative of x with respect to y is equal to zero. Because your y can change, but as your y changes, your x does not change. So can we use this little insight here on vertical lines to think about the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical? Well, before, they told us, they gave us the curve of the equation, and they also told us what dy dx is."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or another way to think about it that's a little bit more in line with that is you could say that your change in x with respect to change in y, notice I took the reciprocal, so now we're talking about a change in the derivative of x with respect to y is equal to zero. Because your y can change, but as your y changes, your x does not change. So can we use this little insight here on vertical lines to think about the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical? Well, before, they told us, they gave us the curve of the equation, and they also told us what dy dx is. Let me just rewrite them again just so we have them there. So we know that y to the third minus xy is equal to two. This is the equation of our curve."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, before, they told us, they gave us the curve of the equation, and they also told us what dy dx is. Let me just rewrite them again just so we have them there. So we know that y to the third minus xy is equal to two. This is the equation of our curve. And we know that dy dx is equal to y over three y squared minus x. So one thing that we could do is, well, let's just figure out what the derivative of x with respect to y is and set that to be equal to zero. So this is the derivative of y with respect to x."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the equation of our curve. And we know that dy dx is equal to y over three y squared minus x. So one thing that we could do is, well, let's just figure out what the derivative of x with respect to y is and set that to be equal to zero. So this is the derivative of y with respect to x. If we take the reciprocal of that, the derivative of x with respect to y, just the reciprocal of this, is going to be the reciprocal of what we have here. So it's going to be three y squared minus x over y. And if we want this to be equal to zero, like we said right over here, well, then that's only going to happen if the numerator is equal to zero."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the derivative of y with respect to x. If we take the reciprocal of that, the derivative of x with respect to y, just the reciprocal of this, is going to be the reciprocal of what we have here. So it's going to be three y squared minus x over y. And if we want this to be equal to zero, like we said right over here, well, then that's only going to happen if the numerator is equal to zero. So we can say, okay, at what xy pair does this numerator equal to zero? Three y squared minus x is equal to zero. You can add x to both sides, and you get three y squared is equal to x."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we want this to be equal to zero, like we said right over here, well, then that's only going to happen if the numerator is equal to zero. So we can say, okay, at what xy pair does this numerator equal to zero? Three y squared minus x is equal to zero. You can add x to both sides, and you get three y squared is equal to x. Now another way you could have thought about it is, well, at what x and y values does the derivative of y with respect to x become undefined? Well, that's going to become undefined if the denominator here is zero. But when you're dealing with things like undefined, it gets a little bit more hand-wavy."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can add x to both sides, and you get three y squared is equal to x. Now another way you could have thought about it is, well, at what x and y values does the derivative of y with respect to x become undefined? Well, that's going to become undefined if the denominator here is zero. But when you're dealing with things like undefined, it gets a little bit more hand-wavy. I like to just think of this as the rate at which x is changing with respect to y is zero. And so that got us to the same conclusion. Well, for that to be true, x has to be equal to three y squared."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "But when you're dealing with things like undefined, it gets a little bit more hand-wavy. I like to just think of this as the rate at which x is changing with respect to y is zero. And so that got us to the same conclusion. Well, for that to be true, x has to be equal to three y squared. And of course, the xy pair has to also satisfy the equation for the curve. So why don't we use both of these constraints, and then we can solve for x and y. And so the easiest thing I can do, I can think of doing, is let's substitute x with three y squared, because they are the same."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, for that to be true, x has to be equal to three y squared. And of course, the xy pair has to also satisfy the equation for the curve. So why don't we use both of these constraints, and then we can solve for x and y. And so the easiest thing I can do, I can think of doing, is let's substitute x with three y squared, because they are the same. That's the second constraint. So if we take our original equation of the curve, we get y to the third minus, instead of writing x, I could write three y squared, three y squared times y, times y is equal to two. And so we get y to the third minus three y to the third is equal to two."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the easiest thing I can do, I can think of doing, is let's substitute x with three y squared, because they are the same. That's the second constraint. So if we take our original equation of the curve, we get y to the third minus, instead of writing x, I could write three y squared, three y squared times y, times y is equal to two. And so we get y to the third minus three y to the third is equal to two. This is negative two y to the third is equal to two. We can divide both sides by negative two. And we get, we get y to the third is equal to negative one, or y is equal to negative one."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we get y to the third minus three y to the third is equal to two. This is negative two y to the third is equal to two. We can divide both sides by negative two. And we get, we get y to the third is equal to negative one, or y is equal to negative one. Negative one to the third power is negative one. Now if y is negative one, what is x? Well, x is going to be equal to three times negative one squared."}, {"video_title": "2015 AP Calculus AB 6b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we get, we get y to the third is equal to negative one, or y is equal to negative one. Negative one to the third power is negative one. Now if y is negative one, what is x? Well, x is going to be equal to three times negative one squared. So negative one squared is just one, so x is going to be equal to three. So the point on that curve at which the tangent line is vertical is going to be the point three, negative one. And we are all done."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over, let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is going to be the square root of 3 times x squared over, let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So a prime, our combined area, the derivative of our combined area as a function of x, is going to be equal to, well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x was the derivative of something squared over 16 with respect to that something."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So a prime, our combined area, the derivative of our combined area as a function of x, is going to be equal to, well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x was the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2 over 16, which is just over 8. And then times, we're just doing the chain rule, times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The derivative of this with respect to x was the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2 over 16, which is just over 8. And then times, we're just doing the chain rule, times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1. So times negative 1. Times, so we'll multiply negative 1 right over here. And so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus, let's see, I could write this as positive x over 8."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The derivative of 100 minus x with respect to x is just negative 1. So times negative 1. Times, so we'll multiply negative 1 right over here. And so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus, let's see, I could write this as positive x over 8. So I could write this as 1 8th x, because negative 1 times negative x is positive x over 8. And then minus 100 over 8, which is negative 12.5. Minus 12.5."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus, let's see, I could write this as positive x over 8. So I could write this as 1 8th x, because negative 1 times negative x is positive x over 8. And then minus 100 over 8, which is negative 12.5. Minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point. If we can find an x, it makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to, if you add the x terms, you get square root of 3 over 18 plus 1 8th x."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we just have to verify that this is going to be a minimum point. If we can find an x, it makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to, if you add the x terms, you get square root of 3 over 18 plus 1 8th x. To solve for x, divide both sides by this business, you get x is equal to 12.5 over square root of 3 over 18 plus 1 over 8. And we are done. At x equals this, our derivative is equal to 0."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So if we add 12.5 to both sides, we get 12.5 is equal to, if you add the x terms, you get square root of 3 over 18 plus 1 8th x. To solve for x, divide both sides by this business, you get x is equal to 12.5 over square root of 3 over 18 plus 1 over 8. And we are done. At x equals this, our derivative is equal to 0. Now, we know, I shouldn't say we're done yet. We don't know whether this is a minimum point. In order to figure out whether this is a minimum point, we have to figure out whether our function is concave upward or concave downward when x is equal to this business."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "At x equals this, our derivative is equal to 0. Now, we know, I shouldn't say we're done yet. We don't know whether this is a minimum point. In order to figure out whether this is a minimum point, we have to figure out whether our function is concave upward or concave downward when x is equal to this business. And to figure that out, let's take the second derivative here. So the second derivative, so let me rewrite the second derivative of all of this business. The second derivative, well, this was the same function as this right over here."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In order to figure out whether this is a minimum point, we have to figure out whether our function is concave upward or concave downward when x is equal to this business. And to figure that out, let's take the second derivative here. So the second derivative, so let me rewrite the second derivative of all of this business. The second derivative, well, this was the same function as this right over here. So let me rewrite it. So a prime, the derivative of our combined area, was equal to the square root of 3 over 18x plus 1 8th x minus 12.5. The second derivative is going to be square root of 3 over 18 plus 1 over 8."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The second derivative, well, this was the same function as this right over here. So let me rewrite it. So a prime, the derivative of our combined area, was equal to the square root of 3 over 18x plus 1 8th x minus 12.5. The second derivative is going to be square root of 3 over 18 plus 1 over 8. So this thing right over here is greater than 0, which means we're concave upward for all x's. Which means for all x's, we're kind of doing a situation like this. So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The second derivative is going to be square root of 3 over 18 plus 1 over 8. So this thing right over here is greater than 0, which means we're concave upward for all x's. Which means for all x's, we're kind of doing a situation like this. So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards. This is concave upwards for all x. So we're going to be at a minimum point. The slope is 0 right over here."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards. This is concave upwards for all x. So we're going to be at a minimum point. The slope is 0 right over here. This right over here will be a minimum point. So once again, this is going to be a minimum point. Now, if we actually had 100 meter wire, this expression isn't too valuable."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope is 0 right over here. This right over here will be a minimum point. So once again, this is going to be a minimum point. Now, if we actually had 100 meter wire, this expression isn't too valuable. We'd want to get a pretty close approximation in terms of where to actually make the cut, an actual decimal number. So let's use a calculator to get that. So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, if we actually had 100 meter wire, this expression isn't too valuable. We'd want to get a pretty close approximation in terms of where to actually make the cut, an actual decimal number. So let's use a calculator to get that. So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8. 1 divided by 8 gives us, and now we deserve our drum roll, this is 56.5. So this is approximately equal to 56.5 meters. So you make this cut roughly 56.5."}, {"video_title": "Minimizing combined area   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8. 1 divided by 8 gives us, and now we deserve our drum roll, this is 56.5. So this is approximately equal to 56.5 meters. So you make this cut roughly 56.5. I'll write roughly. I'll make it little. 56.5 meters from the left-hand side."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And what we want to do is take its first and second derivatives and use as much of our techniques as we have at our disposal to attempt to graph it without a graphing calculator. If we have time, I'll take out the graphing calculator and see if our answer matches up. So a good place to start is to take the first derivative of this. So the derivative of f, well, you take the derivative of the inside, so take the derivative of that right there, which is 4x to the third. And then multiply it times the derivative of the outside with respect to the inside. So the derivative of natural log of x is 1 over x. So the derivative of this whole thing with respect to this inside expression is going to be times 1 over x to the fourth plus 27."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the derivative of f, well, you take the derivative of the inside, so take the derivative of that right there, which is 4x to the third. And then multiply it times the derivative of the outside with respect to the inside. So the derivative of natural log of x is 1 over x. So the derivative of this whole thing with respect to this inside expression is going to be times 1 over x to the fourth plus 27. If you found that confusing, you might want to re-watch the chain rule videos. But that's the first derivative of our function. And I could rewrite this, this is equal to 4x to the third over x to the fourth plus 27, or I could write it as 4x to the third times x to the fourth plus 27 to the negative 1."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the derivative of this whole thing with respect to this inside expression is going to be times 1 over x to the fourth plus 27. If you found that confusing, you might want to re-watch the chain rule videos. But that's the first derivative of our function. And I could rewrite this, this is equal to 4x to the third over x to the fourth plus 27, or I could write it as 4x to the third times x to the fourth plus 27 to the negative 1. All three of these expressions are equivalent. I'm just writing, I've multiplied it out, or I could write this as a negative exponent, or I could write this as a fraction with this in the denominator. They're all equivalent."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And I could rewrite this, this is equal to 4x to the third over x to the fourth plus 27, or I could write it as 4x to the third times x to the fourth plus 27 to the negative 1. All three of these expressions are equivalent. I'm just writing, I've multiplied it out, or I could write this as a negative exponent, or I could write this as a fraction with this in the denominator. They're all equivalent. So that's our first derivative. Let's do our second derivative. Our second derivative, this looks like it'll get a little bit hairier."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "They're all equivalent. So that's our first derivative. Let's do our second derivative. Our second derivative, this looks like it'll get a little bit hairier. So our second derivative is the derivative of this. So it's equal to, we can now use the product rule, it's the derivative of this first expression times the second expression. So the derivative of this first expression, 3 times 4 is 12."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Our second derivative, this looks like it'll get a little bit hairier. So our second derivative is the derivative of this. So it's equal to, we can now use the product rule, it's the derivative of this first expression times the second expression. So the derivative of this first expression, 3 times 4 is 12. 12x squared, we just decrement the 3 by 1, times the second expression, times x to the fourth plus 27 to the minus 1. And then to that we want to add just the first expression, not its derivative, so just 4x to the third, times the derivative of the second expression. So the second expression, we could take the derivative of the inside, which is just 4x to the third, this derivative of 27 is just 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the derivative of this first expression, 3 times 4 is 12. 12x squared, we just decrement the 3 by 1, times the second expression, times x to the fourth plus 27 to the minus 1. And then to that we want to add just the first expression, not its derivative, so just 4x to the third, times the derivative of the second expression. So the second expression, we could take the derivative of the inside, which is just 4x to the third, this derivative of 27 is just 0. So times 4x to the third, times the derivative of this whole thing with respect to the inside. So times, so you take this exponent, put it out front. So times minus 1, times this whole thing, x to the fourth plus 27, to the, we decrement this by 1 more, so minus 2."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the second expression, we could take the derivative of the inside, which is just 4x to the third, this derivative of 27 is just 0. So times 4x to the third, times the derivative of this whole thing with respect to the inside. So times, so you take this exponent, put it out front. So times minus 1, times this whole thing, x to the fourth plus 27, to the, we decrement this by 1 more, so minus 2. So let's see if I can simplify this expression a little bit. So this is equal to, so this right here is equal to 12x squared over this thing, x to the fourth plus 27. And then let's see, we're going to have a minus here, so it's minus, you multiply these two guys, 4 times 4 is 16."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So times minus 1, times this whole thing, x to the fourth plus 27, to the, we decrement this by 1 more, so minus 2. So let's see if I can simplify this expression a little bit. So this is equal to, so this right here is equal to 12x squared over this thing, x to the fourth plus 27. And then let's see, we're going to have a minus here, so it's minus, you multiply these two guys, 4 times 4 is 16. 16x to the third times x to the third is x to the sixth, over this thing squared, over x to the fourth plus 27 squared. That's just another way to rewrite that expression right there, right? To the minus 2, you can just put it in the denominator and make it to a positive 2 in the denominator."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then let's see, we're going to have a minus here, so it's minus, you multiply these two guys, 4 times 4 is 16. 16x to the third times x to the third is x to the sixth, over this thing squared, over x to the fourth plus 27 squared. That's just another way to rewrite that expression right there, right? To the minus 2, you can just put it in the denominator and make it to a positive 2 in the denominator. Same thing. Now, if you've seen these problems in the past, we always want to set these things equal to 0. We want to solve for x equals 0, so it'll be useful to have this expressed as just one fraction, instead of the difference or the sum of two fractions."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "To the minus 2, you can just put it in the denominator and make it to a positive 2 in the denominator. Same thing. Now, if you've seen these problems in the past, we always want to set these things equal to 0. We want to solve for x equals 0, so it'll be useful to have this expressed as just one fraction, instead of the difference or the sum of two fractions. So what we could do is to have, we could have a common denominator, so we could multiply both the numerator and denominator of this expression by x to the fourth plus 27, and what do we get? So this is equal to, so if we multiply this first expression times x to the fourth plus 27, we get 12x squared times x to the fourth plus 27, and then in the denominator, you have x to the fourth plus 27 squared. All I did, I multiplied this numerator and this denominator by x to the fourth plus 27, I didn't change it."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We want to solve for x equals 0, so it'll be useful to have this expressed as just one fraction, instead of the difference or the sum of two fractions. So what we could do is to have, we could have a common denominator, so we could multiply both the numerator and denominator of this expression by x to the fourth plus 27, and what do we get? So this is equal to, so if we multiply this first expression times x to the fourth plus 27, we get 12x squared times x to the fourth plus 27, and then in the denominator, you have x to the fourth plus 27 squared. All I did, I multiplied this numerator and this denominator by x to the fourth plus 27, I didn't change it. And then we have that second term, minus 16x to the sixth over x to the fourth plus 27 squared. The whole reason why I did that, now I have a common denominator, now I can just add the numerators. So this is going to be equal to, this is going to be equal to, let's see, let's multiply, well, the denominator, we know what the denominator is."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "All I did, I multiplied this numerator and this denominator by x to the fourth plus 27, I didn't change it. And then we have that second term, minus 16x to the sixth over x to the fourth plus 27 squared. The whole reason why I did that, now I have a common denominator, now I can just add the numerators. So this is going to be equal to, this is going to be equal to, let's see, let's multiply, well, the denominator, we know what the denominator is. It is x to the fourth plus 27 squared, that's our denominator, and then we can multiply this out. This is 12x squared times x to the fourth, so that's 12x to the sixth plus 27 times 12, I don't even feel like multiplying 27 times 12, so I'll just write that out. So plus 27 times 12x squared, I just multiply the 12x squared times 27, and then minus 16x to the sixth."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is going to be equal to, let's see, let's multiply, well, the denominator, we know what the denominator is. It is x to the fourth plus 27 squared, that's our denominator, and then we can multiply this out. This is 12x squared times x to the fourth, so that's 12x to the sixth plus 27 times 12, I don't even feel like multiplying 27 times 12, so I'll just write that out. So plus 27 times 12x squared, I just multiply the 12x squared times 27, and then minus 16x to the sixth. And this simplifies to, let's see if I can simplify this even further. So I have an x to the sixth here, x to the sixth here, so this is equal to, I'll do this in pink, this is equal to the 27 times 12x squared, I don't feel like figuring that out right now, times 12x squared, and then you have minus 16x to the sixth and plus 12x to the sixth. So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So plus 27 times 12x squared, I just multiply the 12x squared times 27, and then minus 16x to the sixth. And this simplifies to, let's see if I can simplify this even further. So I have an x to the sixth here, x to the sixth here, so this is equal to, I'll do this in pink, this is equal to the 27 times 12x squared, I don't feel like figuring that out right now, times 12x squared, and then you have minus 16x to the sixth and plus 12x to the sixth. So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared. And that is our second derivative. Now, we've done all of the derivatives, and this was actually a pretty hairy problem, and now we can solve for when the first and the second derivatives equal 0, and we'll have our candidate, well we'll know our critical points, and then we'll have our candidate inflection points and see if we can make any headway from there. So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared. And that is our second derivative. Now, we've done all of the derivatives, and this was actually a pretty hairy problem, and now we can solve for when the first and the second derivatives equal 0, and we'll have our candidate, well we'll know our critical points, and then we'll have our candidate inflection points and see if we can make any headway from there. So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined. So this is equal to 0, if we want to set, the only place that this can equal to 0 is if this numerator is equal to 0. This denominator actually, if we are assuming we're dealing with real numbers, this term right here is always going to be greater than or equal to 0 for any value of x, because it's an even exponent, so this thing can never equal 0, right, because you're adding 27 to something that's non-negative, so this will never equal 0, so this will also never be undefined. So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined. So this is equal to 0, if we want to set, the only place that this can equal to 0 is if this numerator is equal to 0. This denominator actually, if we are assuming we're dealing with real numbers, this term right here is always going to be greater than or equal to 0 for any value of x, because it's an even exponent, so this thing can never equal 0, right, because you're adding 27 to something that's non-negative, so this will never equal 0, so this will also never be undefined. So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily. If we wanted to set this equal to 0, we just say 4x to the third is equal to 0, and we know what x value will make that equal to 0, x has to be equal to 0. 4 times something to the third is equal to 0, that something has to be 0. x to the third has to be 0, x has to be 0. So we can write f prime of 0 is equal to 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily. If we wanted to set this equal to 0, we just say 4x to the third is equal to 0, and we know what x value will make that equal to 0, x has to be equal to 0. 4 times something to the third is equal to 0, that something has to be 0. x to the third has to be 0, x has to be 0. So we can write f prime of 0 is equal to 0. So 0 is a critical point. The slope at 0 is 0, we don't know if it's a maximum or a minimum or an inflection point yet, or it could be, you know, we'll explore it a little bit more. And actually just so we get the coordinate, what's the coordinate?"}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we can write f prime of 0 is equal to 0. So 0 is a critical point. The slope at 0 is 0, we don't know if it's a maximum or a minimum or an inflection point yet, or it could be, you know, we'll explore it a little bit more. And actually just so we get the coordinate, what's the coordinate? The coordinate x is 0, and then y is the natural log of x is 0, this just turns out, so it's the natural log of 27. So it's the natural log of, let me figure out what that is and get the calculator out. I said I wouldn't use a graphing calculator, but I could use a regular calculator."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And actually just so we get the coordinate, what's the coordinate? The coordinate x is 0, and then y is the natural log of x is 0, this just turns out, so it's the natural log of 27. So it's the natural log of, let me figure out what that is and get the calculator out. I said I wouldn't use a graphing calculator, but I could use a regular calculator. 27, if I were to take the natural log of that, it's like 3 point, well for our purpose this is called 3.3, we're just trying to get the general shape of the graph. So 3.3, 3 point, well we could just say 2.9 and it kept going. So this is a critical point right here, the slope is 0 here."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I said I wouldn't use a graphing calculator, but I could use a regular calculator. 27, if I were to take the natural log of that, it's like 3 point, well for our purpose this is called 3.3, we're just trying to get the general shape of the graph. So 3.3, 3 point, well we could just say 2.9 and it kept going. So this is a critical point right here, the slope is 0 here. Slope is equal to 0 at x is equal to 0. So this is one thing we want to block off. And let's see if we can find any candidate inflection points."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is a critical point right here, the slope is 0 here. Slope is equal to 0 at x is equal to 0. So this is one thing we want to block off. And let's see if we can find any candidate inflection points. And remember, candidate inflection points are where the second derivative equals 0. Now if the second derivative equals 0, that doesn't tell us that those are definitely inflection points. Let me make this very clear."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And let's see if we can find any candidate inflection points. And remember, candidate inflection points are where the second derivative equals 0. Now if the second derivative equals 0, that doesn't tell us that those are definitely inflection points. Let me make this very clear. If x is inflection, then the second derivative at x is going to be equal to 0. Because you're having a change in the concavity, I always say con-ca-ti-vity, but it's the con-cavity. You have a change in the slope goes from either increasing to decreasing or from decreasing to increasing."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me make this very clear. If x is inflection, then the second derivative at x is going to be equal to 0. Because you're having a change in the concavity, I always say con-ca-ti-vity, but it's the con-cavity. You have a change in the slope goes from either increasing to decreasing or from decreasing to increasing. But if the derivative is equal to 0, you cannot assume that it's an inflection point. So what we're going to do is, we're going to find all of the points at which this is true, and then see if we actually do have a sign change in the second derivative at that point. And only if you have a sign change, then you can say it's an inflection point."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You have a change in the slope goes from either increasing to decreasing or from decreasing to increasing. But if the derivative is equal to 0, you cannot assume that it's an inflection point. So what we're going to do is, we're going to find all of the points at which this is true, and then see if we actually do have a sign change in the second derivative at that point. And only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because the second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go below that x, the second derivative has to actually change signs."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because the second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go below that x, the second derivative has to actually change signs. Only then. So we can say if f' changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely going to be 0 right at x."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It has to have a second derivative of 0, and when you go below that x, the second derivative has to actually change signs. Only then. So we can say if f' changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely going to be 0 right at x. But you have to actually see that if it's negative before x, it has to be positive after x, or if it's positive before x, it has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And if it's changing signs around x, then it's definitely going to be 0 right at x. But you have to actually see that if it's negative before x, it has to be positive after x, or if it's positive before x, it has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points and then see if this is true, that the sign actually changes. So we want to find where this thing over here is equal to 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points and then see if this is true, that the sign actually changes. So we want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where our numerator can be equal to 0 for the second derivative."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12 x squared minus 4x to the 6th is equal to 0. Remember, that's just the numerator of our second derivative."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's see where our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12 x squared minus 4x to the 6th is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So let's factor out a 4x squared. Now we'll have 27 times, if we factor a 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the 4th is equal to 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So let's factor out a 4x squared. Now we'll have 27 times, if we factor a 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the 4th is equal to 0. So the x's that will make this equal to 0 will satisfy either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head, that's 81, 20 times 3 is 60, 7 times 3 is 21, 60 plus 21 is 81. Or 81 minus x to the 4th is equal to 0. Any x that satisfies either of these will make this entire expression equal to 0, so if this thing is 0, the whole thing is going to be equal to 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Now we'll have 27 times, if we factor a 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the 4th is equal to 0. So the x's that will make this equal to 0 will satisfy either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head, that's 81, 20 times 3 is 60, 7 times 3 is 21, 60 plus 21 is 81. Or 81 minus x to the 4th is equal to 0. Any x that satisfies either of these will make this entire expression equal to 0, so if this thing is 0, the whole thing is going to be equal to 0. If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Any x that satisfies either of these will make this entire expression equal to 0, so if this thing is 0, the whole thing is going to be equal to 0. If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0 itself. This is going to be equal to 0 when, let's see, if we add x to the 4th to both sides, you get x to the 4th is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus 3."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's solve this. This is going to be 0 when x is equal to 0 itself. This is going to be equal to 0 when, let's see, if we add x to the 4th to both sides, you get x to the 4th is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus 3. These are our candidate inflection points. x is equal to 0, x is equal to plus 3, or x is equal to minus 3. What we have to do now is to see whether the second derivative changes signs around these points in order to be able to label them inflection points."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus 3. These are our candidate inflection points. x is equal to 0, x is equal to plus 3, or x is equal to minus 3. What we have to do now is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. What happens when x is slightly below 0? Let's take the situation, let's do all the scenarios. What happens when x is slightly below 0?"}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What we have to do now is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. What happens when x is slightly below 0? Let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them necessarily, but if x is like 0.1, what is the second derivative going to be doing? If x is minus 0.1, this term right here is going to be positive, and then this is going to be 81 minus 0.1 to the 4th. That's going to be a very small number."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What happens when x is slightly below 0? Not all of them necessarily, but if x is like 0.1, what is the second derivative going to be doing? If x is minus 0.1, this term right here is going to be positive, and then this is going to be 81 minus 0.1 to the 4th. That's going to be a very small number. It's going to be some positive number times 81 minus a small number, so it's going to be a positive number. When x is less than 0, or just slightly less than 0, our second derivative is positive. What happens when x is slightly larger?"}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's going to be a very small number. It's going to be some positive number times 81 minus a small number, so it's going to be a positive number. When x is less than 0, or just slightly less than 0, our second derivative is positive. What happens when x is slightly larger? When I write this notation, I want to be careful. I mean really just right below 0. When x is right above 0, what happens?"}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What happens when x is slightly larger? When I write this notation, I want to be careful. I mean really just right below 0. When x is right above 0, what happens? Let's say x was 0.01, or 0.1, positive 0.1. It's going to be the same thing, because in both cases we're squaring and we're taking the 4th, so you're kind of losing your sign information. If x is 0.1, this thing is going to be a small positive number."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "When x is right above 0, what happens? Let's say x was 0.01, or 0.1, positive 0.1. It's going to be the same thing, because in both cases we're squaring and we're taking the 4th, so you're kind of losing your sign information. If x is 0.1, this thing is going to be a small positive number. You're going to be subtracting a very small positive number from 81, but 81 minus a small number is still going to be positive. You're going to have a positive times a positive, so your second derivative is still going to be greater than 0. Something interesting here."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If x is 0.1, this thing is going to be a small positive number. You're going to be subtracting a very small positive number from 81, but 81 minus a small number is still going to be positive. You're going to have a positive times a positive, so your second derivative is still going to be greater than 0. Something interesting here. Your second derivative is 0 when x is equal to 0, but it is not an inflection point, because notice the concavity did not change around 0. Our second derivative is positive as we approach 0 from the left, and it's positive as we approach 0 from the right. In general, at 0, as we're near 0 from either direction, we're going to be concave upwards."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Something interesting here. Your second derivative is 0 when x is equal to 0, but it is not an inflection point, because notice the concavity did not change around 0. Our second derivative is positive as we approach 0 from the left, and it's positive as we approach 0 from the right. In general, at 0, as we're near 0 from either direction, we're going to be concave upwards. The fact that 0 is a critical point, and that we're always concave upward as we approach 0 from either side, this tells us that this is a minimum point. Because we're concave upwards all around 0. 0 is not an inflection point."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "In general, at 0, as we're near 0 from either direction, we're going to be concave upwards. The fact that 0 is a critical point, and that we're always concave upward as we approach 0 from either side, this tells us that this is a minimum point. Because we're concave upwards all around 0. 0 is not an inflection point. Let's see if positive and negative 3 are inflection points. If you study this equation, let me write our... Actually, I just want to be clear. I've just been using the numerator of the second derivative."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "0 is not an inflection point. Let's see if positive and negative 3 are inflection points. If you study this equation, let me write our... Actually, I just want to be clear. I've just been using the numerator of the second derivative. The whole second derivative is this thing right here, but I've been ignoring the denominator, because the denominator is always positive. If we're trying to understand whether things are positive or negative, we just really have to determine whether the numerator is positive or negative, because this expression right here is always positive. It's something to the second power."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I've just been using the numerator of the second derivative. The whole second derivative is this thing right here, but I've been ignoring the denominator, because the denominator is always positive. If we're trying to understand whether things are positive or negative, we just really have to determine whether the numerator is positive or negative, because this expression right here is always positive. It's something to the second power. Let's test whether we have a change in concavity around x is equal to positive or negative 3. Remember, the numerator of our... Let me just rewrite our second derivative, just so you see it here. f prime prime of x."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's something to the second power. Let's test whether we have a change in concavity around x is equal to positive or negative 3. Remember, the numerator of our... Let me just rewrite our second derivative, just so you see it here. f prime prime of x. The numerator is this thing right here. It's 4x squared times 81 minus x to the fourth. The denominator was up here, x to the fourth plus 27 squared."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "f prime prime of x. The numerator is this thing right here. It's 4x squared times 81 minus x to the fourth. The denominator was up here, x to the fourth plus 27 squared. x to the fourth plus 27 squared. That was our second derivative. Let's see if this changes signs around positive or negative 3."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The denominator was up here, x to the fourth plus 27 squared. x to the fourth plus 27 squared. That was our second derivative. Let's see if this changes signs around positive or negative 3. Actually, we should get the same answer, because regardless of whether we put positive or negative 3 here, you lose all your sign information, because you're taking it to the fourth power. You're taking it to the second power. Obviously, anything to the fourth power is always going to be positive."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's see if this changes signs around positive or negative 3. Actually, we should get the same answer, because regardless of whether we put positive or negative 3 here, you lose all your sign information, because you're taking it to the fourth power. You're taking it to the second power. Obviously, anything to the fourth power is always going to be positive. Anything to the second power is always going to be negative. When we do our test, if it's true for positive 3, it's probably going to be true for negative 3 as well. Let's just try it out."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Obviously, anything to the fourth power is always going to be positive. Anything to the second power is always going to be negative. When we do our test, if it's true for positive 3, it's probably going to be true for negative 3 as well. Let's just try it out. When x is just a little bit less than positive 3, what's the sign of f prime prime of x? It's going to be 4 times 9. It's going to be 4 times a positive number."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's just try it out. When x is just a little bit less than positive 3, what's the sign of f prime prime of x? It's going to be 4 times 9. It's going to be 4 times a positive number. It might be like 2.999, but this is still going to be positive. This is going to be positive when x is approaching 3. Then this is going to be, well, if x is 3, this is 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's going to be 4 times a positive number. It might be like 2.999, but this is still going to be positive. This is going to be positive when x is approaching 3. Then this is going to be, well, if x is 3, this is 0. If x is a little bit less than 3, if it's like 2.9999, this number is going to be less than 81. This is also going to be positive. Of course, the denominator is always positive."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Then this is going to be, well, if x is 3, this is 0. If x is a little bit less than 3, if it's like 2.9999, this number is going to be less than 81. This is also going to be positive. Of course, the denominator is always positive. As x is less than 3, as it's approaching from the left, we are concave upwards. This thing is going to be a positive. f prime prime is greater than 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Of course, the denominator is always positive. As x is less than 3, as it's approaching from the left, we are concave upwards. This thing is going to be a positive. f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? This first term is still going to be positive, but if x is just larger than 3, x to the fourth is going to be just larger than 81."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? This first term is still going to be positive, but if x is just larger than 3, x to the fourth is going to be just larger than 81. This second term is going to be negative in that situation. It's going to be negative. Let me do it in a new color."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This first term is still going to be positive, but if x is just larger than 3, x to the fourth is going to be just larger than 81. This second term is going to be negative in that situation. It's going to be negative. Let me do it in a new color. It's going to be negative when x is larger than 3 because this is going to be larger than 81. If this is negative and this is positive, then the whole thing is going to be negative because this denominator is still going to be positive. Then f prime prime is going to be less than 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do it in a new color. It's going to be negative when x is larger than 3 because this is going to be larger than 81. If this is negative and this is positive, then the whole thing is going to be negative because this denominator is still going to be positive. Then f prime prime is going to be less than 0. We're going to be concave downwards. One last one. What happens when x is just greater than minus 3?"}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Then f prime prime is going to be less than 0. We're going to be concave downwards. One last one. What happens when x is just greater than minus 3? Just being greater than minus 3, that's like minus 2.9999. When you take minus 2.99 and square it, you're going to get a positive number. This is going to be positive."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What happens when x is just greater than minus 3? Just being greater than minus 3, that's like minus 2.9999. When you take minus 2.99 and square it, you're going to get a positive number. This is going to be positive. If you take minus 2.99 to the fourth, that's going to be a little bit less than 81 because 2.99 to the fourth is a little bit less than 81. This is still going to be positive. You have positive times a positive divided by a positive."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is going to be positive. If you take minus 2.99 to the fourth, that's going to be a little bit less than 81 because 2.99 to the fourth is a little bit less than 81. This is still going to be positive. You have positive times a positive divided by a positive. You're going to be concave upwards because your second derivative is going to be greater than 0. Concave upwards. Finally, when x is just less than negative 3, remember when I write this, I don't mean for all x's larger than negative 3 or all x's smaller than negative 3."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You have positive times a positive divided by a positive. You're going to be concave upwards because your second derivative is going to be greater than 0. Concave upwards. Finally, when x is just less than negative 3, remember when I write this, I don't mean for all x's larger than negative 3 or all x's smaller than negative 3. There's actually no notation that would say just as we just approach 3, in this case, from the left. What happens if we just go to minus 3.11 or 3.01? This term right here is going to be positive."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Finally, when x is just less than negative 3, remember when I write this, I don't mean for all x's larger than negative 3 or all x's smaller than negative 3. There's actually no notation that would say just as we just approach 3, in this case, from the left. What happens if we just go to minus 3.11 or 3.01? This term right here is going to be positive. If we take minus 3.1 to the fourth, that's going to be larger than positive 81. The sign will become positive. It will be larger than 81, so this will become negative."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This term right here is going to be positive. If we take minus 3.1 to the fourth, that's going to be larger than positive 81. The sign will become positive. It will be larger than 81, so this will become negative. In that case as well, we'll have a positive times a negative divided by a positive. Then our second derivative is going to be negative. We're going to be downwards."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It will be larger than 81, so this will become negative. In that case as well, we'll have a positive times a negative divided by a positive. Then our second derivative is going to be negative. We're going to be downwards. I think we're ready to plot. First of all, is x plus or minus 3 inflection points? Sure."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We're going to be downwards. I think we're ready to plot. First of all, is x plus or minus 3 inflection points? Sure. As we approach x is equal to 3 from the left, we are concave upwards. Then as we cross 3, the second derivative is 0. The second derivative is 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Sure. As we approach x is equal to 3 from the left, we are concave upwards. Then as we cross 3, the second derivative is 0. The second derivative is 0. I lost it up here. The second derivative is 0. Then as we go to the right of 3, we become concave downwards."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The second derivative is 0. I lost it up here. The second derivative is 0. Then as we go to the right of 3, we become concave downwards. We got our sign change in the second derivative. x is equal to 3. This 3 is definitely an inflection point."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Then as we go to the right of 3, we become concave downwards. We got our sign change in the second derivative. x is equal to 3. This 3 is definitely an inflection point. The same argument can be made for negative 3. We switch signs as we cross 3. These definitely are inflection points."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This 3 is definitely an inflection point. The same argument can be made for negative 3. We switch signs as we cross 3. These definitely are inflection points. Just so we get the exact coordinates, let's figure out what f of 3 is, or f of positive and negative 3. Then we're ready to graph. First of all, we know that the point 0,3.29, that this was a minimum."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "These definitely are inflection points. Just so we get the exact coordinates, let's figure out what f of 3 is, or f of positive and negative 3. Then we're ready to graph. First of all, we know that the point 0,3.29, that this was a minimum. Because 0 is a critical point, the slope is 0 there. Because it's concave upwards, all around 0. So 0 is definitely not an inflection point."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "First of all, we know that the point 0,3.29, that this was a minimum. Because 0 is a critical point, the slope is 0 there. Because it's concave upwards, all around 0. So 0 is definitely not an inflection point. Then we know that the points positive 3 and minus 3 are inflection points. In order to figure out their y coordinates, we can just evaluate them. They're actually going to have the same y coordinates."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So 0 is definitely not an inflection point. Then we know that the points positive 3 and minus 3 are inflection points. In order to figure out their y coordinates, we can just evaluate them. They're actually going to have the same y coordinates. Because if you put a minus 3 or positive 3 and take it to the fourth power, you're going to get the same thing. Let's figure out what they are. If we take 3 to the fourth power, that's 81."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "They're actually going to have the same y coordinates. Because if you put a minus 3 or positive 3 and take it to the fourth power, you're going to get the same thing. Let's figure out what they are. If we take 3 to the fourth power, that's 81. 81 plus 27 is equal to 108. Then we want to take the natural log of it. We want to take the natural log."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If we take 3 to the fourth power, that's 81. 81 plus 27 is equal to 108. Then we want to take the natural log of it. We want to take the natural log. It's like 4.7, just to get a rough idea. That's true of whether we do positive or negative 3, because we took to the fourth power. So it's 4.7."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We want to take the natural log. It's like 4.7, just to get a rough idea. That's true of whether we do positive or negative 3, because we took to the fourth power. So it's 4.7. These are both inflection points. We should be ready to graph it. Let's graph it."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's 4.7. These are both inflection points. We should be ready to graph it. Let's graph it. Let me draw my axis. Just like that. This is my y-axis."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's graph it. Let me draw my axis. Just like that. This is my y-axis. This is my x-axis. This is y. You can even call it the f of x axis, if you like."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is my y-axis. This is my x-axis. This is y. You can even call it the f of x axis, if you like. This is x. The point 0, 3.29. Let's say this is 1, 2, 3, 4, 5."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You can even call it the f of x axis, if you like. This is x. The point 0, 3.29. Let's say this is 1, 2, 3, 4, 5. The point 0, 3.29. That's 0, 1, 2, 3. A little bit above 3 is right there."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say this is 1, 2, 3, 4, 5. The point 0, 3.29. That's 0, 1, 2, 3. A little bit above 3 is right there. That's a minimum point. Then we're concave. The slope is 0 right there."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "A little bit above 3 is right there. That's a minimum point. Then we're concave. The slope is 0 right there. We figured that out, because the first derivative was 0 there. It's a critical point. It's concave upwards around there."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope is 0 right there. We figured that out, because the first derivative was 0 there. It's a critical point. It's concave upwards around there. That told us we're at a minimum point right there. Then at positive 3. 1, 2, 3."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's concave upwards around there. That told us we're at a minimum point right there. Then at positive 3. 1, 2, 3. At positive 3, 4.7. 4.7 will look something like that. We have an inflection point."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "1, 2, 3. At positive 3, 4.7. 4.7 will look something like that. We have an inflection point. Before that, we're concave upwards. Then after that, we're concave downwards. It looks something like this."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We have an inflection point. Before that, we're concave upwards. Then after that, we're concave downwards. It looks something like this. We're concave upwards up to that point. Maybe, actually, you should let me ignore that yellow thing I drew before. Let me get rid of that."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It looks something like this. We're concave upwards up to that point. Maybe, actually, you should let me ignore that yellow thing I drew before. Let me get rid of that. Let me draw it like this. 1, 2, 3. 3, 4.7 looks like that."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me get rid of that. Let me draw it like this. 1, 2, 3. 3, 4.7 looks like that. Minus 3, 4.7. 1, 2, 3. 4.7 looks like that."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "3, 4.7 looks like that. Minus 3, 4.7. 1, 2, 3. 4.7 looks like that. We know at 0, we have a slope of 0. We're concave upwards. We look like this."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "4.7 looks like that. We know at 0, we have a slope of 0. We're concave upwards. We look like this. We're concave upwards until x is equal to 3. Then at x equals 3, we become concave downwards. We become concave downwards."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We look like this. We're concave upwards until x is equal to 3. Then at x equals 3, we become concave downwards. We become concave downwards. Let me try my best to draw it well. We go off like that. Then we're concave upwards around 0."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We become concave downwards. Let me try my best to draw it well. We go off like that. Then we're concave upwards around 0. Until we get, we're concave upwards as long as x is greater than minus 3. Then at minus 3, we become concave downwards again. Maybe, I should do it in that color."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Then we're concave upwards around 0. Until we get, we're concave upwards as long as x is greater than minus 3. Then at minus 3, we become concave downwards again. Maybe, I should do it in that color. This concave downwards right here, that's this right here. That's that right there. This concave downwards right here."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Maybe, I should do it in that color. This concave downwards right here, that's this right here. That's that right there. This concave downwards right here. Sorry, I meant to do it in that red color. This concave downwards right here is this right there. Then the concave upwards around 0 is right there."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This concave downwards right here. Sorry, I meant to do it in that red color. This concave downwards right here is this right there. Then the concave upwards around 0 is right there. You could even imagine this concave upwards that we measured. That's this concave upwards. This concave upwards is that."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Then the concave upwards around 0 is right there. You could even imagine this concave upwards that we measured. That's this concave upwards. This concave upwards is that. Then around 0, we're always upwards. This is my sense of what the graph will look like. Maybe, it turns into, you could think about what it does."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This concave upwards is that. Then around 0, we're always upwards. This is my sense of what the graph will look like. Maybe, it turns into, you could think about what it does. As x approaches positive or negative infinity, some of the terms. I won't go into that. Let's test whether we've graphed it correctly using a graphing calculator."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Maybe, it turns into, you could think about what it does. As x approaches positive or negative infinity, some of the terms. I won't go into that. Let's test whether we've graphed it correctly using a graphing calculator. Let me get out my trusty TI-85 and let's graph this sucker. Press graph, y equals, it was the natural log of x to the fourth plus 27. I want to get that graph there."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's test whether we've graphed it correctly using a graphing calculator. Let me get out my trusty TI-85 and let's graph this sucker. Press graph, y equals, it was the natural log of x to the fourth plus 27. I want to get that graph there. I do second graph. Let's cross our fingers. It looks pretty good."}, {"video_title": "Another example graphing with derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I want to get that graph there. I do second graph. Let's cross our fingers. It looks pretty good. It looks almost exactly like what we drew. I think our mathematics was correct. This was actually very satisfying."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first thing, let's see if we can take the antiderivative of nine sine of x. So we could use some of our integration properties to simplify this a little bit. So this is going to be equal to, this is the same thing as nine times the integral from 11 pi over two to six pi of sine of x dx. And what's the antiderivative of sine of x? Well we know from our derivatives that the derivative with respect to x of cosine of x is equal to negative sine of x. Negative sine of x. So can we construct this in some way so this is a negative sine of x?"}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what's the antiderivative of sine of x? Well we know from our derivatives that the derivative with respect to x of cosine of x is equal to negative sine of x. Negative sine of x. So can we construct this in some way so this is a negative sine of x? Well what if I multiplied on the inside, what if I multiplied it by negative one? Well I can't just multiply it only one place by negative one I need to multiply by negative one twice so I'm not changing its value. So what if I said negative nine times negative sine of x?"}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So can we construct this in some way so this is a negative sine of x? Well what if I multiplied on the inside, what if I multiplied it by negative one? Well I can't just multiply it only one place by negative one I need to multiply by negative one twice so I'm not changing its value. So what if I said negative nine times negative sine of x? Well this is still gonna be nine sine of x. If you took negative nine times negative sine of x it is nine sine of x. And I did it this way because now negative sine of x it matches the derivative of cosine of x."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what if I said negative nine times negative sine of x? Well this is still gonna be nine sine of x. If you took negative nine times negative sine of x it is nine sine of x. And I did it this way because now negative sine of x it matches the derivative of cosine of x. So we could say that this is all going to be equal to, it's all going to be equal to, you have your negative nine out front, negative nine times, times, and I'll put it in brackets, negative nine times the antiderivative of negative sine of x. Well that is just going to be cosine of x. Cosine of x. And we're going to evaluate it at its bounds."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I did it this way because now negative sine of x it matches the derivative of cosine of x. So we could say that this is all going to be equal to, it's all going to be equal to, you have your negative nine out front, negative nine times, times, and I'll put it in brackets, negative nine times the antiderivative of negative sine of x. Well that is just going to be cosine of x. Cosine of x. And we're going to evaluate it at its bounds. We're going to evaluate it at six pi. Let me do that in a color I haven't used yet. We're gonna do that at six pi."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to evaluate it at its bounds. We're going to evaluate it at six pi. Let me do that in a color I haven't used yet. We're gonna do that at six pi. And we're also going to do that at 11 pi over two. 11 pi over two. And so this is going to be equal to, this is equal to negative nine times, I'm gonna create some space here."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna do that at six pi. And we're also going to do that at 11 pi over two. 11 pi over two. And so this is going to be equal to, this is equal to negative nine times, I'm gonna create some space here. So, actually that's probably more space than I need. It's going to be cosine of six pi. Cosine of six, six pi."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to, this is equal to negative nine times, I'm gonna create some space here. So, actually that's probably more space than I need. It's going to be cosine of six pi. Cosine of six, six pi. Cosine of six pi minus, minus cosine of 11 pi over two. Cosine of 11 pi over two. Well what is cosine of six pi going to be?"}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of six, six pi. Cosine of six pi minus, minus cosine of 11 pi over two. Cosine of 11 pi over two. Well what is cosine of six pi going to be? Well, cosine of any multiple of two pi is going to be equal to one. You could use six pi as we're going around the unit circle three times. So, this is the same thing as cosine of two pi or the same thing as cosine of zero."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well what is cosine of six pi going to be? Well, cosine of any multiple of two pi is going to be equal to one. You could use six pi as we're going around the unit circle three times. So, this is the same thing as cosine of two pi or the same thing as cosine of zero. So that is going to be equal to one. If that seems unfamiliar to you, I encourage you to review the unit circle definition of cosine. And what is cosine of 11 pi over two?"}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, this is the same thing as cosine of two pi or the same thing as cosine of zero. So that is going to be equal to one. If that seems unfamiliar to you, I encourage you to review the unit circle definition of cosine. And what is cosine of 11 pi over two? Let's see, let's subtract some, let's subtract some multiple of two pi here to put it in values that we can understand better. So this is, so let me write it here. Cosine of 11 pi over two."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what is cosine of 11 pi over two? Let's see, let's subtract some, let's subtract some multiple of two pi here to put it in values that we can understand better. So this is, so let me write it here. Cosine of 11 pi over two. That is the same thing as, let's see, if we were to subtract, this is the same thing as cosine of 11 pi over two minus, let's see, this is the same thing as five and 1 1\u20442 pi. Right? Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of 11 pi over two. That is the same thing as, let's see, if we were to subtract, this is the same thing as cosine of 11 pi over two minus, let's see, this is the same thing as five and 1 1\u20442 pi. Right? Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two. In fact, no, let's subtract five. Let's subtract, no, let's subtract four pi, which is eight pi over two. So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two. In fact, no, let's subtract five. Let's subtract, no, let's subtract four pi, which is eight pi over two. So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine. And so this is going to be equal to cosine of three pi over two. And if we imagine the unit circle, let me draw the unit circle here. So it's my y-axis, my x-axis, and then I have the unit circle."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine. And so this is going to be equal to cosine of three pi over two. And if we imagine the unit circle, let me draw the unit circle here. So it's my y-axis, my x-axis, and then I have the unit circle. So, whoops, all right, the unit circle, just like that. So if we start at, this is zero, then you go to pi over two, then you go to pi, then you go to three pi over two. So that's this point on the unit circle."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's my y-axis, my x-axis, and then I have the unit circle. So, whoops, all right, the unit circle, just like that. So if we start at, this is zero, then you go to pi over two, then you go to pi, then you go to three pi over two. So that's this point on the unit circle. So the cosine is the x-coordinate, so this is going to be zero. This is zero, so this is zero. And so we get one minus zero, so everything in the brackets evaluates out to one."}, {"video_title": "Definite integral of trig function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's this point on the unit circle. So the cosine is the x-coordinate, so this is going to be zero. This is zero, so this is zero. And so we get one minus zero, so everything in the brackets evaluates out to one. And so we are left with, so let me do that. So all of this is equal to one. And so you have negative nine times one, which of course is just negative nine, is what this definite integral evaluates to."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be the same thing as the limit as x approaches negative one of x plus one over, over the limit, the limit as x approaches negative one of square root of x plus five minus two. And then we could say, all right, this thing up here, x plus one, if we think about the graph y equals x plus one, it's continuous everywhere, especially at x equals negative one, and so to evaluate this limit, we just have to evaluate this expression at x equals negative one. So this numerator is just going to evaluate to negative one plus one. And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to?"}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. Now when you see that, you might be tempted to give up."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. So let's do that."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one?"}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. While f of x, f of x would have the same graph, except it wouldn't have, it wouldn't have the gap."}, {"video_title": "Limits by rationalizing   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. While f of x, f of x would have the same graph, except it wouldn't have, it wouldn't have the gap. And so if you're trying to find the limit, it seems completely reasonable. Well, let's just use f of x and evaluate what f of x would be to kind of fill that gap at x equals negative one. So hopefully this graphical version helps a little bit, or if it confuses you, ignore it."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we were to take the derivative with respect to time, so if we were to take the derivative with respect to time of this function s, what are we going to get? Well, we're going to get ds dt, or the rate at which position changes with respect to time, and what's another word for that? The rate at which position changes with respect to time? Well, that's just velocity. So that we could write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time? So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's just velocity. So that we could write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time? So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. So this is going to be, we can write this as dv dt, the rate at which velocity is changing with respect to time, and what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. So this is going to be, we can write this as dv dt, the rate at which velocity is changing with respect to time, and what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time. So you start with the position as a function of time, take its derivative with respect to time, you get velocity. Take that derivative with respect to time, you get acceleration. Well, you could go the other way around."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be our acceleration as a function of time. So you start with the position as a function of time, take its derivative with respect to time, you get velocity. Take that derivative with respect to time, you get acceleration. Well, you could go the other way around. If you started with acceleration, and you were to take the antiderivative of it, an antiderivative of it is going to be, actually, let me just write it this way. So an antiderivative, I'll just use the integral symbol to show that I'm taking the antiderivative, is going to be the integral of the antiderivative of a of t, and this is going to give you some expression with a plus c, and we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you could go the other way around. If you started with acceleration, and you were to take the antiderivative of it, an antiderivative of it is going to be, actually, let me just write it this way. So an antiderivative, I'll just use the integral symbol to show that I'm taking the antiderivative, is going to be the integral of the antiderivative of a of t, and this is going to give you some expression with a plus c, and we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. The position is a function of time. Just like this, we would have some plus c here, but if we know our position at a given time, we could solve for that c. Now that we've reviewed it a little bit, but we've rewritten it in, I guess you could say, thinking of it not just from the differential point of view or from the derivative point of view, but thinking of it from the antiderivative point of view, let's see if we can solve an interesting problem. Let's say that we know that the acceleration of a particle as a function of time is equal to 1."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. The position is a function of time. Just like this, we would have some plus c here, but if we know our position at a given time, we could solve for that c. Now that we've reviewed it a little bit, but we've rewritten it in, I guess you could say, thinking of it not just from the differential point of view or from the derivative point of view, but thinking of it from the antiderivative point of view, let's see if we can solve an interesting problem. Let's say that we know that the acceleration of a particle as a function of time is equal to 1. It's always accelerating at 1 unit per, and I'm not giving you time, but let's just say that we're thinking in terms of meters and seconds. This is 1 meter per second squared right over here. That's our acceleration as a function of time."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that we know that the acceleration of a particle as a function of time is equal to 1. It's always accelerating at 1 unit per, and I'm not giving you time, but let's just say that we're thinking in terms of meters and seconds. This is 1 meter per second squared right over here. That's our acceleration as a function of time. Let's say we don't know the velocity expressions, but we know the velocity at a particular time, and we don't know the position expressions, but we know the position at a particular time. Let's say we know that the velocity at time 3, let's say 3 seconds, is negative 3 meters per second. Actually, I'm going to write the units here and make it a little bit."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's our acceleration as a function of time. Let's say we don't know the velocity expressions, but we know the velocity at a particular time, and we don't know the position expressions, but we know the position at a particular time. Let's say we know that the velocity at time 3, let's say 3 seconds, is negative 3 meters per second. Actually, I'm going to write the units here and make it a little bit. This is meters per second squared. That's going to be our unit for acceleration. This is our unit for velocity."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, I'm going to write the units here and make it a little bit. This is meters per second squared. That's going to be our unit for acceleration. This is our unit for velocity. Let's say that we know that the position at time 2, at 2 seconds, is equal to negative 10 meters. If we're thinking in one dimension, if we're just moving along the number line, it's 10 to the left of the origin. Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time?"}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our unit for velocity. Let's say that we know that the position at time 2, at 2 seconds, is equal to negative 10 meters. If we're thinking in one dimension, if we're just moving along the number line, it's 10 to the left of the origin. Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? Not just velocity at time 3, but velocity generally as a function of time and position as a function of time. I encourage you to pause this video right now and try to figure it out on your own. Let's just work through this."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? Not just velocity at time 3, but velocity generally as a function of time and position as a function of time. I encourage you to pause this video right now and try to figure it out on your own. Let's just work through this. What is, we know that velocity as a function of time is going to be the antiderivative of our acceleration as a function of time. Our acceleration is just 1. This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just work through this. What is, we know that velocity as a function of time is going to be the antiderivative of our acceleration as a function of time. Our acceleration is just 1. This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. Let's just write that down. V of 3 is going to be equal to 3, 3 plus C. Every place where I saw the T, or every place where I have the T, I replace it with this 3 right over here. Actually, let me make it a little bit clearer."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. Let's just write that down. V of 3 is going to be equal to 3, 3 plus C. Every place where I saw the T, or every place where I have the T, I replace it with this 3 right over here. Actually, let me make it a little bit clearer. V of 3 is equal to 3 plus C, and they tell us that that's equal to negative 3. That is equal to negative 3. What's C going to be?"}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me make it a little bit clearer. V of 3 is equal to 3 plus C, and they tell us that that's equal to negative 3. That is equal to negative 3. What's C going to be? If we just look at this part of the equation, just this equation right over here, if you subtract 3 from both sides, you get C is equal to negative 6. Now we know the exact expression that defines velocity as a function of time. V of T is equal to T, T plus negative 6, or T minus 6."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's C going to be? If we just look at this part of the equation, just this equation right over here, if you subtract 3 from both sides, you get C is equal to negative 6. Now we know the exact expression that defines velocity as a function of time. V of T is equal to T, T plus negative 6, or T minus 6. We can verify that. The derivative of this with respect to time is just 1, and when time is equal to 3, 3 minus 6 is indeed negative 3. We've been able to figure out velocity as a function of time."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "V of T is equal to T, T plus negative 6, or T minus 6. We can verify that. The derivative of this with respect to time is just 1, and when time is equal to 3, 3 minus 6 is indeed negative 3. We've been able to figure out velocity as a function of time. Now let's do a similar thing to figure out position as a function of time. We know that position is going to be an antiderivative of the velocity function. Let's write that down."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've been able to figure out velocity as a function of time. Now let's do a similar thing to figure out position as a function of time. We know that position is going to be an antiderivative of the velocity function. Let's write that down. Position as a function of time is going to be equal to the antiderivative of V of T, dT, which is equal to the antiderivative of T minus 6, dT, which is equal to, well, the antiderivative of T is T squared over 2. T squared over 2, we've seen that before. The antiderivative of negative 6 is negative 6T."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's write that down. Position as a function of time is going to be equal to the antiderivative of V of T, dT, which is equal to the antiderivative of T minus 6, dT, which is equal to, well, the antiderivative of T is T squared over 2. T squared over 2, we've seen that before. The antiderivative of negative 6 is negative 6T. Of course, we can't forget our constant. So plus C. This is what S of T is equal to. S of T is equal to all of this business right over here."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of negative 6 is negative 6T. Of course, we can't forget our constant. So plus C. This is what S of T is equal to. S of T is equal to all of this business right over here. Now we can try to solve for our constant. We do that using this information right over here. At 2 seconds, our position is negative 10 meters."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "S of T is equal to all of this business right over here. Now we can try to solve for our constant. We do that using this information right over here. At 2 seconds, our position is negative 10 meters. S of 2, or I could just write it this way. Let me write it this way. S of 2 at 2 seconds is going to be equal to 2 squared over 2."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "At 2 seconds, our position is negative 10 meters. S of 2, or I could just write it this way. Let me write it this way. S of 2 at 2 seconds is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2. That's going to be 2 minus 6 times 2. So minus 12 plus C is equal to negative 10."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "S of 2 at 2 seconds is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2. That's going to be 2 minus 6 times 2. So minus 12 plus C is equal to negative 10. Is equal to negative 10. Let's see, we get 2 minus 12 is negative 10, plus C is equal to negative 10. You add 10 to both sides, you get C, in this case, is equal to 0."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So minus 12 plus C is equal to negative 10. Is equal to negative 10. Let's see, we get 2 minus 12 is negative 10, plus C is equal to negative 10. You add 10 to both sides, you get C, in this case, is equal to 0. So we've figured out what our position function is as well. This C right over here is just going to be 0. So our position, as a function of time, is equal to T squared over 2 minus 6T."}, {"video_title": "Worked example motion problems (with definite integrals)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You add 10 to both sides, you get C, in this case, is equal to 0. So we've figured out what our position function is as well. This C right over here is just going to be 0. So our position, as a function of time, is equal to T squared over 2 minus 6T. And you can verify. When T is equal to 2, 2 squared over 2 is 2, minus 12 is negative 10. You take the derivative here, you get T minus 6, and you can see, and we already verified, that V of 3 is negative 3."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is try to get a graphical appreciation for inflection points, which we also cover in some detail in other videos. So the first thing to appreciate is an inflection point is a point on our graph where our slope goes from decreasing to increasing or from increasing to decreasing. So right over here I have the graph of some function, and let me draw the slope of a tangent line at different points. So when x is equal to negative two, that is what the tangent line looks like, and you can see its slope. And then as we increase x, we can see that the slope is positive, but it is decreasing. Then it goes to zero, and then it goes negative, and the slope keeps decreasing all the way until we get, it looks like, we get to about x equals negative one, and then our slope begins to increase again. So something interesting happened right at x equals negative one."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x is equal to negative two, that is what the tangent line looks like, and you can see its slope. And then as we increase x, we can see that the slope is positive, but it is decreasing. Then it goes to zero, and then it goes negative, and the slope keeps decreasing all the way until we get, it looks like, we get to about x equals negative one, and then our slope begins to increase again. So something interesting happened right at x equals negative one. And so that's a pretty good indication. We're just doing it graphically here. We're not proving it."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So something interesting happened right at x equals negative one. And so that's a pretty good indication. We're just doing it graphically here. We're not proving it. But that at this point right over here, we have an inflection point. So let me write that down. So let me show you that again now that the point is labeled."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're not proving it. But that at this point right over here, we have an inflection point. So let me write that down. So let me show you that again now that the point is labeled. For x at negative two, we have a positive slope. It decreases, decreases, decreases. It's negative."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me show you that again now that the point is labeled. For x at negative two, we have a positive slope. It decreases, decreases, decreases. It's negative. It still decreases. X equals negative one, and then our slope begins increasing again. So that's how you could tell it just from the function itself, but you could also tell inflection points by looking at your first derivative."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's negative. It still decreases. X equals negative one, and then our slope begins increasing again. So that's how you could tell it just from the function itself, but you could also tell inflection points by looking at your first derivative. Remember, an inflection point is when our slope goes from increasing to decreasing or from decreasing to increasing. The derivative is just the slope of the tangent line. So this right over here, this is the derivative of our original blue function."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's how you could tell it just from the function itself, but you could also tell inflection points by looking at your first derivative. Remember, an inflection point is when our slope goes from increasing to decreasing or from decreasing to increasing. The derivative is just the slope of the tangent line. So this right over here, this is the derivative of our original blue function. So here we can see the interesting parts. And so notice what's happening. On the derivative, the derivative is decreasing, which means the slope of our tangent line of our original function is decreasing, and we saw that."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here, this is the derivative of our original blue function. So here we can see the interesting parts. And so notice what's happening. On the derivative, the derivative is decreasing, which means the slope of our tangent line of our original function is decreasing, and we saw that. Notice, while the derivative is decreasing right over here, our slope will be decreasing. Our slope is positive, our slope is positive, but decreasing, then it becomes negative, but decreasing all the way until this point, which is at x equals negative one. So let's do that again."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "On the derivative, the derivative is decreasing, which means the slope of our tangent line of our original function is decreasing, and we saw that. Notice, while the derivative is decreasing right over here, our slope will be decreasing. Our slope is positive, our slope is positive, but decreasing, then it becomes negative, but decreasing all the way until this point, which is at x equals negative one. So let's do that again. So our slope is positive and decreasing, and then right over about there, right over here, our slope keeps decreasing, but then it actually turns negative, and it keeps decreasing all the way until x equals negative one. And then our slope begins increasing again. So the derivative begins increasing, which means the slope of our tangent line of our original function begins increasing."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that again. So our slope is positive and decreasing, and then right over about there, right over here, our slope keeps decreasing, but then it actually turns negative, and it keeps decreasing all the way until x equals negative one. And then our slope begins increasing again. So the derivative begins increasing, which means the slope of our tangent line of our original function begins increasing. So that point is interesting. An inflection point, one way to identify an inflection point from the first derivative is to look at a minimum point or to look at a maximum point, because that shows a place where your derivative is changing direction. It's going from increasing to decreasing, or in this case, from decreasing to increasing, which tells you that this is likely an inflection point."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative begins increasing, which means the slope of our tangent line of our original function begins increasing. So that point is interesting. An inflection point, one way to identify an inflection point from the first derivative is to look at a minimum point or to look at a maximum point, because that shows a place where your derivative is changing direction. It's going from increasing to decreasing, or in this case, from decreasing to increasing, which tells you that this is likely an inflection point. Now let's think about the second derivative. So right over here, this is the derivative of the derivative, and I could zoom out to look at the whole thing. You actually can't see the whole thing right over here."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going from increasing to decreasing, or in this case, from decreasing to increasing, which tells you that this is likely an inflection point. Now let's think about the second derivative. So right over here, this is the derivative of the derivative, and I could zoom out to look at the whole thing. You actually can't see the whole thing right over here. Actually, I can zoom out a little bit more so that you can really see what's going on. And so what's interesting here? Well, it looks like right at x equals negative one, we cross, our second derivative crosses the x-axis."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You actually can't see the whole thing right over here. Actually, I can zoom out a little bit more so that you can really see what's going on. And so what's interesting here? Well, it looks like right at x equals negative one, we cross, our second derivative crosses the x-axis. So let me label that. So right over there, we cross the x-axis, which is exactly where we have the inflection point. And that makes sense, because if our second derivative goes from being negative to positive, that means our first derivative goes from being decreasing to increasing, which means the slope of our tangent line of our function goes from decreasing to increasing."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it looks like right at x equals negative one, we cross, our second derivative crosses the x-axis. So let me label that. So right over there, we cross the x-axis, which is exactly where we have the inflection point. And that makes sense, because if our second derivative goes from being negative to positive, that means our first derivative goes from being decreasing to increasing, which means the slope of our tangent line of our function goes from decreasing to increasing. We've seen that over and over, decreasing to increasing right over here. Now it's important to realize the second derivative doesn't need to just touch the x-axis. It needs to cross it."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that makes sense, because if our second derivative goes from being negative to positive, that means our first derivative goes from being decreasing to increasing, which means the slope of our tangent line of our function goes from decreasing to increasing. We've seen that over and over, decreasing to increasing right over here. Now it's important to realize the second derivative doesn't need to just touch the x-axis. It needs to cross it. So you might say, well, what about this point right over here, two comma zero? The second derivative touches the x-axis there, but it doesn't cross it. So we never go from our derivative increasing to our derivative decreasing."}, {"video_title": "Inflection points from graphs of function & derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It needs to cross it. So you might say, well, what about this point right over here, two comma zero? The second derivative touches the x-axis there, but it doesn't cross it. So we never go from our derivative increasing to our derivative decreasing. So big takeaways, you can figure out the inflection point from either the graph of the function, from the graph of the derivative, or the graph of the second derivative. On the function itself, you just wanna inspect the slopes of the tangent line and think about where does it go from decreasing to increasing, or the other way around, from increasing to decreasing. If you're looking at the first derivative, you really just wanna look at minimum or maximum points."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I want to think about the limit of f of x as x approaches positive infinity and the limit of f of x as x approaches negative infinity. So let's think about what these are going to be. Well, once again, and I'm not doing this in an ultra rigorous way, but more in an intuitive way, is to think about what this function approximately equals as we get larger and larger and larger x's. This is the case if we're getting very positive x's, very positive infinity direction, or very negative. It's still the absolute value of those x's that are very, very, very large as we approach positive infinity or negative infinity. Well, in the numerator we only have one term. We have this x term."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the case if we're getting very positive x's, very positive infinity direction, or very negative. It's still the absolute value of those x's that are very, very, very large as we approach positive infinity or negative infinity. Well, in the numerator we only have one term. We have this x term. But in the denominator we have two terms under the radical here. And as x gets larger and larger and larger, either in the positive or the negative direction, this x squared term is going to really dominate this one. You can imagine when x is a million, you're going to have a million squared plus 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have this x term. But in the denominator we have two terms under the radical here. And as x gets larger and larger and larger, either in the positive or the negative direction, this x squared term is going to really dominate this one. You can imagine when x is a million, you're going to have a million squared plus 1. The value of the denominator is going to be dictated by this x squared term. So this is going to be approximately equal to x over the square root of x squared. This term right over here, the 1, isn't going to matter so much when we get very, very, very large x's."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can imagine when x is a million, you're going to have a million squared plus 1. The value of the denominator is going to be dictated by this x squared term. So this is going to be approximately equal to x over the square root of x squared. This term right over here, the 1, isn't going to matter so much when we get very, very, very large x's. And this right over here, x over the square root of x squared, or x over the principal root of x squared, this is going to be equal to x over. If I square something and then take the principal root, remember the principal root is the positive square root of something, then I'm essentially taking the absolute value of x. It's going to be equal to x over the absolute value of x for x approaches infinity or for x approaches negative infinity."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This term right over here, the 1, isn't going to matter so much when we get very, very, very large x's. And this right over here, x over the square root of x squared, or x over the principal root of x squared, this is going to be equal to x over. If I square something and then take the principal root, remember the principal root is the positive square root of something, then I'm essentially taking the absolute value of x. It's going to be equal to x over the absolute value of x for x approaches infinity or for x approaches negative infinity. So another way to say this, another way to restate these limits, is as we approach infinity, this limit, we can restate it as the limit, this is going to be equal to the limit as x approaches infinity of x over the absolute value of x. Now, for positive x's, the absolute value of x is just going to be x. This is going to be x divided by x."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be equal to x over the absolute value of x for x approaches infinity or for x approaches negative infinity. So another way to say this, another way to restate these limits, is as we approach infinity, this limit, we can restate it as the limit, this is going to be equal to the limit as x approaches infinity of x over the absolute value of x. Now, for positive x's, the absolute value of x is just going to be x. This is going to be x divided by x. So this is just going to be 1. Similarly, right over here, we're taking the limit as we go to negative infinity. This is going to be the limit of x over the absolute value of x as x approaches negative infinity."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be x divided by x. So this is just going to be 1. Similarly, right over here, we're taking the limit as we go to negative infinity. This is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason why I was able to make this statement is that f of x and this thing right over here become very, very similar, or you can kind of say converge to each other, as x gets very, very, very large or x gets very, very, very, very negative. Now, for negative values of x, the absolute value of x is going to be positive. x is obviously going to be negative, and we're just going to get negative 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason why I was able to make this statement is that f of x and this thing right over here become very, very similar, or you can kind of say converge to each other, as x gets very, very, very large or x gets very, very, very, very negative. Now, for negative values of x, the absolute value of x is going to be positive. x is obviously going to be negative, and we're just going to get negative 1. So using this, we can actually try to graph our function. So let's try to do that. So let's say that is my y-axis."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "x is obviously going to be negative, and we're just going to get negative 1. So using this, we can actually try to graph our function. So let's try to do that. So let's say that is my y-axis. This is my x-axis. And we see that we have two horizontal asymptotes. We have one horizontal asymptote at y is equal to 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that is my y-axis. This is my x-axis. And we see that we have two horizontal asymptotes. We have one horizontal asymptote at y is equal to 1. So let's say this right over here is y is equal to 1. Let me draw that line as a dotted line. We're going to approach this thing."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have one horizontal asymptote at y is equal to 1. So let's say this right over here is y is equal to 1. Let me draw that line as a dotted line. We're going to approach this thing. And then we have another horizontal asymptote at y is equal to negative 1. So that might be right over there. y is equal to negative 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to approach this thing. And then we have another horizontal asymptote at y is equal to negative 1. So that might be right over there. y is equal to negative 1. And if we want to plot at least one point, we could think about what does f of 0 equal. So f of 0 is going to be equal to 0 over the square root of 0 plus 1, or 0 squared plus 1. Well, that's all just going to be equal to 0."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "y is equal to negative 1. And if we want to plot at least one point, we could think about what does f of 0 equal. So f of 0 is going to be equal to 0 over the square root of 0 plus 1, or 0 squared plus 1. Well, that's all just going to be equal to 0. So we have this point right over here. And we know that as x approaches infinity, we're approaching this blue horizontal asymptote. So it might look something like this."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's all just going to be equal to 0. So we have this point right over here. And we know that as x approaches infinity, we're approaching this blue horizontal asymptote. So it might look something like this. Let me do it a little bit differently. Let me do it a little bit. There you go."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it might look something like this. Let me do it a little bit differently. Let me do it a little bit. There you go. Clean this up. So it might look something like this. That's not the color I wanted to use."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "There you go. Clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger. And then like this."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger. And then like this. We get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y is equal to f of x."}, {"video_title": "Limits at infinity of quotients with square roots (odd power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then like this. We get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y is equal to f of x. And you can verify this by taking a calculator, trying to plot more points, or using some type of graphing calculator or something. But anyway, I just wanted to tackle another situation where we're approaching infinity or negative infinity, and we're trying to determine the horizontal asymptotes. And remember, the key is just to say what terms dominate as x approaches positive infinity or negative infinity to say, well, what is that function going to approach?"}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to figure out f prime of x, and as we will see, the chain rule is going to be very useful here. And what I'm gonna do is I'm going to first just apply the chain rule, and then maybe dig into it a little bit to make sure we draw the connection between what we're doing here and then what you might see in maybe some of your calculus textbooks that explain the chain rule. So if we have a function that is defined as essentially a composite function, notice this expression right here, we are taking something to the third power. It isn't just an x that we're taking to the third power, we are taking a cosine of x to the third power. So we're taking a function, you could view it this way, we're taking the function cosine of x, and then we're inputting it into another function that takes it to the third power. So let me put it this way. If you viewed, if you say, look, we could take an x, we put it into one function, that is, that first function is cosine of x, so first we evaluate the cosine, and so that's going to produce cosine of x, cosine of x, and then we're going to input it into a function that just takes things to the third power."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "It isn't just an x that we're taking to the third power, we are taking a cosine of x to the third power. So we're taking a function, you could view it this way, we're taking the function cosine of x, and then we're inputting it into another function that takes it to the third power. So let me put it this way. If you viewed, if you say, look, we could take an x, we put it into one function, that is, that first function is cosine of x, so first we evaluate the cosine, and so that's going to produce cosine of x, cosine of x, and then we're going to input it into a function that just takes things to the third power. So it just takes things to the third power, and so what are you going to end up with? Well, you're going to end up with, what are you taking to the third power? You're taking cosine of x. Cosine of x to the third power."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you viewed, if you say, look, we could take an x, we put it into one function, that is, that first function is cosine of x, so first we evaluate the cosine, and so that's going to produce cosine of x, cosine of x, and then we're going to input it into a function that just takes things to the third power. So it just takes things to the third power, and so what are you going to end up with? Well, you're going to end up with, what are you taking to the third power? You're taking cosine of x. Cosine of x to the third power. This is a composite function. You could view this, you could view this as the function, let's call this blue one the function v, and let's call this the function u, and so if we're taking x and into u, this is u of x, and then if we're taking u of x into the input, or as the input into the function v, then this output right over here, this is going to be v of, well, what was inputted? V of u of x. V of u of x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're taking cosine of x. Cosine of x to the third power. This is a composite function. You could view this, you could view this as the function, let's call this blue one the function v, and let's call this the function u, and so if we're taking x and into u, this is u of x, and then if we're taking u of x into the input, or as the input into the function v, then this output right over here, this is going to be v of, well, what was inputted? V of u of x. V of u of x. Or, another way of writing it, I'm going to write it multiple ways, it's the same thing as v of cosine of x. V of cosine of x. And so v, whatever you input into it, it just takes it to the third power. If you were to write v of x, it would be x to the third power."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "V of u of x. V of u of x. Or, another way of writing it, I'm going to write it multiple ways, it's the same thing as v of cosine of x. V of cosine of x. And so v, whatever you input into it, it just takes it to the third power. If you were to write v of x, it would be x to the third power. So the chain rule tells us, or the chain rule is what our brain should say, hey, it becomes applicable if we're going to take the derivative of a function that can be expressed as a composite function like this. So just to be clear, we can write f of x, f of x is equal to v of u of x. I know I'm essentially saying the same thing over and over again, but I'm saying it in slightly different ways because the first time you learn this, it can be a little bit hard to grok, or really deeply understand. So I'm going to try to write it in different ways."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you were to write v of x, it would be x to the third power. So the chain rule tells us, or the chain rule is what our brain should say, hey, it becomes applicable if we're going to take the derivative of a function that can be expressed as a composite function like this. So just to be clear, we can write f of x, f of x is equal to v of u of x. I know I'm essentially saying the same thing over and over again, but I'm saying it in slightly different ways because the first time you learn this, it can be a little bit hard to grok, or really deeply understand. So I'm going to try to write it in different ways. And the chain rule tells us that if you have a situation like this, then the derivative, f prime of x, and this is something that you will see in your textbooks, so this is going to be the derivative of this whole thing with respect to u of x, so we could write that as v prime of u of x, v prime of u of x, times the derivative of u with respect to x, times u prime of x. This right over here, this is one expression of the chain rule. And so how do we evaluate it in this case?"}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm going to try to write it in different ways. And the chain rule tells us that if you have a situation like this, then the derivative, f prime of x, and this is something that you will see in your textbooks, so this is going to be the derivative of this whole thing with respect to u of x, so we could write that as v prime of u of x, v prime of u of x, times the derivative of u with respect to x, times u prime of x. This right over here, this is one expression of the chain rule. And so how do we evaluate it in this case? Let me color code it in a similar way. So the v function, this outer thing that just takes things to the third power, I'll put in blue. So f prime of x, another way of expressing it, and I'll use it with more of the differential notation."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so how do we evaluate it in this case? Let me color code it in a similar way. So the v function, this outer thing that just takes things to the third power, I'll put in blue. So f prime of x, another way of expressing it, and I'll use it with more of the differential notation. You could view this as the derivative of, well, I'll write it a couple of different ways. You could view it as the derivative of v, the derivative of v with respect to u, I want to get the colors right, the derivative of v with respect to u, that's what this thing is right over here, times the derivative of u with respect to x. So times the derivative of u with respect to x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f prime of x, another way of expressing it, and I'll use it with more of the differential notation. You could view this as the derivative of, well, I'll write it a couple of different ways. You could view it as the derivative of v, the derivative of v with respect to u, I want to get the colors right, the derivative of v with respect to u, that's what this thing is right over here, times the derivative of u with respect to x. So times the derivative of u with respect to x. And just to make clear, so you're familiar with the different notations you'll see in different textbooks, this is this right over here, just using different notations, and this is this right over here. So let's actually evaluate these things. You're probably tired of just talking in the abstract."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So times the derivative of u with respect to x. And just to make clear, so you're familiar with the different notations you'll see in different textbooks, this is this right over here, just using different notations, and this is this right over here. So let's actually evaluate these things. You're probably tired of just talking in the abstract. So this is going to be equal to, this is going to be equal to, and I'm gonna write it out again, this is the derivative, instead of just writing v and u, I'm gonna write it, let me write it this way, this is going to be, I keep wanting to be using the wrong colors, this is going to be the derivative of, I'm gonna leave some space, times the derivative of something else with respect to something else. So we're gonna first take the derivative of v. Well v is cosine of x to the third power, cosine of x. We're gonna take the derivative of that with respect to u, which is just cosine of x, and we're gonna multiply that times the derivative of u, which is cosine of x, with respect to x, so this one we have good, we've seen this before."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're probably tired of just talking in the abstract. So this is going to be equal to, this is going to be equal to, and I'm gonna write it out again, this is the derivative, instead of just writing v and u, I'm gonna write it, let me write it this way, this is going to be, I keep wanting to be using the wrong colors, this is going to be the derivative of, I'm gonna leave some space, times the derivative of something else with respect to something else. So we're gonna first take the derivative of v. Well v is cosine of x to the third power, cosine of x. We're gonna take the derivative of that with respect to u, which is just cosine of x, and we're gonna multiply that times the derivative of u, which is cosine of x, with respect to x, so this one we have good, we've seen this before. We know that the derivative with respect to x of cosine of x, cosine, let me use it in that same color, the derivative of cosine of x, well that's equal to negative sine of x. So this one right over here, that is negative sine of x. You might be more familiar with seeing the derivative operator this way, but in theory, you won't see this as often, but this helps my brain really grok what we're doing."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna take the derivative of that with respect to u, which is just cosine of x, and we're gonna multiply that times the derivative of u, which is cosine of x, with respect to x, so this one we have good, we've seen this before. We know that the derivative with respect to x of cosine of x, cosine, let me use it in that same color, the derivative of cosine of x, well that's equal to negative sine of x. So this one right over here, that is negative sine of x. You might be more familiar with seeing the derivative operator this way, but in theory, you won't see this as often, but this helps my brain really grok what we're doing. We're taking the derivative of cosine of x with respect to x, well that's gonna be negative sine of x. Well what about taking the derivative of cosine of x to the third power with respect to cosine of x? What does this thing over here mean?"}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You might be more familiar with seeing the derivative operator this way, but in theory, you won't see this as often, but this helps my brain really grok what we're doing. We're taking the derivative of cosine of x with respect to x, well that's gonna be negative sine of x. Well what about taking the derivative of cosine of x to the third power with respect to cosine of x? What does this thing over here mean? Well, if I was taking the derivative, if I was taking the derivative of, let me write it this way, if I was taking the derivative of x to the third power, x to the third power with respect to x, if it was like that, well this is just going to be, and let me put some brackets here to make it a little bit clearer. If I'm taking the derivative of that, that is going to be, that is going to be, we bring the exponent out front, it's going to be three, three times x, three times x to the second power. Three times x to the second power."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "What does this thing over here mean? Well, if I was taking the derivative, if I was taking the derivative of, let me write it this way, if I was taking the derivative of x to the third power, x to the third power with respect to x, if it was like that, well this is just going to be, and let me put some brackets here to make it a little bit clearer. If I'm taking the derivative of that, that is going to be, that is going to be, we bring the exponent out front, it's going to be three, three times x, three times x to the second power. Three times x to the second power. So the general notion here is, if I'm taking the derivative of something, whatever this something happens to be, let me just add a new color. It could be, I'm doing the derivative of orange circle to the third power with respect to orange circle, well that's just going to be three times orange or yellow circle. Let me make it an actual orange circle."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Three times x to the second power. So the general notion here is, if I'm taking the derivative of something, whatever this something happens to be, let me just add a new color. It could be, I'm doing the derivative of orange circle to the third power with respect to orange circle, well that's just going to be three times orange or yellow circle. Let me make it an actual orange circle. So the derivative of orange circle to the third power with respect to orange circle, that's going to be three times the orange circle squared. So if I'm taking the derivative of cosine of x to the third power with respect to cosine of x, well that's just going to be, this is just going to be three times cosine of x, cosine of x to the second power. To the second power."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make it an actual orange circle. So the derivative of orange circle to the third power with respect to orange circle, that's going to be three times the orange circle squared. So if I'm taking the derivative of cosine of x to the third power with respect to cosine of x, well that's just going to be, this is just going to be three times cosine of x, cosine of x to the second power. To the second power. Notice, we're just, one way to think about it, is taking the derivative of this outside function with respect to the inside. So I would do the same thing as taking the derivative of x to the third power, but instead of an x, I have a cosine of x. So instead of it being three x squared, it is three cosine of x squared."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "To the second power. Notice, we're just, one way to think about it, is taking the derivative of this outside function with respect to the inside. So I would do the same thing as taking the derivative of x to the third power, but instead of an x, I have a cosine of x. So instead of it being three x squared, it is three cosine of x squared. And then the chain rule says, if we want to finally get the derivative with respect to x, we then take the derivative of cosine of x with respect to x. Now that's a big mouthful, but we are at the home stretch. We've now figured out the derivative."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So instead of it being three x squared, it is three cosine of x squared. And then the chain rule says, if we want to finally get the derivative with respect to x, we then take the derivative of cosine of x with respect to x. Now that's a big mouthful, but we are at the home stretch. We've now figured out the derivative. It's going to be this times this. So let's see, that's going to be negative three, negative three times sine of x times cosine squared of x. And I know that was kind of a long way of saying it."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've now figured out the derivative. It's going to be this times this. So let's see, that's going to be negative three, negative three times sine of x times cosine squared of x. And I know that was kind of a long way of saying it. I'm trying to explain the chain rule at the same time. But once you get the hang of it, you're just going to say, all right, well let me take the derivative of the outside of something to the third power with respect to the inside. Let me just treat that cosine of x like as if it was an x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I know that was kind of a long way of saying it. I'm trying to explain the chain rule at the same time. But once you get the hang of it, you're just going to say, all right, well let me take the derivative of the outside of something to the third power with respect to the inside. Let me just treat that cosine of x like as if it was an x. Well that's going to be, the second I did that, that's going to be three cosine squared of x. So that's that part and that part. And then let me take the derivative of the inside with respect to x."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the point is to really show you where that formula that you might see in a calculus book actually comes from, that it just comes from the exact same principles that we did in the last video. It's not for you to memorize that formula. I highly recommend against that because then you really won't know what's going on. It's really just to do it, it's better to do it from first principles where you find the volume of each of these disks and think of it that way. But let's just generalize what we saw in the last video. So instead of saying that this is y is equal to x squared, let's say that this is the graph, this function right over here, this function right over here, let's just generalize it and call it y is equal to f of x. And instead of saying we're going between 0 and 2, let's just say we're going between a and b."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's really just to do it, it's better to do it from first principles where you find the volume of each of these disks and think of it that way. But let's just generalize what we saw in the last video. So instead of saying that this is y is equal to x squared, let's say that this is the graph, this function right over here, this function right over here, let's just generalize it and call it y is equal to f of x. And instead of saying we're going between 0 and 2, let's just say we're going between a and b. So these are just two endpoints along the x-axis. So how would we find the volume of this? Well, just like the last video, we would still take a disk just like this, but what is the height of the disk now?"}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And instead of saying we're going between 0 and 2, let's just say we're going between a and b. So these are just two endpoints along the x-axis. So how would we find the volume of this? Well, just like the last video, we would still take a disk just like this, but what is the height of the disk now? Well, the height of the disk is not just x squared, we've generalized it. It is going to be whatever the value of our function is at that point. So the height of the disk is going to be f of x."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, just like the last video, we would still take a disk just like this, but what is the height of the disk now? Well, the height of the disk is not just x squared, we've generalized it. It is going to be whatever the value of our function is at that point. So the height of the disk is going to be f of x. The area of the space of this disk is going to be pi times our radius squared. So our radius is f of x, and we are just going to square it. That's the area of this face right over here."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the height of the disk is going to be f of x. The area of the space of this disk is going to be pi times our radius squared. So our radius is f of x, and we are just going to square it. That's the area of this face right over here. What is the volume of our disk? We're just going to have to multiply that times our depth. So it's going to be that times dx."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the area of this face right over here. What is the volume of our disk? We're just going to have to multiply that times our depth. So it's going to be that times dx. And we want to take the sum of all of these disks from a to b. And we're going to take the sum of them and then take the limit as the dx's get smaller and smaller and smaller, and we have an infinite number of these disks. And that means we're just going to take a definite integral."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be that times dx. And we want to take the sum of all of these disks from a to b. And we're going to take the sum of them and then take the limit as the dx's get smaller and smaller and smaller, and we have an infinite number of these disks. And that means we're just going to take a definite integral. So we're going to take the definite integral of this from a to b. And this right here is the formula that you will often see in a calculus book for using the disk method when you're rotating around the x-axis. But I just wanted to show you that it comes out of the common sense of finding the volume of this disk."}, {"video_title": "Generalizing disc method around x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that means we're just going to take a definite integral. So we're going to take the definite integral of this from a to b. And this right here is the formula that you will often see in a calculus book for using the disk method when you're rotating around the x-axis. But I just wanted to show you that it comes out of the common sense of finding the volume of this disk. The f of x right over here is just the radius of the disk. So this part right over here is just really pi r squared. We multiply it times the depth, and then we take the sum from a to b."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're told, consider the sum two plus five plus eight plus 11. Which expression is equal to the sum above? And they tell us choose all answers that apply. So like always, pause the video and see if you can work through this on your own. So when you look at the sum, it's clear you're starting at two, you're adding three each time, and we also are dealing with you have four total terms. Now we could try to construct an equation here or an expression using sigma notation, but instead what I like to do is look at our options. We really only have to look at these two options here and expand them out."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So like always, pause the video and see if you can work through this on your own. So when you look at the sum, it's clear you're starting at two, you're adding three each time, and we also are dealing with you have four total terms. Now we could try to construct an equation here or an expression using sigma notation, but instead what I like to do is look at our options. We really only have to look at these two options here and expand them out. What sum would each of these be? Well this is going to be the same thing as. We're starting at n equals one."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We really only have to look at these two options here and expand them out. What sum would each of these be? Well this is going to be the same thing as. We're starting at n equals one. So this is going to be three times when n equals one, three times one minus one, and then plus, then we'll go to n equals two, three times two minus one. Then we're gonna go to n equals three, so plus three times three minus one. And then finally we're gonna go to n equals four."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're starting at n equals one. So this is going to be three times when n equals one, three times one minus one, and then plus, then we'll go to n equals two, three times two minus one. Then we're gonna go to n equals three, so plus three times three minus one. And then finally we're gonna go to n equals four. So plus three times, when n equals four, this is three times four minus one. So just to be clear, this is what we did when n equals one. Let me write it down, n equals one."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally we're gonna go to n equals four. So plus three times, when n equals four, this is three times four minus one. So just to be clear, this is what we did when n equals one. Let me write it down, n equals one. This is what we got when n equals two. This is what we got when n equals three. And this is what we got when n is equal to four."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write it down, n equals one. This is what we got when n equals two. This is what we got when n equals three. And this is what we got when n is equal to four. And we stopped at n equals four because it tells us right over there. We start at n equals one and we go all the way to n equals four. So what does this equal?"}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is what we got when n is equal to four. And we stopped at n equals four because it tells us right over there. We start at n equals one and we go all the way to n equals four. So what does this equal? So let's see, three times one minus one is two, so this is looking good so far. Three times two minus one, that's six minus one, that is five, so still looking good. Three times three minus one, that's nine minus one."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what does this equal? So let's see, three times one minus one is two, so this is looking good so far. Three times two minus one, that's six minus one, that is five, so still looking good. Three times three minus one, that's nine minus one. Once again, still looking good. Three times four minus one, that is 11. So we like this choice, so I would definitely select this one."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Three times three minus one, that's nine minus one. Once again, still looking good. Three times four minus one, that is 11. So we like this choice, so I would definitely select this one. Now let's do the same thing over here. When n is equal to zero, it's going to be two plus three times zero, so that's just going to be two. And then plus, when n is equal to one, it's going to be two plus three times one, which is five, this is starting to look good."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we like this choice, so I would definitely select this one. Now let's do the same thing over here. When n is equal to zero, it's going to be two plus three times zero, so that's just going to be two. And then plus, when n is equal to one, it's going to be two plus three times one, which is five, this is starting to look good. Now when n is equal to, this was n equals zero, this was n equals one, so now we're at n equals two. So at n equals two, two plus three times two is two plus six, which is eight. And this makes sense."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then plus, when n is equal to one, it's going to be two plus three times one, which is five, this is starting to look good. Now when n is equal to, this was n equals zero, this was n equals one, so now we're at n equals two. So at n equals two, two plus three times two is two plus six, which is eight. And this makes sense. Every time we increase n by one, we are adding another three, which is consistent with what we saw there. We start at two when n equals zero. This is just, that three n is just zero, so you start at two, and then you keep, every time you increase n by one, you are just adding three again."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this makes sense. Every time we increase n by one, we are adding another three, which is consistent with what we saw there. We start at two when n equals zero. This is just, that three n is just zero, so you start at two, and then you keep, every time you increase n by one, you are just adding three again. So finally, when n is equal to three, two plus three times three is two plus, or is 11, I should say. And so this also is exactly the same sum. So I feel good about both of these."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is just, that three n is just zero, so you start at two, and then you keep, every time you increase n by one, you are just adding three again. So finally, when n is equal to three, two plus three times three is two plus, or is 11, I should say. And so this also is exactly the same sum. So I feel good about both of these. Let's do one more example. So we're here, we're given the sum, and we're saying choose one answer. Which of these is equivalent to this sum right over here?"}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I feel good about both of these. Let's do one more example. So we're here, we're given the sum, and we're saying choose one answer. Which of these is equivalent to this sum right over here? Well, like we did before, let's just expand it out. And what's different here is that we just have a variable, but that shouldn't make it too much more difficult. So let's do the situation, and I'll write it out."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Which of these is equivalent to this sum right over here? Well, like we did before, let's just expand it out. And what's different here is that we just have a variable, but that shouldn't make it too much more difficult. So let's do the situation, and I'll write it out. Let's do the situation when n is equal to one. When n is equal to one, it's gonna be k over one plus one. K over one plus one, and I'll write it out."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do the situation, and I'll write it out. Let's do the situation when n is equal to one. When n is equal to one, it's gonna be k over one plus one. K over one plus one, and I'll write it out. This is n is equal to one. And then plus, when n is equal to two, it's gonna be k over two plus one. This is n is equal to two."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "K over one plus one, and I'll write it out. This is n is equal to one. And then plus, when n is equal to two, it's gonna be k over two plus one. This is n is equal to two. And then we're gonna keep going. When n is equal to three, it's gonna be k over three plus one as n is equal to three. And then plus, finally, because we stop right over here at n equals four."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is n is equal to two. And then we're gonna keep going. When n is equal to three, it's gonna be k over three plus one as n is equal to three. And then plus, finally, because we stop right over here at n equals four. When n equals four, it's going to be k over four plus one. That's when n is equal to four. So this is all going to be equal."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then plus, finally, because we stop right over here at n equals four. When n equals four, it's going to be k over four plus one. That's when n is equal to four. So this is all going to be equal. I'll just write it over here. This is equal to k over two plus k over three plus k over four plus k over five, which is exactly this choice right over here. And actually, if I had looked at the choices ahead of time, I might have even been able to save even more time by just saying, well, look, actually, if you just try to compute the first term, when n is equal to one, it would be k over two."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is all going to be equal. I'll just write it over here. This is equal to k over two plus k over three plus k over four plus k over five, which is exactly this choice right over here. And actually, if I had looked at the choices ahead of time, I might have even been able to save even more time by just saying, well, look, actually, if you just try to compute the first term, when n is equal to one, it would be k over two. Well, only this one is starting with a k over two. This one has no k's here, which is sketchy. They're trying to look at the error where you try to replace the k with the number as well, not just the n. So that's what they're trying to do here."}, {"video_title": "Worked examples Summation notation   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually, if I had looked at the choices ahead of time, I might have even been able to save even more time by just saying, well, look, actually, if you just try to compute the first term, when n is equal to one, it would be k over two. Well, only this one is starting with a k over two. This one has no k's here, which is sketchy. They're trying to look at the error where you try to replace the k with the number as well, not just the n. So that's what they're trying to do here. Here, they're trying to, let's see, well, this isn't as obvious what they're even trying to do, right over here where they put the k in the denominator. And here, if you swapped the n and the k's, then you would have gotten this thing right over here. So we definitely feel good about choice A."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "So you probably realize that some of the traditional techniques that we've already had in our toolkits don't seem to be directly applicable. You substitution and others. And the key here to realize is we have a rational expression here where the numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. And whenever you see something like that, it's probably a good idea to divide the denominator into the numerator. That's what this rational expression could be interpreted as, x minus five divided by negative two x plus two. So let's do a little bit of algebraic long division to actually divide negative two x plus two into x minus five to see if we can rewrite this in a way that where we can evaluate the integral."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "In this case, the numerator and the denominator have the same degree. And whenever you see something like that, it's probably a good idea to divide the denominator into the numerator. That's what this rational expression could be interpreted as, x minus five divided by negative two x plus two. So let's do a little bit of algebraic long division to actually divide negative two x plus two into x minus five to see if we can rewrite this in a way that where we can evaluate the integral. So let's do that. We're gonna take x minus five. So x minus five."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "So let's do a little bit of algebraic long division to actually divide negative two x plus two into x minus five to see if we can rewrite this in a way that where we can evaluate the integral. So let's do that. We're gonna take x minus five. So x minus five. And divide negative two x plus two into that. So negative two x plus two. So look at the highest degree terms."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "So x minus five. And divide negative two x plus two into that. So negative two x plus two. So look at the highest degree terms. How many times does negative two x go into x? Well, it's gonna go negative 1 1 2 times. Negative 1 1 2 times two is negative one."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "So look at the highest degree terms. How many times does negative two x go into x? Well, it's gonna go negative 1 1 2 times. Negative 1 1 2 times two is negative one. Negative 1 1 2 times negative two x is just going to be positive x, just like that. And now we want to subtract this yellow expression from this blue expression. And so let's just, let me just take the negative of this and then add."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "Negative 1 1 2 times two is negative one. Negative 1 1 2 times negative two x is just going to be positive x, just like that. And now we want to subtract this yellow expression from this blue expression. And so let's just, let me just take the negative of this and then add. So I'm just gonna take the negative of it and add. And so we are left with negative five plus one is negative four. So you could say negative two x plus two goes into x minus five negative 1 1 2 times with negative four left over."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "And so let's just, let me just take the negative of this and then add. So I'm just gonna take the negative of it and add. And so we are left with negative five plus one is negative four. So you could say negative two x plus two goes into x minus five negative 1 1 2 times with negative four left over. And we can rewrite this integral, our original integral, as, we can rewrite it as negative 1 1 2 minus four over negative two x plus two dx. Now let's see, it looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two, all of these terms are divisible by two."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "So you could say negative two x plus two goes into x minus five negative 1 1 2 times with negative four left over. And we can rewrite this integral, our original integral, as, we can rewrite it as negative 1 1 2 minus four over negative two x plus two dx. Now let's see, it looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two, all of these terms are divisible by two. Actually we have all these negatives, that always unnecessarily complicates things. So let's actually divide the numerator and the denominator by negative two. So what are we gonna have then?"}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "The numerator and the denominator, they're both divisible by two, all of these terms are divisible by two. Actually we have all these negatives, that always unnecessarily complicates things. So let's actually divide the numerator and the denominator by negative two. So what are we gonna have then? So if we divide the numerator by negative two, if this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's just going to become x. And then two divided by negative two is going to be minus one."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "So what are we gonna have then? So if we divide the numerator by negative two, if this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's just going to become x. And then two divided by negative two is going to be minus one. So our original integral, once again, this is just algebra, everything we've done so far is algebra. We've just rewritten it using a little bit of algebraic long division. Our original integral has simplified to negative 1 1 2."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "And then two divided by negative two is going to be minus one. So our original integral, once again, this is just algebra, everything we've done so far is algebra. We've just rewritten it using a little bit of algebraic long division. Our original integral has simplified to negative 1 1 2. And some might argue it's not simplified, but it's actually much more useful for finding the integral. Negative 1 1 2 plus two over x minus one dx. Now, how do we evaluate this?"}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "Our original integral has simplified to negative 1 1 2. And some might argue it's not simplified, but it's actually much more useful for finding the integral. Negative 1 1 2 plus two over x minus one dx. Now, how do we evaluate this? Well, the antiderivative of negative 1 1 2 is pretty straightforward. That's just going to be negative 1 1 2 x. Negative 1 1 2 x."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "Now, how do we evaluate this? Well, the antiderivative of negative 1 1 2 is pretty straightforward. That's just going to be negative 1 1 2 x. Negative 1 1 2 x. What's the antiderivative of two over x minus one? And you might be able to do this in your head. The derivative of x minus one is just one."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "Negative 1 1 2 x. What's the antiderivative of two over x minus one? And you might be able to do this in your head. The derivative of x minus one is just one. So you could say that the derivative is sitting there. And so we can essentially do u substitution in our heads and say, okay, let's just take the antiderivative, I guess you could say with respect to x minus one, which would be the natural log of the absolute value of x minus one. If all of that sounds really confusing, I'll actually do the u substitution."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "The derivative of x minus one is just one. So you could say that the derivative is sitting there. And so we can essentially do u substitution in our heads and say, okay, let's just take the antiderivative, I guess you could say with respect to x minus one, which would be the natural log of the absolute value of x minus one. If all of that sounds really confusing, I'll actually do the u substitution. So if I were just trying to evaluate, if I were just trying to evaluate the integral two over x minus one dx, I could see, okay, the derivative of x minus one is just one. So I could say u is equal to x minus one. And then du is going to be equal to dx."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "If all of that sounds really confusing, I'll actually do the u substitution. So if I were just trying to evaluate, if I were just trying to evaluate the integral two over x minus one dx, I could see, okay, the derivative of x minus one is just one. So I could say u is equal to x minus one. And then du is going to be equal to dx. And so this is going to be, we can rewrite in terms of u as two, I'll just take the constant out, two times the integral of one over u du, which we know is two times the natural log of the absolute value of u plus c. And in this case, we know that u is x minus one. So this is equal to two times the natural log of x minus one plus c. And so that's what we're gonna have right over here. So plus two times the natural log of the absolute value of x minus one plus c. And the plus c doesn't just come from this one."}, {"video_title": "Dividing expressions to evaluate integral   AP Calculus BC   Khan Academy.mp3", "Sentence": "And then du is going to be equal to dx. And so this is going to be, we can rewrite in terms of u as two, I'll just take the constant out, two times the integral of one over u du, which we know is two times the natural log of the absolute value of u plus c. And in this case, we know that u is x minus one. So this is equal to two times the natural log of x minus one plus c. And so that's what we're gonna have right over here. So plus two times the natural log of the absolute value of x minus one plus c. And the plus c doesn't just come from this one. This, in general, when we're taking the integral of the whole thing, there could be some constant, because obviously if we go the other way, we take the derivative, the constant will go away. So let me just put the plus c right over there. And we are done."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me graph a potential s of t right over here. We have a horizontal axis as the time axis. And let me just graph something. I'll draw it kind of parabola looking. Although I could have done it general, but just to make things a little bit simpler for me. So I'll draw it kind of parabola looking. So that is, if we call this a y-axis, we could even call this y equals s of t as a reasonable way to graph our position as a function of time function."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll draw it kind of parabola looking. Although I could have done it general, but just to make things a little bit simpler for me. So I'll draw it kind of parabola looking. So that is, if we call this a y-axis, we could even call this y equals s of t as a reasonable way to graph our position as a function of time function. And now let's think about what happens if we want to think about the change in position between two times. Let's say between time a. Let's say that's time a right over there."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is, if we call this a y-axis, we could even call this y equals s of t as a reasonable way to graph our position as a function of time function. And now let's think about what happens if we want to think about the change in position between two times. Let's say between time a. Let's say that's time a right over there. And then this right over here is time b. So time b is right over here. So what would be the change in position between time a and between time b?"}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that's time a right over there. And then this right over here is time b. So time b is right over here. So what would be the change in position between time a and between time b? Well, at time b, we are at s of b. We are at s of b position. And at time a, we were at s of a position."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what would be the change in position between time a and between time b? Well, at time b, we are at s of b. We are at s of b position. And at time a, we were at s of a position. So the change in position between time a and time b, let me write this down, the change in position between, and this might be obvious to you, but I'll write it down, between times a and b is going to be equal to s of b, this position, s of b minus this position, minus s of a. So nothing earth shattering so far. But now let's think about what happens if we take the derivative of this function right over here."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And at time a, we were at s of a position. So the change in position between time a and time b, let me write this down, the change in position between, and this might be obvious to you, but I'll write it down, between times a and b is going to be equal to s of b, this position, s of b minus this position, minus s of a. So nothing earth shattering so far. But now let's think about what happens if we take the derivative of this function right over here. So what happens when we take the derivative of a position as a function of time? So remember, the derivative gives us the slope of the tangent line at any point. So let's say we're looking at a point right over there."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now let's think about what happens if we take the derivative of this function right over here. So what happens when we take the derivative of a position as a function of time? So remember, the derivative gives us the slope of the tangent line at any point. So let's say we're looking at a point right over there. The slope of the tangent line, it tells us for a very small change in t, I'm exaggerating it visually, for a very, very small change in t, how much are we changing in position? How much are we changing in position? So we write that as ds dt is the derivative of our position function at any given time."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say we're looking at a point right over there. The slope of the tangent line, it tells us for a very small change in t, I'm exaggerating it visually, for a very, very small change in t, how much are we changing in position? How much are we changing in position? So we write that as ds dt is the derivative of our position function at any given time. So when we're talking about how the rate at which position changes with respect to time, what is that? Well, that is equal to velocity. So this is equal to velocity."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we write that as ds dt is the derivative of our position function at any given time. So when we're talking about how the rate at which position changes with respect to time, what is that? Well, that is equal to velocity. So this is equal to velocity. But let me write this in different notations. So this itself is going to be a function of time. So we could write this, this is equal to s prime of t. These are just two different ways of writing the derivative of s with respect to t. This makes it a little bit clearer that this itself is a function of time."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to velocity. But let me write this in different notations. So this itself is going to be a function of time. So we could write this, this is equal to s prime of t. These are just two different ways of writing the derivative of s with respect to t. This makes it a little bit clearer that this itself is a function of time. And we know that this is the exact same thing as velocity as function of time, which we will write as v of t. So let's graph what v of t might look like down here. Let's graph it. So let me put another axis down here that looks pretty close to the original."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could write this, this is equal to s prime of t. These are just two different ways of writing the derivative of s with respect to t. This makes it a little bit clearer that this itself is a function of time. And we know that this is the exact same thing as velocity as function of time, which we will write as v of t. So let's graph what v of t might look like down here. Let's graph it. So let me put another axis down here that looks pretty close to the original. Give myself some real estate. So that looks pretty good. And then let me try to graph v of t. So once again, if this is my y-axis, this is my t-axis."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me put another axis down here that looks pretty close to the original. Give myself some real estate. So that looks pretty good. And then let me try to graph v of t. So once again, if this is my y-axis, this is my t-axis. And I'm going to graph y is equal to v of t. And if this really is a parabola, then the slope over here is 0. The slope, the rate of change is 0. And then it keeps increasing."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let me try to graph v of t. So once again, if this is my y-axis, this is my t-axis. And I'm going to graph y is equal to v of t. And if this really is a parabola, then the slope over here is 0. The slope, the rate of change is 0. And then it keeps increasing. The slope gets steeper and steeper and steeper. And so v of t might look something like this. So this is the graph of y is equal to v of t. Now, using this graph, let's think if we can conceptualize the distance or the change in position between time a and between time b."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it keeps increasing. The slope gets steeper and steeper and steeper. And so v of t might look something like this. So this is the graph of y is equal to v of t. Now, using this graph, let's think if we can conceptualize the distance or the change in position between time a and between time b. Well, let's go back to our Riemann sums. Let's think about what an area of a very small rectangle would represent. So let's divide this into a bunch of rectangles."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the graph of y is equal to v of t. Now, using this graph, let's think if we can conceptualize the distance or the change in position between time a and between time b. Well, let's go back to our Riemann sums. Let's think about what an area of a very small rectangle would represent. So let's divide this into a bunch of rectangles. So I'll do fairly large rectangles so we have some space to work with. You can imagine much smaller ones. And I'm going to do a left Riemann sum here, just because we've done those a bunch."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's divide this into a bunch of rectangles. So I'll do fairly large rectangles so we have some space to work with. You can imagine much smaller ones. And I'm going to do a left Riemann sum here, just because we've done those a bunch. But we could do a right Riemann sum. We could do a trapezoidal sum. We could do anything we want."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm going to do a left Riemann sum here, just because we've done those a bunch. But we could do a right Riemann sum. We could do a trapezoidal sum. We could do anything we want. So then we could keep going all the way. Actually, let me just do 3 right now. Let me just do 3 right over here."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could do anything we want. So then we could keep going all the way. Actually, let me just do 3 right now. Let me just do 3 right over here. And so this is actually a very rough approximation. But you can imagine it might get closer. But what is the area of each of these rectangles trying?"}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just do 3 right over here. And so this is actually a very rough approximation. But you can imagine it might get closer. But what is the area of each of these rectangles trying? What is it an approximation for? Well, this one right over here, you have f of a, or I should say v of a. So your velocity at time a is the height right over here."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what is the area of each of these rectangles trying? What is it an approximation for? Well, this one right over here, you have f of a, or I should say v of a. So your velocity at time a is the height right over here. And then this distance right over here is a change in time, times delta t. So the area for that rectangle is your velocity at that moment times your change in time. What is the velocity at that moment times your change in time? Well, that's going to be your change in position."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So your velocity at time a is the height right over here. And then this distance right over here is a change in time, times delta t. So the area for that rectangle is your velocity at that moment times your change in time. What is the velocity at that moment times your change in time? Well, that's going to be your change in position. So this will tell you this is an approximation of your change in position over this time. Then this rectangle, the area of this rectangle, is another approximation for your change in position over the next delta t. And then you can imagine this right over here is an approximation for your change in position for the next delta t. So if you really wanted to figure out your change in position between a and b, you might want to just do a Riemann sum if you wanted to approximate it. You would want to take the sum from i equals 1 to i equals n of v of, and I'll do a left Riemann sum."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's going to be your change in position. So this will tell you this is an approximation of your change in position over this time. Then this rectangle, the area of this rectangle, is another approximation for your change in position over the next delta t. And then you can imagine this right over here is an approximation for your change in position for the next delta t. So if you really wanted to figure out your change in position between a and b, you might want to just do a Riemann sum if you wanted to approximate it. You would want to take the sum from i equals 1 to i equals n of v of, and I'll do a left Riemann sum. But once again, we could use a midpoint. We could do trapezoids. We could do the right Riemann sum."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "You would want to take the sum from i equals 1 to i equals n of v of, and I'll do a left Riemann sum. But once again, we could use a midpoint. We could do trapezoids. We could do the right Riemann sum. But I'll just do a left one because that's what I depicted right here. v of t of i minus 1. So if this would be t0, it would be a."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could do the right Riemann sum. But I'll just do a left one because that's what I depicted right here. v of t of i minus 1. So if this would be t0, it would be a. So this is the first rectangle. So the first rectangle, you use the function evaluated at t0. For the second rectangle, you use the function evaluated at t1."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if this would be t0, it would be a. So this is the first rectangle. So the first rectangle, you use the function evaluated at t0. For the second rectangle, you use the function evaluated at t1. We've done this in multiple videos already. And then we multiply it times each of the changes in time. This will be an approximation for our total."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "For the second rectangle, you use the function evaluated at t1. We've done this in multiple videos already. And then we multiply it times each of the changes in time. This will be an approximation for our total. And let me make it clear. Where delta t is equal to b minus a over the number of intervals we have. We already know from many, many videos when we looked at Riemann sums that this will be an approximation."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "This will be an approximation for our total. And let me make it clear. Where delta t is equal to b minus a over the number of intervals we have. We already know from many, many videos when we looked at Riemann sums that this will be an approximation. Well, it will be an approximation for two things. We just talked about it will be an approximation for our change in position. But it's also an approximation for our area."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already know from many, many videos when we looked at Riemann sums that this will be an approximation. Well, it will be an approximation for two things. We just talked about it will be an approximation for our change in position. But it's also an approximation for our area. So this right over here. So we're trying to approximate change in position. And this is also approximate of the area under the curve."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it's also an approximation for our area. So this right over here. So we're trying to approximate change in position. And this is also approximate of the area under the curve. So hopefully this satisfies you that if you are able to calculate the area under the curve. And actually, this one's pretty easy because it's a trapezoid. But even if this was a function, if it was kind of a wacky function, it would still apply."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is also approximate of the area under the curve. So hopefully this satisfies you that if you are able to calculate the area under the curve. And actually, this one's pretty easy because it's a trapezoid. But even if this was a function, if it was kind of a wacky function, it would still apply. That when you're calculating the area under the curve of the velocity function, you are actually figuring out the change in position. These are the two things. Well, we already know what could we do to get the exact area under the curve or to get the exact change in position."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But even if this was a function, if it was kind of a wacky function, it would still apply. That when you're calculating the area under the curve of the velocity function, you are actually figuring out the change in position. These are the two things. Well, we already know what could we do to get the exact area under the curve or to get the exact change in position. Well, we just have a ton of rectangles. We take the limit as the number of rectangles we have approaches infinity. We take the limit as n approaches infinity."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we already know what could we do to get the exact area under the curve or to get the exact change in position. Well, we just have a ton of rectangles. We take the limit as the number of rectangles we have approaches infinity. We take the limit as n approaches infinity. And as n approaches infinity, because delta t is b minus a divided by n, delta t is going to become infinitely small. It's going to turn into dt. This is one way to think about it."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We take the limit as n approaches infinity. And as n approaches infinity, because delta t is b minus a divided by n, delta t is going to become infinitely small. It's going to turn into dt. This is one way to think about it. And we already have notation for this. This is one way to think about a Riemann integral. We just use the left Riemann sum."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is one way to think about it. And we already have notation for this. This is one way to think about a Riemann integral. We just use the left Riemann sum. Once again, we could use the right Riemann sum, et cetera, et cetera. We could have used a more general Riemann sum, but this one will work. So this will be equal to the definite integral from a to b of v of t dt."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We just use the left Riemann sum. Once again, we could use the right Riemann sum, et cetera, et cetera. We could have used a more general Riemann sum, but this one will work. So this will be equal to the definite integral from a to b of v of t dt. So this right over here is one way of saying, look, if we want the exact area under the curve of the velocity curve, which is going to be the exact change in position between a and b, we can denote it this way. It's the limit of this Riemann sum as n approaches infinity or the definite integral from a to b of v of t dt. But what did we just figure out?"}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this will be equal to the definite integral from a to b of v of t dt. So this right over here is one way of saying, look, if we want the exact area under the curve of the velocity curve, which is going to be the exact change in position between a and b, we can denote it this way. It's the limit of this Riemann sum as n approaches infinity or the definite integral from a to b of v of t dt. But what did we just figure out? So remember, this is another. We could call this the exact change in position between times a and b. But we already figured out what the exact change in positions between times a and b are."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what did we just figure out? So remember, this is another. We could call this the exact change in position between times a and b. But we already figured out what the exact change in positions between times a and b are. It's this thing right over here. And so this gets interesting. We now have a way of evaluating this definite integral."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we already figured out what the exact change in positions between times a and b are. It's this thing right over here. And so this gets interesting. We now have a way of evaluating this definite integral. Conceptually, we knew that this is the exact change in position between a and b, but we already figured out a way to figure out the exact change in position between a and b. So let me write all this down. We have that the definite integral between a and b of v of t dt is equal to s of b minus s of a, where s of t is the anti-derivative of v of t. And this notion, although I've written in a very nontraditional used position velocity, this is the second fundamental theorem of calculus."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We now have a way of evaluating this definite integral. Conceptually, we knew that this is the exact change in position between a and b, but we already figured out a way to figure out the exact change in position between a and b. So let me write all this down. We have that the definite integral between a and b of v of t dt is equal to s of b minus s of a, where s of t is the anti-derivative of v of t. And this notion, although I've written in a very nontraditional used position velocity, this is the second fundamental theorem of calculus. And you're probably wondering about the first. We'll talk about that in another video. But this is a super useful way of evaluating definite integrals and finding the area under a curve."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have that the definite integral between a and b of v of t dt is equal to s of b minus s of a, where s of t is the anti-derivative of v of t. And this notion, although I've written in a very nontraditional used position velocity, this is the second fundamental theorem of calculus. And you're probably wondering about the first. We'll talk about that in another video. But this is a super useful way of evaluating definite integrals and finding the area under a curve. Second fundamental theorem of calculus, very closely tied to the first fundamental theorem, which we won't talk about now. So why is this such a big deal? Well, let me write it in a more general notation, the way that you might be used to seeing it in your calculus book."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this is a super useful way of evaluating definite integrals and finding the area under a curve. Second fundamental theorem of calculus, very closely tied to the first fundamental theorem, which we won't talk about now. So why is this such a big deal? Well, let me write it in a more general notation, the way that you might be used to seeing it in your calculus book. It's telling us that if we want the area under the curve between two points, a and b, between two x points, a and b, of f of x. And so this is how we would denote the area under the curve between those two intervals. So let me draw that just to make it clear what I'm talking about in general terms."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let me write it in a more general notation, the way that you might be used to seeing it in your calculus book. It's telling us that if we want the area under the curve between two points, a and b, between two x points, a and b, of f of x. And so this is how we would denote the area under the curve between those two intervals. So let me draw that just to make it clear what I'm talking about in general terms. So this right over here could be f of x. And we care about the area under the curve between a and b. If we want to find the exact area under the curve, we can figure it out by taking the antiderivative of f. And let's just say that capital F of x is the antiderivative, or is an antiderivative, because you can have multiple that are shifted by constants, is an antiderivative of f. Then you just have to evaluate the antiderivative at the end points and take the difference."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw that just to make it clear what I'm talking about in general terms. So this right over here could be f of x. And we care about the area under the curve between a and b. If we want to find the exact area under the curve, we can figure it out by taking the antiderivative of f. And let's just say that capital F of x is the antiderivative, or is an antiderivative, because you can have multiple that are shifted by constants, is an antiderivative of f. Then you just have to evaluate the antiderivative at the end points and take the difference. So you take the end point first. You subtract the antiderivative evaluated at the starting point from the antiderivative evaluated at the end point. So you get capital F of b minus capital F of a."}, {"video_title": "Intuition for second part of fundamental theorem of calculus   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we want to find the exact area under the curve, we can figure it out by taking the antiderivative of f. And let's just say that capital F of x is the antiderivative, or is an antiderivative, because you can have multiple that are shifted by constants, is an antiderivative of f. Then you just have to evaluate the antiderivative at the end points and take the difference. So you take the end point first. You subtract the antiderivative evaluated at the starting point from the antiderivative evaluated at the end point. So you get capital F of b minus capital F of a. So if you want to figure out the exact area under the curve, you take the antiderivative of it and evaluate that at the end point. And from that, you subtract the starting point. So hopefully that makes sense."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Using correct units, explain the meaning of the expression in the context of this problem. So we know that if we just want to evaluate this definite integral from zero to 10 of h prime of t dt, this is just the same thing as evaluating the antiderivative of this thing right over here, which is just h of t at 10, and subtract from that the antiderivative of this evaluated at zero. This is the second fundamental theorem of calculus. This is exactly how we evaluate definite integrals. And when you look at it from this, you just see that it's, when you evaluate it, it just gives us the difference in temperature from zero minutes to 10 minutes. We're taking the temperature at 10 minutes, and from that we're subtracting from that the temperature at zero minutes. So this is really our change in temperature over the first, over those 10 minutes."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is exactly how we evaluate definite integrals. And when you look at it from this, you just see that it's, when you evaluate it, it just gives us the difference in temperature from zero minutes to 10 minutes. We're taking the temperature at 10 minutes, and from that we're subtracting from that the temperature at zero minutes. So this is really our change in temperature over the first, over those 10 minutes. And we can actually evaluate it. We know what our temperature was after 10 minutes. H of 10 is 43 degrees Celsius."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is really our change in temperature over the first, over those 10 minutes. And we can actually evaluate it. We know what our temperature was after 10 minutes. H of 10 is 43 degrees Celsius. So this is 43 right over here. And from that, we're going to subtract our initial temperature, our temperature at zero minutes, which is 66 degrees Celsius. We're gonna subtract 66 degrees Celsius."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "H of 10 is 43 degrees Celsius. So this is 43 right over here. And from that, we're going to subtract our initial temperature, our temperature at zero minutes, which is 66 degrees Celsius. We're gonna subtract 66 degrees Celsius. This gives us negative 23 degrees Celsius. So our change in temperature is negative 23 degrees, or our temperature has gone down 23 degrees Celsius over the course of the first 10 minutes. So that is part, that is part C right over there."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna subtract 66 degrees Celsius. This gives us negative 23 degrees Celsius. So our change in temperature is negative 23 degrees, or our temperature has gone down 23 degrees Celsius over the course of the first 10 minutes. So that is part, that is part C right over there. Now let's do part D, part D, D. At time t equals zero, biscuits with temperature 100 degrees were removed from an oven. So now we're talking about biscuits. We started with t, now biscuits."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is part, that is part C right over there. Now let's do part D, part D, D. At time t equals zero, biscuits with temperature 100 degrees were removed from an oven. So now we're talking about biscuits. We started with t, now biscuits. The temperature of the biscuits at time t is modeled by a differentiable function B for which it is known that B prime of t is equal to this business right over here. Using the given models at time t equals 10, how much cooler are the biscuits than the t? Well, we know what the temperature of the t is, so we just have to figure out the temperature of the biscuits to figure out how much cooler they are than the t. And to figure out the temperature of the biscuits, we can essentially just use the exact same idea."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We started with t, now biscuits. The temperature of the biscuits at time t is modeled by a differentiable function B for which it is known that B prime of t is equal to this business right over here. Using the given models at time t equals 10, how much cooler are the biscuits than the t? Well, we know what the temperature of the t is, so we just have to figure out the temperature of the biscuits to figure out how much cooler they are than the t. And to figure out the temperature of the biscuits, we can essentially just use the exact same idea. We can say how much did the, we know that the biscuits started off at 100 degrees Celsius, and we can say, well, how much did they cool down, or what was their change in temperature over the 10 minutes? If we know the change in temperature and we know it started at 100 degrees, then we can use that information to get what its temperature is at 10 minutes, and then we can answer their question. So the change in temperature over those 10 minutes is just the definite integral from zero to 10 of this business right over here."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we know what the temperature of the t is, so we just have to figure out the temperature of the biscuits to figure out how much cooler they are than the t. And to figure out the temperature of the biscuits, we can essentially just use the exact same idea. We can say how much did the, we know that the biscuits started off at 100 degrees Celsius, and we can say, well, how much did they cool down, or what was their change in temperature over the 10 minutes? If we know the change in temperature and we know it started at 100 degrees, then we can use that information to get what its temperature is at 10 minutes, and then we can answer their question. So the change in temperature over those 10 minutes is just the definite integral from zero to 10 of this business right over here. Of, let me just write it so you see it's the exact same pattern as what we saw over here. This was for the cookies, now we're talking about the biscuits. And this right over here is going to be the definite integral from zero to 10."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the change in temperature over those 10 minutes is just the definite integral from zero to 10 of this business right over here. Of, let me just write it so you see it's the exact same pattern as what we saw over here. This was for the cookies, now we're talking about the biscuits. And this right over here is going to be the definite integral from zero to 10. B prime of t, they give it to us right over here, is negative 13.84 e to the negative 0.173 t dt. And now we just evaluate this. So we can, let's take the, well, we can evaluate this."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this right over here is going to be the definite integral from zero to 10. B prime of t, they give it to us right over here, is negative 13.84 e to the negative 0.173 t dt. And now we just evaluate this. So we can, let's take the, well, we can evaluate this. What we have to do is, because we know that the derivative, we know the derivative of e to the ax is a e to the x. Let me write that. The derivative of e to the ax is equal to a e to the ax, just from the chain rule, derivative of the inside is just a, and then we multiply that times the derivative of the entire thing, and derivative of e to the x is just e to the x."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can, let's take the, well, we can evaluate this. What we have to do is, because we know that the derivative, we know the derivative of e to the ax is a e to the x. Let me write that. The derivative of e to the ax is equal to a e to the ax, just from the chain rule, derivative of the inside is just a, and then we multiply that times the derivative of the entire thing, and derivative of e to the x is just e to the x. Or we could say that the integral of e to the ax dx is equal to one over a e to the ax plus c. And you could take the derivative of this to see that you would get this right over here. So using that same idea, the antiderivative of this over here is just going to be the negative 13.84, and we're going to divide by this coefficient on the t right over here. So negative 0.173, and you could use your calculator."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of e to the ax is equal to a e to the ax, just from the chain rule, derivative of the inside is just a, and then we multiply that times the derivative of the entire thing, and derivative of e to the x is just e to the x. Or we could say that the integral of e to the ax dx is equal to one over a e to the ax plus c. And you could take the derivative of this to see that you would get this right over here. So using that same idea, the antiderivative of this over here is just going to be the negative 13.84, and we're going to divide by this coefficient on the t right over here. So negative 0.173, and you could use your calculator. Calculators are allowed for this part of the problem, but we could do this analytically. Times e to the negative 0.173 times t, and we're going to evaluate that. We're going to evaluate that from zero to 10."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative 0.173, and you could use your calculator. Calculators are allowed for this part of the problem, but we could do this analytically. Times e to the negative 0.173 times t, and we're going to evaluate that. We're going to evaluate that from zero to 10. So we're going to evaluate it at 10 and subtract from that this thing evaluated at zero. So this is going to be, let me just factor out this part right over here. So it's going to be negative 13.84 over negative 0.173 times this evaluated at when t is 10."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to evaluate that from zero to 10. So we're going to evaluate it at 10 and subtract from that this thing evaluated at zero. So this is going to be, let me just factor out this part right over here. So it's going to be negative 13.84 over negative 0.173 times this evaluated at when t is 10. So e to the, if we multiply this times 10, this is negative 1.73. That's when I evaluated at 10. And from that, we want to subtract when this is evaluated at zero."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be negative 13.84 over negative 0.173 times this evaluated at when t is 10. So e to the, if we multiply this times 10, this is negative 1.73. That's when I evaluated at 10. And from that, we want to subtract when this is evaluated at zero. If the exponent here is zero, if t is zero, the whole exponent is zero. So e to the zero power is just equal to one. And now we can get our calculator out to evaluate this."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And from that, we want to subtract when this is evaluated at zero. If the exponent here is zero, if t is zero, the whole exponent is zero. So e to the zero power is just equal to one. And now we can get our calculator out to evaluate this. So get my TI-85 out, and I get, I'm just going to evaluate this inner expression right over here. So e to the negative 1.73, and from that, I want to subtract one. So that gives me negative 0.822."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we can get our calculator out to evaluate this. So get my TI-85 out, and I get, I'm just going to evaluate this inner expression right over here. So e to the negative 1.73, and from that, I want to subtract one. So that gives me negative 0.822. So that's this part in parentheses. And then I want to multiply it times what I have out front. So my previous answer times, and I won't write the negative just because the negative divided by negative is going to be a positive."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that gives me negative 0.822. So that's this part in parentheses. And then I want to multiply it times what I have out front. So my previous answer times, and I won't write the negative just because the negative divided by negative is going to be a positive. So it's just 13.84 divided by 0.173. Because I didn't write the negatives because they just cancel out. And that gives me negative 65.817."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So my previous answer times, and I won't write the negative just because the negative divided by negative is going to be a positive. So it's just 13.84 divided by 0.173. Because I didn't write the negatives because they just cancel out. And that gives me negative 65.817. So this is our change in temperature for the biscuits. It's our change in temperature. Let me write it down."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that gives me negative 65.817. So this is our change in temperature for the biscuits. It's our change in temperature. Let me write it down. Negative 65.81 or 82 degrees. So this is negative 65.82 degrees Celsius. This is our change in temperature."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write it down. Negative 65.81 or 82 degrees. So this is negative 65.82 degrees Celsius. This is our change in temperature. This is our change in temp for the biscuits. Now we know that they started off at 100 degrees, and they went down by 65.82 degrees over the 10 minutes. So let's just, they started at 100, and then they went down by this previous amount."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our change in temperature. This is our change in temp for the biscuits. Now we know that they started off at 100 degrees, and they went down by 65.82 degrees over the 10 minutes. So let's just, they started at 100, and then they went down by this previous amount. So, whoops, I want to subtract the answer. So minus the answer. Oh, sorry."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just, they started at 100, and then they went down by this previous amount. So, whoops, I want to subtract the answer. So minus the answer. Oh, sorry. I want to actually add the change in temperature. I want to be very careful here. It went down by 65 degrees."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Oh, sorry. I want to actually add the change in temperature. I want to be very careful here. It went down by 65 degrees. So I really should just say, I really should say 100 minus 65.817. Yeah, I could keep adding digits if I want. 247, that's enough."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It went down by 65 degrees. So I really should just say, I really should say 100 minus 65.817. Yeah, I could keep adding digits if I want. 247, that's enough. So after 10 minutes, the biscuits are 34.18 degrees. And we already know, we already knew what the change in temperature for the, or what the temperature of the tea is after 10 minutes. We already knew it was 43 degrees."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "247, that's enough. So after 10 minutes, the biscuits are 34.18 degrees. And we already know, we already knew what the change in temperature for the, or what the temperature of the tea is after 10 minutes. We already knew it was 43 degrees. So the biscuits are at 34 degrees. The tea is at, sorry, the biscuits are at 34 degrees. The tea is at 43 degrees."}, {"video_title": "2011 Calculus AB free response #2 (c & d)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already knew it was 43 degrees. So the biscuits are at 34 degrees. The tea is at, sorry, the biscuits are at 34 degrees. The tea is at 43 degrees. So if we do 43 is where the tea is, and we subtract the temperature of the biscuits, we see that the biscuits are 8.82 degrees cooler. 8.82 degrees Celsius cooler. So we can say the biscuits, the biscuits are 8.82 degrees Celsius cooler than the tea."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "On the left-hand side, it says, Avery tried to find the derivative of seven minus five x using basic differentiation rules. Here is her work. And on the right-hand side, it says, Hannah tried to find the derivative of negative three plus eight x using basic differentiation rules. Here is her work. And these are two different examples from differentiation rules exercise on Khan Academy. And I thought I would just do them side by side because we can kind of think about what each of these people are doing correct or incorrect. So these are similar expressions."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here is her work. And these are two different examples from differentiation rules exercise on Khan Academy. And I thought I would just do them side by side because we can kind of think about what each of these people are doing correct or incorrect. So these are similar expressions. We have a constant, and then we have a first-degree term, a constant, and then a first-degree term. So they're gonna take the derivative. So let's see, step one for Avery."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So these are similar expressions. We have a constant, and then we have a first-degree term, a constant, and then a first-degree term. So they're gonna take the derivative. So let's see, step one for Avery. She took, she's separately taking the derivative of seven and separately taking the derivative of five x. So this is, my spider sense is already going off here. Because what happened to this negative right over here?"}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, step one for Avery. She took, she's separately taking the derivative of seven and separately taking the derivative of five x. So this is, my spider sense is already going off here. Because what happened to this negative right over here? So it would have made sense for her to do the derivative of seven, and she could have said minus the derivative of five x. That's one possibility that she could have done. The derivative of a difference is equal to the difference of the derivatives."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because what happened to this negative right over here? So it would have made sense for her to do the derivative of seven, and she could have said minus the derivative of five x. That's one possibility that she could have done. The derivative of a difference is equal to the difference of the derivatives. We've seen that property. Or she could have said the derivative, she could have said this is equal to the derivative of seven plus the derivative with respect to x of negative five x. These two things would have been equivalent to this one."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of a difference is equal to the difference of the derivatives. We've seen that property. Or she could have said the derivative, she could have said this is equal to the derivative of seven plus the derivative with respect to x of negative five x. These two things would have been equivalent to this one. But for this one, she somehow forgot to include the negative. So I think she had a problem right at step one. Now if you just follow her logic after step one, let's see if she makes any more mistakes."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "These two things would have been equivalent to this one. But for this one, she somehow forgot to include the negative. So I think she had a problem right at step one. Now if you just follow her logic after step one, let's see if she makes any more mistakes. So she takes the derivative of a constant. So a constant isn't going to change with respect to x. So that makes sense that that derivative is zero."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now if you just follow her logic after step one, let's see if she makes any more mistakes. So she takes the derivative of a constant. So a constant isn't going to change with respect to x. So that makes sense that that derivative is zero. And so we still have the derivative of five x. And remember, it should have been negative five x or minus the derivative of five x. And let's see what she does here."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that makes sense that that derivative is zero. And so we still have the derivative of five x. And remember, it should have been negative five x or minus the derivative of five x. And let's see what she does here. So that zero disappears, and now she takes the constant out. And that's true. The derivative of a constant times something is equal to the constant times the derivative of that something."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see what she does here. So that zero disappears, and now she takes the constant out. And that's true. The derivative of a constant times something is equal to the constant times the derivative of that something. And then she finds that the derivative with respect to x of x is one. And that's true. So the slope, if you had the graph of y equals x, the slope there is one."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of a constant times something is equal to the constant times the derivative of that something. And then she finds that the derivative with respect to x of x is one. And that's true. So the slope, if you had the graph of y equals x, the slope there is one. Or what's the rate of change at which x changes with respect to x? Well, it's going to be one for one. So the slope here is one."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope, if you had the graph of y equals x, the slope there is one. Or what's the rate of change at which x changes with respect to x? Well, it's going to be one for one. So the slope here is one. So this is going to be five times one, which is equal to five. And at the end, they just say, what step did Avery make a mistake? So she clearly made a mistake at step one."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope here is one. So this is going to be five times one, which is equal to five. And at the end, they just say, what step did Avery make a mistake? So she clearly made a mistake at step one. This right over here should have been a negative. If that's a negative, then that would have been a negative. Then this would have been a negative."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So she clearly made a mistake at step one. This right over here should have been a negative. If that's a negative, then that would have been a negative. Then this would have been a negative. Then that would have been a negative. And then her final answer should have been, should have been a negative five. Now let's go back to Hannah to see if she made any mistakes and where."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then this would have been a negative. Then that would have been a negative. And then her final answer should have been, should have been a negative five. Now let's go back to Hannah to see if she made any mistakes and where. So she's differentiating a similar expression. So first she takes the derivative of the constant plus the derivative of the first degree term. Derivative of a constant is zero."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's go back to Hannah to see if she made any mistakes and where. So she's differentiating a similar expression. So first she takes the derivative of the constant plus the derivative of the first degree term. Derivative of a constant is zero. That looks good. So you get the zero. And then you have the derivative of the first degree term."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative of a constant is zero. That looks good. So you get the zero. And then you have the derivative of the first degree term. That's what she's trying to figure out. And then, let's see. She's taking, let's see."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you have the derivative of the first degree term. That's what she's trying to figure out. And then, let's see. She's taking, let's see. So this seems off. She is assuming that the derivative of a product is equal to the product of the derivatives. That is not the case."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "She's taking, let's see. So this seems off. She is assuming that the derivative of a product is equal to the product of the derivatives. That is not the case. And especially, and it's, if you have a constant here, there's actually a much simpler way of thinking about it, frankly, the way that Avery thought about it. Avery had made a mistake at step one. But this is actually going to be equal to the derivative of a constant times an expression is equal to the same thing as the constant times the derivative of the expression."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is not the case. And especially, and it's, if you have a constant here, there's actually a much simpler way of thinking about it, frankly, the way that Avery thought about it. Avery had made a mistake at step one. But this is actually going to be equal to the derivative of a constant times an expression is equal to the same thing as the constant times the derivative of the expression. So this would have been the correct way to go. And then the derivative of x with respect to x, well, that's just going to be one. So this should have all simplified to eight."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this is actually going to be equal to the derivative of a constant times an expression is equal to the same thing as the constant times the derivative of the expression. So this would have been the correct way to go. And then the derivative of x with respect to x, well, that's just going to be one. So this should have all simplified to eight. What she did is, she is assumed, she tried to take the derivative of eight and multiply that times the derivative of x. That is not the way it works. In the future, you will learn something called the product rule."}, {"video_title": "Basic derivative rules find the error   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this should have all simplified to eight. What she did is, she is assumed, she tried to take the derivative of eight and multiply that times the derivative of x. That is not the way it works. In the future, you will learn something called the product rule. But you won't even have to apply that here because one of these components, I guess you could say, is a constant. So this is the wrong step. This is where Hannah makes a mistake."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if we can take the derivative with respect to x of the fourth root of x to the third power plus four x squared plus seven. And at first you might say, alright, how do I take the derivative of a fourth root of something? It looks like I have a composite function. I'm taking the fourth root of another expression here, and you'd be right. And if you're dealing with a composite function, the chain rule should be front of mind. But first, let's just make this fourth root a little bit more tractable for us, and just realize that this fourth root is really nothing but a fractional exponent. So this is the same thing as the derivative with respect to x of x to the third plus four x squared plus seven to the 1 4th power."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm taking the fourth root of another expression here, and you'd be right. And if you're dealing with a composite function, the chain rule should be front of mind. But first, let's just make this fourth root a little bit more tractable for us, and just realize that this fourth root is really nothing but a fractional exponent. So this is the same thing as the derivative with respect to x of x to the third plus four x squared plus seven to the 1 4th power. To the 1 4th power. Now, how do we take the derivative of this? Well, we can view this, as I said a few seconds ago, we can view this as a composite function."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the same thing as the derivative with respect to x of x to the third plus four x squared plus seven to the 1 4th power. To the 1 4th power. Now, how do we take the derivative of this? Well, we can view this, as I said a few seconds ago, we can view this as a composite function. What do we do first with our x? Well, we do all of this business. And we could call this u of x."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can view this, as I said a few seconds ago, we can view this as a composite function. What do we do first with our x? Well, we do all of this business. And we could call this u of x. And then, whatever we get for u of x, we raise that to the 4th power. So the way that we would take the derivative, we would take the derivative of this, you could view it as the outer function with respect to u of x, and then multiply that times the derivative of u with respect to x. So let's do that."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could call this u of x. And then, whatever we get for u of x, we raise that to the 4th power. So the way that we would take the derivative, we would take the derivative of this, you could view it as the outer function with respect to u of x, and then multiply that times the derivative of u with respect to x. So let's do that. So what this is going to be, this is going to be equal to, so we're gonna take our outside function, which I'm highlighting in green now, so where I take something to the 1 4th, I'm gonna take the derivative of that with respect to the inside, with respect to u of x. Well, I'm just gonna use the power rule here. I'm just gonna bring that 1 4th out front."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So what this is going to be, this is going to be equal to, so we're gonna take our outside function, which I'm highlighting in green now, so where I take something to the 1 4th, I'm gonna take the derivative of that with respect to the inside, with respect to u of x. Well, I'm just gonna use the power rule here. I'm just gonna bring that 1 4th out front. So it's going to be 1 4th times whatever I'm taking the derivative with respect to, to the 1 4th minus one power. Look, all I did is use the power rule here. I didn't have an x here."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm just gonna bring that 1 4th out front. So it's going to be 1 4th times whatever I'm taking the derivative with respect to, to the 1 4th minus one power. Look, all I did is use the power rule here. I didn't have an x here. Now I'm taking the derivative with respect to u of x, with respect to this polynomial expression here. So I could just throw the u of x in here if I like. Actually, let me just do that."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I didn't have an x here. Now I'm taking the derivative with respect to u of x, with respect to this polynomial expression here. So I could just throw the u of x in here if I like. Actually, let me just do that. So this is going to be x to the 3rd plus four x squared plus seven. And then I wanna multiply that, and this is the chain rule. I took the derivative of the outside with respect to the inside, and I'm gonna multiply that times the derivative of the inside."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me just do that. So this is going to be x to the 3rd plus four x squared plus seven. And then I wanna multiply that, and this is the chain rule. I took the derivative of the outside with respect to the inside, and I'm gonna multiply that times the derivative of the inside. So what's the derivative of u of x? u prime of x. Let's see, we're just gonna use the power rule a bunch of times."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I took the derivative of the outside with respect to the inside, and I'm gonna multiply that times the derivative of the inside. So what's the derivative of u of x? u prime of x. Let's see, we're just gonna use the power rule a bunch of times. It's going to be three x squared plus two times four is eight. X to the two minus one is just one power, first power. So that's just, I can just write that as eight x."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see, we're just gonna use the power rule a bunch of times. It's going to be three x squared plus two times four is eight. X to the two minus one is just one power, first power. So that's just, I can just write that as eight x. And then the derivative with respect to x of seven, well, derivative with respect to x of a constant is just gonna be zero. So that's u prime of x. So then I'm just gonna multiply by u prime of x, which is three x squared plus, three x squared plus eight x."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's just, I can just write that as eight x. And then the derivative with respect to x of seven, well, derivative with respect to x of a constant is just gonna be zero. So that's u prime of x. So then I'm just gonna multiply by u prime of x, which is three x squared plus, three x squared plus eight x. And so, well, I can clean this up a little bit. So this would be equal to, this would be equal to, actually, let me just rewrite that exponent there. So this 1 4th minus one, I can rewrite it, 1 4th minus one is negative 3 4ths."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So then I'm just gonna multiply by u prime of x, which is three x squared plus, three x squared plus eight x. And so, well, I can clean this up a little bit. So this would be equal to, this would be equal to, actually, let me just rewrite that exponent there. So this 1 4th minus one, I can rewrite it, 1 4th minus one is negative 3 4ths. Negative 3 4ths. Negative 3 4ths power. And you could manipulate this in different ways if you like, but the key is to just recognize that this is an application of the chain rule."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this 1 4th minus one, I can rewrite it, 1 4th minus one is negative 3 4ths. Negative 3 4ths. Negative 3 4ths power. And you could manipulate this in different ways if you like, but the key is to just recognize that this is an application of the chain rule. Derivative of the outside, well, actually, the first thing to realize is the fourth root is the same thing as taking something to the 1 4th power, basic exponent property. And then realize that, okay, I have a composite function here. So I can take the derivative of the outside with respect to the inside, that's what we did here, times the derivative of the inside with respect to x."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could manipulate this in different ways if you like, but the key is to just recognize that this is an application of the chain rule. Derivative of the outside, well, actually, the first thing to realize is the fourth root is the same thing as taking something to the 1 4th power, basic exponent property. And then realize that, okay, I have a composite function here. So I can take the derivative of the outside with respect to the inside, that's what we did here, times the derivative of the inside with respect to x. And so if someone were to tell you, if someone were to say, all right, f of x, f of x is equal to the fourth root of x to the third plus four x squared plus seven. And then they said, well, what is f prime of, I don't know, negative three? Well, you would evaluate this at negative three."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I can take the derivative of the outside with respect to the inside, that's what we did here, times the derivative of the inside with respect to x. And so if someone were to tell you, if someone were to say, all right, f of x, f of x is equal to the fourth root of x to the third plus four x squared plus seven. And then they said, well, what is f prime of, I don't know, negative three? Well, you would evaluate this at negative three. So let me just do that. So it's 1 4th times, see, you have negative 27. I hope this works out reasonably well, plus 36, plus 36, plus seven to the negative 3 4ths."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you would evaluate this at negative three. So let me just do that. So it's 1 4th times, see, you have negative 27. I hope this works out reasonably well, plus 36, plus 36, plus seven to the negative 3 4ths. What does this result to? This is going to be equal to, this right over here is 16, right? Negative 27 plus seven is negative 20 plus 36."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I hope this works out reasonably well, plus 36, plus 36, plus seven to the negative 3 4ths. What does this result to? This is going to be equal to, this right over here is 16, right? Negative 27 plus seven is negative 20 plus 36. So this is 16. I think this is gonna work out nicely. And then times, three times negative, so three times nine, which is 27 minus 24."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative 27 plus seven is negative 20 plus 36. So this is 16. I think this is gonna work out nicely. And then times, three times negative, so three times nine, which is 27 minus 24. So this is going to be right over here. That is going to be three. Now what is 16 to the negative 3 4ths?"}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then times, three times negative, so three times nine, which is 27 minus 24. So this is going to be right over here. That is going to be three. Now what is 16 to the negative 3 4ths? So let me, this is 1 4th times, so 16 to the 1 4th is two, and then you raise that to the, let me, actually, I don't wanna skip steps here. But this is, at this point, we are dealing with algebra, or maybe even pre-algebra. So this is going to be times, times 16 to the 1 4th, and then we're gonna raise that to the negative three, times that three out front."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is 16 to the negative 3 4ths? So let me, this is 1 4th times, so 16 to the 1 4th is two, and then you raise that to the, let me, actually, I don't wanna skip steps here. But this is, at this point, we are dealing with algebra, or maybe even pre-algebra. So this is going to be times, times 16 to the 1 4th, and then we're gonna raise that to the negative three, times that three out front. So we could put that three there. 16 to the 1 4th is two. Two to the 3rd is eight."}, {"video_title": "Radical functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be times, times 16 to the 1 4th, and then we're gonna raise that to the negative three, times that three out front. So we could put that three there. 16 to the 1 4th is two. Two to the 3rd is eight. So two to the negative 3rd power is 1 8th. So that is 1 8th. So we have 3 4ths times 1 8th, which is equal to three over 32."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let f be a function such that f of negative one is equal to three, f prime of negative one is equal to five. Let g be the function g of x is equal to two x to the third power. Let capital F be a function defined as, so capital F is defined as lowercase f of x divided by lowercase g of x. And they want us to evaluate the derivative of capital F at x equals negative one. So the way that we can do that is let's just take the derivative of capital F and then evaluate it as at x equals one. And the way they've set up capital F, this function definition, we can see that it is the quotient of two functions. So if we want to take its derivative, you might say, well maybe the quotient rule is important here."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they want us to evaluate the derivative of capital F at x equals negative one. So the way that we can do that is let's just take the derivative of capital F and then evaluate it as at x equals one. And the way they've set up capital F, this function definition, we can see that it is the quotient of two functions. So if we want to take its derivative, you might say, well maybe the quotient rule is important here. And I'll always give you my aside. The quotient rule, I'm gonna state it right now, and it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. But let me just state the quotient rule right now."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we want to take its derivative, you might say, well maybe the quotient rule is important here. And I'll always give you my aside. The quotient rule, I'm gonna state it right now, and it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. But let me just state the quotient rule right now. So if you have some function defined as some function in a numerator divided by some function in the denominator, we can say its derivative, and this is really just a restatement of the quotient rule, its derivative is going to be the derivative of the function in the numerator, so d dx f of x times the function in the denominator, so times g of x minus, minus the function in the numerator, minus f of x, not taking its derivative, times the derivative of the function in the denominator, d dx g of x, all of that over, so all of this is going to be over the function in the denominator squared. So this g of x squared, g of x, g of x squared. And you could use different types of notation here, you could say, instead of writing this with the derivative operator, you could say this is the same thing as g prime of x, and likewise, you could say, well, that is the same thing as f prime of x."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let me just state the quotient rule right now. So if you have some function defined as some function in a numerator divided by some function in the denominator, we can say its derivative, and this is really just a restatement of the quotient rule, its derivative is going to be the derivative of the function in the numerator, so d dx f of x times the function in the denominator, so times g of x minus, minus the function in the numerator, minus f of x, not taking its derivative, times the derivative of the function in the denominator, d dx g of x, all of that over, so all of this is going to be over the function in the denominator squared. So this g of x squared, g of x, g of x squared. And you could use different types of notation here, you could say, instead of writing this with the derivative operator, you could say this is the same thing as g prime of x, and likewise, you could say, well, that is the same thing as f prime of x. And so now we just want to evaluate this thing, and you might say, well, how do I evaluate this thing? Well, let's just try it. Let's just say, well, we want to evaluate f prime when x is equal to negative one."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could use different types of notation here, you could say, instead of writing this with the derivative operator, you could say this is the same thing as g prime of x, and likewise, you could say, well, that is the same thing as f prime of x. And so now we just want to evaluate this thing, and you might say, well, how do I evaluate this thing? Well, let's just try it. Let's just say, well, we want to evaluate f prime when x is equal to negative one. So we can write f prime of negative one is equal to, well, everywhere we see an x, let's put a negative one here. It's going to be f prime of negative one, f prime, lowercase f prime, it's a little confusing, lowercase f prime of negative one times g of negative one, g of negative one, minus f of negative one, f of negative one, times g prime of negative one, g prime of negative one, all of that over, let me do it in that same color, so I take my colors seriously. All right, all of that over, g of negative one squared, g of negative one squared."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just say, well, we want to evaluate f prime when x is equal to negative one. So we can write f prime of negative one is equal to, well, everywhere we see an x, let's put a negative one here. It's going to be f prime of negative one, f prime, lowercase f prime, it's a little confusing, lowercase f prime of negative one times g of negative one, g of negative one, minus f of negative one, f of negative one, times g prime of negative one, g prime of negative one, all of that over, let me do it in that same color, so I take my colors seriously. All right, all of that over, g of negative one squared, g of negative one squared. Now, can we figure out what f prime of negative one, f of negative one, g of negative one, and g prime of negative one, what they are? Well, some of them they tell us outright. They tell us f and f prime at negative one, and for g, we can actually solve for those."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, all of that over, g of negative one squared, g of negative one squared. Now, can we figure out what f prime of negative one, f of negative one, g of negative one, and g prime of negative one, what they are? Well, some of them they tell us outright. They tell us f and f prime at negative one, and for g, we can actually solve for those. So let's see, if this is, well, let's just first evaluate g of negative one. g of negative one is going to be two times negative one to the third power. Well, negative one to the third power is just negative one times two, so this is negative two, and g prime of x, I'll do it here, g prime of x, which use the power rule, bring that three out front, three times two is six, x, decrement that exponent, three minus one is two, and so g prime of negative one is equal to six times negative one squared."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "They tell us f and f prime at negative one, and for g, we can actually solve for those. So let's see, if this is, well, let's just first evaluate g of negative one. g of negative one is going to be two times negative one to the third power. Well, negative one to the third power is just negative one times two, so this is negative two, and g prime of x, I'll do it here, g prime of x, which use the power rule, bring that three out front, three times two is six, x, decrement that exponent, three minus one is two, and so g prime of negative one is equal to six times negative one squared. Well, negative one squared is just one, so this is going to be equal to six. So we actually know what all of these values are now. We know, so first we want to figure out f prime of negative one."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, negative one to the third power is just negative one times two, so this is negative two, and g prime of x, I'll do it here, g prime of x, which use the power rule, bring that three out front, three times two is six, x, decrement that exponent, three minus one is two, and so g prime of negative one is equal to six times negative one squared. Well, negative one squared is just one, so this is going to be equal to six. So we actually know what all of these values are now. We know, so first we want to figure out f prime of negative one. Well, they tell us that right over here. f prime of negative one is equal to five, so that is five. g of negative one, well, we figure that right here."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know, so first we want to figure out f prime of negative one. Well, they tell us that right over here. f prime of negative one is equal to five, so that is five. g of negative one, well, we figure that right here. g of negative one is negative two, so this is negative two. f of negative one, so f of negative one, they tell us that right over there. That is equal to three, and then g prime of negative one, let me just circle it in this green color, g prime of negative one, we figured it out."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "g of negative one, well, we figure that right here. g of negative one is negative two, so this is negative two. f of negative one, so f of negative one, they tell us that right over there. That is equal to three, and then g prime of negative one, let me just circle it in this green color, g prime of negative one, we figured it out. It is equal to six, so this is equal to six, and then finally g of negative one right over here, we already figured that out. That was equal to negative two, so negative two. So this is all going to simplify to, so you have five times negative two, which is negative 10, minus three times six, which is 18, all of that over negative two squared."}, {"video_title": "Worked example Quotient rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is equal to three, and then g prime of negative one, let me just circle it in this green color, g prime of negative one, we figured it out. It is equal to six, so this is equal to six, and then finally g of negative one right over here, we already figured that out. That was equal to negative two, so negative two. So this is all going to simplify to, so you have five times negative two, which is negative 10, minus three times six, which is 18, all of that over negative two squared. Well, negative two squared is just going to be positive four, so this is going to be equal to negative 28 over positive four, which is equal to negative seven. And there you have it. It looks intimidating at first, but if you just say, okay, look, I can use the quotient rule right over here, and then once I apply the quotient rule, I can actually just directly figure out what g of negative one, g prime of negative one, and they gave us f of negative one and f prime of negative one, so hopefully you find that helpful."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's equal to the definite integral between t is equal to negative three and t is equal to x of our lower case g of t, g of t dt. So given how we have just defined capital G of x, let's see if we can find some values. So let's evaluate the function capital G at x is equal to four. Let's also evaluate the function capital G at x is equal to eight. And I encourage you to pause the video right now and try to think about these on your own, and then we can work through them together. So let's tackle capital G of four first. So this is going to be, well, if x is equal to four, this top boundary is going to be four."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's also evaluate the function capital G at x is equal to eight. And I encourage you to pause the video right now and try to think about these on your own, and then we can work through them together. So let's tackle capital G of four first. So this is going to be, well, if x is equal to four, this top boundary is going to be four. So this is going to be the definite integral of g of t between t is equal to negative three and t is equal to four, g of t, g of t dt. Now what's that going to be equal to? Well, let's see."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be, well, if x is equal to four, this top boundary is going to be four. So this is going to be the definite integral of g of t between t is equal to negative three and t is equal to four, g of t, g of t dt. Now what's that going to be equal to? Well, let's see. Let's look at this graph here. So we have negative three. So this t is equal to negative three, which is right over here."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see. Let's look at this graph here. So we have negative three. So this t is equal to negative three, which is right over here. t is equal to negative three. And we're going to go all the way to t is equal to four. Let me circle that in orange."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this t is equal to negative three, which is right over here. t is equal to negative three. And we're going to go all the way to t is equal to four. Let me circle that in orange. All the way to t is equal to four, which is right over here. So this is, one way to think about this is that this is going to be the area above the t axis and below the graph of g. So it's going to be this area right over here that is above the t axis and below the graph of g of t. But we're not going to add to it this area because this area over here, I'll shade it in yellow, this area I'm shading in yellow over here, this is going to be negative. Why is it going to be negative?"}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me circle that in orange. All the way to t is equal to four, which is right over here. So this is, one way to think about this is that this is going to be the area above the t axis and below the graph of g. So it's going to be this area right over here that is above the t axis and below the graph of g of t. But we're not going to add to it this area because this area over here, I'll shade it in yellow, this area I'm shading in yellow over here, this is going to be negative. Why is it going to be negative? Because we want the area that is above t and below g. This is the other way around. This is below the t axis and above the graph of g. So one way to think about it is we could split it up. So we could, actually let me just clear this out so we have more space."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Why is it going to be negative? Because we want the area that is above t and below g. This is the other way around. This is below the t axis and above the graph of g. So one way to think about it is we could split it up. So we could, actually let me just clear this out so we have more space. So this right over here is going to be equal to the integral, I'll do that in this purple color, the integral from t equals negative three to zero of g of t dt plus, I'll do this in the yellow color, plus the integral from zero, t equals zero, to t is equal to four of g of t dt. Now what are each of these going to be? Well this is just a triangle where the base is three, the base has length three, the height is length three, so it's going to be three times three, which is nine times one half, because three times three would give us the area of this entire square, but this triangle is half of that."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could, actually let me just clear this out so we have more space. So this right over here is going to be equal to the integral, I'll do that in this purple color, the integral from t equals negative three to zero of g of t dt plus, I'll do this in the yellow color, plus the integral from zero, t equals zero, to t is equal to four of g of t dt. Now what are each of these going to be? Well this is just a triangle where the base is three, the base has length three, the height is length three, so it's going to be three times three, which is nine times one half, because three times three would give us the area of this entire square, but this triangle is half of that. So this right over here is going to be 4.5. And then what's this area in yellow? Well let's see, we have a triangle, its width here is four, its height is four, four times four is 16, which would be the area of this entire square, we take half of that, it's eight."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well this is just a triangle where the base is three, the base has length three, the height is length three, so it's going to be three times three, which is nine times one half, because three times three would give us the area of this entire square, but this triangle is half of that. So this right over here is going to be 4.5. And then what's this area in yellow? Well let's see, we have a triangle, its width here is four, its height is four, four times four is 16, which would be the area of this entire square, we take half of that, it's eight. Now we don't just add it to it, because once again, this is going to be negative area, the graph over here is below the t axis. So this we would say, this integral is going to evaluate to negative eight. Once again, why is it negative eight?"}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well let's see, we have a triangle, its width here is four, its height is four, four times four is 16, which would be the area of this entire square, we take half of that, it's eight. Now we don't just add it to it, because once again, this is going to be negative area, the graph over here is below the t axis. So this we would say, this integral is going to evaluate to negative eight. Once again, why is it negative eight? Because the graph is below the t axis. And so what do we get? We get g of four, which is this area, essentially minus this area, 4.5 minus eight is going to be, let's see, four minus eight is negative four, add a.5, that's negative 3.5."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, why is it negative eight? Because the graph is below the t axis. And so what do we get? We get g of four, which is this area, essentially minus this area, 4.5 minus eight is going to be, let's see, four minus eight is negative four, add a.5, that's negative 3.5. Now let's try to figure out what g of eight is. So g of eight. And if you couldn't figure it out the first time around, try to pause the video again, and now that we've figured out g of four, try to figure out what g of eight is."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get g of four, which is this area, essentially minus this area, 4.5 minus eight is going to be, let's see, four minus eight is negative four, add a.5, that's negative 3.5. Now let's try to figure out what g of eight is. So g of eight. And if you couldn't figure it out the first time around, try to pause the video again, and now that we've figured out g of four, try to figure out what g of eight is. Well g of eight is, one way to think about it, it's going to be that minus that area, and then we're going to add, and then we're going to figure out the area, and then we're going to have two more areas to think about. We're going to go all the way to eight, so actually let me draw the line there. So we're going to have to think about, we're going to think about this whole area now, this whole area now, and then we're going to think about this one."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you couldn't figure it out the first time around, try to pause the video again, and now that we've figured out g of four, try to figure out what g of eight is. Well g of eight is, one way to think about it, it's going to be that minus that area, and then we're going to add, and then we're going to figure out the area, and then we're going to have two more areas to think about. We're going to go all the way to eight, so actually let me draw the line there. So we're going to have to think about, we're going to think about this whole area now, this whole area now, and then we're going to think about this one. So we could write, this is going to be equal to the integral between, let me do that purple color. So it's going to be this integral, which is the integral between negative t equals negative three and zero, g of t dt, plus this entire yellow region right now, the part that we looked at before, plus this part over here, so I'll just write that as plus the definite integral between t is equal to zero and six, g of t dt, and then finally, plus the definite integral between t is equal to six and t is equal to eight, g of t dt. Now, we already know that this first one evaluates to 4.5, so that one is 4.5."}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to have to think about, we're going to think about this whole area now, this whole area now, and then we're going to think about this one. So we could write, this is going to be equal to the integral between, let me do that purple color. So it's going to be this integral, which is the integral between negative t equals negative three and zero, g of t dt, plus this entire yellow region right now, the part that we looked at before, plus this part over here, so I'll just write that as plus the definite integral between t is equal to zero and six, g of t dt, and then finally, plus the definite integral between t is equal to six and t is equal to eight, g of t dt. Now, we already know that this first one evaluates to 4.5, so that one is 4.5. Now what's this one going to be? We have a triangle whose base is length six, height is four, six times four is 24, times 1.5 is going to get us 12. So this is going to be, that over there is 12, and then finally, what is this area?"}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, we already know that this first one evaluates to 4.5, so that one is 4.5. Now what's this one going to be? We have a triangle whose base is length six, height is four, six times four is 24, times 1.5 is going to get us 12. So this is going to be, that over there is 12, and then finally, what is this area? Oh, and we have to be careful. This is below the t-axis and above the graph, so this is going to be a negative 12, and then finally we have this area, which is once again going to be a positive area, it's below the graph and above the t-axis, and so let's see, two times four is eight, times 1.5 is four, so we're going to have plus four. Plus four, and so what do we have?"}, {"video_title": "Functions defined by integrals   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be, that over there is 12, and then finally, what is this area? Oh, and we have to be careful. This is below the t-axis and above the graph, so this is going to be a negative 12, and then finally we have this area, which is once again going to be a positive area, it's below the graph and above the t-axis, and so let's see, two times four is eight, times 1.5 is four, so we're going to have plus four. Plus four, and so what do we have? We end up with 4.5 plus four is 8.5 minus 12, so this is going to be equal to, 8.8 minus 12 would be negative four plus five, it is negative 3.5 again. Now why did these two things end up being the same? Well, think about what happened here."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's take the derivative of x to the n. And now that we know the binomial theorem, we have the tools to do it. So how do we take the derivative? Well, what's the classic definition of the derivative? It is the limit as delta x approaches 0 of f of x plus delta x, right? So f of x plus delta x in this situation is x plus delta x to the nth power, right? Minus f of x. Well, f of x here is just x to the n. All of that over delta x."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It is the limit as delta x approaches 0 of f of x plus delta x, right? So f of x plus delta x in this situation is x plus delta x to the nth power, right? Minus f of x. Well, f of x here is just x to the n. All of that over delta x. And now that we know the binomial theorem, we can figure out what the expansion of x plus delta x is to the nth power. And if you don't know the binomial theorem, go to my pre-calculus playlist and watch the videos on the binomial theorem. So the binomial theorem tells us that this is equal to, and let me, I'm going to need some space for this one, the limit as delta x approaches 0."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, f of x here is just x to the n. All of that over delta x. And now that we know the binomial theorem, we can figure out what the expansion of x plus delta x is to the nth power. And if you don't know the binomial theorem, go to my pre-calculus playlist and watch the videos on the binomial theorem. So the binomial theorem tells us that this is equal to, and let me, I'm going to need some space for this one, the limit as delta x approaches 0. And what's the binomial theorem tell us? This is going to be equal to, I'm just going to do the numerator, x to the n plus n choose 1. And once again, review the binomial theorem if this looks like Latin to you and you don't know Latin."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the binomial theorem tells us that this is equal to, and let me, I'm going to need some space for this one, the limit as delta x approaches 0. And what's the binomial theorem tell us? This is going to be equal to, I'm just going to do the numerator, x to the n plus n choose 1. And once again, review the binomial theorem if this looks like Latin to you and you don't know Latin. n choose 1 of x to the n minus 1 delta x plus n choose 2 x to the n minus 2 delta x squared. And then plus, and we have a bunch of digits. And in this proof, we don't have to go through all the digits, but the binomial theorem tells us what they are."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And once again, review the binomial theorem if this looks like Latin to you and you don't know Latin. n choose 1 of x to the n minus 1 delta x plus n choose 2 x to the n minus 2 delta x squared. And then plus, and we have a bunch of digits. And in this proof, we don't have to go through all the digits, but the binomial theorem tells us what they are. And of course, the last digit, we just keep adding, is going to be 1, well, it would be n choose n, which is 1. Let me just write that down. n choose n is going to be x to the 0 times delta x to the n. So that's the binomial expansion."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And in this proof, we don't have to go through all the digits, but the binomial theorem tells us what they are. And of course, the last digit, we just keep adding, is going to be 1, well, it would be n choose n, which is 1. Let me just write that down. n choose n is going to be x to the 0 times delta x to the n. So that's the binomial expansion. And let me switch back to minus. So all of that, that green, that's x plus delta x to the n, so minus x to the n. Minus x to the n power, that's x to the n, I know I squashed it there. All of that over delta x."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "n choose n is going to be x to the 0 times delta x to the n. So that's the binomial expansion. And let me switch back to minus. So all of that, that green, that's x plus delta x to the n, so minus x to the n. Minus x to the n power, that's x to the n, I know I squashed it there. All of that over delta x. Let's see if we can simplify. So first of all, we have an x to the n here, and at the very end, we subtract out an x to the n, so these two cancel out. And then if we look at every term here, every term in the numerator has a delta x, right?"}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "All of that over delta x. Let's see if we can simplify. So first of all, we have an x to the n here, and at the very end, we subtract out an x to the n, so these two cancel out. And then if we look at every term here, every term in the numerator has a delta x, right? So we can divide the numerator and the denominator, essentially, by delta x. This is the same thing as 1 over delta x times this whole thing. So that is equal to, let me, the limit as delta x approaches 0 of, so we divide the top and the bottom by delta x, or we multiply the numerator times 1 over delta x, we get n choose 1 x to the n minus 1."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then if we look at every term here, every term in the numerator has a delta x, right? So we can divide the numerator and the denominator, essentially, by delta x. This is the same thing as 1 over delta x times this whole thing. So that is equal to, let me, the limit as delta x approaches 0 of, so we divide the top and the bottom by delta x, or we multiply the numerator times 1 over delta x, we get n choose 1 x to the n minus 1. What's delta x divided by delta x? Well, that's just 1, right? Plus n choose 2 x to the n minus 2."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So that is equal to, let me, the limit as delta x approaches 0 of, so we divide the top and the bottom by delta x, or we multiply the numerator times 1 over delta x, we get n choose 1 x to the n minus 1. What's delta x divided by delta x? Well, that's just 1, right? Plus n choose 2 x to the n minus 2. This is delta x squared, but when we divide by delta x, we just get a delta x here. And then we keep having a bunch of terms. We're going to divide all of them by delta x, right?"}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Plus n choose 2 x to the n minus 2. This is delta x squared, but when we divide by delta x, we just get a delta x here. And then we keep having a bunch of terms. We're going to divide all of them by delta x, right? And then the last term is delta x to the n. But now we're going to divide that by delta x. So the last term becomes n choose n. x to the 0 is 1, so we can ignore that. Delta x to the n divided by delta x."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We're going to divide all of them by delta x, right? And then the last term is delta x to the n. But now we're going to divide that by delta x. So the last term becomes n choose n. x to the 0 is 1, so we can ignore that. Delta x to the n divided by delta x. Well, that's delta x to the n minus 1, right? And then what are we doing now? Well, remember, we're taking the limit as delta x approaches 0."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Delta x to the n divided by delta x. Well, that's delta x to the n minus 1, right? And then what are we doing now? Well, remember, we're taking the limit as delta x approaches 0. So as delta x approaches 0, pretty much every term that has a delta x in it, it becomes 0, right? We just, when you multiply by 0, you get 0. So every term, this first term has no delta x in it, but every other term does."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, remember, we're taking the limit as delta x approaches 0. So as delta x approaches 0, pretty much every term that has a delta x in it, it becomes 0, right? We just, when you multiply by 0, you get 0. So every term, this first term has no delta x in it, but every other term does. Every other term, even after we divide it by delta x, has a delta x in it. So that's a 0. Every term is 0."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So every term, this first term has no delta x in it, but every other term does. Every other term, even after we divide it by delta x, has a delta x in it. So that's a 0. Every term is 0. All of the other, I don't know, what is it, n minus 1 terms, they're all 0. So all we're left with is that this is equal to n choose 1 of x to the n minus 1. And what's n choose 1?"}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Every term is 0. All of the other, I don't know, what is it, n minus 1 terms, they're all 0. So all we're left with is that this is equal to n choose 1 of x to the n minus 1. And what's n choose 1? That equals n factorial over 1 factorial divided by n minus 1 factorial times x to the n minus 1. 1 factorial is 1. And if I have 7 factorial divided by 6 factorial, that's just 7."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And what's n choose 1? That equals n factorial over 1 factorial divided by n minus 1 factorial times x to the n minus 1. 1 factorial is 1. And if I have 7 factorial divided by 6 factorial, that's just 7. Or if I have 3 factorial divided by 2 factorial, that's just 3. You can work it out. 10 factorial divided by 9 factorial, that's 10."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And if I have 7 factorial divided by 6 factorial, that's just 7. Or if I have 3 factorial divided by 2 factorial, that's just 3. You can work it out. 10 factorial divided by 9 factorial, that's 10. So n factorial divided by n minus 1 factorial, that's just equal to n. So this is equal to n times x to the n minus 1. So that's the derivative of x to the n. n times x to the n minus 1. So we just proved the derivative for any positive integer, x to the power n, where n is any positive integer."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "10 factorial divided by 9 factorial, that's 10. So n factorial divided by n minus 1 factorial, that's just equal to n. So this is equal to n times x to the n minus 1. So that's the derivative of x to the n. n times x to the n minus 1. So we just proved the derivative for any positive integer, x to the power n, where n is any positive integer. And we see later it actually works for all, actually, real numbers in the exponent. So I will see you in a future video. Oh, and another thing I wanted to point out is, I said that we had to know the binomial theorem."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we just proved the derivative for any positive integer, x to the power n, where n is any positive integer. And we see later it actually works for all, actually, real numbers in the exponent. So I will see you in a future video. Oh, and another thing I wanted to point out is, I said that we had to know the binomial theorem. But if you think about it, we really didn't even have to know the binomial theorem. Because we knew in any binomial expansion, I mean, you'd have to know a little bit. But if you did a little experimentation, you would realize that whenever you expand a plus b to the nth power, the first term is going to be a to the n. And then the second term is going to be plus n a to the n minus 1 b."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Oh, and another thing I wanted to point out is, I said that we had to know the binomial theorem. But if you think about it, we really didn't even have to know the binomial theorem. Because we knew in any binomial expansion, I mean, you'd have to know a little bit. But if you did a little experimentation, you would realize that whenever you expand a plus b to the nth power, the first term is going to be a to the n. And then the second term is going to be plus n a to the n minus 1 b. And then you could keep having a bunch of terms. But these are the only terms that are relevant in this proof, because all the other terms get canceled out when delta x approaches 0. So if you just knew that, you could have done this."}, {"video_title": "Proof d dx(x^n)   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But if you did a little experimentation, you would realize that whenever you expand a plus b to the nth power, the first term is going to be a to the n. And then the second term is going to be plus n a to the n minus 1 b. And then you could keep having a bunch of terms. But these are the only terms that are relevant in this proof, because all the other terms get canceled out when delta x approaches 0. So if you just knew that, you could have done this. But it's much better to do it with the binomial theorem. Ignore what I just said if it confused you. I'm just saying that we could have just said, oh, the rest of these terms just all go to 0."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do using our powers of calculus is find the area of this yellow region. And if at any point you get inspired, I always encourage you to pause the video and try to work through it on your own. So the key here is you might recognize, hey, this is an area between curves. A definite integral might be useful. So I'll just set up the definite integral sign. And so first we need to think about what are our left and right boundaries of our region? Well, it looks like the left boundary is where the two graphs intersect right over here, and the right boundary is where they intersect right over there."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "A definite integral might be useful. So I'll just set up the definite integral sign. And so first we need to think about what are our left and right boundaries of our region? Well, it looks like the left boundary is where the two graphs intersect right over here, and the right boundary is where they intersect right over there. Well, what is this point of intersection? It looks like it is negative one comma negative two. We can verify that."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it looks like the left boundary is where the two graphs intersect right over here, and the right boundary is where they intersect right over there. Well, what is this point of intersection? It looks like it is negative one comma negative two. We can verify that. In this red curve, if x is negative one, let's see, you square that, you'll get one minus three. You do indeed get y is equal to negative two. And in this blue function, when x is equal to negative one, you get one minus four plus one."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can verify that. In this red curve, if x is negative one, let's see, you square that, you'll get one minus three. You do indeed get y is equal to negative two. And in this blue function, when x is equal to negative one, you get one minus four plus one. Once again, you do indeed get y is equal to negative two. And the same thing is true when x is equal to one. One minus three, negative two."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in this blue function, when x is equal to negative one, you get one minus four plus one. Once again, you do indeed get y is equal to negative two. And the same thing is true when x is equal to one. One minus three, negative two. One minus four plus one, negative two. So our bounds are indeed, we're going from x equals negative one to x equals positive one. And now let's think about our upper and lower bounds."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "One minus three, negative two. One minus four plus one, negative two. So our bounds are indeed, we're going from x equals negative one to x equals positive one. And now let's think about our upper and lower bounds. Over that interval, this blue graph is our upper bound. And so we would subtract the lower bound from the upper bound. So we would have x to the fourth minus four x squared plus one."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now let's think about our upper and lower bounds. Over that interval, this blue graph is our upper bound. And so we would subtract the lower bound from the upper bound. So we would have x to the fourth minus four x squared plus one. And from that, we will subtract x squared minus three dx. And in many other videos, we have talked about why you do this, why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them. But now we just have to evaluate this definite integral."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would have x to the fourth minus four x squared plus one. And from that, we will subtract x squared minus three dx. And in many other videos, we have talked about why you do this, why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them. But now we just have to evaluate this definite integral. So let's just get down to business. All right. So we have the integral from negative one to one."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now we just have to evaluate this definite integral. So let's just get down to business. All right. So we have the integral from negative one to one. And so we have x to the fourth, x to the fourth. And now we have minus four x squared. And then when you distribute this negative sign, you're gonna subtract another x squared."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have the integral from negative one to one. And so we have x to the fourth, x to the fourth. And now we have minus four x squared. And then when you distribute this negative sign, you're gonna subtract another x squared. So you're gonna have minus five x squared. And then you have plus one. And then you're gonna subtract a negative three."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then when you distribute this negative sign, you're gonna subtract another x squared. So you're gonna have minus five x squared. And then you have plus one. And then you're gonna subtract a negative three. So it's gonna be one plus three. So it's gonna be plus four dx. And just to be clear, I should put parentheses right over there because it's really the dx is being multiplied by this entire expression."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you're gonna subtract a negative three. So it's gonna be one plus three. So it's gonna be plus four dx. And just to be clear, I should put parentheses right over there because it's really the dx is being multiplied by this entire expression. And so let's see. Let's find the antiderivative of this. This should be pretty straightforward."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just to be clear, I should put parentheses right over there because it's really the dx is being multiplied by this entire expression. And so let's see. Let's find the antiderivative of this. This should be pretty straightforward. We're just gonna use the reverse power rule multiple times. So this is going to be, the antiderivative of x to the fourth is x to the fifth over five. We just incremented the exponent and divided by that incremented exponent minus, same idea here, five x to the third over three plus four x."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "This should be pretty straightforward. We're just gonna use the reverse power rule multiple times. So this is going to be, the antiderivative of x to the fourth is x to the fifth over five. We just incremented the exponent and divided by that incremented exponent minus, same idea here, five x to the third over three plus four x. And then we are going to evaluate it at one and then subtract from that it evaluated at negative one. So let's first evaluate it at one. We're gonna get 1 5th minus 5 3rds plus four."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "We just incremented the exponent and divided by that incremented exponent minus, same idea here, five x to the third over three plus four x. And then we are going to evaluate it at one and then subtract from that it evaluated at negative one. So let's first evaluate it at one. We're gonna get 1 5th minus 5 3rds plus four. And now let us evaluate it at negative one. So minus, let's see, if this is negative one, we're gonna negative 1 5th, negative 1 5th, and this is gonna be plus 5 3rds, plus 5 3rds, and then this is going to be minus four, minus four. But then when you distribute the negative sign, this, so we're gonna distribute this over all of these terms."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna get 1 5th minus 5 3rds plus four. And now let us evaluate it at negative one. So minus, let's see, if this is negative one, we're gonna negative 1 5th, negative 1 5th, and this is gonna be plus 5 3rds, plus 5 3rds, and then this is going to be minus four, minus four. But then when you distribute the negative sign, this, so we're gonna distribute this over all of these terms. And so this is going to be, if we make this positive, this will be positive. This one will be negative, and then this one will be positive. So you have 1 5th plus 1 5th, which is going to be 2 5ths, that and that, and then minus 5 3rds, minus 5 3rds."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then when you distribute the negative sign, this, so we're gonna distribute this over all of these terms. And so this is going to be, if we make this positive, this will be positive. This one will be negative, and then this one will be positive. So you have 1 5th plus 1 5th, which is going to be 2 5ths, that and that, and then minus 5 3rds, minus 5 3rds. So minus 10 over three, and then four plus four. So plus eight. And also we just need to simplify this."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you have 1 5th plus 1 5th, which is going to be 2 5ths, that and that, and then minus 5 3rds, minus 5 3rds. So minus 10 over three, and then four plus four. So plus eight. And also we just need to simplify this. This is going to be, let's see, it's gonna be eight, and then if I write, so plus, I'm gonna write these two with a denominator of 15, because that's the common denominator of three and five. Let's see, 2 5ths is 6 15ths. Yeah, that's right, five times three is 15, two times three is six."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "And also we just need to simplify this. This is going to be, let's see, it's gonna be eight, and then if I write, so plus, I'm gonna write these two with a denominator of 15, because that's the common denominator of three and five. Let's see, 2 5ths is 6 15ths. Yeah, that's right, five times three is 15, two times three is six. And then 10 3rds, let's see, if we multiply the denominator times five, we have to multiply the numerator times five, so this is gonna be 50 15ths. And so what's 6 15ths minus 50 15ths? So this is going to be equal to eight minus six minus 50 is minus 44."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yeah, that's right, five times three is 15, two times three is six. And then 10 3rds, let's see, if we multiply the denominator times five, we have to multiply the numerator times five, so this is gonna be 50 15ths. And so what's 6 15ths minus 50 15ths? So this is going to be equal to eight minus six minus 50 is minus 44. Minus 44 over 15. And so what is 44 over 15? 44 over 15 is equal to two and 14 15ths."}, {"video_title": "Worked example area between curves   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to eight minus six minus 50 is minus 44. Minus 44 over 15. And so what is 44 over 15? 44 over 15 is equal to two and 14 15ths. So that's really what we're subtracting. We're gonna subtract two and 14 15ths. So if you subtract two from this, you would get six minus 14 over 15, because we still have to subtract the 14 15ths."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are told the population of a town grows at a rate of e to the 1.2t power minus 2t people per year, where t is the number of years. At t equals two years, the town has 1,500 people. So first they ask us, approximately by how many people does the population grow between t equals two and t equals five, and then what is the town's population at t equals five years? And if we actually figure out this first question, the second question is actually pretty straightforward. We figure out the amount that it grows and then add it to what we were at at t equals two, add it to 1,500. So pause this video and see if you can figure it out. So the key here is to appreciate that this right over here is expressing the rate of how fast the population is growing and we've been seen in multiple videos now, let me just draw, do a quick review of this notion of a rate curve."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we actually figure out this first question, the second question is actually pretty straightforward. We figure out the amount that it grows and then add it to what we were at at t equals two, add it to 1,500. So pause this video and see if you can figure it out. So the key here is to appreciate that this right over here is expressing the rate of how fast the population is growing and we've been seen in multiple videos now, let me just draw, do a quick review of this notion of a rate curve. So those are my axes and this is my t axis, my time axis. And so this is showing me how my rate of change changes as a function of time. So let's say it's something like this."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the key here is to appreciate that this right over here is expressing the rate of how fast the population is growing and we've been seen in multiple videos now, let me just draw, do a quick review of this notion of a rate curve. So those are my axes and this is my t axis, my time axis. And so this is showing me how my rate of change changes as a function of time. So let's say it's something like this. So once again, if I said at this time right over here, this is my rate, this doesn't tell me, for example, what my population is, this tells me what is my rate of change of a population. And we have seen in previous videos that if you wanna figure out the change in the thing that the rate is, that the rate is the rate of change of, say, the change in population, you would find the area under the rate curve between those two appropriate times. And why does that make sense?"}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say it's something like this. So once again, if I said at this time right over here, this is my rate, this doesn't tell me, for example, what my population is, this tells me what is my rate of change of a population. And we have seen in previous videos that if you wanna figure out the change in the thing that the rate is, that the rate is the rate of change of, say, the change in population, you would find the area under the rate curve between those two appropriate times. And why does that make sense? Well, imagine a very small change in time right over here. If you have a very small change in time and if you assume that your rate is approximately constant over that very small change in time, well then your change in, let's say we're measuring the rate of change of population here, your accumulation, you could say, is going to be your rate times your change in time, which would be the area of this rectangle. And so that would be the, roughly, that would be the area under the curve over that very, very small change in time."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And why does that make sense? Well, imagine a very small change in time right over here. If you have a very small change in time and if you assume that your rate is approximately constant over that very small change in time, well then your change in, let's say we're measuring the rate of change of population here, your accumulation, you could say, is going to be your rate times your change in time, which would be the area of this rectangle. And so that would be the, roughly, that would be the area under the curve over that very, very small change in time. So what we really wanna do is find the area under this curve from t equals two to t equals five. And we have seen multiple times in calculus how to express that. So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that would be the, roughly, that would be the area under the curve over that very, very small change in time. So what we really wanna do is find the area under this curve from t equals two to t equals five. And we have seen multiple times in calculus how to express that. So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt. So if we just evaluate that, that will be the answer to this first question. So what is this going to be? Well, let's actually work it out."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt. So if we just evaluate that, that will be the answer to this first question. So what is this going to be? Well, let's actually work it out. So what is the antiderivative of e to the 1.2t? Well, let me just try to do it over here. So if I am trying to calculate, let me write it as e to the five, or actually 6 5ths t, 12 tenths is the same thing as 6 5ths, 6 5ths t dt."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's actually work it out. So what is the antiderivative of e to the 1.2t? Well, let me just try to do it over here. So if I am trying to calculate, let me write it as e to the five, or actually 6 5ths t, 12 tenths is the same thing as 6 5ths, 6 5ths t dt. So this is an indefinite integral. I'm just trying to figure out the antiderivative here. Well, if I had a 6 5ths right over here, then u substitution, or sometimes you would say the inverse chain rule, would be very appropriate."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I am trying to calculate, let me write it as e to the five, or actually 6 5ths t, 12 tenths is the same thing as 6 5ths, 6 5ths t dt. So this is an indefinite integral. I'm just trying to figure out the antiderivative here. Well, if I had a 6 5ths right over here, then u substitution, or sometimes you would say the inverse chain rule, would be very appropriate. Well, we could put a 6 5ths there if we write a 5 6ths right over here, 5 6ths times 6 5ths, and we can take constants in and out of the integral like this, scaling constants, I should say. Well, now, so this is going to be equal to 5 6ths, this 5 6ths right over here. This antiderivative is pretty straightforward."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if I had a 6 5ths right over here, then u substitution, or sometimes you would say the inverse chain rule, would be very appropriate. Well, we could put a 6 5ths there if we write a 5 6ths right over here, 5 6ths times 6 5ths, and we can take constants in and out of the integral like this, scaling constants, I should say. Well, now, so this is going to be equal to 5 6ths, this 5 6ths right over here. This antiderivative is pretty straightforward. Since I have the derivative of 6 5ths t right over here, I can find the antiderivative with respect to 6 5ths t, which is essentially I'm doing u substitution. If you had to do u substitution, you would make that right over there u, and then that, and that would be your du. But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This antiderivative is pretty straightforward. Since I have the derivative of 6 5ths t right over here, I can find the antiderivative with respect to 6 5ths t, which is essentially I'm doing u substitution. If you had to do u substitution, you would make that right over there u, and then that, and that would be your du. But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference. So let's evaluate this at when t is equal to five. Well, you are going to, let me color code this a little bit. When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference. So let's evaluate this at when t is equal to five. Well, you are going to, let me color code this a little bit. When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract. When we evaluate it at two, we get 5 6ths e to the 6 5ths times two is the same thing as 12 5ths, or we could say that's 2.4, 2.4 minus four. Two squared is four, and so what do we get? Well, there's a couple of ways that we could do this."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract. When we evaluate it at two, we get 5 6ths e to the 6 5ths times two is the same thing as 12 5ths, or we could say that's 2.4, 2.4 minus four. Two squared is four, and so what do we get? Well, there's a couple of ways that we could do this. So we could write this as, let me write it this way. We could write this so we have a 5 6ths and a 5 6ths, so we could write this as 5 6ths times e to the 6th minus e to the 2.4 minus, because we distribute that negative sign, e to the 2.4, e to the 2.4 power, and then we have minus 25. Let me just send another color to keep track of it."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, there's a couple of ways that we could do this. So we could write this as, let me write it this way. We could write this so we have a 5 6ths and a 5 6ths, so we could write this as 5 6ths times e to the 6th minus e to the 2.4 minus, because we distribute that negative sign, e to the 2.4, e to the 2.4 power, and then we have minus 25. Let me just send another color to keep track of it. We have minus 25, and then you have minus negative four, so that would be plus four, so that would be minus 21, and I would need a calculator to figure this out, so let me do that. Let me get my calculator on this computer, and there we go, and so let's see. If we want to find e to the 6th power, that's 430, okay, so now let me figure out, so minus, I would say 22.4, that looks like a 24."}, {"video_title": "Worked example problem involving definite integral (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just send another color to keep track of it. We have minus 25, and then you have minus negative four, so that would be plus four, so that would be minus 21, and I would need a calculator to figure this out, so let me do that. Let me get my calculator on this computer, and there we go, and so let's see. If we want to find e to the 6th power, that's 430, okay, so now let me figure out, so minus, I would say 22.4, that looks like a 24. I'll correct it as soon as I get back to that screen. E to that power, e to the 2.4 power, and I get equals, so what's in parentheses is this number right here, so times 5 6ths, times five, divided by six, is equal to that minus 21, minus 21, is equal to this, so if I round to the nearest hundredth, it's going to be approximately 306.00, so this is approximately 306.00, so approximately by how many people does a population grow between t equals two and t equals five? Well, by approximately 306 people."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let h be the vertical distance between the graphs of f and g in region S. Find the rate at which h changes with respect to x when x is equal to 1.8. So we have region S right over here. You can't see it that well since I drew over it. And what you see in region S, the function f is greater than the function g. It's above the function g. And so we can write h of x. We could write h of x as being equal to f of x minus g of x. And what we want to do is we want to find the rate at which h changes with respect to x. And so we could write that as h prime of, we could say h prime of x, but we want the rate when x is equal to 1.8."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what you see in region S, the function f is greater than the function g. It's above the function g. And so we can write h of x. We could write h of x as being equal to f of x minus g of x. And what we want to do is we want to find the rate at which h changes with respect to x. And so we could write that as h prime of, we could say h prime of x, but we want the rate when x is equal to 1.8. So h prime of 1.8 is what we want to figure out. Now we could evaluate f prime of 1.8 and g prime of 1.8, and to do that we would take the derivatives of each of these things and we know how to do that. It's within our capabilities."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we could write that as h prime of, we could say h prime of x, but we want the rate when x is equal to 1.8. So h prime of 1.8 is what we want to figure out. Now we could evaluate f prime of 1.8 and g prime of 1.8, and to do that we would take the derivatives of each of these things and we know how to do that. It's within our capabilities. But it's important to realize when you're taking the AP test that you have a calculator at your disposal. And a calculator can numerically integrate and it can numerically evaluate derivatives. And so whenever they actually want us to find the area or evaluate an integral where they give the endpoints or evaluate a derivative at a point, well that's a pretty good sign that you could probably use your calculator here."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's within our capabilities. But it's important to realize when you're taking the AP test that you have a calculator at your disposal. And a calculator can numerically integrate and it can numerically evaluate derivatives. And so whenever they actually want us to find the area or evaluate an integral where they give the endpoints or evaluate a derivative at a point, well that's a pretty good sign that you could probably use your calculator here. And what's extra good about this is we have already essentially inputted h of x in the previous steps and really in part a. I had defined this function here and this function is essentially h of x. I took the absolute value of it so it's always positive over either region, but I could delete the absolute value if we want. So let me delete that absolute value. Then I have to get rid of that parentheses at the end."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so whenever they actually want us to find the area or evaluate an integral where they give the endpoints or evaluate a derivative at a point, well that's a pretty good sign that you could probably use your calculator here. And what's extra good about this is we have already essentially inputted h of x in the previous steps and really in part a. I had defined this function here and this function is essentially h of x. I took the absolute value of it so it's always positive over either region, but I could delete the absolute value if we want. So let me delete that absolute value. Then I have to get rid of that parentheses at the end. So let me delete that. And so notice, this is h of x. We have our f of x, which is one plus x plus e to the x squared minus two x."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then I have to get rid of that parentheses at the end. So let me delete that. And so notice, this is h of x. We have our f of x, which is one plus x plus e to the x squared minus two x. And then from that we subtract g of x. So we have g of x was a positive x to the fourth, but we're subtracting, so negative x to the fourth. Let me show you g of x right over here."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have our f of x, which is one plus x plus e to the x squared minus two x. And then from that we subtract g of x. So we have g of x was a positive x to the fourth, but we're subtracting, so negative x to the fourth. Let me show you g of x right over here. G of x is right over here. And notice, we are subtracting it. So the y one, as I've defined in my calculator, and I just press this y equals button right over here, that is my h of x."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me show you g of x right over here. G of x is right over here. And notice, we are subtracting it. So the y one, as I've defined in my calculator, and I just press this y equals button right over here, that is my h of x. And so now I can go back to the other screen and I can evaluate its derivative when x is equal to 1.8. I go to math. I scroll down."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the y one, as I've defined in my calculator, and I just press this y equals button right over here, that is my h of x. And so now I can go back to the other screen and I can evaluate its derivative when x is equal to 1.8. I go to math. I scroll down. We have n derivative right here. And so click enter there. And what am I gonna take the derivative of?"}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I scroll down. We have n derivative right here. And so click enter there. And what am I gonna take the derivative of? Well, the function y sub one that I've already defined in my calculator. I can go to variables, y variables. It's already selected function, so I'll just press enter."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what am I gonna take the derivative of? Well, the function y sub one that I've already defined in my calculator. I can go to variables, y variables. It's already selected function, so I'll just press enter. And I'll select the function y sub one that I've already defined. So I'm taking the derivative of y sub one. I'm taking the derivative with respect to x."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's already selected function, so I'll just press enter. And I'll select the function y sub one that I've already defined. So I'm taking the derivative of y sub one. I'm taking the derivative with respect to x. And I'm going to evaluate that derivative when x is equal to 1.8. That simple. And then I click enter."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm taking the derivative with respect to x. And I'm going to evaluate that derivative when x is equal to 1.8. That simple. And then I click enter. And there you have it. It's approximately negative 3.812. So approximately negative 3.812."}, {"video_title": "2015 AP Calculus 2c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I click enter. And there you have it. It's approximately negative 3.812. So approximately negative 3.812. And we're done. And one thing that you might appreciate from this entire question, and even the question one, is they really want to make sure that you understand the underlying conceptual ideas behind derivatives and integrals. And if you understand the conceptual ideas of how do you use them to solve problems, and you have your calculator at disposal, that these are not too hairy, that these can be done fairly quickly."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it, and so there's a couple of interesting things here. The first thing, at least that my brain does, it says, well, I'm used to taking derivatives and antiderivatives of e to the x. Not some other base to the x. So we know that the derivative with respect to x of e to the x is e to the x, or we could say that the antiderivative of e to the x is equal to e to the x plus c. So since I'm dealing with something raised to a, this particular situation, something raised to a function of x, it seems like I might wanna put some, I might wanna change the base here. But how do I do that? Well, the way I would do that is re-express two in terms of e. So what would be two in terms of e? Well, two is equal to e, is equal to e raised to the power that you need to raise e to to get to two."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that the derivative with respect to x of e to the x is e to the x, or we could say that the antiderivative of e to the x is equal to e to the x plus c. So since I'm dealing with something raised to a, this particular situation, something raised to a function of x, it seems like I might wanna put some, I might wanna change the base here. But how do I do that? Well, the way I would do that is re-express two in terms of e. So what would be two in terms of e? Well, two is equal to e, is equal to e raised to the power that you need to raise e to to get to two. Well, what's the power that you have to raise e to to get to two? Well, that's the natural log of two. Once again, the natural log of two is the exponent that you have to raise e to to get to two."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, two is equal to e, is equal to e raised to the power that you need to raise e to to get to two. Well, what's the power that you have to raise e to to get to two? Well, that's the natural log of two. Once again, the natural log of two is the exponent that you have to raise e to to get to two. So if you actually raise e to it, you're going to get two. So this is what two is. Now what is two to the x to the third?"}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, the natural log of two is the exponent that you have to raise e to to get to two. So if you actually raise e to it, you're going to get two. So this is what two is. Now what is two to the x to the third? Well, if we raise both sides of this to the x to the third power, if we raise both sides to the x to the third power, two to the x to the third is equal to, if I raise something to an exponent and then raise that to an exponent, it's going to be equal to e to the x to the third, x to the third times the natural log of two, times the natural log of two. So that already seems pretty interesting. So let's rewrite this."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is two to the x to the third? Well, if we raise both sides of this to the x to the third power, if we raise both sides to the x to the third power, two to the x to the third is equal to, if I raise something to an exponent and then raise that to an exponent, it's going to be equal to e to the x to the third, x to the third times the natural log of two, times the natural log of two. So that already seems pretty interesting. So let's rewrite this. And actually, what I'm going to do is let's just focus on the indefinite integral first, see if we can figure that out, and then we can apply, and then we can evaluate the definite ones. So let's just think about this. Let's think about the indefinite integral of x squared times two to the x to the third power dx."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's rewrite this. And actually, what I'm going to do is let's just focus on the indefinite integral first, see if we can figure that out, and then we can apply, and then we can evaluate the definite ones. So let's just think about this. Let's think about the indefinite integral of x squared times two to the x to the third power dx. I really want to find the antiderivative of this. Well, this is going to be the exact same thing as the integral of, so I'll write my x squared still, but instead of two to the x to the third, I'm going to write all of this business. Let me just copy and paste that."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's think about the indefinite integral of x squared times two to the x to the third power dx. I really want to find the antiderivative of this. Well, this is going to be the exact same thing as the integral of, so I'll write my x squared still, but instead of two to the x to the third, I'm going to write all of this business. Let me just copy and paste that. We already established this is the same thing as two to the x to the third power. Copy and paste it just like that. And then let me just close it with a dx."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just copy and paste that. We already established this is the same thing as two to the x to the third power. Copy and paste it just like that. And then let me just close it with a dx. So I was able to get it in terms of e as a base. That makes me a little bit more comfortable, but it still seems pretty complicated. But you might be saying, well, okay, look, maybe u substitution could be at play here because I have this kind of crazy expression, x to the third times the natural log of two, but what's the derivative of that?"}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let me just close it with a dx. So I was able to get it in terms of e as a base. That makes me a little bit more comfortable, but it still seems pretty complicated. But you might be saying, well, okay, look, maybe u substitution could be at play here because I have this kind of crazy expression, x to the third times the natural log of two, but what's the derivative of that? Well, that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. Well, that's just a constant times x squared. We already have a x squared here, and so maybe we can engineer this a little bit to have the constant there as well."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "But you might be saying, well, okay, look, maybe u substitution could be at play here because I have this kind of crazy expression, x to the third times the natural log of two, but what's the derivative of that? Well, that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. Well, that's just a constant times x squared. We already have a x squared here, and so maybe we can engineer this a little bit to have the constant there as well. So let's think about that. So if we made this, if we defined this as u, so if we said u is equal to x to the third times the natural log of two, what is du going to be? Well, du is going to be, it's going to be, well, natural log of two is just a constant, so it's going to be three x squared times the natural log of two."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already have a x squared here, and so maybe we can engineer this a little bit to have the constant there as well. So let's think about that. So if we made this, if we defined this as u, so if we said u is equal to x to the third times the natural log of two, what is du going to be? Well, du is going to be, it's going to be, well, natural log of two is just a constant, so it's going to be three x squared times the natural log of two. And we can actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing, just using logarithm properties, as x squared times the natural log of two to the third power. Three natural log of two is the same thing as the natural log of two to the third power."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, du is going to be, it's going to be, well, natural log of two is just a constant, so it's going to be three x squared times the natural log of two. And we can actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing, just using logarithm properties, as x squared times the natural log of two to the third power. Three natural log of two is the same thing as the natural log of two to the third power. So this is equal to x squared times the natural log of eight. So let's see, if this is u, where is du? Oh, and of course, we can't forget the dx."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Three natural log of two is the same thing as the natural log of two to the third power. So this is equal to x squared times the natural log of eight. So let's see, if this is u, where is du? Oh, and of course, we can't forget the dx. This is a dx right over here. dx, dx, dx. So where is the du?"}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Oh, and of course, we can't forget the dx. This is a dx right over here. dx, dx, dx. So where is the du? Well, we have a dx. Let me circle things. So you have a dx here."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So where is the du? Well, we have a dx. Let me circle things. So you have a dx here. You have a dx there. You have an x squared here. You have an x squared here."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you have a dx here. You have a dx there. You have an x squared here. You have an x squared here. So really, all we need is, all we need here is the natural log of eight. So if we, ideally, we would have a natural log of eight right over here. And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by a natural log of eight."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have an x squared here. So really, all we need is, all we need here is the natural log of eight. So if we, ideally, we would have a natural log of eight right over here. And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by a natural log of eight. And so we could do it right over here. We could divide by natural log of eight. But we know that the antiderivative of some constant times a function is the same thing as the constant times the antiderivative of that function."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by a natural log of eight. And so we could do it right over here. We could divide by natural log of eight. But we know that the antiderivative of some constant times a function is the same thing as the constant times the antiderivative of that function. So we could just take that on the outside. So it's one over the natural log of eight. So let's write this in terms of u and du."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we know that the antiderivative of some constant times a function is the same thing as the constant times the antiderivative of that function. So we could just take that on the outside. So it's one over the natural log of eight. So let's write this in terms of u and du. This simplifies to one over the natural log of eight times the antiderivative of e, e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's write this in terms of u and du. This simplifies to one over the natural log of eight times the antiderivative of e, e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward. We know what this is going to be. This is going to be equal to, so let me just write the one over natural log of eight out here. One over natural log of eight times e to the u."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is straightforward. We know what this is going to be. This is going to be equal to, so let me just write the one over natural log of eight out here. One over natural log of eight times e to the u. Times e to the u, e to the u. And of course, if we're thinking in terms of just antiderivative, there would be some constant out there. And then we would just reverse the substitution."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "One over natural log of eight times e to the u. Times e to the u, e to the u. And of course, if we're thinking in terms of just antiderivative, there would be some constant out there. And then we would just reverse the substitution. We already know what u is. So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two. And of course, we could put a plus c there."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we would just reverse the substitution. We already know what u is. So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two. And of course, we could put a plus c there. Now, going back to the original problem. We just need to evaluate the antiderivative of this at each of these points. So let's rewrite this."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And of course, we could put a plus c there. Now, going back to the original problem. We just need to evaluate the antiderivative of this at each of these points. So let's rewrite this. So given what we just figured out, so let me copy and paste that. This is just going to be equal to, it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated at zero. We don't have to worry about the constants because those will cancel out."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's rewrite this. So given what we just figured out, so let me copy and paste that. This is just going to be equal to, it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated at zero. We don't have to worry about the constants because those will cancel out. And so we are going to get, we are going to get one, let me evaluate it first at one. So you're gonna get one over the natural log of eight times e to the one to the third power, which is just one, times the natural log of two. Natural log of two, that's evaluated at one."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't have to worry about the constants because those will cancel out. And so we are going to get, we are going to get one, let me evaluate it first at one. So you're gonna get one over the natural log of eight times e to the one to the third power, which is just one, times the natural log of two. Natural log of two, that's evaluated at one. And then we're gonna have minus it evaluated at zero. So it's going to be one over the natural log of eight times e to the, well, when x is zero, this whole thing is going to be zero. Well, e to the zero is just one."}, {"video_title": "_-substitution definite integral of exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Natural log of two, that's evaluated at one. And then we're gonna have minus it evaluated at zero. So it's going to be one over the natural log of eight times e to the, well, when x is zero, this whole thing is going to be zero. Well, e to the zero is just one. And e to the natural log of two, well, that's just going to be two. We already established that early on. This is just going to be equal to two."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we've got a very interesting scenario here. I have this conical, thimble-like cup that is 4 centimeters high. And also, the diameter of the top of the cup is also 4 centimeters. And I'm pouring water into this cup right now. And I'm pouring the water at a rate of 1 cubic centimeter per second. And right at this moment, there is a height of 2 centimeters of water in the cup right now. So the height right now from the bottom of the cup to this point right over here is 2 centimeters."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm pouring water into this cup right now. And I'm pouring the water at a rate of 1 cubic centimeter per second. And right at this moment, there is a height of 2 centimeters of water in the cup right now. So the height right now from the bottom of the cup to this point right over here is 2 centimeters. So my question to you is, at what rate? We know the rate at which the water is flowing into the cup. We're being given a volume per time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the height right now from the bottom of the cup to this point right over here is 2 centimeters. So my question to you is, at what rate? We know the rate at which the water is flowing into the cup. We're being given a volume per time. My question to you is, right at this moment, right when we are filling our cup at 1 cubic centimeter per second, and we have exactly 2 centimeters of water in the cup, 2 centimeters deep of water in the cup, what is the rate at which the height of the water is changing? What is the rate at which this height right over here is actually changing? We know it's 2 centimeters, but how fast is it changing?"}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're being given a volume per time. My question to you is, right at this moment, right when we are filling our cup at 1 cubic centimeter per second, and we have exactly 2 centimeters of water in the cup, 2 centimeters deep of water in the cup, what is the rate at which the height of the water is changing? What is the rate at which this height right over here is actually changing? We know it's 2 centimeters, but how fast is it changing? Well, let's think about this a little bit. What are we being given? We're being given the rate at which the volume of the water is changing with respect to time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know it's 2 centimeters, but how fast is it changing? Well, let's think about this a little bit. What are we being given? We're being given the rate at which the volume of the water is changing with respect to time. So let's write that down. We're being given the rate at which the volume of the water is changing with respect to time. And we're being told."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're being given the rate at which the volume of the water is changing with respect to time. So let's write that down. We're being given the rate at which the volume of the water is changing with respect to time. And we're being told. We're told that this is 1 cubic centimeter per second. And what are we trying to figure out? Well, we're trying to figure out how fast the height of the water is changing with respect to time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're being told. We're told that this is 1 cubic centimeter per second. And what are we trying to figure out? Well, we're trying to figure out how fast the height of the water is changing with respect to time. We know that the height right now is 2 centimeters, but what we want to figure out is the rate at which the height is changing with respect to time. If we can figure out this, then we have essentially answered the question. So one way that we can do this is that we can come up with a relationship between the volume at any moment in time and the height at any moment in time, and then maybe take the derivative of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're trying to figure out how fast the height of the water is changing with respect to time. We know that the height right now is 2 centimeters, but what we want to figure out is the rate at which the height is changing with respect to time. If we can figure out this, then we have essentially answered the question. So one way that we can do this is that we can come up with a relationship between the volume at any moment in time and the height at any moment in time, and then maybe take the derivative of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing. So let's try to do it step by step. So first of all, can we come up with a relationship between the volume and the height at any given moment? Well, we have also been given the formula for the volume of a cone right over here."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one way that we can do this is that we can come up with a relationship between the volume at any moment in time and the height at any moment in time, and then maybe take the derivative of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing. So let's try to do it step by step. So first of all, can we come up with a relationship between the volume and the height at any given moment? Well, we have also been given the formula for the volume of a cone right over here. The volume of a cone is 1 third times the area of the base of the cone times the height. And we won't prove it here, although we could prove it later on, especially when we start doing solids of revolutions in integral calculus. But we'll just take it on faith right now that this is how we can figure out the volume of a cone."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we have also been given the formula for the volume of a cone right over here. The volume of a cone is 1 third times the area of the base of the cone times the height. And we won't prove it here, although we could prove it later on, especially when we start doing solids of revolutions in integral calculus. But we'll just take it on faith right now that this is how we can figure out the volume of a cone. So given this, can we figure out an expression that relates volume to the height of the cone? Well, we could say that volume, and I'll do it in this blue color, the volume of water is what we really care about. The volume of water is going to be equal to 1 third times the area of the surface of the water, area of water surface times our height of the water, so times h. So how can we figure out the area of the water surface, preferably in terms of h?"}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we'll just take it on faith right now that this is how we can figure out the volume of a cone. So given this, can we figure out an expression that relates volume to the height of the cone? Well, we could say that volume, and I'll do it in this blue color, the volume of water is what we really care about. The volume of water is going to be equal to 1 third times the area of the surface of the water, area of water surface times our height of the water, so times h. So how can we figure out the area of the water surface, preferably in terms of h? Well, we see right over here the diameter across the top of the cone is 4 centimeters, and the height of the whole cup is 4 centimeters. And so that ratio is going to be true at any depth of water. It's always going to have the same ratio between the diameter across the top and the height, because these are lines right over here."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "The volume of water is going to be equal to 1 third times the area of the surface of the water, area of water surface times our height of the water, so times h. So how can we figure out the area of the water surface, preferably in terms of h? Well, we see right over here the diameter across the top of the cone is 4 centimeters, and the height of the whole cup is 4 centimeters. And so that ratio is going to be true at any depth of water. It's always going to have the same ratio between the diameter across the top and the height, because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water, if the depth is h, the diameter across the surface of the water is also going to be h. And so from that, we can figure out what the radius is going to be. The radius is going to be h over 2."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's always going to have the same ratio between the diameter across the top and the height, because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water, if the depth is h, the diameter across the surface of the water is also going to be h. And so from that, we can figure out what the radius is going to be. The radius is going to be h over 2. And so the area of the water surface is going to be pi r squared, pi times the radius squared, h over 2 squared. That's the area of the surface of the water. And of course, we still have the 1 third out here, and we're still multiplying by this h over here."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "The radius is going to be h over 2. And so the area of the water surface is going to be pi r squared, pi times the radius squared, h over 2 squared. That's the area of the surface of the water. And of course, we still have the 1 third out here, and we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1 third times pi h squared over 4 times another h, which is equal to, we have pi h to the third power over 12. So that is our volume."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "And of course, we still have the 1 third out here, and we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1 third times pi h squared over 4 times another h, which is equal to, we have pi h to the third power over 12. So that is our volume. Now what we want to do is relate how fast the volume is changing with respect to time and how fast the height is changing with respect to time. So we care with respect to time. Since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is our volume. Now what we want to do is relate how fast the volume is changing with respect to time and how fast the height is changing with respect to time. So we care with respect to time. Since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. And just so I have enough space to do that, let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "Since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. And just so I have enough space to do that, let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well, the derivative with respect to time of our volume, we can just rewrite that as dv dt, this thing right over here. This is dv dt."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well, the derivative with respect to time of our volume, we can just rewrite that as dv dt, this thing right over here. This is dv dt. And this is going to be equal to, well, we could take the constants out of this. This is going to be equal to pi over 12 times the derivative with respect to t of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is dv dt. And this is going to be equal to, well, we could take the constants out of this. This is going to be equal to pi over 12 times the derivative with respect to t of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. In fact, it's definitely a function of time. As time goes on, the height will change because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power, just to make it clear that this is a function of t. h of t to the third power."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. In fact, it's definitely a function of time. As time goes on, the height will change because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power, just to make it clear that this is a function of t. h of t to the third power. Now, what is the derivative with respect to t of h of t to the third power? Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power, just to make it clear that this is a function of t. h of t to the third power. Now, what is the derivative with respect to t of h of t to the third power? Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us, let me rewrite everything else, dv with respect to t is going to be equal to pi over 12 times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be, let me write this in a different color, maybe an orange."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about the chain rule. The chain rule tells us, let me rewrite everything else, dv with respect to t is going to be equal to pi over 12 times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be, let me write this in a different color, maybe an orange. So that's going to be 3 times our something squared times the derivative of that something with respect to t times dh, let me, I've already used that pink, times dh dt. Let's just be very clear. This orange term right over here, and I'm just using the chain rule."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's going to be, let me write this in a different color, maybe an orange. So that's going to be 3 times our something squared times the derivative of that something with respect to t times dh, let me, I've already used that pink, times dh dt. Let's just be very clear. This orange term right over here, and I'm just using the chain rule. This is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect to t. This will give us the derivative of h of t to the third power with respect to t, which is exactly what we want to do when we apply this operator. How fast is this changing? How is this changing with respect to time?"}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "This orange term right over here, and I'm just using the chain rule. This is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect to t. This will give us the derivative of h of t to the third power with respect to t, which is exactly what we want to do when we apply this operator. How fast is this changing? How is this changing with respect to time? So we can just rewrite this, just so it gets a little bit cleaner. Let me rewrite everything I've done. So we've got dv, the rate at which our volume is changing with respect to time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "How is this changing with respect to time? So we can just rewrite this, just so it gets a little bit cleaner. Let me rewrite everything I've done. So we've got dv, the rate at which our volume is changing with respect to time. The rate at which our volume is changing with respect to time is equal to pi over 12 times 3h of t squared. Or I could just write that as 3h squared. Times the rate at which the height is changing with respect to time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we've got dv, the rate at which our volume is changing with respect to time. The rate at which our volume is changing with respect to time is equal to pi over 12 times 3h of t squared. Or I could just write that as 3h squared. Times the rate at which the height is changing with respect to time. Times dh dt. And you might be a little confused. You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "Times the rate at which the height is changing with respect to time. Times dh dt. And you might be a little confused. You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time. So we're assuming we did express volume as a function of height, but we're saying that height itself is a function of time. So we're taking the derivative of everything with respect to time. So that's why the chain rule came into play when we were taking the derivative of h or the derivative of h of t, because we're assuming that h is a function of time."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time. So we're assuming we did express volume as a function of height, but we're saying that height itself is a function of time. So we're taking the derivative of everything with respect to time. So that's why the chain rule came into play when we were taking the derivative of h or the derivative of h of t, because we're assuming that h is a function of time. Now, what does this thing right over here get us? Well, we're telling us at the exact moment that we set up this problem, we know what dv dt is. We know that it's 1 centimeter cubed per second."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's why the chain rule came into play when we were taking the derivative of h or the derivative of h of t, because we're assuming that h is a function of time. Now, what does this thing right over here get us? Well, we're telling us at the exact moment that we set up this problem, we know what dv dt is. We know that it's 1 centimeter cubed per second. We know that this right over here is 1 centimeter cubed per second. We know what our height is right at this moment. We were told it is 2 centimeters."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that it's 1 centimeter cubed per second. We know that this right over here is 1 centimeter cubed per second. We know what our height is right at this moment. We were told it is 2 centimeters. Our height right over here, we know it is 2 centimeters. So the only unknown we have over here is a rate at which our height is changing with respect to time, which is exactly what we needed to figure out in the first place. So we just have to solve for that."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "We were told it is 2 centimeters. Our height right over here, we know it is 2 centimeters. So the only unknown we have over here is a rate at which our height is changing with respect to time, which is exactly what we needed to figure out in the first place. So we just have to solve for that. So we get 1 cubic centimeter. Let me make it clear. We get 1 cubic centimeter per second."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to solve for that. So we get 1 cubic centimeter. Let me make it clear. We get 1 cubic centimeter per second. I won't write the units to save some space. Is equal to pi over 2. And I'll write this in a neutral color."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get 1 cubic centimeter per second. I won't write the units to save some space. Is equal to pi over 2. And I'll write this in a neutral color. Actually, let me write it in the same color. Is equal to pi over 2 times 3 times h squared. h is 2, so you're going to get 4 squared centimeters."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll write this in a neutral color. Actually, let me write it in the same color. Is equal to pi over 2 times 3 times h squared. h is 2, so you're going to get 4 squared centimeters. We kept the units. So 3 times 4. Let me be careful."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "h is 2, so you're going to get 4 squared centimeters. We kept the units. So 3 times 4. Let me be careful. That wasn't pi over 2. That was pi over 12. This is a pi over 12 right over here."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me be careful. That wasn't pi over 2. That was pi over 12. This is a pi over 12 right over here. Pi over 12. So you get pi over 12 times 3 times 2 squared times dh dt. All of this is equal to 1."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is a pi over 12 right over here. Pi over 12. So you get pi over 12 times 3 times 2 squared times dh dt. All of this is equal to 1. So now I'll switch to a neutral color. We get 1 is equal to 3 times 4 is 12. Cancels out with that 12."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of this is equal to 1. So now I'll switch to a neutral color. We get 1 is equal to 3 times 4 is 12. Cancels out with that 12. We get 1 is equal to pi times dh dt. To solve for dh dt, divide both sides by pi. And we get our drum roll now."}, {"video_title": "Related rates water pouring into a cone   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cancels out with that 12. We get 1 is equal to pi times dh dt. To solve for dh dt, divide both sides by pi. And we get our drum roll now. The rate at which our height is changing with respect to time, as we're putting 1 cubic centimeter of water per second in, and right when our height is 2 centimeters high, the rate at which this height is changing with respect to time is 1 over pi. And I haven't done the dimensional analysis, but this is going to be in centimeters per second. You could work through the dimensional analysis if you like by putting the dimensions right over here."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's now take the definite integral from two to four of six plus x squared over x to the third power dx. At first this might seem pretty daunting. I have this rational expression, but if we just rewrite this, it might jump out at you how this could be a little bit simpler. So this is equal to the integral from two to four of six over x to the third power plus x squared over x to the third power dx. I just separated this numerator out. I just divided each of those terms by x to the third power. And this I could rewrite."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to the integral from two to four of six over x to the third power plus x squared over x to the third power dx. I just separated this numerator out. I just divided each of those terms by x to the third power. And this I could rewrite. This is equal to the integral from two to four of six x to the negative three power. That's that first term there. And x squared divided by x to the third, well that is going to be one over x."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this I could rewrite. This is equal to the integral from two to four of six x to the negative three power. That's that first term there. And x squared divided by x to the third, well that is going to be one over x. So plus one over x dx. Now this is going to be equal to, let's take the antiderivative of the different parts, and we're going to evaluate that at four, and we're going to evaluate that at two. And we're going to find the difference between this expression, the antiderivative evaluated at four and at two."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And x squared divided by x to the third, well that is going to be one over x. So plus one over x dx. Now this is going to be equal to, let's take the antiderivative of the different parts, and we're going to evaluate that at four, and we're going to evaluate that at two. And we're going to find the difference between this expression, the antiderivative evaluated at four and at two. Now what is the antiderivative of six x to the negative three? Well here once again we can just use, we could use as a power rule for taking the antiderivative, or it's the reverse of the derivative power rule. We know that if we're taking the integral of x to the n dx, the antiderivative of that is going to be x to the n plus one over n plus one."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to find the difference between this expression, the antiderivative evaluated at four and at two. Now what is the antiderivative of six x to the negative three? Well here once again we can just use, we could use as a power rule for taking the antiderivative, or it's the reverse of the derivative power rule. We know that if we're taking the integral of x to the n dx, the antiderivative of that is going to be x to the n plus one over n plus one. And if we were just taking an indefinite integral, there would be some plus c. The reason why we don't put the plus c's here is when you evaluated both bounds of integration, the c would cancel out regardless of what it is. So we don't really think about the c much when we're taking indefinite integrals. But let's apply that to six x to the negative third power."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that if we're taking the integral of x to the n dx, the antiderivative of that is going to be x to the n plus one over n plus one. And if we were just taking an indefinite integral, there would be some plus c. The reason why we don't put the plus c's here is when you evaluated both bounds of integration, the c would cancel out regardless of what it is. So we don't really think about the c much when we're taking indefinite integrals. But let's apply that to six x to the negative third power. So it's going to be, we're going to take x to the negative three plus one, so it's x to the negative two, and so we're going to divide by negative two as well. And of course we had that six out front from the get-go, so that's the antiderivative six x to the negative three power. And what's the antiderivative one over x?"}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's apply that to six x to the negative third power. So it's going to be, we're going to take x to the negative three plus one, so it's x to the negative two, and so we're going to divide by negative two as well. And of course we had that six out front from the get-go, so that's the antiderivative six x to the negative three power. And what's the antiderivative one over x? You might be tempted to use this same idea right over here. You might be tempted to say, all right, well the antiderivative of x to the negative one, which is the same thing as one over x, would be equal to x to the negative one plus one over negative one plus one. But what is negative one plus one?"}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what's the antiderivative one over x? You might be tempted to use this same idea right over here. You might be tempted to say, all right, well the antiderivative of x to the negative one, which is the same thing as one over x, would be equal to x to the negative one plus one over negative one plus one. But what is negative one plus one? It is zero. So this doesn't fit this property right over here. But lucky for us, there is another property."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what is negative one plus one? It is zero. So this doesn't fit this property right over here. But lucky for us, there is another property. And we went the other way when we were first taking derivatives of natural log functions. The antiderivative of one over x, or x to the negative one, is equal to, sometimes you'll see it written as natural log of x plus c. And sometimes, and I actually prefer this one because you could actually evaluate it for negative values, is to say the absolute value, the natural log of the absolute value of x. And this is useful because this is defined for negative values, not just positive values."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "But lucky for us, there is another property. And we went the other way when we were first taking derivatives of natural log functions. The antiderivative of one over x, or x to the negative one, is equal to, sometimes you'll see it written as natural log of x plus c. And sometimes, and I actually prefer this one because you could actually evaluate it for negative values, is to say the absolute value, the natural log of the absolute value of x. And this is useful because this is defined for negative values, not just positive values. The natural log of x is only defined for positive values of x. But when you take the absolute value, now it could be negative or positive values of x. And it works, the derivative of this is indeed one over x."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is useful because this is defined for negative values, not just positive values. The natural log of x is only defined for positive values of x. But when you take the absolute value, now it could be negative or positive values of x. And it works, the derivative of this is indeed one over x. Now it's not so relevant here because our bounds of integration are both positive. But if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is the natural log of absolute value of x. So this, we could say, is plus the natural log of the absolute value of x."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it works, the derivative of this is indeed one over x. Now it's not so relevant here because our bounds of integration are both positive. But if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is the natural log of absolute value of x. So this, we could say, is plus the natural log of the absolute value of x. It's not a bad habit to do it, and if everything's positive, well, the absolute value of x is equal to x. And so what is this going to be equal to? This is equal to, let's evaluate everything at four."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this, we could say, is plus the natural log of the absolute value of x. It's not a bad habit to do it, and if everything's positive, well, the absolute value of x is equal to x. And so what is this going to be equal to? This is equal to, let's evaluate everything at four. And actually, before I even evaluate at four, what's six divided by negative two? That's negative three. So if we evaluate it at four, it's going to be negative three over four squared."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is equal to, let's evaluate everything at four. And actually, before I even evaluate at four, what's six divided by negative two? That's negative three. So if we evaluate it at four, it's going to be negative three over four squared. Four to the negative two is one over four squared. And then plus the natural log of the, we could say the absolute value of four, but the absolute value of four is just four. So the natural log of four."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we evaluate it at four, it's going to be negative three over four squared. Four to the negative two is one over four squared. And then plus the natural log of the, we could say the absolute value of four, but the absolute value of four is just four. So the natural log of four. And from that, we're going to subtract everything evaluated at two. So let's do that. So if we evaluated at two, it's going to be negative three over two squared."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the natural log of four. And from that, we're going to subtract everything evaluated at two. So let's do that. So if we evaluated at two, it's going to be negative three over two squared. So two to the negative two is one over two squared. Over two squared plus the natural log of, the absolute value of positive two is once again, is just two. And so what does this give us?"}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we evaluated at two, it's going to be negative three over two squared. So two to the negative two is one over two squared. Over two squared plus the natural log of, the absolute value of positive two is once again, is just two. And so what does this give us? So let's try to simplify it a little bit. So this is negative 3 16ths. Let me do that same color."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what does this give us? So let's try to simplify it a little bit. So this is negative 3 16ths. Let me do that same color. So this is going to be equal to negative three, sorry, not negative 3 16ths, gotta be very careful. Oh, sorry, yes. Sorry, it is negative 3 16ths."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do that same color. So this is going to be equal to negative three, sorry, not negative 3 16ths, gotta be very careful. Oh, sorry, yes. Sorry, it is negative 3 16ths. For some reason my brain started thinking four to the third power. Negative 3 16ths plus natural log of four. And then this right over here is negative 3 4ths."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sorry, it is negative 3 16ths. For some reason my brain started thinking four to the third power. Negative 3 16ths plus natural log of four. And then this right over here is negative 3 4ths. Negative 3 4ths, do that same color. This right over here is negative 3 4ths. We have this negative sign out front that we're going to have to distribute."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this right over here is negative 3 4ths. Negative 3 4ths, do that same color. This right over here is negative 3 4ths. We have this negative sign out front that we're going to have to distribute. So the negative of negative 3 4ths is plus 3 4ths. Plus 3 4ths. And then we're going to subtract, remember we're distributing this negative sign, the natural log, the natural log of two."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have this negative sign out front that we're going to have to distribute. So the negative of negative 3 4ths is plus 3 4ths. Plus 3 4ths. And then we're going to subtract, remember we're distributing this negative sign, the natural log, the natural log of two. And what is this equal to? All right, so this is going to be equal to, and I'm now going to switch to a neutral color. So let's add these two terms that don't involve the natural log."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're going to subtract, remember we're distributing this negative sign, the natural log, the natural log of two. And what is this equal to? All right, so this is going to be equal to, and I'm now going to switch to a neutral color. So let's add these two terms that don't involve the natural log. And let's see, if we have a common denominator, three over four is the same thing. That is the same thing as we multiply the numerator and denominator by four. That is 12 over 16."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's add these two terms that don't involve the natural log. And let's see, if we have a common denominator, three over four is the same thing. That is the same thing as we multiply the numerator and denominator by four. That is 12 over 16. And so you have negative 3 16ths, negative 3 16ths plus 12 16ths will give you 9 16ths. 9 16ths. And then we're going to have the ones that do involve the natural log."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is 12 over 16. And so you have negative 3 16ths, negative 3 16ths plus 12 16ths will give you 9 16ths. 9 16ths. And then we're going to have the ones that do involve the natural log. Natural log of four minus the natural log of two. So we could write this plus the natural log of four minus the natural log of two. And you might remember from your logarithm properties that this over here, this is the same thing as the natural log of four divided by two."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're going to have the ones that do involve the natural log. Natural log of four minus the natural log of two. So we could write this plus the natural log of four minus the natural log of two. And you might remember from your logarithm properties that this over here, this is the same thing as the natural log of four divided by two. This comes straight out of your logarithm properties. And so this is going to be the natural log of two. Natural log of two."}, {"video_title": "Definite integral involving natural log   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you might remember from your logarithm properties that this over here, this is the same thing as the natural log of four divided by two. This comes straight out of your logarithm properties. And so this is going to be the natural log of two. Natural log of two. So we deserve a little bit of a drum roll now. This is all going to be equal to, this is going to be equal to the natural log, sorry, nine over 16 plus the natural log of two. Plus the natural log of two."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And one of the reasons that it's one of my most favorite theorems in mathematics is that it has the word squeeze in it, a word that you don't see showing up in a lot of mathematics, but it is appropriately named. This is oftentimes also called the Sandwich Theorem, which is also an appropriate name, as we'll see in a second. And since it can be called the Sandwich Theorem, let's first just think about an analogy to kind of get the intuition behind the Squeeze or the Sandwich Theorem. Let's say that there are three people. Let's say that there is Imran. Imran. Let's say there's Diya."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that there are three people. Let's say that there is Imran. Imran. Let's say there's Diya. And let's say there is Sal. And let's say that Imran, on any given day, he always has the fewest amount of calories, and Sal, on any given day, always has the most number of calories. So on a given day, we can always say Diya eats at least as much as Imran."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say there's Diya. And let's say there is Sal. And let's say that Imran, on any given day, he always has the fewest amount of calories, and Sal, on any given day, always has the most number of calories. So on a given day, we can always say Diya eats at least as much as Imran. And then we can say Sal eats at least as much, not just to repeat those words, as Diya. And so we could set up a little inequality here. On a given day, we could write that Imran's calories on a given day are going to be less than or equal to Diya's calories on that same day, which is going to be less than or equal to Sal's calories on that same day."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So on a given day, we can always say Diya eats at least as much as Imran. And then we can say Sal eats at least as much, not just to repeat those words, as Diya. And so we could set up a little inequality here. On a given day, we could write that Imran's calories on a given day are going to be less than or equal to Diya's calories on that same day, which is going to be less than or equal to Sal's calories on that same day. Now let's say that it's Tuesday. Let's say on Tuesday, you find out that Imran ate 1,500 calories. And on that same day, Sal also ate 1,500 calories."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "On a given day, we could write that Imran's calories on a given day are going to be less than or equal to Diya's calories on that same day, which is going to be less than or equal to Sal's calories on that same day. Now let's say that it's Tuesday. Let's say on Tuesday, you find out that Imran ate 1,500 calories. And on that same day, Sal also ate 1,500 calories. So based on this, how many calories must Diya have eaten that day? Well, she always eats at least as many as Imran's. So she ate 1,500 calories or more."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And on that same day, Sal also ate 1,500 calories. So based on this, how many calories must Diya have eaten that day? Well, she always eats at least as many as Imran's. So she ate 1,500 calories or more. But she always has less than or equal to the number of calories Sal eats. So it must be less than or equal to 1,500. Well, there's only one number that is greater than or equal to 1,500 and less than or equal to 1,500, and that is 1,500 calories."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So she ate 1,500 calories or more. But she always has less than or equal to the number of calories Sal eats. So it must be less than or equal to 1,500. Well, there's only one number that is greater than or equal to 1,500 and less than or equal to 1,500, and that is 1,500 calories. So Diya must have eaten 1,500 calories. This is common sense. So this Diya must have had 1,500 calories."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, there's only one number that is greater than or equal to 1,500 and less than or equal to 1,500, and that is 1,500 calories. So Diya must have eaten 1,500 calories. This is common sense. So this Diya must have had 1,500 calories. And the squeeze theorem is essentially the mathematical version of this for functions. And you could even view this is Imran's calories as a function of the day, Sal's calories as a function of the day, and Diya's calories as a function of the day is always going to be in between those. So now let's make this a little bit more mathematical."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So this Diya must have had 1,500 calories. And the squeeze theorem is essentially the mathematical version of this for functions. And you could even view this is Imran's calories as a function of the day, Sal's calories as a function of the day, and Diya's calories as a function of the day is always going to be in between those. So now let's make this a little bit more mathematical. So let me clear this out so we can have some space to do some math in. So let's say that we have the same analogy. So let's say that we have three functions."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So now let's make this a little bit more mathematical. So let me clear this out so we can have some space to do some math in. So let's say that we have the same analogy. So let's say that we have three functions. Let's say f of x over some interval is always less than or equal to g of x over that same interval, which is always less than or equal to h of x over that same interval. So let me depict this graphically. So let's depict it graphically right over here."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say that we have three functions. Let's say f of x over some interval is always less than or equal to g of x over that same interval, which is always less than or equal to h of x over that same interval. So let me depict this graphically. So let's depict it graphically right over here. So that is my y-axis. This is my x-axis. And I'll just depict some interval in the x-axis right over here."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's depict it graphically right over here. So that is my y-axis. This is my x-axis. And I'll just depict some interval in the x-axis right over here. So let's say h of x looks something like that. Let me make it more interesting. Whoops."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And I'll just depict some interval in the x-axis right over here. So let's say h of x looks something like that. Let me make it more interesting. Whoops. This is the x-axis. So let's say h of x looks something like this. So that's my h of x."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Whoops. This is the x-axis. So let's say h of x looks something like this. So that's my h of x. Let's say f of x looks something like this. Maybe it does some interesting things. And then it comes in, and then it goes up like this."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's my h of x. Let's say f of x looks something like this. Maybe it does some interesting things. And then it comes in, and then it goes up like this. So f of x looks something like that. And then g of x, for any x value, g of x is always in between these two. So g of x is always in between this."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And then it comes in, and then it goes up like this. So f of x looks something like that. And then g of x, for any x value, g of x is always in between these two. So g of x is always in between this. And I think you see where the squeeze is happening and where the sandwich is happening. So this looks like if h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread. So it would look something like this."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So g of x is always in between this. And I think you see where the squeeze is happening and where the sandwich is happening. So this looks like if h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread. So it would look something like this. Now let's say that we know this is analogous to saying on a particular day, Sal and Imran ate the same amount. Let's say for a particular x value, the limit as f and h approach that x value, they approach the same limit. So let's take this x value right over here."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So it would look something like this. Now let's say that we know this is analogous to saying on a particular day, Sal and Imran ate the same amount. Let's say for a particular x value, the limit as f and h approach that x value, they approach the same limit. So let's take this x value right over here. Let's say the x value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's take this x value right over here. Let's say the x value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it. But over this interval, this has to be true. And if these limits right over here are defined, because we know that g of x is always sandwiched in between these two functions, therefore, on that day, or for that x value, I should get out of that food eating analogy, this tells us, if all of this is true over this interval, this tells us that the limit as x approaches c of g of x must also be equal to L. And once again, this is common sense. f of x is approaching L. h of x is approaching L. g of x is sandwiched in between it."}, {"video_title": "Squeeze theorem or sandwich theorem   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it. But over this interval, this has to be true. And if these limits right over here are defined, because we know that g of x is always sandwiched in between these two functions, therefore, on that day, or for that x value, I should get out of that food eating analogy, this tells us, if all of this is true over this interval, this tells us that the limit as x approaches c of g of x must also be equal to L. And once again, this is common sense. f of x is approaching L. h of x is approaching L. g of x is sandwiched in between it. So it also has to be approaching L. And you might say, well, this is common sense. Why is this useful? Well, as you'll see, this is useful for finding the limits of some wacky functions."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In fact, the definition of a derivative uses the notion of a limit. It's a slope around the point as we take the limit of points closer and closer to the point in question. You've seen that many, many, many times over. In this video, we're going to do it in the opposite direction. We're going to use derivatives to figure out limits. In particular, limits that end up in indeterminate form. When I say by indeterminate form, I mean that when we just take the limit as it is, we end up with something like 0 over 0 or infinity over infinity or negative infinity over infinity."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In this video, we're going to do it in the opposite direction. We're going to use derivatives to figure out limits. In particular, limits that end up in indeterminate form. When I say by indeterminate form, I mean that when we just take the limit as it is, we end up with something like 0 over 0 or infinity over infinity or negative infinity over infinity. All of these are indeterminate, undefined forms. To do that, we're going to use L'Hopital's Rule. In this video, I'm just going to show you what L'Hopital's Rule says and how to apply it."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "When I say by indeterminate form, I mean that when we just take the limit as it is, we end up with something like 0 over 0 or infinity over infinity or negative infinity over infinity. All of these are indeterminate, undefined forms. To do that, we're going to use L'Hopital's Rule. In this video, I'm just going to show you what L'Hopital's Rule says and how to apply it. It's fairly straightforward. It's actually a very useful tool sometimes if you're in some type of a math competition. They ask you to find a difficult limit that when you just plug the numbers in, you get something like this."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In this video, I'm just going to show you what L'Hopital's Rule says and how to apply it. It's fairly straightforward. It's actually a very useful tool sometimes if you're in some type of a math competition. They ask you to find a difficult limit that when you just plug the numbers in, you get something like this. L'Hopital's Rule is normally what they are testing you for. In a future video, I might prove it, but that gets a little bit more involved. The application is actually reasonably straightforward."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "They ask you to find a difficult limit that when you just plug the numbers in, you get something like this. L'Hopital's Rule is normally what they are testing you for. In a future video, I might prove it, but that gets a little bit more involved. The application is actually reasonably straightforward. What L'Hopital's Rule tells us is that if we have, and I'll do it in abstract form first, but I think when I show you the example, it will all be made clear. If the limit as x approaches c of f of x is equal to 0, and the limit as x approaches c of g of x is equal to 0, and the limit as x approaches c of f prime of x over g prime of x exists and it equals L, then all of these conditions have to be met. This is the indeterminate form of 0 over 0, so this is the first case."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The application is actually reasonably straightforward. What L'Hopital's Rule tells us is that if we have, and I'll do it in abstract form first, but I think when I show you the example, it will all be made clear. If the limit as x approaches c of f of x is equal to 0, and the limit as x approaches c of g of x is equal to 0, and the limit as x approaches c of f prime of x over g prime of x exists and it equals L, then all of these conditions have to be met. This is the indeterminate form of 0 over 0, so this is the first case. Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L. This might seem a little bit bizarre to you right now, and I'm actually going to write the other case, and then I'll do an example. We'll do multiple examples, and the examples are going to make it all clear. This is the first case, and the example we're going to do is actually going to be an example of this case."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the indeterminate form of 0 over 0, so this is the first case. Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L. This might seem a little bit bizarre to you right now, and I'm actually going to write the other case, and then I'll do an example. We'll do multiple examples, and the examples are going to make it all clear. This is the first case, and the example we're going to do is actually going to be an example of this case. The other case is if the limit as x approaches c of f of x is equal to positive or negative infinity, and the limit as x approaches c of g of x is equal to positive or negative infinity, and the limit of, I guess you could say the quotient of the derivatives exists, and the limit as x approaches c of f prime of x over g prime of x is equal to L, then we can make this same statement again. Then we can make this exact same statement. Let me just copy that out."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the first case, and the example we're going to do is actually going to be an example of this case. The other case is if the limit as x approaches c of f of x is equal to positive or negative infinity, and the limit as x approaches c of g of x is equal to positive or negative infinity, and the limit of, I guess you could say the quotient of the derivatives exists, and the limit as x approaches c of f prime of x over g prime of x is equal to L, then we can make this same statement again. Then we can make this exact same statement. Let me just copy that out. Then this again. Edit, copy, and then let me paste it. In either of these two situations, just to make sure you understand what you're looking at, this is a situation where if you just tried to evaluate this limit right here, you're going to get f of c, which is 0, or the limit as x approaches c of f of x over the limit as x approaches c of g of x, and that's going to give you 0 over 0."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me just copy that out. Then this again. Edit, copy, and then let me paste it. In either of these two situations, just to make sure you understand what you're looking at, this is a situation where if you just tried to evaluate this limit right here, you're going to get f of c, which is 0, or the limit as x approaches c of f of x over the limit as x approaches c of g of x, and that's going to give you 0 over 0. You say, hey, I don't know what that limit is, but this says, well, look, if this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit. If I get a number, if that exists, then they're going to be the same limit. This is a situation where when we take the limit, we get infinity over infinity, or negative infinity, or positive infinity over positive or negative infinity."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In either of these two situations, just to make sure you understand what you're looking at, this is a situation where if you just tried to evaluate this limit right here, you're going to get f of c, which is 0, or the limit as x approaches c of f of x over the limit as x approaches c of g of x, and that's going to give you 0 over 0. You say, hey, I don't know what that limit is, but this says, well, look, if this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit. If I get a number, if that exists, then they're going to be the same limit. This is a situation where when we take the limit, we get infinity over infinity, or negative infinity, or positive infinity over positive or negative infinity. These are the two indeterminate forms. To make it all clear, let me just show you an example, because I think this will make things a lot more clear. Let's say we are trying to find the limit."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is a situation where when we take the limit, we get infinity over infinity, or negative infinity, or positive infinity over positive or negative infinity. These are the two indeterminate forms. To make it all clear, let me just show you an example, because I think this will make things a lot more clear. Let's say we are trying to find the limit. Notice in a new color. Let me do it in this purplish color. Let's say we wanted to find the limit as x approaches 0 of sine of x over x."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say we are trying to find the limit. Notice in a new color. Let me do it in this purplish color. Let's say we wanted to find the limit as x approaches 0 of sine of x over x. Now, if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0 over 0. Sine of 0 is 0, or the limit as x approaches 0 of sine of x is 0, and obviously as x approaches 0 of x, that's also going to be 0. This is our indeterminate form."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say we wanted to find the limit as x approaches 0 of sine of x over x. Now, if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0 over 0. Sine of 0 is 0, or the limit as x approaches 0 of sine of x is 0, and obviously as x approaches 0 of x, that's also going to be 0. This is our indeterminate form. If you want to think about it, this is our f of x. That f of x right there is sine of x. Our g of x, this g of x right there for this first case, is the x. g of x is equal to x, and f of x is equal to sine of x."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is our indeterminate form. If you want to think about it, this is our f of x. That f of x right there is sine of x. Our g of x, this g of x right there for this first case, is the x. g of x is equal to x, and f of x is equal to sine of x. Notice, we definitely know that this meets the first two constraints. The limit as x, and in this case, c is 0. The limit as x approaches 0 of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Our g of x, this g of x right there for this first case, is the x. g of x is equal to x, and f of x is equal to sine of x. Notice, we definitely know that this meets the first two constraints. The limit as x, and in this case, c is 0. The limit as x approaches 0 of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0. We get our indeterminate form. Let's see at least whether this limit even exists. If we take the derivative of f of x and we put that over the derivative of g of x and take that the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The limit as x approaches 0 of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0. We get our indeterminate form. Let's see at least whether this limit even exists. If we take the derivative of f of x and we put that over the derivative of g of x and take that the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. I'll do that in the blue. Let me write the derivatives of the two functions. f prime of x, if f of x is sine of x, what's f prime of x?"}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "If we take the derivative of f of x and we put that over the derivative of g of x and take that the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. I'll do that in the blue. Let me write the derivatives of the two functions. f prime of x, if f of x is sine of x, what's f prime of x? It's just cosine of x. We've learned that many times. If g of x is x, what is g prime of x?"}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "f prime of x, if f of x is sine of x, what's f prime of x? It's just cosine of x. We've learned that many times. If g of x is x, what is g prime of x? That's super easy. The derivative of x is just 1. Let's try to take the limit as x approaches 0 of f prime of x over g prime of x, over their derivatives."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "If g of x is x, what is g prime of x? That's super easy. The derivative of x is just 1. Let's try to take the limit as x approaches 0 of f prime of x over g prime of x, over their derivatives. That's going to be the limit as x approaches 0 of cosine of x over 1. I wrote that 1 a little strange. Over 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's try to take the limit as x approaches 0 of f prime of x over g prime of x, over their derivatives. That's going to be the limit as x approaches 0 of cosine of x over 1. I wrote that 1 a little strange. Over 1. This is pretty straightforward. What is this going to be? As x approaches 0 of cosine of x, that's going to be equal to 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Over 1. This is pretty straightforward. What is this going to be? As x approaches 0 of cosine of x, that's going to be equal to 1. That's equal to 1. Obviously, the limit as x approaches 0 of 1, that's also going to be equal to 1. In this situation, we just saw that the limit as x approaches, our c in this case is 0, as x approaches 0 of f prime of x over g prime of x is equal to 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "As x approaches 0 of cosine of x, that's going to be equal to 1. That's equal to 1. Obviously, the limit as x approaches 0 of 1, that's also going to be equal to 1. In this situation, we just saw that the limit as x approaches, our c in this case is 0, as x approaches 0 of f prime of x over g prime of x is equal to 1. This limit exists and it equals 1. We've met all of the conditions. This is the case we're dealing with."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In this situation, we just saw that the limit as x approaches, our c in this case is 0, as x approaches 0 of f prime of x over g prime of x is equal to 1. This limit exists and it equals 1. We've met all of the conditions. This is the case we're dealing with. Limit as x approaches 0 of sine of x is equal to 0. Limit as x approaches 0 of x is also equal to 0. The limit of the derivative of sine of x over the derivative of x, which is cosine of x over 1, we found this to be equal to 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the case we're dealing with. Limit as x approaches 0 of sine of x is equal to 0. Limit as x approaches 0 of x is also equal to 0. The limit of the derivative of sine of x over the derivative of x, which is cosine of x over 1, we found this to be equal to 1. All of these top conditions are met. Then we know this must be the case, that the limit as x approaches 0 of sine of x over x must be equal to 1. It must be the same limit as this value right here, where we take the derivative of the f of x and of the g of x. I'll do more examples in the next few videos, and I think it will make it a lot more concrete."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "At the point negative one comma one, I would draw a short segment of slope blank. And like always, pause this video and see if you can fill out these three blanks. Well, when you're, the short segments that you're trying to draw to construct this slope field, you figure out their slope based on the differential equation. So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field. So the point negative one comma one, negative one comma one, a short segment of slope three."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field. So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that."}, {"video_title": "Worked example forming a slope field   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that. And then at the point two comma three, at two comma three, a short segment of slope negative one. So two comma three, a segment of slope negative one. It would look something like that."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "Divya received the following problem. A particle moves in a straight line with velocity v of t is equal to the square root of three t minus one meters per second, where t is time in seconds. At t equals two, the particle's distance from the starting point was eight meters in the positive direction. What is the particle's position at t equals seven seconds? Which expression should Divya use to solve the problem? So pause this video and have a go at it. All right, now let's do this together."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the particle's position at t equals seven seconds? Which expression should Divya use to solve the problem? So pause this video and have a go at it. All right, now let's do this together. So we wanna know the particle's position at t is equal to seven. So what we could do, they tell us what our position is at t equals two. So what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now let's do this together. So we wanna know the particle's position at t is equal to seven. So what we could do, they tell us what our position is at t equals two. So what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven. And there's another word for this. You could also call this your displacement from t equals two to t equals seven. And we know how to think about displacement."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven. And there's another word for this. You could also call this your displacement from t equals two to t equals seven. And we know how to think about displacement. Velocity is your rate of change of displacement. And so if you wanna figure out your displacement, you're between two times, you would integrate the velocity function, so this is going to be the integral from t equals two to t equals seven of our velocity function, v of t dt. This would be our displacement from time two to time seven."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know how to think about displacement. Velocity is your rate of change of displacement. And so if you wanna figure out your displacement, you're between two times, you would integrate the velocity function, so this is going to be the integral from t equals two to t equals seven of our velocity function, v of t dt. This would be our displacement from time two to time seven. So if they said what is our change in position from time two to time seven, it would be just this expression. But they want us to figure out, or they want Divya to figure out, what is the particle's position at t equals seven seconds? So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "This would be our displacement from time two to time seven. So if they said what is our change in position from time two to time seven, it would be just this expression. But they want us to figure out, or they want Divya to figure out, what is the particle's position at t equals seven seconds? So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is. It was eight meters in the positive direction, so we could just call that positive eight meters. So it's going to be eight plus your change in position, which is going to be your displacement. And let's see, we can see this choice right over there, and that's what we would pick."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is. It was eight meters in the positive direction, so we could just call that positive eight meters. So it's going to be eight plus your change in position, which is going to be your displacement. And let's see, we can see this choice right over there, and that's what we would pick. This first option, v of seven, that just gives us our velocity at time seven, or exactly at seven seconds. Or another way is our rate of change of displacement at seven seconds, so that's not what we want. This one right over here, you have your position at t equals two, but then you have your change in position from t equals zero to t equals seven."}, {"video_title": "Analyzing motion problems position   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we can see this choice right over there, and that's what we would pick. This first option, v of seven, that just gives us our velocity at time seven, or exactly at seven seconds. Or another way is our rate of change of displacement at seven seconds, so that's not what we want. This one right over here, you have your position at t equals two, but then you have your change in position from t equals zero to t equals seven. So this doesn't seem, this isn't right. And this is your position at time two plus your v prime, the derivative of velocity, is the acceleration, plus your acceleration at time seven. So that's definitely not gonna give you the particle's position, so we like that second choice."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "On which intervals is g increasing? Well at first you might say, well they don't even give us g. How do we figure out when g is increasing? Well, the answer is, all we need is g prime, which they do give us. And saying on which intervals is g increasing, that's equivalent to saying on which intervals is the first derivative with respect to x, on which intervals is that going to be greater than zero? If your rate of change with respect to x is greater than zero, if it's positive, then your function itself is going to be increasing. And so there's a couple of ways that we could approach this. You might just want to inspect kind of the structure of this expression and think about, well when is that going to be greater than zero?"}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And saying on which intervals is g increasing, that's equivalent to saying on which intervals is the first derivative with respect to x, on which intervals is that going to be greater than zero? If your rate of change with respect to x is greater than zero, if it's positive, then your function itself is going to be increasing. And so there's a couple of ways that we could approach this. You might just want to inspect kind of the structure of this expression and think about, well when is that going to be greater than zero? Or we could do it a little bit more methodically. We could say, well let's look at the critical points or the critical values for g. So critical, critical points for g. And just to remind ourselves what critical points are, that is when g prime of x is equal to zero or g prime of x is undefined, is undefined. And we have videos on critical points or critical values."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "You might just want to inspect kind of the structure of this expression and think about, well when is that going to be greater than zero? Or we could do it a little bit more methodically. We could say, well let's look at the critical points or the critical values for g. So critical, critical points for g. And just to remind ourselves what critical points are, that is when g prime of x is equal to zero or g prime of x is undefined, is undefined. And we have videos on critical points or critical values. And why those are relevant is those are the places, those are the possible places where the sign could change, the sign of g prime could change. So when is g prime of x equal to zero? Well the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we have videos on critical points or critical values. And why those are relevant is those are the places, those are the possible places where the sign could change, the sign of g prime could change. So when is g prime of x equal to zero? Well the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero. So that's the only place where g prime of x is equal to zero. And where is g prime of x undefined? Well it's going to be undefined if the denominator becomes undefined."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero. So that's the only place where g prime of x is equal to zero. And where is g prime of x undefined? Well it's going to be undefined if the denominator becomes undefined. The denominator becomes undefined if the denominator is zero. And so that's going to happen if x minus two is equal to zero x minus two is equal to zero or x is equal to two. So we have two critical points or critical values here."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well it's going to be undefined if the denominator becomes undefined. The denominator becomes undefined if the denominator is zero. And so that's going to happen if x minus two is equal to zero x minus two is equal to zero or x is equal to two. So we have two critical points or critical values here. And what I'm gonna do is I'm gonna graph them, let's put them on a number line. And let's just think about what g prime is doing in the intervals between the critical points. So let's start at zero, one, two, three, and then let's go to negative one."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have two critical points or critical values here. And what I'm gonna do is I'm gonna graph them, let's put them on a number line. And let's just think about what g prime is doing in the intervals between the critical points. So let's start at zero, one, two, three, and then let's go to negative one. And we have a critical point at, let me do that in magenta. We have a critical point at x equals zero, right over there. And we have a critical point at x equals two, right over there."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's start at zero, one, two, three, and then let's go to negative one. And we have a critical point at, let me do that in magenta. We have a critical point at x equals zero, right over there. And we have a critical point at x equals two, right over there. And so let's think about what g prime is doing in the intervals between the critical values or on either side of the critical values. So let's think about, let's first think about this interval. Let me do it in this purple color."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we have a critical point at x equals two, right over there. And so let's think about what g prime is doing in the intervals between the critical values or on either side of the critical values. So let's think about, let's first think about this interval. Let me do it in this purple color. Let's think about the interval between negative infinity and zero. So if we think about this interval, so negative infinity and zero, that open interval. Well, if we look at g prime, the numerator is still going to be positive."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do it in this purple color. Let's think about the interval between negative infinity and zero. So if we think about this interval, so negative infinity and zero, that open interval. Well, if we look at g prime, the numerator is still going to be positive. You take any negative value, you square it, you're gonna get a positive value. So this is going to be positive. Now what about the denominator?"}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if we look at g prime, the numerator is still going to be positive. You take any negative value, you square it, you're gonna get a positive value. So this is going to be positive. Now what about the denominator? You take a negative number, you subtract two from it, you're still gonna get a negative number, and then you take it to the third power. Well, a negative number to the third power is gonna be a negative number. So that right over there is going to be negative."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what about the denominator? You take a negative number, you subtract two from it, you're still gonna get a negative number, and then you take it to the third power. Well, a negative number to the third power is gonna be a negative number. So that right over there is going to be negative. So you're gonna have a positive divided by a negative. So g prime is going to be negative. So let me write that down."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that right over there is going to be negative. So you're gonna have a positive divided by a negative. So g prime is going to be negative. So let me write that down. So on this interval, on this interval, I'll write it like this, g prime of x is less than zero. Or if we cared, if we wanted to know when it's decreasing, we would know it's definitely decreasing over that interval. Now let's take the interval between zero and two, right over here."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write that down. So on this interval, on this interval, I'll write it like this, g prime of x is less than zero. Or if we cared, if we wanted to know when it's decreasing, we would know it's definitely decreasing over that interval. Now let's take the interval between zero and two, right over here. So this is the interval from zero to two, the open interval. So what's gonna go on with g prime of x here? Well, once again, x squared, anything greater than zero, and this is, we're not including zero in this interval, well, this is for sure going to be positive."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's take the interval between zero and two, right over here. So this is the interval from zero to two, the open interval. So what's gonna go on with g prime of x here? Well, once again, x squared, anything greater than zero, and this is, we're not including zero in this interval, well, this is for sure going to be positive. And so let's see, if we have x minus two, where x is greater than zero but less than two. So if x, we could just say, for example, if x was one, one minus two is negative one. We're still gonna get negative values in this denominator right over here."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, once again, x squared, anything greater than zero, and this is, we're not including zero in this interval, well, this is for sure going to be positive. And so let's see, if we have x minus two, where x is greater than zero but less than two. So if x, we could just say, for example, if x was one, one minus two is negative one. We're still gonna get negative values in this denominator right over here. So since we're still gonna get negative values in this denominator, the denominator is still going to be, take a negative value to the third power, well, you're gonna still get a negative value. So this is going to be negative. Negative."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're still gonna get negative values in this denominator right over here. So since we're still gonna get negative values in this denominator, the denominator is still going to be, take a negative value to the third power, well, you're gonna still get a negative value. So this is going to be negative. Negative. So you're still going to have g prime as less than zero. So let me write that down. So you still have g prime of x is less than zero."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative. So you're still going to have g prime as less than zero. So let me write that down. So you still have g prime of x is less than zero. And then let's take the interval above. Let's take the interval from two to infinity. Two to infinity."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you still have g prime of x is less than zero. And then let's take the interval above. Let's take the interval from two to infinity. Two to infinity. Well, the numerator is positive, it's always gonna be positive for any x not being equal to zero. And this denominator, you're taking values greater than two, subtracting two from it, which is still gonna give you a positive value. You take the third power, it's all gonna be positive."}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two to infinity. Well, the numerator is positive, it's always gonna be positive for any x not being equal to zero. And this denominator, you're taking values greater than two, subtracting two from it, which is still gonna give you a positive value. You take the third power, it's all gonna be positive. It is all going to be positive. So this is the interval where g prime of x is greater than zero. So on which intervals is g increasing?"}, {"video_title": "Finding increasing interval given the derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "You take the third power, it's all gonna be positive. It is all going to be positive. So this is the interval where g prime of x is greater than zero. So on which intervals is g increasing? Well, that's where g prime of x is greater than zero. So it's going to be from two, from two to infinity. Or we could just write it like this."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Part B, for x is not equal to zero, express f prime of x as a piecewise defined function. Find the value of x for which f prime of x is equal to negative 3. So the first thing you might be wondering is why did we even have to take out x is equal to zero from the, or why is the derivative not going to be defined there? And that's just because you're going to see that the derivative is going to be something different when we approach x is equal to zero from the left versus when we approach x is equal to zero from the right. And that's why they just took it out of there for us. So let's just figure out what that derivative is for all the other values of x. So f prime of x is equal to, so for x is less than zero, we're going to take the derivative of this first case."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's just because you're going to see that the derivative is going to be something different when we approach x is equal to zero from the left versus when we approach x is equal to zero from the right. And that's why they just took it out of there for us. So let's just figure out what that derivative is for all the other values of x. So f prime of x is equal to, so for x is less than zero, we're going to take the derivative of this first case. So the derivative of one is just zero. The derivative of negative two sine of x, well derivative of sine of x is just cosine of x. So it's going to be negative two cosine of x."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f prime of x is equal to, so for x is less than zero, we're going to take the derivative of this first case. So the derivative of one is just zero. The derivative of negative two sine of x, well derivative of sine of x is just cosine of x. So it's going to be negative two cosine of x. Negative two cosine of x for x is less than zero. And then for x is greater than zero, I'll do this in another color, I'll do it in orange. We have this case right over here."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be negative two cosine of x. Negative two cosine of x for x is less than zero. And then for x is greater than zero, I'll do this in another color, I'll do it in orange. We have this case right over here. And we'll just do the chain rule, derivative of negative four x with respect to x is negative four. And derivative of e to the negative four x with respect to e, the derivative of e to the negative four x with respect to negative four x is just e to the negative four x. Sometimes you can say this is derivative of the inside times the derivative of the outside with respect to the inside."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have this case right over here. And we'll just do the chain rule, derivative of negative four x with respect to x is negative four. And derivative of e to the negative four x with respect to e, the derivative of e to the negative four x with respect to negative four x is just e to the negative four x. Sometimes you can say this is derivative of the inside times the derivative of the outside with respect to the inside. So either way, it's negative four e to the negative four x for x is greater than zero. So we did the first part, we expressed f prime of x as a piecewise defined function. We didn't define the derivative, actually I forgot a parenthesis here."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sometimes you can say this is derivative of the inside times the derivative of the outside with respect to the inside. So either way, it's negative four e to the negative four x for x is greater than zero. So we did the first part, we expressed f prime of x as a piecewise defined function. We didn't define the derivative, actually I forgot a parenthesis here. We didn't define the derivative when x is equal to zero because it's actually not going to be defined there. Now let's do the second part. Find the value of x for which f prime of x is equal to negative three."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "We didn't define the derivative, actually I forgot a parenthesis here. We didn't define the derivative when x is equal to zero because it's actually not going to be defined there. Now let's do the second part. Find the value of x for which f prime of x is equal to negative three. And so if this wasn't piecewise defined, you'd just very simply just say, oh look, f prime of x is equal to negative three. You would take whatever f prime of x is equal to and you'd do some algebra to solve for it. But here you're like, well which case do I use?"}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Find the value of x for which f prime of x is equal to negative three. And so if this wasn't piecewise defined, you'd just very simply just say, oh look, f prime of x is equal to negative three. You would take whatever f prime of x is equal to and you'd do some algebra to solve for it. But here you're like, well which case do I use? I don't know if the x that gets us to negative three is going to be less than zero or I don't know if it's going to be greater than zero, so I don't know which case to use. And one thing to realize is to look at these functions a little bit and realize that cosine of x is a bounded function. Cosine of x can only go between positive one and negative one."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "But here you're like, well which case do I use? I don't know if the x that gets us to negative three is going to be less than zero or I don't know if it's going to be greater than zero, so I don't know which case to use. And one thing to realize is to look at these functions a little bit and realize that cosine of x is a bounded function. Cosine of x can only go between positive one and negative one. So negative two cosine of x can only go between positive two and negative two. So this can only go between positive two and negative two. So it can never get to negative three."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of x can only go between positive one and negative one. So negative two cosine of x can only go between positive two and negative two. So this can only go between positive two and negative two. So it can never get to negative three. So if anything's ever going to get to negative three, it's going to have to be this part of the derivative or this part of the derivative definition. So it's going to have to be this thing right over here. And hopefully there's some values of x greater than zero where this thing right over here is equal to negative three."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it can never get to negative three. So if anything's ever going to get to negative three, it's going to have to be this part of the derivative or this part of the derivative definition. So it's going to have to be this thing right over here. And hopefully there's some values of x greater than zero where this thing right over here is equal to negative three. So let's try it out. Negative four e to the negative four x needs to be equal to negative three. We can divide both sides by negative four."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And hopefully there's some values of x greater than zero where this thing right over here is equal to negative three. So let's try it out. Negative four e to the negative four x needs to be equal to negative three. We can divide both sides by negative four. We get e to the negative four x is equal to negative three-fourths divided by negative four is three-fourths. We can take the natural log of both sides and we will get negative four x is equal to the natural log of three-fourths. And just to be clear what I did here, you literally could put the natural log here, natural log there, and you could put the natural log there as well to see that step."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can divide both sides by negative four. We get e to the negative four x is equal to negative three-fourths divided by negative four is three-fourths. We can take the natural log of both sides and we will get negative four x is equal to the natural log of three-fourths. And just to be clear what I did here, you literally could put the natural log here, natural log there, and you could put the natural log there as well to see that step. This is saying what power do I have to raise e to to get e to the negative four x? Well, obviously I just need to raise e to the negative four x over there. So this power is negative four x."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just to be clear what I did here, you literally could put the natural log here, natural log there, and you could put the natural log there as well to see that step. This is saying what power do I have to raise e to to get e to the negative four x? Well, obviously I just need to raise e to the negative four x over there. So this power is negative four x. And then we just took the natural log of the right-hand side as well. And then to solve for x, we can divide both sides by negative four. So you get x is equal to, or we could multiply both sides by negative one-fourth either way."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this power is negative four x. And then we just took the natural log of the right-hand side as well. And then to solve for x, we can divide both sides by negative four. So you get x is equal to, or we could multiply both sides by negative one-fourth either way. Negative one-fourth natural log of three-fourths. And what we need to do is verify that this x, so we used this case right over here. We used that case, but we have to make sure that this x, that we can use this case, that this x is greater than zero."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you get x is equal to, or we could multiply both sides by negative one-fourth either way. Negative one-fourth natural log of three-fourths. And what we need to do is verify that this x, so we used this case right over here. We used that case, but we have to make sure that this x, that we can use this case, that this x is greater than zero. And we might be tempted right when we look at this to say, wait, wait, this looks like a negative number. But we have to remind ourselves that the natural log of three-fourths, since three-fourths is less than e, the natural log of three-fourths is going to be a negative number. It's going to be e to some negative exponent."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "We used that case, but we have to make sure that this x, that we can use this case, that this x is greater than zero. And we might be tempted right when we look at this to say, wait, wait, this looks like a negative number. But we have to remind ourselves that the natural log of three-fourths, since three-fourths is less than e, the natural log of three-fourths is going to be a negative number. It's going to be e to some negative exponent. So since this is negative and this part right over here is negative, so that is also negative. You have a negative times a negative, so this right over here is going to be positive. So this is a positive value right over here."}, {"video_title": "2011 Calculus AB free response #6b   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be e to some negative exponent. So since this is negative and this part right over here is negative, so that is also negative. You have a negative times a negative, so this right over here is going to be positive. So this is a positive value right over here. So you would use this case right over here. So that's our answer. x is equal to negative one-fourth natural log of three-fourths."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say I run some type of a factory and I've studied my operations and I'm able to figure out how my cost varies as a function of quantity over a week, on a weekly period. And so to visualize that, let me draw it, draw this cost function. So this is my cost axis. This right over here could be my quantity axis. So that's quantity or Q, my Q axis. Let me just call that Q. That's my Q axis."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here could be my quantity axis. So that's quantity or Q, my Q axis. Let me just call that Q. That's my Q axis. And my function might look something like this. My function might look something like this. It seems reasonable to me."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's my Q axis. And my function might look something like this. My function might look something like this. It seems reasonable to me. Even if I produce nothing, I still have fixed costs. I have to pay rent on the factory. I have to probably pay people even if we produce nothing."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It seems reasonable to me. Even if I produce nothing, I still have fixed costs. I have to pay rent on the factory. I have to probably pay people even if we produce nothing. And so let's say that fixed cost in the week is $1,000. And then as my quantity increases, so do my costs. So if I produce 100 units right over here, then my cost goes up to 1,300."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I have to probably pay people even if we produce nothing. And so let's say that fixed cost in the week is $1,000. And then as my quantity increases, so do my costs. So if I produce 100 units right over here, then my cost goes up to 1,300. 1,300. If I produce more than that, you see my costs increase and they increase at an ever faster rate. Now, I go into a lot more depth on things like cost functions in the economics playlist."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I produce 100 units right over here, then my cost goes up to 1,300. 1,300. If I produce more than that, you see my costs increase and they increase at an ever faster rate. Now, I go into a lot more depth on things like cost functions in the economics playlist. But what I wanna think about in the calculus context is what would the derivative of this represent? What would the derivative of C with respect to Q, which could be also written as C prime of Q, what does that represent? Well, if we think about it visually, we know that we can think about the derivative as the slope of the tangent line."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, I go into a lot more depth on things like cost functions in the economics playlist. But what I wanna think about in the calculus context is what would the derivative of this represent? What would the derivative of C with respect to Q, which could be also written as C prime of Q, what does that represent? Well, if we think about it visually, we know that we can think about the derivative as the slope of the tangent line. So for example, that's the tangent line when Q is equal to 100. So the slope of that tangent line is, you could view as C prime, or it is C prime of 100. But what is that slope telling us?"}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if we think about it visually, we know that we can think about the derivative as the slope of the tangent line. So for example, that's the tangent line when Q is equal to 100. So the slope of that tangent line is, you could view as C prime, or it is C prime of 100. But what is that slope telling us? Well, the slope is the change in our cost divided by the change in our quantity. And it's the slope of the tangent line. This is what we first learned in calculus."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what is that slope telling us? Well, the slope is the change in our cost divided by the change in our quantity. And it's the slope of the tangent line. This is what we first learned in calculus. As we get to smaller and smaller and smaller changes in quantity, we essentially take the limit as our change in quantity approaches zero. That's how we get the instantaneous change. So one way to think about it, one way to think about it is this is the instantaneous, this is the rate right on the margin at which our cost is changing with respect to quantity."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is what we first learned in calculus. As we get to smaller and smaller and smaller changes in quantity, we essentially take the limit as our change in quantity approaches zero. That's how we get the instantaneous change. So one way to think about it, one way to think about it is this is the instantaneous, this is the rate right on the margin at which our cost is changing with respect to quantity. So if I were to produce just another drop, another atom of whatever I'm producing, at what rate is my cost going to increase? And the reason why I'm saying it right on the margin is we see that it's not constant. If our cost function were a line, we would have a constant slope."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one way to think about it, one way to think about it is this is the instantaneous, this is the rate right on the margin at which our cost is changing with respect to quantity. So if I were to produce just another drop, another atom of whatever I'm producing, at what rate is my cost going to increase? And the reason why I'm saying it right on the margin is we see that it's not constant. If our cost function were a line, we would have a constant slope. The tangent line would essentially be the cost function. But we see it changes right over here. The incremental atom to produce here costs less than the incremental atom right over here."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If our cost function were a line, we would have a constant slope. The tangent line would essentially be the cost function. But we see it changes right over here. The incremental atom to produce here costs less than the incremental atom right over here. The slope has gone up. And it might make sense. Maybe I'm using some raw material out there in the world, and as I use more and more of it, it becomes more and more scarce, and so the market price of it goes up and up and up."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The incremental atom to produce here costs less than the incremental atom right over here. The slope has gone up. And it might make sense. Maybe I'm using some raw material out there in the world, and as I use more and more of it, it becomes more and more scarce, and so the market price of it goes up and up and up. But you might say, well, why do I even care about the rate at which my costs are increasing, the rate at which my costs are increasing on the margin, which is why this is called marginal cost. Well, the reason why you care about it is you might be trying to figure out when do I stop producing? Let's say this is orange juice."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe I'm using some raw material out there in the world, and as I use more and more of it, it becomes more and more scarce, and so the market price of it goes up and up and up. But you might say, well, why do I even care about the rate at which my costs are increasing, the rate at which my costs are increasing on the margin, which is why this is called marginal cost. Well, the reason why you care about it is you might be trying to figure out when do I stop producing? Let's say this is orange juice. If I know that next gallon is going to cost me $5 to produce, and I can sell it for $6, then I'm gonna do it. But if that next gallon, if I'm up here, I've already produced a lot, and I'm taking all the oranges off the market, and now I have to transport oranges from the other side of the planet or whatever it might be, and now that incremental gallon of oranges or a gallon of orange juice costs me $10 to produce, and I'm not going to be able to sell it for more than $6, it doesn't make sense for me to produce it anymore. So in a calculus context, or I guess you say in an economics context, if you can model your cost as a function of quantity, the derivative of that is the marginal cost."}, {"video_title": "Marginal cost & differential calculus   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say this is orange juice. If I know that next gallon is going to cost me $5 to produce, and I can sell it for $6, then I'm gonna do it. But if that next gallon, if I'm up here, I've already produced a lot, and I'm taking all the oranges off the market, and now I have to transport oranges from the other side of the planet or whatever it might be, and now that incremental gallon of oranges or a gallon of orange juice costs me $10 to produce, and I'm not going to be able to sell it for more than $6, it doesn't make sense for me to produce it anymore. So in a calculus context, or I guess you say in an economics context, if you can model your cost as a function of quantity, the derivative of that is the marginal cost. It's the rate at which costs are increasing for that incremental unit. And there's other similar ideas. If we modeled our profit as a function of quantity, if we took the derivative, that would be our marginal profit."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's get some practice rewriting definite integrals as the limit of a Riemann sum. So let's say I wanted to take the definite integral from pi to two pi of cosine of x dx. And what I want to do is I want to write it as the limit as n approaches infinity of a Riemann sum. So it's gonna take the form of the limit as n approaches infinity, and we can have our sigma notation right over here. And I would say from, let's say, i is equal to one all the way to n. Let me scroll down a little bit so it doesn't get all squenched up at the top. Of, and so let me draw what's actually going on so that we can get a better sense of what to write here within the sigma notation. So, let me do it large."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's gonna take the form of the limit as n approaches infinity, and we can have our sigma notation right over here. And I would say from, let's say, i is equal to one all the way to n. Let me scroll down a little bit so it doesn't get all squenched up at the top. Of, and so let me draw what's actually going on so that we can get a better sense of what to write here within the sigma notation. So, let me do it large. So if this is pi right over here, pi right over here, this would be three pi over two, and this would be two pi right over here, two pi. Now what does the graph of cosine of x do? Well, at pi, cosine of pi is negative one."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, let me do it large. So if this is pi right over here, pi right over here, this would be three pi over two, and this would be two pi right over here, two pi. Now what does the graph of cosine of x do? Well, at pi, cosine of pi is negative one. We'll assume that's negative one there. And cosine of two pi is one. And so the graph is gonna do something like this."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, at pi, cosine of pi is negative one. We'll assume that's negative one there. And cosine of two pi is one. And so the graph is gonna do something like this. And this is obviously just a hand-drawn version of it. You have seen cosine functions before. This is just part of it."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the graph is gonna do something like this. And this is obviously just a hand-drawn version of it. You have seen cosine functions before. This is just part of it. And so this definite integral represents the area from pi to two pi between the curve and the x-axis. And you might already know that this area is going to be, or this part of the definite integral would be negative, and this would be positive and they'll cancel out, and this would all actually end up being zero in this case. But this exercise for this video is to rewrite this as a limit as n approaches infinity of a Riemann sum."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is just part of it. And so this definite integral represents the area from pi to two pi between the curve and the x-axis. And you might already know that this area is going to be, or this part of the definite integral would be negative, and this would be positive and they'll cancel out, and this would all actually end up being zero in this case. But this exercise for this video is to rewrite this as a limit as n approaches infinity of a Riemann sum. So as a Riemann sum, what we wanna do is think about breaking this up into a bunch of rectangles. So let's say, or I should say n rectangles. So that's our first one right over there."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this exercise for this video is to rewrite this as a limit as n approaches infinity of a Riemann sum. So as a Riemann sum, what we wanna do is think about breaking this up into a bunch of rectangles. So let's say, or I should say n rectangles. So that's our first one right over there. Then this might be our second one. And let's do right Riemann sum, where the right boundary of our rectangle, what the value of the function is at that point, that's what defines the height. So that's our second one, all the way until this one right over here is going to be our nth one."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's our first one right over there. Then this might be our second one. And let's do right Riemann sum, where the right boundary of our rectangle, what the value of the function is at that point, that's what defines the height. So that's our second one, all the way until this one right over here is going to be our nth one. So this is one, let me write it this way. This is i is equal to one, this is i is equal to two, all the way until we get to i is equal to n. And then if we take the limit as n approaches infinity, the sum of the areas of these rectangles are going to get better and better and better. And so let's first think about it."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's our second one, all the way until this one right over here is going to be our nth one. So this is one, let me write it this way. This is i is equal to one, this is i is equal to two, all the way until we get to i is equal to n. And then if we take the limit as n approaches infinity, the sum of the areas of these rectangles are going to get better and better and better. And so let's first think about it. What is the width of each of these rectangles going to be? Well, I am taking this interval from pi to two pi, and I'm gonna divide it into n equal intervals. So the width of each of these, the width of each of these, is going to be two pi minus pi."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's first think about it. What is the width of each of these rectangles going to be? Well, I am taking this interval from pi to two pi, and I'm gonna divide it into n equal intervals. So the width of each of these, the width of each of these, is going to be two pi minus pi. So I'm just taking the difference between my bounds of integration, and I am dividing by n, which is equal to pi over n. So that's the width of each of these, that's pi over n. This is pi over n, this is pi over n. And what's the height of each of these rectangles? And remember, this is a right Riemann sum, so it's going to be the right end of our rectangle is going to define the height. So this right over here, what would this height be?"}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the width of each of these, the width of each of these, is going to be two pi minus pi. So I'm just taking the difference between my bounds of integration, and I am dividing by n, which is equal to pi over n. So that's the width of each of these, that's pi over n. This is pi over n, this is pi over n. And what's the height of each of these rectangles? And remember, this is a right Riemann sum, so it's going to be the right end of our rectangle is going to define the height. So this right over here, what would this height be? Well, this height, this value, I should say, this is going to be equal to f of what? Well, this was pi, and this is going to be pi plus the length of our interval right over here, the base of the rectangle. So we started at pi, so it's going to be pi plus, this one's going to be pi over n, I could say, times one."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here, what would this height be? Well, this height, this value, I should say, this is going to be equal to f of what? Well, this was pi, and this is going to be pi plus the length of our interval right over here, the base of the rectangle. So we started at pi, so it's going to be pi plus, this one's going to be pi over n, I could say, times one. That's this height right over here. What's this one going to be right over here? Well, this one is going to be f of pi, our first start, plus pi over n times what?"}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we started at pi, so it's going to be pi plus, this one's going to be pi over n, I could say, times one. That's this height right over here. What's this one going to be right over here? Well, this one is going to be f of pi, our first start, plus pi over n times what? We're going to add pi over n two times. Pi over n times two. So the general form of the right boundary is going to be, so for example, this height right over here, this is going to be f of, we started at pi, plus, we're doing the right Riemann sum, so we're gonna add pi over n n times by this point."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this one is going to be f of pi, our first start, plus pi over n times what? We're going to add pi over n two times. Pi over n times two. So the general form of the right boundary is going to be, so for example, this height right over here, this is going to be f of, we started at pi, plus, we're doing the right Riemann sum, so we're gonna add pi over n n times by this point. Pi over n times n. Or, if we wanted to say it generally, if we're talking about the ith rectangle, remember, we're gonna sum them all up, what's their height? Well, the height is going to be, in this case, it's going to be cosine of pi plus, if we're with the ith rectangle, we are going to add pi over n i times. Pi over n times i."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the general form of the right boundary is going to be, so for example, this height right over here, this is going to be f of, we started at pi, plus, we're doing the right Riemann sum, so we're gonna add pi over n n times by this point. Pi over n times n. Or, if we wanted to say it generally, if we're talking about the ith rectangle, remember, we're gonna sum them all up, what's their height? Well, the height is going to be, in this case, it's going to be cosine of pi plus, if we're with the ith rectangle, we are going to add pi over n i times. Pi over n times i. And then, that's the height of each of our rectangles. And then what's the width? Well, we already figured that out."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's now tackle part C, which tells us for T is less than 10, an alternate model for the internal temperature of the potato at time T minutes is the function G that satisfies the differential equation, the derivative of G with respect to T is equal to the negative of G minus 27 to the 2 3rds power, where G of T is measured in degrees Celsius, and G of zero is equal to 91. Find an expression for G of T based on this model. What is the internal temperature of the potato at time T is equal to three? So they gave us a differential equation. They want us to find an expression for G of T, so essentially find a solution to this differential equation, and then they want us to use that solution to find the internal temperature at time T equals three. So the first thing to appreciate, this is in an AP calculus class, and so if they're asking us to solve a differential equation or find a solution to a differential equation, it is unlikely to be a really strange differential equation. It's likely to be a separable differential equation, and then once we find that solution, we just have to evaluate it at time T is equal to three."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they gave us a differential equation. They want us to find an expression for G of T, so essentially find a solution to this differential equation, and then they want us to use that solution to find the internal temperature at time T equals three. So the first thing to appreciate, this is in an AP calculus class, and so if they're asking us to solve a differential equation or find a solution to a differential equation, it is unlikely to be a really strange differential equation. It's likely to be a separable differential equation, and then once we find that solution, we just have to evaluate it at time T is equal to three. So let's rewrite the differential equation, and then let's try to evaluate. Let's see if we can find a solution, and then we'll evaluate. The derivative of G with respect to T is equal to the negative of G minus 27 to the 2 3rds power."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's likely to be a separable differential equation, and then once we find that solution, we just have to evaluate it at time T is equal to three. So let's rewrite the differential equation, and then let's try to evaluate. Let's see if we can find a solution, and then we'll evaluate. The derivative of G with respect to T is equal to the negative of G minus 27 to the 2 3rds power. So if this is going to be a separable differential equation, I want to separate the DG and the DT. So I'm gonna treat my differential like numbers, so, or variables. So I'm gonna multiply both sides times the T differential, and so then I'm going to have the capital G differential, DG is equal to negative times G minus 27 to the 2 3rds power DT, and the whole notion here, I'm trying to get all of the G things on the side with the DG, and then all of the things that involve T on the side with the DT."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of G with respect to T is equal to the negative of G minus 27 to the 2 3rds power. So if this is going to be a separable differential equation, I want to separate the DG and the DT. So I'm gonna treat my differential like numbers, so, or variables. So I'm gonna multiply both sides times the T differential, and so then I'm going to have the capital G differential, DG is equal to negative times G minus 27 to the 2 3rds power DT, and the whole notion here, I'm trying to get all of the G things on the side with the DG, and then all of the things that involve T on the side with the DT. So we don't see any Ts here, so really we just have to get the Gs over on this side while leaving the DT there, and then we can integrate both sides. So let's see. If we were to divide both sides by G minus 27 to the 2 3rds, G minus 27 to the 2 3rds, G minus 27 to the 2 3rds, what do we have?"}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm gonna multiply both sides times the T differential, and so then I'm going to have the capital G differential, DG is equal to negative times G minus 27 to the 2 3rds power DT, and the whole notion here, I'm trying to get all of the G things on the side with the DG, and then all of the things that involve T on the side with the DT. So we don't see any Ts here, so really we just have to get the Gs over on this side while leaving the DT there, and then we can integrate both sides. So let's see. If we were to divide both sides by G minus 27 to the 2 3rds, G minus 27 to the 2 3rds, G minus 27 to the 2 3rds, what do we have? Well we can rewrite the left side as G minus 27 to the negative 2 3rds, negative 2 3rds, DG is equal to, I'm left with just this negative DT, is equal to negative DT. Now let's see. Now we can, or actually just to simplify, or just to make it a little bit more obvious, I could write that as negative one DT, and now we just have to integrate both sides."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we were to divide both sides by G minus 27 to the 2 3rds, G minus 27 to the 2 3rds, G minus 27 to the 2 3rds, what do we have? Well we can rewrite the left side as G minus 27 to the negative 2 3rds, negative 2 3rds, DG is equal to, I'm left with just this negative DT, is equal to negative DT. Now let's see. Now we can, or actually just to simplify, or just to make it a little bit more obvious, I could write that as negative one DT, and now we just have to integrate both sides. So we could rewrite this as, let me write an integral sign. I'm going to integrate, I'm going to integrate both sides here, and so what is this going to give me? Well on the left side here, you could try to do some U substitution, saying U is equal to G minus 27, and then DU would be DG, or you might recognize that look, the derivative of G minus 27 is just DG."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we can, or actually just to simplify, or just to make it a little bit more obvious, I could write that as negative one DT, and now we just have to integrate both sides. So we could rewrite this as, let me write an integral sign. I'm going to integrate, I'm going to integrate both sides here, and so what is this going to give me? Well on the left side here, you could try to do some U substitution, saying U is equal to G minus 27, and then DU would be DG, or you might recognize that look, the derivative of G minus 27 is just DG. The derivative of G minus 27 is just going to be equal to one so you could even say, look I have the derivative there, so I could really integrate with respect to G minus 27. And so really I would just use the reverse power rule. I would take my G minus 27, I would increment this exponent by one, so let's see negative 2 3rds plus one is positive 1 3rd, so positive 1 3rd power, and then I would divide by this new exponent, so dividing by 1 3rd is the same thing as multiplying by three, so that's the left hand side."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well on the left side here, you could try to do some U substitution, saying U is equal to G minus 27, and then DU would be DG, or you might recognize that look, the derivative of G minus 27 is just DG. The derivative of G minus 27 is just going to be equal to one so you could even say, look I have the derivative there, so I could really integrate with respect to G minus 27. And so really I would just use the reverse power rule. I would take my G minus 27, I would increment this exponent by one, so let's see negative 2 3rds plus one is positive 1 3rd, so positive 1 3rd power, and then I would divide by this new exponent, so dividing by 1 3rd is the same thing as multiplying by three, so that's the left hand side. This is going to be equal to, the right hand side, this is just going to be negative T, and for good measure, I'm going to have a plus C right over here. So how can we solve for C? Well they give us some information."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "I would take my G minus 27, I would increment this exponent by one, so let's see negative 2 3rds plus one is positive 1 3rd, so positive 1 3rd power, and then I would divide by this new exponent, so dividing by 1 3rd is the same thing as multiplying by three, so that's the left hand side. This is going to be equal to, the right hand side, this is just going to be negative T, and for good measure, I'm going to have a plus C right over here. So how can we solve for C? Well they give us some information. They say G of zero is equal to 91, so let's write that down. So when T is zero, G is 91, so we can write, let me write it over here. So three times, so when T is zero, G is 91."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well they give us some information. They say G of zero is equal to 91, so let's write that down. So when T is zero, G is 91, so we can write, let me write it over here. So three times, so when T is zero, G is 91. 91 minus 27 to the 1 3rd power is equal to negative T. Well now we're saying T is zero, so we could write negative zero there, or we could just not write it, and then plus C. So that's what C is going to be equal to. Let's see, 91 minus 27 is 64. 64 to the 1 3rd power is positive four, and so we have C is equal to 12."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So three times, so when T is zero, G is 91. 91 minus 27 to the 1 3rd power is equal to negative T. Well now we're saying T is zero, so we could write negative zero there, or we could just not write it, and then plus C. So that's what C is going to be equal to. Let's see, 91 minus 27 is 64. 64 to the 1 3rd power is positive four, and so we have C is equal to 12. C is equal to 12, and let's see. We want to write an expression for G of T, so now let's just manipulate this and solve for G. So if we, so let me just take this and go right over here. So if we divide both sides by three, we are going to get G minus 27 to the 1 3rd power is equal to negative T over three plus four."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "64 to the 1 3rd power is positive four, and so we have C is equal to 12. C is equal to 12, and let's see. We want to write an expression for G of T, so now let's just manipulate this and solve for G. So if we, so let me just take this and go right over here. So if we divide both sides by three, we are going to get G minus 27 to the 1 3rd power is equal to negative T over three plus four. I just divided both sides by three. Now I can take the cube of both sides, and I would get G minus 27 is equal to negative T over three plus four to the 3rd power, and now I just have to add 27 to both sides, and I'll make it clear. G is a function of T is equal to negative T over three plus four."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we divide both sides by three, we are going to get G minus 27 to the 1 3rd power is equal to negative T over three plus four. I just divided both sides by three. Now I can take the cube of both sides, and I would get G minus 27 is equal to negative T over three plus four to the 3rd power, and now I just have to add 27 to both sides, and I'll make it clear. G is a function of T is equal to negative T over three plus four. All of that to the 3rd power plus 27. So I did the first part. This is, well, let me not cross this."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "G is a function of T is equal to negative T over three plus four. All of that to the 3rd power plus 27. So I did the first part. This is, well, let me not cross this. This right over here is an expression for G of T. It was indeed a separable differential equation. It took a little bit of algebraic manipulation to get us there, but we were able to do it, and we were not only able to solve for the general solution. We were able to find the particular solution using this initial condition right over here, that G of zero is equal to 91."}, {"video_title": "2017 AP Calculus AB BC 4c   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is, well, let me not cross this. This right over here is an expression for G of T. It was indeed a separable differential equation. It took a little bit of algebraic manipulation to get us there, but we were able to do it, and we were not only able to solve for the general solution. We were able to find the particular solution using this initial condition right over here, that G of zero is equal to 91. Now we'll do the easy part. Based on this model, what is the internal temperature of the potato at time T equals three? So G of three is equal to negative three over three plus four, let me just write it out."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say I have two functions that are the inverse of each other. So I have f of x, and then I also have g of x, which is equal to the inverse of f of x, and f of x would be the inverse of g of x as well. If the notion of an inverse function is completely unfamiliar to you, I encourage you to review inverse functions on Khan Academy. Now one of the properties of inverse functions are that if I were to take g of f of x, g of f of x, or I could say the f inverse of f of x, that this is just going to be equal to x. And it comes straight out of what an inverse of a function is. If this is x right over here, the function f would map to some value f of x, so that's f of x right over there, and then the function g, or f inverse, if you input f of x into it, it would take you back, it would take you back to x. So that would be f inverse, or we're saying g is the same thing as f inverse."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now one of the properties of inverse functions are that if I were to take g of f of x, g of f of x, or I could say the f inverse of f of x, that this is just going to be equal to x. And it comes straight out of what an inverse of a function is. If this is x right over here, the function f would map to some value f of x, so that's f of x right over there, and then the function g, or f inverse, if you input f of x into it, it would take you back, it would take you back to x. So that would be f inverse, or we're saying g is the same thing as f inverse. So all of that so far is a review of inverse functions, but now we're going to apply a little bit of calculus to it using the chain rule. And we're gonna get a pretty interesting result. What I wanna do is take the derivative of both sides of this equation right over here."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would be f inverse, or we're saying g is the same thing as f inverse. So all of that so far is a review of inverse functions, but now we're going to apply a little bit of calculus to it using the chain rule. And we're gonna get a pretty interesting result. What I wanna do is take the derivative of both sides of this equation right over here. So let's apply the derivative operator, d dx on the left-hand side, d dx on the right-hand side, and what are we going to get? Well, on the left-hand side, we would apply the chain rule. So this is going to be the derivative of g with respect to f of x, so that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x, and then that is going to be equal to what?"}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I wanna do is take the derivative of both sides of this equation right over here. So let's apply the derivative operator, d dx on the left-hand side, d dx on the right-hand side, and what are we going to get? Well, on the left-hand side, we would apply the chain rule. So this is going to be the derivative of g with respect to f of x, so that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x, and then that is going to be equal to what? Well, the derivative with respect to x of x, that's just equal to one. And this is where we get our interesting result. All we did so far is we used something we knew about inverse functions, and we used the chain rule to take the derivative of the left-hand side, but if you divide both sides by g prime of f of x, what are you going to get?"}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be the derivative of g with respect to f of x, so that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x, and then that is going to be equal to what? Well, the derivative with respect to x of x, that's just equal to one. And this is where we get our interesting result. All we did so far is we used something we knew about inverse functions, and we used the chain rule to take the derivative of the left-hand side, but if you divide both sides by g prime of f of x, what are you going to get? You're going to get a relationship between the derivative of a function and the derivative of its inverse. So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x. And this is really neat, because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "All we did so far is we used something we knew about inverse functions, and we used the chain rule to take the derivative of the left-hand side, but if you divide both sides by g prime of f of x, what are you going to get? You're going to get a relationship between the derivative of a function and the derivative of its inverse. So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x. And this is really neat, because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse. And we can actually see this is true with some classic functions. So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f, so f inverse, which is, what's the inverse of e to the x? Well, one way to think about it is if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is really neat, because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse. And we can actually see this is true with some classic functions. So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f, so f inverse, which is, what's the inverse of e to the x? Well, one way to think about it is if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again. So you'd get x is equal to e to the y. You take the natural log of both sides, you get natural log of x is equal to y. So the inverse of e to the x is natural log of x."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, one way to think about it is if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again. So you'd get x is equal to e to the y. You take the natural log of both sides, you get natural log of x is equal to y. So the inverse of e to the x is natural log of x. And once again, that's all review of inverse functions. If that's unfamiliar, review it on Khan Academy. So g of x is going to be equal to the natural log of x."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the inverse of e to the x is natural log of x. And once again, that's all review of inverse functions. If that's unfamiliar, review it on Khan Academy. So g of x is going to be equal to the natural log of x. Now, let's see if this holds true for these two functions. Well, what is f prime of x going to be? Well, this is one of those amazing results in calculus, one of these neat things about the number e is that the derivative of e to the x is e to the x."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g of x is going to be equal to the natural log of x. Now, let's see if this holds true for these two functions. Well, what is f prime of x going to be? Well, this is one of those amazing results in calculus, one of these neat things about the number e is that the derivative of e to the x is e to the x. And in other videos, we also saw that the derivative of the natural log of x is one over x. So let's see if this holds out. So we should get a result, f prime of x, e to the x, should be equal to one over g prime of f of x."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is one of those amazing results in calculus, one of these neat things about the number e is that the derivative of e to the x is e to the x. And in other videos, we also saw that the derivative of the natural log of x is one over x. So let's see if this holds out. So we should get a result, f prime of x, e to the x, should be equal to one over g prime of f of x. So g prime of f of x, so g prime is one over our f of x, and f of x is e to the x, one over e to the x. Is this indeed true? Yes, it is."}, {"video_title": "Derivatives of inverse functions   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we should get a result, f prime of x, e to the x, should be equal to one over g prime of f of x. So g prime of f of x, so g prime is one over our f of x, and f of x is e to the x, one over e to the x. Is this indeed true? Yes, it is. One over one over e to the x is just going to be e to the x. So it all checks out. And you could do it the other way because these are inverses of each other."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "you are likely already familiar with finding the area between curves, and in fact, if you're not, I encourage you to review that on Khan Academy. For example, we could find this yellow area using a definite integral. But what we're going to do in this video is do something even more interesting. We're gonna find the volume of shapes where the base is defined in some way by the area between two curves. And in this video, we're gonna think about a shape, and I'm gonna draw it in three dimensions. So let me draw this over again, but with a little bit of perspective. So let's make this the y-axis."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna find the volume of shapes where the base is defined in some way by the area between two curves. And in this video, we're gonna think about a shape, and I'm gonna draw it in three dimensions. So let me draw this over again, but with a little bit of perspective. So let's make this the y-axis. So that's the y-axis. This is my x-axis. That is my x-axis."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's make this the y-axis. So that's the y-axis. This is my x-axis. That is my x-axis. This is the line y is equal to six, right over there. Y is equal to six. This dotted line, we could just draw it like this."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is my x-axis. This is the line y is equal to six, right over there. Y is equal to six. This dotted line, we could just draw it like this. And so this would be the point x equals two. And then the graph of y is equal to four times the natural log of three minus x would look something like this. Look something like this."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "This dotted line, we could just draw it like this. And so this would be the point x equals two. And then the graph of y is equal to four times the natural log of three minus x would look something like this. Look something like this. And so this region is this region. But it's going to be the base of a three-dimensional shape where any cross-section, if I were to take a cross-section right over here, is going to be a square. So whatever this length is, we also go that much high."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Look something like this. And so this region is this region. But it's going to be the base of a three-dimensional shape where any cross-section, if I were to take a cross-section right over here, is going to be a square. So whatever this length is, we also go that much high. And so the cross-section is a square right over there. The cross-section right over here is going to be a square, whatever the difference between these two functions is, that's also how high we are going to go. This length, which is six at this point, this is also going to be the height."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So whatever this length is, we also go that much high. And so the cross-section is a square right over there. The cross-section right over here is going to be a square, whatever the difference between these two functions is, that's also how high we are going to go. This length, which is six at this point, this is also going to be the height. It is going to be a square. It's going to be quite big. I'm gonna have to scroll down so we can draw the whole thing roughly at the right proportion."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "This length, which is six at this point, this is also going to be the height. It is going to be a square. It's going to be quite big. I'm gonna have to scroll down so we can draw the whole thing roughly at the right proportion. So it'll look something like this. This should be a square. It's gonna look something like this."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm gonna have to scroll down so we can draw the whole thing roughly at the right proportion. So it'll look something like this. This should be a square. It's gonna look something like this. And so the whole shape would look something like this. Would look something like that. Try to shade that in a little bit so that you can appreciate it a little bit more."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's gonna look something like this. And so the whole shape would look something like this. Would look something like that. Try to shade that in a little bit so that you can appreciate it a little bit more. But hopefully you get the idea. And some of you might be excited and some of you might be a little intimidated. Well, hey, I've been dealing with the two dimensions for so long."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Try to shade that in a little bit so that you can appreciate it a little bit more. But hopefully you get the idea. And some of you might be excited and some of you might be a little intimidated. Well, hey, I've been dealing with the two dimensions for so long. What's going on with these three dimensions? But you'll quickly appreciate that you already have the powers of integration to solve this and to do that, we just have to break up the shape into a bunch of these, you could view them as these little square tiles that have some depth to them. So let's make that into a little tile, this one into it that also has some depth to it."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, hey, I've been dealing with the two dimensions for so long. What's going on with these three dimensions? But you'll quickly appreciate that you already have the powers of integration to solve this and to do that, we just have to break up the shape into a bunch of these, you could view them as these little square tiles that have some depth to them. So let's make that into a little tile, this one into it that also has some depth to it. You could even, I could draw it multiple places. You could view it as a, break it up into these things that have a very small depth that we could call dx. And we know how to figure out what their volume is."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's make that into a little tile, this one into it that also has some depth to it. You could even, I could draw it multiple places. You could view it as a, break it up into these things that have a very small depth that we could call dx. And we know how to figure out what their volume is. What would be the volume of one of these things? Well, it would be the depth times the area, times the surface area of this cross section right over here. Let me do that in a different color."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know how to figure out what their volume is. What would be the volume of one of these things? Well, it would be the depth times the area, times the surface area of this cross section right over here. Let me do that in a different color. So what would be the area that I am shading in in pink right over here? Well, that area is going to be the base length squared. What's the base length?"}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do that in a different color. So what would be the area that I am shading in in pink right over here? Well, that area is going to be the base length squared. What's the base length? What's the difference between these two functions? It is going to be six minus, our bottom function is four times the natural log of three minus x. And so that would just give us that length."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's the base length? What's the difference between these two functions? It is going to be six minus, our bottom function is four times the natural log of three minus x. And so that would just give us that length. But if we square it, we get this entire area. We get that entire area, you square it. And then you multiply it times the depth."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that would just give us that length. But if we square it, we get this entire area. We get that entire area, you square it. And then you multiply it times the depth. You multiply it times the depth. Now you have the volume of just this little section right over here. And I think you might see where this is going."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you multiply it times the depth. You multiply it times the depth. Now you have the volume of just this little section right over here. And I think you might see where this is going. Now what if you were to add up all of these from x equals zero to x equals two? Well, then you would have the volume of the entire thing. This is the power of the definite integral."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I think you might see where this is going. Now what if you were to add up all of these from x equals zero to x equals two? Well, then you would have the volume of the entire thing. This is the power of the definite integral. So we could just integrate from x equals zero to x equals two from x equals zero to x equals two. If you drew where these intersect our base, you would say, all right, this thing right over here would be this thing right over here, where it's dx. Instead of just multiplying dx times the difference between these functions, we're going to square the difference of these functions because we're visualizing this three-dimensional shape, the surface area of this three-dimensional shape as opposed to just the height of this little rectangle."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the power of the definite integral. So we could just integrate from x equals zero to x equals two from x equals zero to x equals two. If you drew where these intersect our base, you would say, all right, this thing right over here would be this thing right over here, where it's dx. Instead of just multiplying dx times the difference between these functions, we're going to square the difference of these functions because we're visualizing this three-dimensional shape, the surface area of this three-dimensional shape as opposed to just the height of this little rectangle. And if you were to evaluate this integral, you would indeed get the volume of this kind of pedestal horn-looking thing. This is not an easy definite integral to evaluate by hand, but we can actually use a calculator for that. And so we can hit Math and then hit choice number nine for definite integral."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of just multiplying dx times the difference between these functions, we're going to square the difference of these functions because we're visualizing this three-dimensional shape, the surface area of this three-dimensional shape as opposed to just the height of this little rectangle. And if you were to evaluate this integral, you would indeed get the volume of this kind of pedestal horn-looking thing. This is not an easy definite integral to evaluate by hand, but we can actually use a calculator for that. And so we can hit Math and then hit choice number nine for definite integral. And then we just have to input everything. We're going from zero till two of, and then we have, let me open parentheses because I'm gonna have to square everything, six minus four times the natural log of x, or actually the natural log of three minus x. And so let me close the parentheses on the natural log part."}, {"video_title": "Volume with cross sections intro   Applications of integration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can hit Math and then hit choice number nine for definite integral. And then we just have to input everything. We're going from zero till two of, and then we have, let me open parentheses because I'm gonna have to square everything, six minus four times the natural log of x, or actually the natural log of three minus x. And so let me close the parentheses on the natural log part. And then if I close the parentheses on this whole thing, I want to then square it. And then I'm integrating with respect to x, Enter. I got approximately 26.27."}, {"video_title": "Switching bounds of definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've already seen one definition of the definite integral, and many of them are closely related to this definition that we've already seen, is, well, look, the definite integral from a to b of f of x, d of x is this area shaded in blue, and we can approximate it by splitting it into n rectangles, so let's say that's the first rectangle, one, that's the second rectangle, two, and you're gonna go all the way to the nth rectangle, so this would be the n minus oneth rectangle, and for the sake of this argument I'm gonna make in this video, I'm gonna assume that they're all the same width, so this is the nth rectangle, and they all have the same width, and we see there are definitions of integration where you don't have to have the same width here, but let's say that each of those widths are delta x, and the way that we calculate delta x is we take b minus a, we take b minus a, and we divide it by n, we divide it by n, which is common sense, or this is what you learned in division, we're just taking this length and dividing it by n to get n equal spacings, which is n equal spacings of delta x, and so if you do this, you say, okay, we see this multiple times, you can approximate it, you can approximate this area using these rectangles as the sum from i equals one to n, so you're summing, you're summing n of these rectangles areas, where the height of each of these rectangles are gonna be f of x sub i, where x sub i is the point at which you're taking the function value to find out its height, so that could be x sub one, x sub two, x sub three, so on and so forth, and you're multiplying that times your delta x, times your delta x, so you take x sub two, f of x sub two is that height right there, f of x sub two is that height right there, you multiply it times delta x, you get the area, and we saw that when we looked at Riemann sums and using that to approximate, and we said, hey, the one definition of the definite integral is that since this is the area, this is going to be the limit as n approaches infinity of this, where delta x is defined as that, so let me just copy and paste that, so copy and paste, where that, so that's one way to think about it. Now, given this definition, what do you think, what do you think this, or maybe another way to think about it, how do you think this expression that I'm writing right over here, based on this definition, should relate to this expression? So notice, all I've done is I've, instead of going from a to b, I'm now going from b to a, I'm now going from b to a. How do you think these two things should relate? And I encourage you to look at all of this to come to that conclusion, and pause the video to do so. Well, let's just think about what's going to happen. This is going to be, if I were to literally just take this, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I am going to want to swap these two."}, {"video_title": "Switching bounds of definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "How do you think these two things should relate? And I encourage you to look at all of this to come to that conclusion, and pause the video to do so. Well, let's just think about what's going to happen. This is going to be, if I were to literally just take this, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I am going to want to swap these two. Instead of b minus a, it's going to be a minus b now. It's going to be, it's going to be a minus b. So each of these are going, this value right over here, let me make these color-coded maybe, so this orange delta x, this orange delta x is going to be the negative of this green, of this green delta x."}, {"video_title": "Switching bounds of definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be, if I were to literally just take this, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I am going to want to swap these two. Instead of b minus a, it's going to be a minus b now. It's going to be, it's going to be a minus b. So each of these are going, this value right over here, let me make these color-coded maybe, so this orange delta x, this orange delta x is going to be the negative of this green, of this green delta x. This is the negative of that right over there. And everything else is the same. So what am I going to end up doing?"}, {"video_title": "Switching bounds of definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "So each of these are going, this value right over here, let me make these color-coded maybe, so this orange delta x, this orange delta x is going to be the negative of this green, of this green delta x. This is the negative of that right over there. And everything else is the same. So what am I going to end up doing? Well, I'm essentially going to end up having the negative value of this. So this is going to be equal to the negative of the integral from a to b of f of x dx. And so this is the result we get, which is another really important integration property that if you swap, if you swap the bounds of integration, and it really just comes from this idea."}, {"video_title": "Switching bounds of definite integral   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what am I going to end up doing? Well, I'm essentially going to end up having the negative value of this. So this is going to be equal to the negative of the integral from a to b of f of x dx. And so this is the result we get, which is another really important integration property that if you swap, if you swap the bounds of integration, and it really just comes from this idea. Instead of delta x being b minus a, if you swap the bounds of integration, it's going to be a minus b. You're going to get the negative delta x, or the negative of your original delta x, which is going to give you the negative of this original value right over here. And once again, this is a really, really useful integration property where you're trying to make sense of some integrals and even sometimes solve some of them."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one way you could think of it, if you set f of x as being equal to sine of x, and g of x being the natural log of x, and let's see, fg, let's say h of x, h of x equaling x squared, then this thing right over here is the exact same thing as trying to take the derivative with respect to x of f of g of h of x. And what I want to do is kind of think about it how I would do it in my head, without having to write all the chain rule notation. So the way I would think about this, if I were doing this in my head, is the derivative of this outer function of f with respect to the level of composition directly below it. So the derivative of sine of x is cosine of x. But instead of it being cosine of x, it's going to be cosine of whatever was inside of here. So it's going to be cosine of natural log. Let me write that in that same color."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of sine of x is cosine of x. But instead of it being cosine of x, it's going to be cosine of whatever was inside of here. So it's going to be cosine of natural log. Let me write that in that same color. Cosine of natural log of x squared. I'm going to do x in that same yellow color. Cosine of x squared."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write that in that same color. Cosine of natural log of x squared. I'm going to do x in that same yellow color. Cosine of x squared. And so you could really view this part, what I just write over here, as f prime. This is f prime of g of h of x. This is f prime of g of h of x."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of x squared. And so you could really view this part, what I just write over here, as f prime. This is f prime of g of h of x. This is f prime of g of h of x. If you want to keep track of things. So I just took the derivative of the outer with respect to whatever was inside of it. And now I have to take the derivative of the inside with respect to x."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is f prime of g of h of x. If you want to keep track of things. So I just took the derivative of the outer with respect to whatever was inside of it. And now I have to take the derivative of the inside with respect to x. But now we have another composite function. So we're going to multiply this times. We're going to do the chain rule again."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now I have to take the derivative of the inside with respect to x. But now we have another composite function. So we're going to multiply this times. We're going to do the chain rule again. The derivative of, we're going to take the derivative of ln with respect to x squared. So the derivative of ln of x is 1 over x. But now we're going to have 1 over not x, but 1 over x squared."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to do the chain rule again. The derivative of, we're going to take the derivative of ln with respect to x squared. So the derivative of ln of x is 1 over x. But now we're going to have 1 over not x, but 1 over x squared. So to be clear, this part right over here is g prime of not x. If it was g prime of x, this would be 1 over x. But instead of an x, we have our h of x there."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now we're going to have 1 over not x, but 1 over x squared. So to be clear, this part right over here is g prime of not x. If it was g prime of x, this would be 1 over x. But instead of an x, we have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we can take the derivative of our inner function."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But instead of an x, we have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we can take the derivative of our inner function. Let me write it. So we could write this as g prime of h of x. And finally, we just have to take the derivative of our innermost function with respect to x."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, we can take the derivative of our inner function. Let me write it. So we could write this as g prime of h of x. And finally, we just have to take the derivative of our innermost function with respect to x. So the derivative of x squared with respect to x is 2x. So times h prime of x. Let me make everything clear."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And finally, we just have to take the derivative of our innermost function with respect to x. So the derivative of x squared with respect to x is 2x. So times h prime of x. Let me make everything clear. So what we have right over here in purple, this, this, and this are the same things. One expressed concretely, one expressed abstractly. This, this, and this are the same thing, expressed concretely and abstractly."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make everything clear. So what we have right over here in purple, this, this, and this are the same things. One expressed concretely, one expressed abstractly. This, this, and this are the same thing, expressed concretely and abstractly. And then finally, this and this are the same thing, expressed concretely and abstractly. But then we're done. All we have to do to be done is to just simplify this."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This, this, and this are the same thing, expressed concretely and abstractly. And then finally, this and this are the same thing, expressed concretely and abstractly. But then we're done. All we have to do to be done is to just simplify this. So if we just change the order in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this 2x over x squared is the same thing as 2 over x."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "All we have to do to be done is to just simplify this. So if we just change the order in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this 2x over x squared is the same thing as 2 over x. And we're multiplying it times all of this business. So we're left with 2 over x. This goes away."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this 2x over x squared is the same thing as 2 over x. And we're multiplying it times all of this business. So we're left with 2 over x. This goes away. 2 over x times the cosine of the natural log of x squared. So it seemed like a very daunting derivative. But we just say, OK, what's the derivative of sine of something with respect to that something?"}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This goes away. 2 over x times the cosine of the natural log of x squared. So it seemed like a very daunting derivative. But we just say, OK, what's the derivative of sine of something with respect to that something? Well, that's cosine of that something. And then we go in one layer. What's the derivative of that something?"}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we just say, OK, what's the derivative of sine of something with respect to that something? Well, that's cosine of that something. And then we go in one layer. What's the derivative of that something? Well, in that something, we have another composition. So the derivative of ln of x or ln of something with respect to another something, well, that's going to be 1 over the something. So we had gotten a 1 over x squared here."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's the derivative of that something? Well, in that something, we have another composition. So the derivative of ln of x or ln of something with respect to another something, well, that's going to be 1 over the something. So we had gotten a 1 over x squared here. That squared got canceled out. And then finally, the derivative of this innermost function. It's like peeling an onion."}, {"video_title": "Derivative of sin(ln(x_))   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we had gotten a 1 over x squared here. That squared got canceled out. And then finally, the derivative of this innermost function. It's like peeling an onion. The derivative of this inner function with respect to x, which was just 2x, which we got right over here. This was 1 over x squared. This was 2x before we did any canceling out."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just think about it in general terms. If we just think about a general function over an interval, where it could have an absolute maximum. So let me draw some axes over here. And I'm speaking in general terms first, and then we can go back to our function g, which is derived from this function f right over here. So let's say that these are my coordinate axes, and let's say we care about some interval here. So let's say this is the interval that I care about. A function could look something like this."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm speaking in general terms first, and then we can go back to our function g, which is derived from this function f right over here. So let's say that these are my coordinate axes, and let's say we care about some interval here. So let's say this is the interval that I care about. A function could look something like this. A function could look something like this. And in this case, its absolute maximum is going to occur at the beginning of the interval. Or a function could look something like this."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "A function could look something like this. A function could look something like this. And in this case, its absolute maximum is going to occur at the beginning of the interval. Or a function could look something like this. It could look something like that. And then the absolute maximum could occur at the end point of the interval, or the other possibility is that the function looks something like this, at which point the maximum would be at this critical point. And I say critical point as opposed to just a point where the slope is zero, because it's possible that the function is not differentiable there."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or a function could look something like this. It could look something like that. And then the absolute maximum could occur at the end point of the interval, or the other possibility is that the function looks something like this, at which point the maximum would be at this critical point. And I say critical point as opposed to just a point where the slope is zero, because it's possible that the function is not differentiable there. You could imagine a function that looks like this. And maybe it wouldn't be differentiable there, but that point there still would be the absolute maximum. So what we really just have to do is evaluate g at the different end points of this interval to see how high it gets or how large of a value we get for the g at the end points."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I say critical point as opposed to just a point where the slope is zero, because it's possible that the function is not differentiable there. You could imagine a function that looks like this. And maybe it wouldn't be differentiable there, but that point there still would be the absolute maximum. So what we really just have to do is evaluate g at the different end points of this interval to see how high it gets or how large of a value we get for the g at the end points. And then we have to see if g has any critical points in between, and then evaluate it there to see if that's a candidate for the global maximum. So let's just evaluate g at the different end points. So let's start off, let's evaluate g at negative 4, at kind of the lowest end or the starting point of our interval."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we really just have to do is evaluate g at the different end points of this interval to see how high it gets or how large of a value we get for the g at the end points. And then we have to see if g has any critical points in between, and then evaluate it there to see if that's a candidate for the global maximum. So let's just evaluate g at the different end points. So let's start off, let's evaluate g at negative 4, at kind of the lowest end or the starting point of our interval. So g of negative 4 is equal to 2 times negative 4, plus the integral from 0 to negative 4 f of t dt. The first part is very easy, 2 times negative 4 is negative 8. Let me do it over here so I have some real estate."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's start off, let's evaluate g at negative 4, at kind of the lowest end or the starting point of our interval. So g of negative 4 is equal to 2 times negative 4, plus the integral from 0 to negative 4 f of t dt. The first part is very easy, 2 times negative 4 is negative 8. Let me do it over here so I have some real estate. So this is equal to negative 8, and instead of leaving this as 0 to negative 4 f of t dt, let's change the bounds of integration here, especially so that we can get the lower number as the lower bound. And that way it becomes a little bit more natural to think of it in terms of areas. So this expression right here can be rewritten as the negative of the integral between negative 4 and 0 f of t dt."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do it over here so I have some real estate. So this is equal to negative 8, and instead of leaving this as 0 to negative 4 f of t dt, let's change the bounds of integration here, especially so that we can get the lower number as the lower bound. And that way it becomes a little bit more natural to think of it in terms of areas. So this expression right here can be rewritten as the negative of the integral between negative 4 and 0 f of t dt. And now this expression right over here is the area under f of t, or in this case f of x, or the area under f, between negative 4 and 0. So it's this area right over here. And we have to be careful because this part over here is below the x-axis, so this we would consider negative area, we would think of it in integration terms, and this would be positive area."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this expression right here can be rewritten as the negative of the integral between negative 4 and 0 f of t dt. And now this expression right over here is the area under f of t, or in this case f of x, or the area under f, between negative 4 and 0. So it's this area right over here. And we have to be careful because this part over here is below the x-axis, so this we would consider negative area, we would think of it in integration terms, and this would be positive area. So the total area here is going to be this positive area minus this area right over here. So let's think about what this is. So this section over here, we did this in part a actually, this is a quarter circle, so it's 1 fourth, so these are both quarter circles, so we could multiply 1 fourth times the area of this circle, the area of this circle, of this entire circle, if we were to draw the entire thing all the way around, it has a radius of 3, so the area of the entire circle would be pi times 3 squared, or it would be 9 pi."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we have to be careful because this part over here is below the x-axis, so this we would consider negative area, we would think of it in integration terms, and this would be positive area. So the total area here is going to be this positive area minus this area right over here. So let's think about what this is. So this section over here, we did this in part a actually, this is a quarter circle, so it's 1 fourth, so these are both quarter circles, so we could multiply 1 fourth times the area of this circle, the area of this circle, of this entire circle, if we were to draw the entire thing all the way around, it has a radius of 3, so the area of the entire circle would be pi times 3 squared, or it would be 9 pi. And of course we're going to divide it by 4, multiply it by 1 fourth to just get this quarter circle right over there. And then this area right over here, the area of the entire circle, the area of this entire circle, we have a radius of 1, so it's going to be pi times 1 squared, so it's going to be pi, and we're going to divide it by 4, we're going to divide it by 4 because it's only 1 fourth of that circle. And we're going to subtract that, so we have negative pi, and we're multiplying it times 1 fourth out here, because it's a quarter circle in either case, and we're subtracting it because the area is below the x axis."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this section over here, we did this in part a actually, this is a quarter circle, so it's 1 fourth, so these are both quarter circles, so we could multiply 1 fourth times the area of this circle, the area of this circle, of this entire circle, if we were to draw the entire thing all the way around, it has a radius of 3, so the area of the entire circle would be pi times 3 squared, or it would be 9 pi. And of course we're going to divide it by 4, multiply it by 1 fourth to just get this quarter circle right over there. And then this area right over here, the area of the entire circle, the area of this entire circle, we have a radius of 1, so it's going to be pi times 1 squared, so it's going to be pi, and we're going to divide it by 4, we're going to divide it by 4 because it's only 1 fourth of that circle. And we're going to subtract that, so we have negative pi, and we're multiplying it times 1 fourth out here, because it's a quarter circle in either case, and we're subtracting it because the area is below the x axis. And so this simplifies to, this is equal to 1 fourth times 8 pi, which is the same thing as 2 pi. Did I do that right? 1 fourth times 8 pi, so this all simplifies to 2 pi."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to subtract that, so we have negative pi, and we're multiplying it times 1 fourth out here, because it's a quarter circle in either case, and we're subtracting it because the area is below the x axis. And so this simplifies to, this is equal to 1 fourth times 8 pi, which is the same thing as 2 pi. Did I do that right? 1 fourth times 8 pi, so this all simplifies to 2 pi. So g of negative 4 is equal to negative 8 minus 2 pi. So clearly a negative number here, more negative than negative 8. So let's try the other bounds."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "1 fourth times 8 pi, so this all simplifies to 2 pi. So g of negative 4 is equal to negative 8 minus 2 pi. So clearly a negative number here, more negative than negative 8. So let's try the other bounds. Let's see what g of positive 3 is. Let's see what g of positive, I'll do it over here so I have more space. So g of positive 3, when x is equal to 3, that, we go back to our definition, that is 2 times 3, plus the integral from 0 to 3, f of t dt."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's try the other bounds. Let's see what g of positive 3 is. Let's see what g of positive, I'll do it over here so I have more space. So g of positive 3, when x is equal to 3, that, we go back to our definition, that is 2 times 3, plus the integral from 0 to 3, f of t dt. And this is going to be equal to 2 times 3 is 6, and the integral from 0 to 3, f of t dt, that's this entire area. So we have positive area over here, and then we have an equal negative area right over here, because it's below the x-axis. So the integral from 0 to 3 is just going to be 0."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g of positive 3, when x is equal to 3, that, we go back to our definition, that is 2 times 3, plus the integral from 0 to 3, f of t dt. And this is going to be equal to 2 times 3 is 6, and the integral from 0 to 3, f of t dt, that's this entire area. So we have positive area over here, and then we have an equal negative area right over here, because it's below the x-axis. So the integral from 0 to 3 is just going to be 0. You're going to have this positive area, and then this negative area right over here is going to completely cancel it out, because it's symmetric right over here. So this thing is going to evaluate to 0. So g of 3 is 6."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the integral from 0 to 3 is just going to be 0. You're going to have this positive area, and then this negative area right over here is going to completely cancel it out, because it's symmetric right over here. So this thing is going to evaluate to 0. So g of 3 is 6. So we already know that our starting point, g of negative 4, that when x is equal to negative 4, that is not where g hits a global maximum, because that's a negative number, but it's the end point where g hits a positive value. So negative 4 is definitely not a candidate. 3, x is equal to 3, is still in the running for the x-coordinate where g has a global absolute maximum."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g of 3 is 6. So we already know that our starting point, g of negative 4, that when x is equal to negative 4, that is not where g hits a global maximum, because that's a negative number, but it's the end point where g hits a positive value. So negative 4 is definitely not a candidate. 3, x is equal to 3, is still in the running for the x-coordinate where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points that it's either undifferentiable or its derivative is equal to 0. So let's look at its derivative."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "3, x is equal to 3, is still in the running for the x-coordinate where g has a global absolute maximum. Now what we have to do is figure out any critical points that g has in between. So points that it's either undifferentiable or its derivative is equal to 0. So let's look at its derivative. So g prime of x, we just take the derivative of this business up here, the derivative of 2x is 2, derivative of this definite integral from 0 to x of f of t dt, we did that in Part A, this is just the fundamental theorem of calculus, this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere over the interval."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at its derivative. So g prime of x, we just take the derivative of this business up here, the derivative of 2x is 2, derivative of this definite integral from 0 to x of f of t dt, we did that in Part A, this is just the fundamental theorem of calculus, this is just going to be plus f of x right over there. So it actually turns out that g is differentiable over the entire interval. You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere over the interval. So you'll get a number here, and obviously 2 is just 2, and you add 2 to it and you get the derivative of g at that point of the interval. So g is actually differentiable throughout the interval. So the only critical points would be where this derivative is equal to 0."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "You give any x value over this interval, we have a value for f of x. f of x isn't differentiable everywhere, but definitely f of x is defined everywhere over the interval. So you'll get a number here, and obviously 2 is just 2, and you add 2 to it and you get the derivative of g at that point of the interval. So g is actually differentiable throughout the interval. So the only critical points would be where this derivative is equal to 0. So let's set this thing equal to 0. So we want to solve the equation, I'll just rewrite it actually. So we want to solve the equation g prime of x is equal to 0, or 2 plus f of x is equal to 0."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the only critical points would be where this derivative is equal to 0. So let's set this thing equal to 0. So we want to solve the equation, I'll just rewrite it actually. So we want to solve the equation g prime of x is equal to 0, or 2 plus f of x is equal to 0. You can subtract 2 from both sides, and you get f of x is equal to negative 2. So any x that satisfies f of x is equal to negative 2 is a point where the derivative of g is equal to 0. And let's see if f of x is equal to negative 2 at any point."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to solve the equation g prime of x is equal to 0, or 2 plus f of x is equal to 0. You can subtract 2 from both sides, and you get f of x is equal to negative 2. So any x that satisfies f of x is equal to negative 2 is a point where the derivative of g is equal to 0. And let's see if f of x is equal to negative 2 at any point. So let me draw a line over here, negative 2. Negative 2, we have to look at it visually, because they've only given us this visual definition of f of x. Doesn't equal negative 2, doesn't equal to negative 2, only equals negative 2 right over there."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see if f of x is equal to negative 2 at any point. So let me draw a line over here, negative 2. Negative 2, we have to look at it visually, because they've only given us this visual definition of f of x. Doesn't equal negative 2, doesn't equal to negative 2, only equals negative 2 right over there. And it looks like we're at about 2 and a half, but let's get exact. Let's actually figure out the slope of the line and figure out what x value actually gives f of x equal to negative 2. And we can figure out the slope of this line fairly visually, or figure out the equation of this line fairly visually."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Doesn't equal negative 2, doesn't equal to negative 2, only equals negative 2 right over there. And it looks like we're at about 2 and a half, but let's get exact. Let's actually figure out the slope of the line and figure out what x value actually gives f of x equal to negative 2. And we can figure out the slope of this line fairly visually, or figure out the equation of this line fairly visually. We can figure out its slope if we run 3. So if our change in x is 3, then our change in y, our rise, is negative 6. Change in y is equal to negative 6."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can figure out the slope of this line fairly visually, or figure out the equation of this line fairly visually. We can figure out its slope if we run 3. So if our change in x is 3, then our change in y, our rise, is negative 6. Change in y is equal to negative 6. Slope is rise over run, or change in y over change in x. So negative 6 divided by 3 is negative 2. It has a slope of negative 2."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Change in y is equal to negative 6. Slope is rise over run, or change in y over change in x. So negative 6 divided by 3 is negative 2. It has a slope of negative 2. I could have done that easier. Where if we go forward 1, we go down by 2. So we could have seen that."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "It has a slope of negative 2. I could have done that easier. Where if we go forward 1, we go down by 2. So we could have seen that. The slope is negative 2. So this part of f of x, we have y is equal to negative 2x, plus, and then the y-intercept is pretty straightforward. This is at 3, 1, 2, 3, plus negative 2x plus 3."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could have seen that. The slope is negative 2. So this part of f of x, we have y is equal to negative 2x, plus, and then the y-intercept is pretty straightforward. This is at 3, 1, 2, 3, plus negative 2x plus 3. So where we're, the part of f of x where clearly it equals negative 2 at some point of that, this part of f of x is defined by this line. Obviously, this part of f of x is not defined by that line. But to figure out the exact value, we just have to figure out when this line is equal to negative 2."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is at 3, 1, 2, 3, plus negative 2x plus 3. So where we're, the part of f of x where clearly it equals negative 2 at some point of that, this part of f of x is defined by this line. Obviously, this part of f of x is not defined by that line. But to figure out the exact value, we just have to figure out when this line is equal to negative 2. So we have negative 2x plus 3 is equal to negative 2. And remember, this is what f of x is equal to over the interval that we care about. If we were talking about f of x over there, we wouldn't be able to put negative 2x plus 3."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "But to figure out the exact value, we just have to figure out when this line is equal to negative 2. So we have negative 2x plus 3 is equal to negative 2. And remember, this is what f of x is equal to over the interval that we care about. If we were talking about f of x over there, we wouldn't be able to put negative 2x plus 3. We'd have to have some form of equation for these circles. But right over here, this is what f of x is, and now we can solve this pretty straightforwardly. So we could subtract 3 from both sides."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we were talking about f of x over there, we wouldn't be able to put negative 2x plus 3. We'd have to have some form of equation for these circles. But right over here, this is what f of x is, and now we can solve this pretty straightforwardly. So we could subtract 3 from both sides. And we get negative 2x is equal to negative 5. Divide both sides by negative 2. You get x is equal to negative 5 over negative 2, which is equal to 5 halves, which is exactly what we thought it was when we looked at it visually."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could subtract 3 from both sides. And we get negative 2x is equal to negative 5. Divide both sides by negative 2. You get x is equal to negative 5 over negative 2, which is equal to 5 halves, which is exactly what we thought it was when we looked at it visually. It looked like we were at about 2 and 1 half, which is the same thing as 5 halves. Now, we don't know what this is. We don't know if this is an inflection point, is this a maximum, is this a minimum?"}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "You get x is equal to negative 5 over negative 2, which is equal to 5 halves, which is exactly what we thought it was when we looked at it visually. It looked like we were at about 2 and 1 half, which is the same thing as 5 halves. Now, we don't know what this is. We don't know if this is an inflection point, is this a maximum, is this a minimum? So really we just want to evaluate g at this point to see if it gets higher than when we evaluate g at 3. So let's evaluate g at 5 halves. So g at 5 halves is going to be equal to 2 times 5 halves plus the integral from 0 to 5 halves of f of t dt."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know if this is an inflection point, is this a maximum, is this a minimum? So really we just want to evaluate g at this point to see if it gets higher than when we evaluate g at 3. So let's evaluate g at 5 halves. So g at 5 halves is going to be equal to 2 times 5 halves plus the integral from 0 to 5 halves of f of t dt. So this first part right over here, the 2's cancel out, so this is going to be equal to 5, and then plus the integral from 0 to 5 halves. Now, you might be able to do it visually, but we know what the value is of f of t over this interval. We already figured out the equation for it over this interval."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g at 5 halves is going to be equal to 2 times 5 halves plus the integral from 0 to 5 halves of f of t dt. So this first part right over here, the 2's cancel out, so this is going to be equal to 5, and then plus the integral from 0 to 5 halves. Now, you might be able to do it visually, but we know what the value is of f of t over this interval. We already figured out the equation for it over this interval. So it's the integral of negative 2x plus 3 dt. And then let's just evaluate this. Let me get some real estate."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already figured out the equation for it over this interval. So it's the integral of negative 2x plus 3 dt. And then let's just evaluate this. Let me get some real estate. So this is, let me draw a line here so we don't get confused. So this is going to be equal to 5 plus, and then take the antiderivative. The antiderivative of negative 2x is negative x squared."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me get some real estate. So this is, let me draw a line here so we don't get confused. So this is going to be equal to 5 plus, and then take the antiderivative. The antiderivative of negative 2x is negative x squared. So negative, let me do it. So we have negative x squared, and the antiderivative of 3 is just going to be 3x. So plus 3x, and we're going to evaluate it from 0 to 5 halves."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of negative 2x is negative x squared. So negative, let me do it. So we have negative x squared, and the antiderivative of 3 is just going to be 3x. So plus 3x, and we're going to evaluate it from 0 to 5 halves. So this is going to be equal to 5 plus, and I'll do all of this stuff right over here. I'll do this stuff in green. So when we evaluate it at 5 halves, this is going to be negative 5 halves squared."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So plus 3x, and we're going to evaluate it from 0 to 5 halves. So this is going to be equal to 5 plus, and I'll do all of this stuff right over here. I'll do this stuff in green. So when we evaluate it at 5 halves, this is going to be negative 5 halves squared. So it's going to be negative 25 over 4 plus 3 times 5 halves, which is 15 over 2. And then from that, we're going to subtract this evaluated at 0. But negative 0 squared plus 3 times 0 is just 0."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when we evaluate it at 5 halves, this is going to be negative 5 halves squared. So it's going to be negative 25 over 4 plus 3 times 5 halves, which is 15 over 2. And then from that, we're going to subtract this evaluated at 0. But negative 0 squared plus 3 times 0 is just 0. So this is what it simplifies to. And so what do we have right over here? So let's get ourselves a common denominator."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "But negative 0 squared plus 3 times 0 is just 0. So this is what it simplifies to. And so what do we have right over here? So let's get ourselves a common denominator. It looks like a common denominator right over here. It could be 4. So this is equal to 5 is the same thing as 20 over 4 minus 25 over 4 and then plus 30 over 4."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's get ourselves a common denominator. It looks like a common denominator right over here. It could be 4. So this is equal to 5 is the same thing as 20 over 4 minus 25 over 4 and then plus 30 over 4. So 20 plus 30 is 50 minus 25 is 25. So this is equal to 25 over 4. And 25 over 4 is the same thing as 6 and 1 fourth."}, {"video_title": "2011 Calculus AB free response #4b   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to 5 is the same thing as 20 over 4 minus 25 over 4 and then plus 30 over 4. So 20 plus 30 is 50 minus 25 is 25. So this is equal to 25 over 4. And 25 over 4 is the same thing as 6 and 1 fourth. So when we evaluate our function at this critical point, at this thing where the slope, where the derivative is equal to 0, we got 6 and 1 fourth, which is higher than 6, which is what g was at this end point. And it's definitely higher than what g was at negative 4. So the x-coordinate of the point at which g has an absolute maximum on the interval negative 4 to 3 is x is equal to 5 halves."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "What happens if n equals 0? So let's just think of the situation. Let's try to take the derivative with respect to x of x to the 0 power. Well, what is x to the 0 power going to be? Well, we can assume that x for this case right over here is not equal to 0. 0 to the 0, weird things happen at that point. But if x does not equal 0, what is x to the 0 power going to be?"}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what is x to the 0 power going to be? Well, we can assume that x for this case right over here is not equal to 0. 0 to the 0, weird things happen at that point. But if x does not equal 0, what is x to the 0 power going to be? Well, this is the same thing as the derivative with respect to x of 1. x to the 0 power is just going to be 1. And so what is the derivative with respect to x of 1? And to answer that question, I'll just graph it."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if x does not equal 0, what is x to the 0 power going to be? Well, this is the same thing as the derivative with respect to x of 1. x to the 0 power is just going to be 1. And so what is the derivative with respect to x of 1? And to answer that question, I'll just graph it. I'll just graph f of x equals 1 to make it a little bit clearer. So that's my y-axis. This is my x-axis."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to answer that question, I'll just graph it. I'll just graph f of x equals 1 to make it a little bit clearer. So that's my y-axis. This is my x-axis. And let me graph y equals 1, or f of x equals 1. So that's 1 right over there. f of x equals 1 is just a horizontal line."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is my x-axis. And let me graph y equals 1, or f of x equals 1. So that's 1 right over there. f of x equals 1 is just a horizontal line. So that right over there is the graph y is equal to f of x, which is equal to 1. Now, remember, the derivative, one way to conceptualize it, is just the slope of the tangent line at any point. So what is the slope of the tangent line at this point?"}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "f of x equals 1 is just a horizontal line. So that right over there is the graph y is equal to f of x, which is equal to 1. Now, remember, the derivative, one way to conceptualize it, is just the slope of the tangent line at any point. So what is the slope of the tangent line at this point? And actually, what's the slope at every point? Well, this is a line, so the slope doesn't change. It has a constant slope, and it's a completely horizontal line."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is the slope of the tangent line at this point? And actually, what's the slope at every point? Well, this is a line, so the slope doesn't change. It has a constant slope, and it's a completely horizontal line. It has a slope of 0. So the slope at every point over here, slope is going to be equal to 0. So the slope of this line at any point is just going to be equal to 0."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It has a constant slope, and it's a completely horizontal line. It has a slope of 0. So the slope at every point over here, slope is going to be equal to 0. So the slope of this line at any point is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say I had f of x is equal to 3. Let's say that that's y is equal to 3."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope of this line at any point is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say I had f of x is equal to 3. Let's say that that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x?"}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. So let me write that. Derivative with respect to x of any constant."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. So let me write that. Derivative with respect to x of any constant. So let's say of a, where this is just a constant, that's going to be equal to 0. So pretty straightforward idea. Now let's explore a few more properties."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative with respect to x of any constant. So let's say of a, where this is just a constant, that's going to be equal to 0. So pretty straightforward idea. Now let's explore a few more properties. Let's say I want to take the derivative with respect to x of, well, let's use the same a. Let's say I have some constant times some function. Well, derivatives work out quite well."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's explore a few more properties. Let's say I want to take the derivative with respect to x of, well, let's use the same a. Let's say I have some constant times some function. Well, derivatives work out quite well. You can actually take this little scalar multiplier, this little constant, and take it out of the derivative. This is going to be equal to a. It's going to be equal to a. I didn't want to do that magenta color."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, derivatives work out quite well. You can actually take this little scalar multiplier, this little constant, and take it out of the derivative. This is going to be equal to a. It's going to be equal to a. I didn't want to do that magenta color. It's going to be equal to a times the derivative of f of x. a times the derivative of f of x. Let me do that in blue color. Of f of x."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be equal to a. I didn't want to do that magenta color. It's going to be equal to a times the derivative of f of x. a times the derivative of f of x. Let me do that in blue color. Of f of x. And the other way to denote the derivative of f of x is to just say that this is the same thing. This is equal to a times, this thing right over here is the exact same thing as f prime of x. Now, this might all look like really fancy notation, but I think if I gave you an example, it might make some sense."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Of f of x. And the other way to denote the derivative of f of x is to just say that this is the same thing. This is equal to a times, this thing right over here is the exact same thing as f prime of x. Now, this might all look like really fancy notation, but I think if I gave you an example, it might make some sense. So what about if I were to ask you the derivative with respect to x of 2 times x to the fifth power? Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative. This is going to be the same thing as 2 times the derivative of x to the fifth."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, this might all look like really fancy notation, but I think if I gave you an example, it might make some sense. So what about if I were to ask you the derivative with respect to x of 2 times x to the fifth power? Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative. This is going to be the same thing as 2 times the derivative of x to the fifth. 2 times the derivative with respect to x of x to the fifth. Essentially, I could just take this scalar multiplier and put it in front of the derivative. So this right here, this is the derivative with respect to x of x to the fifth."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be the same thing as 2 times the derivative of x to the fifth. 2 times the derivative with respect to x of x to the fifth. Essentially, I could just take this scalar multiplier and put it in front of the derivative. So this right here, this is the derivative with respect to x of x to the fifth. And we know how to do that using the power rule. This is going to be equal to 2 times, let me write that, I want to keep it consistent with the colors. This is going to be 2 times derivative of x to the fifth."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right here, this is the derivative with respect to x of x to the fifth. And we know how to do that using the power rule. This is going to be equal to 2 times, let me write that, I want to keep it consistent with the colors. This is going to be 2 times derivative of x to the fifth. Well, the power rule tells us n is 5. It's going to be 5x to the 5 minus 1, or 5x to the fourth power. So it's going to be 5x to the fourth power, which is going to be equal to 2 times 5 is 10x to the fourth."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be 2 times derivative of x to the fifth. Well, the power rule tells us n is 5. It's going to be 5x to the 5 minus 1, or 5x to the fourth power. So it's going to be 5x to the fourth power, which is going to be equal to 2 times 5 is 10x to the fourth. So 2x to the fifth, you can literally just say, OK, the power rule tells me the derivative of that is 5x to the fourth. 5 times 2 is 10. So that simplifies our life a good bit."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be 5x to the fourth power, which is going to be equal to 2 times 5 is 10x to the fourth. So 2x to the fifth, you can literally just say, OK, the power rule tells me the derivative of that is 5x to the fourth. 5 times 2 is 10. So that simplifies our life a good bit. We can now, using the power rule and this one property, take the derivative of anything that takes the form ax to the n power. Now let's think about another very useful derivative property. And these don't just apply to the power rule."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that simplifies our life a good bit. We can now, using the power rule and this one property, take the derivative of anything that takes the form ax to the n power. Now let's think about another very useful derivative property. And these don't just apply to the power rule. They apply to any derivative, but they are especially useful for the power rule because it allows us to construct polynomials and take the derivatives of them. If I were to take the derivative of the sum of two functions, so the derivative of, let's say, one function is f of x, and then the other function is g of x, it's lucky for us that this ends up being the same thing as the derivative of f of x plus the derivative of g of x. So this is the same thing as f. Actually, let me use that derivative operator just to make it clear."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And these don't just apply to the power rule. They apply to any derivative, but they are especially useful for the power rule because it allows us to construct polynomials and take the derivatives of them. If I were to take the derivative of the sum of two functions, so the derivative of, let's say, one function is f of x, and then the other function is g of x, it's lucky for us that this ends up being the same thing as the derivative of f of x plus the derivative of g of x. So this is the same thing as f. Actually, let me use that derivative operator just to make it clear. It's the same thing as the derivative with respect to x of f of x plus the derivative with respect to x of g of x. So we put f of x right over here and put g of x right over there. And so with the other notation, we can say this is going to be the same thing."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the same thing as f. Actually, let me use that derivative operator just to make it clear. It's the same thing as the derivative with respect to x of f of x plus the derivative with respect to x of g of x. So we put f of x right over here and put g of x right over there. And so with the other notation, we can say this is going to be the same thing. Derivative with respect to x of f of x, we can write as f prime of x. And the derivative with respect to x of g of x, we can write as g prime of x. Now, once again, this might look like kind of fancy notation to you, but when you see an example, it'll make it pretty clear."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so with the other notation, we can say this is going to be the same thing. Derivative with respect to x of f of x, we can write as f prime of x. And the derivative with respect to x of g of x, we can write as g prime of x. Now, once again, this might look like kind of fancy notation to you, but when you see an example, it'll make it pretty clear. If I want to take the derivative with respect to x of, let's say, x to the third power plus x to the negative 4 power, this just tells us that the derivative of the sum is just the sum of the derivatives. So we can take the derivative of this term using the power rule. So it's going to be 3x squared."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, once again, this might look like kind of fancy notation to you, but when you see an example, it'll make it pretty clear. If I want to take the derivative with respect to x of, let's say, x to the third power plus x to the negative 4 power, this just tells us that the derivative of the sum is just the sum of the derivatives. So we can take the derivative of this term using the power rule. So it's going to be 3x squared. And to that, we can add the derivative of this thing right over here. So it's going to be plus, that's a different shade of blue, plus, and over here is negative 4, so it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power. So we have, and I can just simplify a little bit, this is going to be equal to 3x squared minus 4x to the negative 5."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be 3x squared. And to that, we can add the derivative of this thing right over here. So it's going to be plus, that's a different shade of blue, plus, and over here is negative 4, so it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power. So we have, and I can just simplify a little bit, this is going to be equal to 3x squared minus 4x to the negative 5. And so now we have all the tools we need in our toolkit to essentially take the derivative of any polynomial. So let's give ourselves a little practice there. So let's say that I have, and I'll do it in white, let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have, and I can just simplify a little bit, this is going to be equal to 3x squared minus 4x to the negative 5. And so now we have all the tools we need in our toolkit to essentially take the derivative of any polynomial. So let's give ourselves a little practice there. So let's say that I have, and I'll do it in white, let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100. What is f prime of x? What is the derivative of f with respect to x going to be? Well, we can use the properties that we just said."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that I have, and I'll do it in white, let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100. What is f prime of x? What is the derivative of f with respect to x going to be? Well, we can use the properties that we just said. The derivative of this is just going to be 2 times the derivative of x to the third. Derivative of x to the third is going to be 3x squared, so it's going to be 2 times 3x squared. What's the derivative of negative 7x squared going to be?"}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can use the properties that we just said. The derivative of this is just going to be 2 times the derivative of x to the third. Derivative of x to the third is going to be 3x squared, so it's going to be 2 times 3x squared. What's the derivative of negative 7x squared going to be? Well, it's just going to be negative 7 times the derivative of x squared, which is 2x. What is the derivative of 3x going to be? Well, it's just going to be 3 times the derivative of x, or 3 times the derivative of x to the first."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's the derivative of negative 7x squared going to be? Well, it's just going to be negative 7 times the derivative of x squared, which is 2x. What is the derivative of 3x going to be? Well, it's just going to be 3 times the derivative of x, or 3 times the derivative of x to the first. The derivative of x to the first is just 1. So this is just going to be plus 3 times, could say 1x to the 0, but that's just 1. And then finally, what's the derivative of a constant going to be?"}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's just going to be 3 times the derivative of x, or 3 times the derivative of x to the first. The derivative of x to the first is just 1. So this is just going to be plus 3 times, could say 1x to the 0, but that's just 1. And then finally, what's the derivative of a constant going to be? Let me do that in a different color. What's the derivative of a constant going to be? Well, we covered that at the beginning of this video."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, what's the derivative of a constant going to be? Let me do that in a different color. What's the derivative of a constant going to be? Well, we covered that at the beginning of this video. The derivative of any constant is just going to be 0. So plus 0. And so now we are ready to simplify."}, {"video_title": "Differentiating polynomials   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we covered that at the beginning of this video. The derivative of any constant is just going to be 0. So plus 0. And so now we are ready to simplify. The derivative of f is going to be 2 times 3x squared is just 6x squared. Negative 7 times 2x is negative 14x plus 3. And we don't have to write the 0 there."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "We want to find the derivative with respect to x of all of this business right over here. And you might guess, and this is definitely a function of x. x is one of the boundaries of integration for this definite integral. And you might say, well, it looks like the fundamental theorem of calculus might apply. But I'm used to seeing the x or the function x as the upper bound, not as the lower bound. How do I deal with this? And the key realization is to realize what happens when you switch bounds for a definite integral. And I'll do a little bit of an aside to review that."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "But I'm used to seeing the x or the function x as the upper bound, not as the lower bound. How do I deal with this? And the key realization is to realize what happens when you switch bounds for a definite integral. And I'll do a little bit of an aside to review that. So if I'm taking the definite integral from a to b of f of t dt, we know that this is capital F, the antiderivative of F evaluated at b, minus the antiderivative of F evaluated at a. This is essentially corollary to the fundamental theorem, or it's the fundamental theorem part two, the second fundamental theorem of calculus. This is how we evaluate definite integrals."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "And I'll do a little bit of an aside to review that. So if I'm taking the definite integral from a to b of f of t dt, we know that this is capital F, the antiderivative of F evaluated at b, minus the antiderivative of F evaluated at a. This is essentially corollary to the fundamental theorem, or it's the fundamental theorem part two, the second fundamental theorem of calculus. This is how we evaluate definite integrals. Now, let's think about what the negative of this is. So the negative of that, of a to b of f of t dt, is just going to be equal to the negative of this, which is equal to the negative of F of b minus F of a, which is equal to capital F of a minus capital F of b. All I did is distribute the negative sign and then switch the two terms."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "This is how we evaluate definite integrals. Now, let's think about what the negative of this is. So the negative of that, of a to b of f of t dt, is just going to be equal to the negative of this, which is equal to the negative of F of b minus F of a, which is equal to capital F of a minus capital F of b. All I did is distribute the negative sign and then switch the two terms. But this right over here is equal to the definite integral from, instead of a to b, but from b to a of F of t dt. So notice, when you put a negative, that's just like switching the signs or switching the boundaries. Or if you switch the boundaries, they're the negatives of each other."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "All I did is distribute the negative sign and then switch the two terms. But this right over here is equal to the definite integral from, instead of a to b, but from b to a of F of t dt. So notice, when you put a negative, that's just like switching the signs or switching the boundaries. Or if you switch the boundaries, they're the negatives of each other. So we can go back to our original problem. We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched. The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "Or if you switch the boundaries, they're the negatives of each other. So we can go back to our original problem. We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched. The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business. I should just copy and paste that. So I'll just copy and paste. Let me paste it."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business. I should just copy and paste that. So I'll just copy and paste. Let me paste it. So times the derivative with respect to x of all of that. And now the fundamental theorem of calculus directly applies. This is going to be equal to, we deserve a drum roll now, this is going to be equal to the negative."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound   Khan Academy.mp3", "Sentence": "Let me paste it. So times the derivative with respect to x of all of that. And now the fundamental theorem of calculus directly applies. This is going to be equal to, we deserve a drum roll now, this is going to be equal to the negative. Can't forget the negative. And the fundamental theorem of calculus tells us that that's just going to be this function as a function of x. So it's going to be negative square root of the absolute value of cosine of not t anymore, but x."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So local linearity is this idea that if we zoom in sufficiently on a point, that even a nonlinear function that is differentiable at that point will actually look linear. So let me show some examples of that. So let's say we had y is equal to x squared. So that's that there, clearly a nonlinear function. But we can zoom in on a point and if we zoom sufficiently in, we will see that it looks roughly linear. So let's say we want to zoom in on the point one comma one. So let's do that."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's that there, clearly a nonlinear function. But we can zoom in on a point and if we zoom sufficiently in, we will see that it looks roughly linear. So let's say we want to zoom in on the point one comma one. So let's do that. So zooming in on the point one comma one, already it is looking roughly linear at that point. And this property of local linearity is very helpful when trying to approximate a function around a point. So for example, we could figure out, we could take the derivative at the point one one, use that as the slope of our tangent line, find the equation of the tangent line, and use that equation to approximate values of our function around x equals one."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So zooming in on the point one comma one, already it is looking roughly linear at that point. And this property of local linearity is very helpful when trying to approximate a function around a point. So for example, we could figure out, we could take the derivative at the point one one, use that as the slope of our tangent line, find the equation of the tangent line, and use that equation to approximate values of our function around x equals one. And you might not need to do that for y is equal to x squared, but it could actually be very, very useful for a more complex function. But the big takeaway here, at the point one one, it is displaying this idea of local linearity. And it is also differentiable at that point."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, we could figure out, we could take the derivative at the point one one, use that as the slope of our tangent line, find the equation of the tangent line, and use that equation to approximate values of our function around x equals one. And you might not need to do that for y is equal to x squared, but it could actually be very, very useful for a more complex function. But the big takeaway here, at the point one one, it is displaying this idea of local linearity. And it is also differentiable at that point. Now let's look at another example of a point on a function where we aren't differentiable and we also don't see the local linearity. So for example, let's do the absolute value of x. And let me shift it over a little bit just so that we don't overlap as much."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it is also differentiable at that point. Now let's look at another example of a point on a function where we aren't differentiable and we also don't see the local linearity. So for example, let's do the absolute value of x. And let me shift it over a little bit just so that we don't overlap as much. All right. So the absolute value of x minus one, it actually is differentiable as long as we're not at this corner right over here, as long as we're not at the point one comma zero. For any other x value, it is differentiable."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me shift it over a little bit just so that we don't overlap as much. All right. So the absolute value of x minus one, it actually is differentiable as long as we're not at this corner right over here, as long as we're not at the point one comma zero. For any other x value, it is differentiable. But right at x equals one, we've talked in other videos how we aren't differentiable there. And then we can use this local linearity idea to test it as well. And once again, this is not rigorous mathematics, but it is to give you an intuition."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "For any other x value, it is differentiable. But right at x equals one, we've talked in other videos how we aren't differentiable there. And then we can use this local linearity idea to test it as well. And once again, this is not rigorous mathematics, but it is to give you an intuition. No matter how far we zoom in, we still see this sharp corner. It would be hard to construct the only tangent line, a unique line that goes through this point one comma zero. I can construct an actual infinite number of lines that go through one comma zero, but that do not go through the rest of the curve."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, this is not rigorous mathematics, but it is to give you an intuition. No matter how far we zoom in, we still see this sharp corner. It would be hard to construct the only tangent line, a unique line that goes through this point one comma zero. I can construct an actual infinite number of lines that go through one comma zero, but that do not go through the rest of the curve. And so notice, wherever you see a hard corner, like we're seeing at one comma zero in this absolute value function, that's a pretty good indication that we are not going to be differentiable at that point. Now let's zoom out a little bit and let's take another function. Let's take a function where the differentiability or the lack of differentiability is not because of a corner, but it's because as we zoom in, it starts to look linear, but it starts to look like a vertical line."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "I can construct an actual infinite number of lines that go through one comma zero, but that do not go through the rest of the curve. And so notice, wherever you see a hard corner, like we're seeing at one comma zero in this absolute value function, that's a pretty good indication that we are not going to be differentiable at that point. Now let's zoom out a little bit and let's take another function. Let's take a function where the differentiability or the lack of differentiability is not because of a corner, but it's because as we zoom in, it starts to look linear, but it starts to look like a vertical line. So a good example of that would be square root of, let's say, four minus x squared. So that's the top half of a circle of radius two. And let's focus on the point two comma zero because right over there, we actually are not differentiable."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's take a function where the differentiability or the lack of differentiability is not because of a corner, but it's because as we zoom in, it starts to look linear, but it starts to look like a vertical line. So a good example of that would be square root of, let's say, four minus x squared. So that's the top half of a circle of radius two. And let's focus on the point two comma zero because right over there, we actually are not differentiable. And if we zoom in far enough, we see right at two comma zero that we are approaching what looks like a vertical line. So once again, we would not be differentiable at two comma zero. Now another thing I wanna point out, all of these, you really didn't have to zoom in too much to appreciate that, hey, I got a corner here on this absolute value function, or at two comma zero or at negative two comma zero, something a little bit stranger than normal is happening there, so maybe I'm not differentiable."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's focus on the point two comma zero because right over there, we actually are not differentiable. And if we zoom in far enough, we see right at two comma zero that we are approaching what looks like a vertical line. So once again, we would not be differentiable at two comma zero. Now another thing I wanna point out, all of these, you really didn't have to zoom in too much to appreciate that, hey, I got a corner here on this absolute value function, or at two comma zero or at negative two comma zero, something a little bit stranger than normal is happening there, so maybe I'm not differentiable. But there are some functions that we don't see as typically in a algebra or precalculus or calculus class, but it can look like a hard corner from a zoomed out perspective. But as we zoom in, once again, we'll see the local linearity, and they're also differentiable at those points. So a good example of that, let me actually get rid of some of these just so we can really zoom in."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now another thing I wanna point out, all of these, you really didn't have to zoom in too much to appreciate that, hey, I got a corner here on this absolute value function, or at two comma zero or at negative two comma zero, something a little bit stranger than normal is happening there, so maybe I'm not differentiable. But there are some functions that we don't see as typically in a algebra or precalculus or calculus class, but it can look like a hard corner from a zoomed out perspective. But as we zoom in, once again, we'll see the local linearity, and they're also differentiable at those points. So a good example of that, let me actually get rid of some of these just so we can really zoom in. Let's say y is equal to x to the, and I'm gonna make a very large exponent here. So x to the 10th power, it's starting to look a little bit like a corner there. Let's make it to the 100th power."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So a good example of that, let me actually get rid of some of these just so we can really zoom in. Let's say y is equal to x to the, and I'm gonna make a very large exponent here. So x to the 10th power, it's starting to look a little bit like a corner there. Let's make it to the 100th power. Well now, it's looking even more like a corner there. Let me go to the 1,000th power just for good measure. So at this scale, it looks like we have a corner at the point one comma zero."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's make it to the 100th power. Well now, it's looking even more like a corner there. Let me go to the 1,000th power just for good measure. So at this scale, it looks like we have a corner at the point one comma zero. Now this curve actually does not go to the point one comma zero. If x is one, then y is going to be one. And we'll see that as we zoom in, this, what looks like a hard corner, is going to soften."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So at this scale, it looks like we have a corner at the point one comma zero. Now this curve actually does not go to the point one comma zero. If x is one, then y is going to be one. And we'll see that as we zoom in, this, what looks like a hard corner, is going to soften. And that's good because this function is actually differentiable at every value of x. It's a little bit more exotic than what we typically see. But as we zoom in, we'll actually see that."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we'll see that as we zoom in, this, what looks like a hard corner, is going to soften. And that's good because this function is actually differentiable at every value of x. It's a little bit more exotic than what we typically see. But as we zoom in, we'll actually see that. Let's just zoom in on what looks like a fairly hard corner. But if we zoom sufficiently enough, even at the part that looks like the hardest part of the corner, the real corner, we'll see that it starts to soften and it curves. And if we zoom in sufficiently, it will actually look like a line."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But as we zoom in, we'll actually see that. Let's just zoom in on what looks like a fairly hard corner. But if we zoom sufficiently enough, even at the part that looks like the hardest part of the corner, the real corner, we'll see that it starts to soften and it curves. And if we zoom in sufficiently, it will actually look like a line. It's hard to believe when you're really zoomed out. And I'm going at the point that really looked like a corner from a distance. But as we zoom in, we see once again this local linearity."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we zoom in sufficiently, it will actually look like a line. It's hard to believe when you're really zoomed out. And I'm going at the point that really looked like a corner from a distance. But as we zoom in, we see once again this local linearity. And it's a non-vertical line. And so once again, this is true at any point on this curve, that we are going to be differentiable. So the whole point here is sometimes you might have to zoom in a lot."}, {"video_title": "Local linearity and differentiability   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But as we zoom in, we see once again this local linearity. And it's a non-vertical line. And so once again, this is true at any point on this curve, that we are going to be differentiable. So the whole point here is sometimes you might have to zoom in a lot. A tool like Desmos, which I'm using right now, is very helpful for doing that. And this isn't rigorous mathematics, but it's to give you an intuitive sense that if you zoom in sufficiently and you start to see a curve looking more and more like a line, good indication that you are differentiable. If you keep zooming in and it still looks like a hard corner, or if you zoom in and it looks like the tangent might be vertical, well then, some questions should arise in your brain."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just remind ourselves a definition of a derivative. And there's multiple ways of writing this. For the sake of this video, I'll write it as the derivative of our function at point C, this is Lagrange notation with this F prime, the derivative of our function F at C, is going to be equal to the limit as X approaches C of F of X minus F of C over X minus C. And at first when you see this formula, and we've seen it before, it looks a little bit strange, but all it is is it's calculating the slope. This is our change in the value of our function, or you could think of it as our change in Y if Y is equal to F of X. And this is our change in X. And we're just trying to see, well, what is that slope as X gets closer and closer to C, as our change in X gets closer and closer to zero? And we talk about that in other videos."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our change in the value of our function, or you could think of it as our change in Y if Y is equal to F of X. And this is our change in X. And we're just trying to see, well, what is that slope as X gets closer and closer to C, as our change in X gets closer and closer to zero? And we talk about that in other videos. So I'm now going to make a few claims in this video. And I'm not going to prove them rigorously. There is another video that will go a little bit more into the proof direction."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we talk about that in other videos. So I'm now going to make a few claims in this video. And I'm not going to prove them rigorously. There is another video that will go a little bit more into the proof direction. But this is more to get an intuition. And so the first claim that I'm going to make is if F is differentiable at X equals C, at X equals C, then F is continuous at X equals C. So I'm saying if we know it's differentiable, if we can find this limit, if we can find this derivative at X equals C, then our function is also continuous at X equals C. It doesn't necessarily mean the other way around. And actually, we'll look at a case where it's not necessarily the case the other way around, that if you're continuous, then you're definitely differentiable."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "There is another video that will go a little bit more into the proof direction. But this is more to get an intuition. And so the first claim that I'm going to make is if F is differentiable at X equals C, at X equals C, then F is continuous at X equals C. So I'm saying if we know it's differentiable, if we can find this limit, if we can find this derivative at X equals C, then our function is also continuous at X equals C. It doesn't necessarily mean the other way around. And actually, we'll look at a case where it's not necessarily the case the other way around, that if you're continuous, then you're definitely differentiable. But another way to interpret what I just wrote down is if you are not continuous, then you definitely will not be differentiable. If F not continuous at X equals C, at X equals C, then F is not differentiable. Differentiable at X is equal to C. So let me give a few examples of a non-continuous function."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually, we'll look at a case where it's not necessarily the case the other way around, that if you're continuous, then you're definitely differentiable. But another way to interpret what I just wrote down is if you are not continuous, then you definitely will not be differentiable. If F not continuous at X equals C, at X equals C, then F is not differentiable. Differentiable at X is equal to C. So let me give a few examples of a non-continuous function. And then think about would we be able to find this limit? So the first is where you have a discontinuity. Our function is defined at C. It's equal to this value."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Differentiable at X is equal to C. So let me give a few examples of a non-continuous function. And then think about would we be able to find this limit? So the first is where you have a discontinuity. Our function is defined at C. It's equal to this value. But you can see as X becomes larger than C, it just jumps down and shifts right over here. So what would happen if you were trying to find this limit? Well, remember, all this is is a slope of a line between when X is some arbitrary value."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our function is defined at C. It's equal to this value. But you can see as X becomes larger than C, it just jumps down and shifts right over here. So what would happen if you were trying to find this limit? Well, remember, all this is is a slope of a line between when X is some arbitrary value. Let's say it's out here. So that would be X. This would be the point X comma F of X."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, remember, all this is is a slope of a line between when X is some arbitrary value. Let's say it's out here. So that would be X. This would be the point X comma F of X. And then this is the point C comma F of C right over here. So this is C comma F of C. So if you find the left-sided limit right over here, you're essentially saying, okay, let's find this slope. And then let me get a little bit closer."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "This would be the point X comma F of X. And then this is the point C comma F of C right over here. So this is C comma F of C. So if you find the left-sided limit right over here, you're essentially saying, okay, let's find this slope. And then let me get a little bit closer. And let's get X a little bit closer, and then let's find this slope. And then let's get X even closer than that and find this slope. And in all of those cases, it would be zero."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let me get a little bit closer. And let's get X a little bit closer, and then let's find this slope. And then let's get X even closer than that and find this slope. And in all of those cases, it would be zero. The slope is zero. So one way to think about it, the derivative, or this limit as we approach from the left, seems to be approaching zero. But what about if we were to take Xs to the right?"}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in all of those cases, it would be zero. The slope is zero. So one way to think about it, the derivative, or this limit as we approach from the left, seems to be approaching zero. But what about if we were to take Xs to the right? So instead of our Xs being there, what if we were to take Xs right over here? Well, for this point, X comma F of X, our slope, if we take F of X minus F of C over X minus C, that would be the slope of this line. If we get X to be even closer, let's say right over here, then this would be the slope of this line."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what about if we were to take Xs to the right? So instead of our Xs being there, what if we were to take Xs right over here? Well, for this point, X comma F of X, our slope, if we take F of X minus F of C over X minus C, that would be the slope of this line. If we get X to be even closer, let's say right over here, then this would be the slope of this line. If we get even closer, then this expression would be the slope of this line. And so as we get closer and closer to X being equal to C, we see that our slope is actually approaching negative infinity. And most importantly, it's approaching a very different value from the right."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we get X to be even closer, let's say right over here, then this would be the slope of this line. If we get even closer, then this expression would be the slope of this line. And so as we get closer and closer to X being equal to C, we see that our slope is actually approaching negative infinity. And most importantly, it's approaching a very different value from the right. This expression is approaching a very different value from the right as it is from the left. And so in this case, this limit up here won't exist. So we can clearly say this is not differentiable."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And most importantly, it's approaching a very different value from the right. This expression is approaching a very different value from the right as it is from the left. And so in this case, this limit up here won't exist. So we can clearly say this is not differentiable. So once again, not a proof here. I'm just getting an intuition for, if something isn't continuous, it's pretty clear, at least in this case, that it's not going to be differentiable. Let's look at another case."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can clearly say this is not differentiable. So once again, not a proof here. I'm just getting an intuition for, if something isn't continuous, it's pretty clear, at least in this case, that it's not going to be differentiable. Let's look at another case. Let's look at a case where we have what's sometimes called a removable discontinuity or a point discontinuity. So once again, let's say we're approaching from the left. This is X, this is the point X comma F of X."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's look at another case. Let's look at a case where we have what's sometimes called a removable discontinuity or a point discontinuity. So once again, let's say we're approaching from the left. This is X, this is the point X comma F of X. Now what's interesting is, where as this expression is the slope of the line connecting X comma F of X and C comma F of C, which is this point, not that point. Remember, we have this removable discontinuity right over here. And so this would be, this expression is calculating the slope of that line."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is X, this is the point X comma F of X. Now what's interesting is, where as this expression is the slope of the line connecting X comma F of X and C comma F of C, which is this point, not that point. Remember, we have this removable discontinuity right over here. And so this would be, this expression is calculating the slope of that line. And then if X gets even closer to C, well then we're gonna be calculating the slope of that line. If X gets even closer to C, we're gonna be calculating the slope of that line. And so as we approach from the left, as X approaches C from the left, we actually have a situation where this expression right over here is going to approach negative infinity."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this would be, this expression is calculating the slope of that line. And then if X gets even closer to C, well then we're gonna be calculating the slope of that line. If X gets even closer to C, we're gonna be calculating the slope of that line. And so as we approach from the left, as X approaches C from the left, we actually have a situation where this expression right over here is going to approach negative infinity. And if we approach from the right, if we approach with X as large as in C, well this is our X comma F of X. So we have a positive slope. And then as we get closer, it gets more positive."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so as we approach from the left, as X approaches C from the left, we actually have a situation where this expression right over here is going to approach negative infinity. And if we approach from the right, if we approach with X as large as in C, well this is our X comma F of X. So we have a positive slope. And then as we get closer, it gets more positive. More positive approaches positive infinity. But either way, it's not approaching a finite value. And one side is approaching positive infinity and the other side is approaching negative infinity."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then as we get closer, it gets more positive. More positive approaches positive infinity. But either way, it's not approaching a finite value. And one side is approaching positive infinity and the other side is approaching negative infinity. This, the limit of this expression is not going to exist. So once again, I'm not doing a rigorous proof here. But try to construct a discontinuous function where you will be able to find this."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And one side is approaching positive infinity and the other side is approaching negative infinity. This, the limit of this expression is not going to exist. So once again, I'm not doing a rigorous proof here. But try to construct a discontinuous function where you will be able to find this. It is very, very hard. And you might say, well what about the situations where F is not even defined at C? Which for sure, you're not gonna be continuous if F is not defined at C. Well if F is not defined at C, then this part of the expression wouldn't even make sense."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But try to construct a discontinuous function where you will be able to find this. It is very, very hard. And you might say, well what about the situations where F is not even defined at C? Which for sure, you're not gonna be continuous if F is not defined at C. Well if F is not defined at C, then this part of the expression wouldn't even make sense. So you definitely wouldn't be differentiable. But now let's ask another thing. I've just given you good arguments for when you're not continuous, you're not going to be differentiable."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Which for sure, you're not gonna be continuous if F is not defined at C. Well if F is not defined at C, then this part of the expression wouldn't even make sense. So you definitely wouldn't be differentiable. But now let's ask another thing. I've just given you good arguments for when you're not continuous, you're not going to be differentiable. But can we make another claim that if you are continuous, then you definitely will be differentiable? Well it turns out that there are for sure many functions, an infinite number of functions, that can be continuous at C but not differentiable. So for example, this could be an absolute value function."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "I've just given you good arguments for when you're not continuous, you're not going to be differentiable. But can we make another claim that if you are continuous, then you definitely will be differentiable? Well it turns out that there are for sure many functions, an infinite number of functions, that can be continuous at C but not differentiable. So for example, this could be an absolute value function. It doesn't have to be an absolute value function, but this could be Y is equal to the absolute value of X minus C. And why is this one not differentiable at C? Well think about what's happening. Think about this expression."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, this could be an absolute value function. It doesn't have to be an absolute value function, but this could be Y is equal to the absolute value of X minus C. And why is this one not differentiable at C? Well think about what's happening. Think about this expression. Remember, this expression, all it's doing is calculating the slope between the point X comma F of X and the point C comma F of C. So if X is say out here, this is X comma F of X, it's going to be calculated, and so as we take the limit as X approaches C from the left, we'll be looking at this slope. And then as we get closer, we'll be looking at this slope, which is actually going to be the same. In this case it would be a negative one."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Think about this expression. Remember, this expression, all it's doing is calculating the slope between the point X comma F of X and the point C comma F of C. So if X is say out here, this is X comma F of X, it's going to be calculated, and so as we take the limit as X approaches C from the left, we'll be looking at this slope. And then as we get closer, we'll be looking at this slope, which is actually going to be the same. In this case it would be a negative one. So as X approaches C from the left, this expression would be negative one. But as X approaches C from the right, this expression is going to be one. The slope of the line that connects these points is one."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "In this case it would be a negative one. So as X approaches C from the left, this expression would be negative one. But as X approaches C from the right, this expression is going to be one. The slope of the line that connects these points is one. The slope of the line that connects these points is one. So the limit of this expression, or I would say the value of this expression, is approaching two different values as X approaches C from the left or the right. From the left, it's approaching negative one, or it's constantly negative one, and so it's approaching negative one, you could say."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope of the line that connects these points is one. The slope of the line that connects these points is one. So the limit of this expression, or I would say the value of this expression, is approaching two different values as X approaches C from the left or the right. From the left, it's approaching negative one, or it's constantly negative one, and so it's approaching negative one, you could say. And from the right, it's one, and it's approaching one the entire time. And so we know if you're approaching two different values from on the left-sided or the right-sided limit, then this limit will not exist. So here, this is not differentiable."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "From the left, it's approaching negative one, or it's constantly negative one, and so it's approaching negative one, you could say. And from the right, it's one, and it's approaching one the entire time. And so we know if you're approaching two different values from on the left-sided or the right-sided limit, then this limit will not exist. So here, this is not differentiable. And even intuitively, we think of the derivative as the slope of the tangent line, and you could actually draw an infinite number of tangent lines here. It's one way to think about it. You could say, well, maybe this is the tangent line right over there, but then why can't I make something like this the tangent line?"}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here, this is not differentiable. And even intuitively, we think of the derivative as the slope of the tangent line, and you could actually draw an infinite number of tangent lines here. It's one way to think about it. You could say, well, maybe this is the tangent line right over there, but then why can't I make something like this the tangent line? That only intersects at the point C comma zero, and then you could keep doing things like that. Why can't that be the tangent line? And you could go on and on and on."}, {"video_title": "Differentiability and continuity   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say, well, maybe this is the tangent line right over there, but then why can't I make something like this the tangent line? That only intersects at the point C comma zero, and then you could keep doing things like that. Why can't that be the tangent line? And you could go on and on and on. So the big takeaways here, at least intuitively, in a future video, I'm going to prove to you that if F is differentiable at C, then it is continuous at C, which can also be interpreted that if you're not continuous at C, then you're not gonna be differentiable. These two examples will hopefully give you some intuition for that, but it's not the case that if something is continuous, that it has to be differentiable. It oftentimes will be differentiable, but it doesn't have to be differentiable, and this absolute value function is an example of a continuous function at C, but it is not differentiable at C."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So with the bottom bound of the horizontal line y is equal to e, and an upper bound with y is equal to e to the third power. So pause this video and see if you can work through it. So one way to think about it, this is just like definite integrals we've done where we're looking between the curve and the x-axis, but now it looks like things are swapped around. We now care about the y-axis. So let's just rewrite our function here, and let's rewrite it in terms of x. So if y is equal to 15 over x, that means if we multiply both sides by x, xy is equal to 15, and if we divide both sides by y, we get x is equal to 15 over y. These right over here are all going to be equivalent."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "We now care about the y-axis. So let's just rewrite our function here, and let's rewrite it in terms of x. So if y is equal to 15 over x, that means if we multiply both sides by x, xy is equal to 15, and if we divide both sides by y, we get x is equal to 15 over y. These right over here are all going to be equivalent. Now how does this right over here help you? Well, think about the area, think about estimating the area as a bunch of little rectangles here. So that's one rectangle, and then another rectangle right over there, and then another rectangle right over there."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "These right over here are all going to be equivalent. Now how does this right over here help you? Well, think about the area, think about estimating the area as a bunch of little rectangles here. So that's one rectangle, and then another rectangle right over there, and then another rectangle right over there. So what's the area of each of those rectangles? So the width here, that is going to be x, but we can express x as a function of y. So that's the width right over there, and we know that that's going to be 15 over y."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's one rectangle, and then another rectangle right over there, and then another rectangle right over there. So what's the area of each of those rectangles? So the width here, that is going to be x, but we can express x as a function of y. So that's the width right over there, and we know that that's going to be 15 over y. And then what's the height gonna be? Well, that's gonna be a very small change in y. The height is going to be dy."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's the width right over there, and we know that that's going to be 15 over y. And then what's the height gonna be? Well, that's gonna be a very small change in y. The height is going to be dy. So the area of one of those little rectangles right over there, say the area of that one right over there, you could view as, let me do it over here, as 15 over y dy. And then we wanna sum all of these little rectangles from y is equal to e all the way to y is equal to e to the third power. So that's what our definite integral does."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "The height is going to be dy. So the area of one of those little rectangles right over there, say the area of that one right over there, you could view as, let me do it over here, as 15 over y dy. And then we wanna sum all of these little rectangles from y is equal to e all the way to y is equal to e to the third power. So that's what our definite integral does. We go from y is equal to e to y is equal to e to the third power. So all we did, we're used to seeing things like this, where this would be 15 over x dx. All we're doing here is, this is 15 over y dy."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's what our definite integral does. We go from y is equal to e to y is equal to e to the third power. So all we did, we're used to seeing things like this, where this would be 15 over x dx. All we're doing here is, this is 15 over y dy. So let's evaluate this. So we take the antiderivative of 15 over y, and then evaluate at these two points. So this is going to be equal to, antiderivative of one over y is going to be the natural log of the absolute value of y."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "All we're doing here is, this is 15 over y dy. So let's evaluate this. So we take the antiderivative of 15 over y, and then evaluate at these two points. So this is going to be equal to, antiderivative of one over y is going to be the natural log of the absolute value of y. So it's 15 times the natural log of the absolute value of y. And then we're going to evaluate that at our endpoints. So we're gonna evaluate it at e to the third and at e. So let's first evaluate at e to the third."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to, antiderivative of one over y is going to be the natural log of the absolute value of y. So it's 15 times the natural log of the absolute value of y. And then we're going to evaluate that at our endpoints. So we're gonna evaluate it at e to the third and at e. So let's first evaluate at e to the third. So that's 15 times the natural log, the absolute time, the natural log of the absolute value of e to the third power minus 15 times the natural log of the absolute value of e. So what does this simplify to? The natural log of e to the third power, what power do I have to raise e to to get to e to the third? Well, that's just going to be three."}, {"video_title": "Area between a curve and and the _-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna evaluate it at e to the third and at e. So let's first evaluate at e to the third. So that's 15 times the natural log, the absolute time, the natural log of the absolute value of e to the third power minus 15 times the natural log of the absolute value of e. So what does this simplify to? The natural log of e to the third power, what power do I have to raise e to to get to e to the third? Well, that's just going to be three. And then the natural log of e, what power do I have to raise e to to get e? Well, that's just one. So this is 15 times three minus 15."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if we can find the limit as theta approaches zero of one minus cosine theta over two sine squared theta. And like always, pause the video and see if you can work through this. All right, well, our first temptation is to say, well, this is going to be the same thing as the limit of one minus cosine theta as x approaches, or not x, as theta approaches zero, of theta as theta approaches zero over the limit as theta approaches zero of two sine squared theta. Now, both of these expressions, which could be used to define a function, they'd be continuous if you graphed them. They'd be continuous at theta equals zero, so the limit is going to be the same thing as just evaluating them at theta equals zero. So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, both of these expressions, which could be used to define a function, they'd be continuous if you graphed them. They'd be continuous at theta equals zero, so the limit is going to be the same thing as just evaluating them at theta equals zero. So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero. So once again, we have that indeterminate form. And once again, this indeterminate form, when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist, it just means, well, maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero. So once again, we have that indeterminate form. And once again, this indeterminate form, when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist, it just means, well, maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist. But let's see what we can do to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is f of x."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist. But let's see what we can do to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is f of x. So f of x is equal to one minus cosine theta over two sine squared theta. And let's see if we can rewrite it in some way, that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that this right over here is f of x. So f of x is equal to one minus cosine theta over two sine squared theta. And let's see if we can rewrite it in some way, that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is a one minus cosine theta, this is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is a one minus cosine theta, this is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have a one minus cosine theta in the numerator, and I have a one minus cosine theta in the denominator. Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have a one minus cosine theta in the numerator, and I have a one minus cosine theta in the denominator. Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta. We could say, well, aren't these the same thing? And we would be almost right, because f of x, this one right over here, this is defined, this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta. We could say, well, aren't these the same thing? And we would be almost right, because f of x, this one right over here, this is defined, this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero. But now let's think about the limit again. Essentially what we want to do is we want to find the limit as theta approaches zero of f of x."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero. But now let's think about the limit again. Essentially what we want to do is we want to find the limit as theta approaches zero of f of x. And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero. f of x is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Essentially what we want to do is we want to find the limit as theta approaches zero of f of x. And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero. f of x is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function. But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero. And so we could say g of x is equal to one over two plus two cosine theta. Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function. But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero. And so we could say g of x is equal to one over two plus two cosine theta. Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero. Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is. But the limits as theta approaches zero are going to be the same, and we've seen that in previous videos. And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta?"}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero. Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is. But the limits as theta approaches zero are going to be the same, and we've seen that in previous videos. And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing. If you do that, if you just cross these two out, and all of a sudden, your expression becomes defined at zero, you are now dealing with a different expression, or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain."}, {"video_title": "Trig limit using pythagorean identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing. If you do that, if you just cross these two out, and all of a sudden, your expression becomes defined at zero, you are now dealing with a different expression, or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain. But lucky for us, we can say if we found another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity out, the limits are going to be equivalent. So the limit as theta approaches zero of g of x, well, that's just going to be, since it's continuous at zero, we could say that's just going to be, we can just substitute, that's going to be equal to g of zero, which is equal to one over two plus cosine two, one over two plus two times cosine of zero. Cosine of zero is one, so it's just one over two plus two, which is equal to, deserve a little bit of a drum roll here, which is equal to 1 4th."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw some axes here. So let's say this is my y-axis. So try to draw a vertical line. So that right over there is my y-axis. And then let's say this is my x-axis. I'll focus on the first quadrant, although I don't have to. So let's say this right over here is my x-axis."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that right over there is my y-axis. And then let's say this is my x-axis. I'll focus on the first quadrant, although I don't have to. So let's say this right over here is my x-axis. And then let me draw a function. So let's say my function looks something like that. It could look like anything, but that seems suitable."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say this right over here is my x-axis. And then let me draw a function. So let's say my function looks something like that. It could look like anything, but that seems suitable. So this is the function y is equal to f of x. And just for the sake of conceptual understanding, I'm going to say it's not defined at a point. I didn't have to do this."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "It could look like anything, but that seems suitable. So this is the function y is equal to f of x. And just for the sake of conceptual understanding, I'm going to say it's not defined at a point. I didn't have to do this. You can find the limit as x approaches a point where the function actually is defined, but it becomes that much more interesting, at least for me, where you start to understand why a limit might be relevant, where a function is not defined at some point. So the way I've drawn it, this function is not defined when x is equal to c. Now, the way that we've thought about a limit is what does f of x approach as x approaches c? So let's think about that a little bit."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "I didn't have to do this. You can find the limit as x approaches a point where the function actually is defined, but it becomes that much more interesting, at least for me, where you start to understand why a limit might be relevant, where a function is not defined at some point. So the way I've drawn it, this function is not defined when x is equal to c. Now, the way that we've thought about a limit is what does f of x approach as x approaches c? So let's think about that a little bit. When x is a reasonable bit lower than c, f of x, for our function that we just drew, is right over here. That's what f of x is going to be equal. y is equal to f of x."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. When x is a reasonable bit lower than c, f of x, for our function that we just drew, is right over here. That's what f of x is going to be equal. y is equal to f of x. When x gets a little bit closer, then our f of x is right over there. When x gets even closer, maybe really almost at c, but not quite at c, then our f of x is right over here. And the way we see it, we see that our f of x seems to be, as x gets closer and closer to c, it looks like our f of x is getting closer and closer to some value."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "y is equal to f of x. When x gets a little bit closer, then our f of x is right over there. When x gets even closer, maybe really almost at c, but not quite at c, then our f of x is right over here. And the way we see it, we see that our f of x seems to be, as x gets closer and closer to c, it looks like our f of x is getting closer and closer to some value. It's getting closer and closer to some value right over there. I'll even draw it, maybe I'll draw it with a more solid line. And that was actually only the case when x was getting closer to c from the left, from values of x less than c. But what happens as we get closer and closer to c from values of x that are larger than c?"}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the way we see it, we see that our f of x seems to be, as x gets closer and closer to c, it looks like our f of x is getting closer and closer to some value. It's getting closer and closer to some value right over there. I'll even draw it, maybe I'll draw it with a more solid line. And that was actually only the case when x was getting closer to c from the left, from values of x less than c. But what happens as we get closer and closer to c from values of x that are larger than c? Well, when x is over here, f of x is right over here. And so that's what f of x is, is right over there. When x gets a little bit closer to c, our f of x is right over there."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that was actually only the case when x was getting closer to c from the left, from values of x less than c. But what happens as we get closer and closer to c from values of x that are larger than c? Well, when x is over here, f of x is right over here. And so that's what f of x is, is right over there. When x gets a little bit closer to c, our f of x is right over there. When x is just very slightly larger than c, then our f of x is right over there. And you see, once again, it seems to be approaching that same value. And we call that value, that value that f of x seems to be approaching as x approaches c, we call that value L, or the limit."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x gets a little bit closer to c, our f of x is right over there. When x is just very slightly larger than c, then our f of x is right over there. And you see, once again, it seems to be approaching that same value. And we call that value, that value that f of x seems to be approaching as x approaches c, we call that value L, or the limit. And so the way we would denote it, is we would call that the limit. We don't have to call it L all the time, but it's referred to as the limit. And the way that we would kind of write that mathematically, is we would say the limit of f of x as x approaches c is equal to L. And this is a fine conceptual understanding of limits."}, {"video_title": "Formal definition of limits Part 1 intuition review   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we call that value, that value that f of x seems to be approaching as x approaches c, we call that value L, or the limit. And so the way we would denote it, is we would call that the limit. We don't have to call it L all the time, but it's referred to as the limit. And the way that we would kind of write that mathematically, is we would say the limit of f of x as x approaches c is equal to L. And this is a fine conceptual understanding of limits. And it really will take you pretty far. And you're ready to progress and start thinking about taking a lot of limits. But this isn't a very mathematically rigorous definition of limits."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What I hope to do in this video is give you a satisfying proof of the product rule. So let's just start with our definition of a derivative. So if I have the function f of x, and if I wanted to take the derivative of it, by definition, by definition, the derivative of f of x is the limit as h approaches zero of f of x plus h minus f of x, all of that over, all of that over h. If we want to think of it visually, this is the slope of the tangent line and all of that. But now I want to do something a little bit more interesting. I want to find the derivative with respect to x, not just of f of x, but the product of two functions, f of x times g of x. And if I can come up with a simple thing for this, that essentially is the product rule. Well, if we just apply the definition of a derivative, that means I'm going to take the limit as h approaches zero."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But now I want to do something a little bit more interesting. I want to find the derivative with respect to x, not just of f of x, but the product of two functions, f of x times g of x. And if I can come up with a simple thing for this, that essentially is the product rule. Well, if we just apply the definition of a derivative, that means I'm going to take the limit as h approaches zero. In the denominator, I'm going to have an h. In the denominator, I'm going to write a big, it's going to be a big rational expression. In the denominator, I'm going to have an h. And then I'm going to evaluate this thing at x plus h. So that's going to be f of x plus h, g of x plus h. And from that, I'm going to subtract this thing, evaluate it at f of x. Or, sorry, this thing evaluate it at x."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, if we just apply the definition of a derivative, that means I'm going to take the limit as h approaches zero. In the denominator, I'm going to have an h. In the denominator, I'm going to write a big, it's going to be a big rational expression. In the denominator, I'm going to have an h. And then I'm going to evaluate this thing at x plus h. So that's going to be f of x plus h, g of x plus h. And from that, I'm going to subtract this thing, evaluate it at f of x. Or, sorry, this thing evaluate it at x. So that's going to be f of x times g of x. And I'm going to put a big, awkward space here, and you're going to see why in a second. So if I just, if I evaluate this at x, this is going to be minus f of x, g of x."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Or, sorry, this thing evaluate it at x. So that's going to be f of x times g of x. And I'm going to put a big, awkward space here, and you're going to see why in a second. So if I just, if I evaluate this at x, this is going to be minus f of x, g of x. All I did so far is I just applied the definition of the derivative. Instead of applying it to f of x, I applied it to f of x times g of x. So you have f of x plus h, g of x plus h, minus f of x, g of x, all of that over h, limit as h approaches zero."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So if I just, if I evaluate this at x, this is going to be minus f of x, g of x. All I did so far is I just applied the definition of the derivative. Instead of applying it to f of x, I applied it to f of x times g of x. So you have f of x plus h, g of x plus h, minus f of x, g of x, all of that over h, limit as h approaches zero. Now why did I put this big, awkward space here? Because just the way I've written it right now, it doesn't seem easy to algebraically manipulate. I don't know how to evaluate this limit."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So you have f of x plus h, g of x plus h, minus f of x, g of x, all of that over h, limit as h approaches zero. Now why did I put this big, awkward space here? Because just the way I've written it right now, it doesn't seem easy to algebraically manipulate. I don't know how to evaluate this limit. There doesn't seem to be anything obvious to do. And what I'm about to show you, I guess you could view it as a little bit of a trick. I can't claim that I would have figured it out on my own, maybe eventually if I were spending hours on it."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I don't know how to evaluate this limit. There doesn't seem to be anything obvious to do. And what I'm about to show you, I guess you could view it as a little bit of a trick. I can't claim that I would have figured it out on my own, maybe eventually if I were spending hours on it. And I'm assuming somebody was fumbling with it long enough they said, oh wait, wait, look, if I just add and subtract the same term here, I can begin to algebraically manipulate it and get it to what we all know is the classic product rule. So what do I add and subtract here? Well, let me give you a clue."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I can't claim that I would have figured it out on my own, maybe eventually if I were spending hours on it. And I'm assuming somebody was fumbling with it long enough they said, oh wait, wait, look, if I just add and subtract the same term here, I can begin to algebraically manipulate it and get it to what we all know is the classic product rule. So what do I add and subtract here? Well, let me give you a clue. So if we have plus, actually let me change this, minus f of x plus h, g of x. I can't just subtract, if I subtract it, I've got to add it too, so I don't change the value of this expression. So plus f of x plus h, g of x. Now I haven't changed the value, I just added and subtracted the same thing."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, let me give you a clue. So if we have plus, actually let me change this, minus f of x plus h, g of x. I can't just subtract, if I subtract it, I've got to add it too, so I don't change the value of this expression. So plus f of x plus h, g of x. Now I haven't changed the value, I just added and subtracted the same thing. But now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well, to keep going, let's just keep exploring this expression."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Now I haven't changed the value, I just added and subtracted the same thing. But now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well, to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as h approaches zero. So the first thing I'm going to do is I'm going to look at this part, this part of the expression. And in particular, let's see, I am going to factor out an f of x plus h. So if you factor out an f of x plus h, this part right over here is going to be f of x plus h, f of x plus h, times, you're going to be left with g of x plus h, g of, it's a slightly shader, different shade of green, g of x plus h, that's that there, minus g of x, minus g of x, whoops, I forgot the parentheses."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as h approaches zero. So the first thing I'm going to do is I'm going to look at this part, this part of the expression. And in particular, let's see, I am going to factor out an f of x plus h. So if you factor out an f of x plus h, this part right over here is going to be f of x plus h, f of x plus h, times, you're going to be left with g of x plus h, g of, it's a slightly shader, different shade of green, g of x plus h, that's that there, minus g of x, minus g of x, whoops, I forgot the parentheses. Oh, it's a different color. I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And in particular, let's see, I am going to factor out an f of x plus h. So if you factor out an f of x plus h, this part right over here is going to be f of x plus h, f of x plus h, times, you're going to be left with g of x plus h, g of, it's a slightly shader, different shade of green, g of x plus h, that's that there, minus g of x, minus g of x, whoops, I forgot the parentheses. Oh, it's a different color. I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. All right, g of x plus h, minus g of x, that's that one right over there, and then all of that over this h, all of that over h, so that's this part here, and then this part over here, this part over here, and actually it's still over h, so let me actually circle it like this. So, this part over here, I can write as, so then we're going to have plus, actually here let me, let me factor out a g of x here. So, plus g of x, plus g of x, times this f of x plus h, times f of x plus h, minus this f of x, minus that f of x, all of that over h, all of that over h. Now, we know from our limit properties, the limit of all of this business, well, that's just going to be the same thing as the limit of this as h approaches zero, plus the limit of this as h approaches zero, and then the limit of the product is going to be the same thing as the product of the limits."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. All right, g of x plus h, minus g of x, that's that one right over there, and then all of that over this h, all of that over h, so that's this part here, and then this part over here, this part over here, and actually it's still over h, so let me actually circle it like this. So, this part over here, I can write as, so then we're going to have plus, actually here let me, let me factor out a g of x here. So, plus g of x, plus g of x, times this f of x plus h, times f of x plus h, minus this f of x, minus that f of x, all of that over h, all of that over h. Now, we know from our limit properties, the limit of all of this business, well, that's just going to be the same thing as the limit of this as h approaches zero, plus the limit of this as h approaches zero, and then the limit of the product is going to be the same thing as the product of the limits. So, if I use both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as h approaches zero of f of x plus h, of f of x plus h, times, times the limit as h approaches zero of all of this business, g of x plus h, minus g of x, minus g of x, all of that over h, I think you might see where this is going, very exciting, all right, plus, plus the limit, let me write that a little bit more clearly, plus the limit as h approaches zero of g of x, our nice brown colored g of x, times, now we're going to have a product here, the limit, the limit as h approaches zero of f of x plus h, of f of x plus h, minus f of x, minus f of x, all of that, all of that over h, and let me put the parentheses where they're appropriate, so that, that, that, that, and all I did here, the limit, the limit of this sum, that's going to be the sum of the limits, that's going to be the limit of this, plus the limit of that, and then the limit of the products is going to be the same thing as the product of the limits, so I just use those limit properties here, but now let's evaluate them. What's the limit, and I'll do it in different colors, what's this thing right over here? The limit as h approaches zero of f of x plus h, well that's just going to be f of x."}, {"video_title": "Product rule proof   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So, plus g of x, plus g of x, times this f of x plus h, times f of x plus h, minus this f of x, minus that f of x, all of that over h, all of that over h. Now, we know from our limit properties, the limit of all of this business, well, that's just going to be the same thing as the limit of this as h approaches zero, plus the limit of this as h approaches zero, and then the limit of the product is going to be the same thing as the product of the limits. So, if I use both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as h approaches zero of f of x plus h, of f of x plus h, times, times the limit as h approaches zero of all of this business, g of x plus h, minus g of x, minus g of x, all of that over h, I think you might see where this is going, very exciting, all right, plus, plus the limit, let me write that a little bit more clearly, plus the limit as h approaches zero of g of x, our nice brown colored g of x, times, now we're going to have a product here, the limit, the limit as h approaches zero of f of x plus h, of f of x plus h, minus f of x, minus f of x, all of that, all of that over h, and let me put the parentheses where they're appropriate, so that, that, that, that, and all I did here, the limit, the limit of this sum, that's going to be the sum of the limits, that's going to be the limit of this, plus the limit of that, and then the limit of the products is going to be the same thing as the product of the limits, so I just use those limit properties here, but now let's evaluate them. What's the limit, and I'll do it in different colors, what's this thing right over here? The limit as h approaches zero of f of x plus h, well that's just going to be f of x. Now, and this is the exciting part, what is this? The limit as h approaches zero of g of x plus h minus g of x over h. Well that's just our, that's the definition of our derivative, that's the derivative of g, so this is going to be, this is going to be the derivative of g of x, which is going to be g prime of x, g prime of x, so you're multiplying these two, and then you're going to have plus, what's the limit as h approaches zero of g of x? Well there's not even any h in here, so this is just going to be g of x, so plus g of x times the limit, so let's see, this one is in brown, and the last one I'll do in yellow, times the limit as h approaches zero, and we're getting very close, deserve, the drum roll should be starting, limit as h approaches zero of f of x plus h minus f of x over h. Well that's the definition of the derivative of f of x, this is f prime of x times f prime of x, so there you have it, the derivative of f of x times g of x is this, and if I wanted to write it in a little bit more condensed form, it is equal to, it is equal to f of x times the derivative of g with respect to x, times the derivative of g with respect to x plus g of x, plus g of x times the derivative of f with respect to x, f with respect to x, or another way to think about it, this is the first function times the derivative of the second, plus the second function times the derivative of the first, this is the proof, or a proof, there's actually others, of the product rule."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first idea is that of displacement. So you might use that word in everyday language, and it literally means your change in position. Your change in position. Now a related idea that sometimes gets confused with displacement is the notion of distance traveled. And you might say, well, isn't that just the same thing as change in position? And you will see shortly, no, it isn't always the same thing. The distance traveled, this is the total length of path."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now a related idea that sometimes gets confused with displacement is the notion of distance traveled. And you might say, well, isn't that just the same thing as change in position? And you will see shortly, no, it isn't always the same thing. The distance traveled, this is the total length of path. Total length of path. So what are we talking about? Well, let's say, and we're going to introduce a little bit of calculus now."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "The distance traveled, this is the total length of path. Total length of path. So what are we talking about? Well, let's say, and we're going to introduce a little bit of calculus now. Let's say that we have a particle's velocity function. And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's say, and we're going to introduce a little bit of calculus now. Let's say that we have a particle's velocity function. And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction. And oftentimes when something's one dimension, people forget, well, that too can be a vector quantity. In fact, this velocity is a vector quantity because you could think of it, if it's positive, it's moving to the right, and if it's negative, it's moving to the left. So it has a direction."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say it's just telling us our velocity in the horizontal direction. And oftentimes when something's one dimension, people forget, well, that too can be a vector quantity. In fact, this velocity is a vector quantity because you could think of it, if it's positive, it's moving to the right, and if it's negative, it's moving to the left. So it has a direction. And so sometimes you will see a vector quantity like this have a little arrow on it. Or you will see it bolded, or you will see it bolded like that. I like to write an arrow in it, although that's not always the convention used in different classes."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it has a direction. And so sometimes you will see a vector quantity like this have a little arrow on it. Or you will see it bolded, or you will see it bolded like that. I like to write an arrow in it, although that's not always the convention used in different classes. Now let's plot what this velocity function actually looks like. And I did that ahead of time. So you can see here, time equals zero."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "I like to write an arrow in it, although that's not always the convention used in different classes. Now let's plot what this velocity function actually looks like. And I did that ahead of time. So you can see here, time equals zero. Let's say time is in seconds. And our velocity is in meters per second. So this is meters per second right over here, and this is seconds in this axis."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you can see here, time equals zero. Let's say time is in seconds. And our velocity is in meters per second. So this is meters per second right over here, and this is seconds in this axis. At exactly time zero, this object is traveling at five meters per second. And we could say to the right. It has a velocity of positive five meters per second."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is meters per second right over here, and this is seconds in this axis. At exactly time zero, this object is traveling at five meters per second. And we could say to the right. It has a velocity of positive five meters per second. But then it keeps decelerating at a constant rate. So five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left. So let's think about a few things."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "It has a velocity of positive five meters per second. But then it keeps decelerating at a constant rate. So five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left. So let's think about a few things. First, let's think about what is the displacement over the first five seconds? Over first five seconds. Well, we've seen already multiple times."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about a few things. First, let's think about what is the displacement over the first five seconds? Over first five seconds. Well, we've seen already multiple times. If you wanna find the change in quantity, you can take the integral of the rate function of it. And so velocity is actually the rate of displacement is one way to think about it. So displacement over the first five seconds, we could take the integral from zero to five, zero to five, of our velocity function, of our velocity function, just like that."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we've seen already multiple times. If you wanna find the change in quantity, you can take the integral of the rate function of it. And so velocity is actually the rate of displacement is one way to think about it. So displacement over the first five seconds, we could take the integral from zero to five, zero to five, of our velocity function, of our velocity function, just like that. And we can even calculate this really fast. That would just be this area right over here, which we could just use a little bit of geometry. This is a five by five triangle."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So displacement over the first five seconds, we could take the integral from zero to five, zero to five, of our velocity function, of our velocity function, just like that. And we can even calculate this really fast. That would just be this area right over here, which we could just use a little bit of geometry. This is a five by five triangle. So five times five is 25, times 1 1\u20442, remember, area of a triangle's 1 1\u20442 base times height. So this is going to be 12.5, and let's see, this is going to be meters per second times seconds, so 12.5 meters. So that's the change in position for that particle over the first five seconds."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is a five by five triangle. So five times five is 25, times 1 1\u20442, remember, area of a triangle's 1 1\u20442 base times height. So this is going to be 12.5, and let's see, this is going to be meters per second times seconds, so 12.5 meters. So that's the change in position for that particle over the first five seconds. Wherever it started, it's now going to be 12.5 meters to the right of it, assuming that positive is to the right. Now what about over the first 10 seconds? Now this gets interesting, and I encourage you to pause your video and think about it."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's the change in position for that particle over the first five seconds. Wherever it started, it's now going to be 12.5 meters to the right of it, assuming that positive is to the right. Now what about over the first 10 seconds? Now this gets interesting, and I encourage you to pause your video and think about it. What would be the displacement over the first 10 seconds? Well, we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt. And so that would be the area from here all the way to right over there, so this entire area."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this gets interesting, and I encourage you to pause your video and think about it. What would be the displacement over the first 10 seconds? Well, we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt. And so that would be the area from here all the way to right over there, so this entire area. But you might appreciate, when you're taking a definite integral, if we are below the t-axis and above the function like this, this is going to be negative area. And in fact, this area and this area are going to exactly cancel out, and you're going to get zero meters. Now you might be saying, how can that be?"}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that would be the area from here all the way to right over there, so this entire area. But you might appreciate, when you're taking a definite integral, if we are below the t-axis and above the function like this, this is going to be negative area. And in fact, this area and this area are going to exactly cancel out, and you're going to get zero meters. Now you might be saying, how can that be? After 10 seconds, how do we, why is our displacement only zero meters? This particle's been moving the entire time. Well, remember what's going on."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now you might be saying, how can that be? After 10 seconds, how do we, why is our displacement only zero meters? This particle's been moving the entire time. Well, remember what's going on. The first five seconds, it's moving to the right, it's decelerating the whole time, and then right at five seconds, it has gone 12.5 meters to the right. But then it starts, its velocity starts becoming negative, and the particle starts moving to the left. And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, remember what's going on. The first five seconds, it's moving to the right, it's decelerating the whole time, and then right at five seconds, it has gone 12.5 meters to the right. But then it starts, its velocity starts becoming negative, and the particle starts moving to the left. And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. And so the particle has gone over 10 seconds, 12.5 meters to the right, and then 12.5 meters to the left, and so its change in position is zero meters. It has not changed. Now you might start to be appreciating what the difference between displacement and distance traveled is."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. And so the particle has gone over 10 seconds, 12.5 meters to the right, and then 12.5 meters to the left, and so its change in position is zero meters. It has not changed. Now you might start to be appreciating what the difference between displacement and distance traveled is. So distance, if you're talking about your total length of path, you don't care as much about direction. And so instead of thinking about velocity, what we would do is think about speed. And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now you might start to be appreciating what the difference between displacement and distance traveled is. So distance, if you're talking about your total length of path, you don't care as much about direction. And so instead of thinking about velocity, what we would do is think about speed. And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity. Later on, when we do multiple dimensions, it would be the magnitude of the velocity function, which is what the absolute value function does in one dimension. So what would this look like if we plotted it? Well, the absolute value of the velocity function would just look like that."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity. Later on, when we do multiple dimensions, it would be the magnitude of the velocity function, which is what the absolute value function does in one dimension. So what would this look like if we plotted it? Well, the absolute value of the velocity function would just look like that. And so if you want the distance, you would find the, the distance traveled, I should say, you would find the integral over the appropriate change in time of the speed function, right over here, which we have graphed. So notice, if we want the distance traveled, so I'll just say, I'll write it out, distance traveled over first five seconds, first five seconds, first five seconds, what would it be? Well, it would be the integral from zero to five of the absolute value of our velocity function, which is, you could just view it as our speed function right over here, dt."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the absolute value of the velocity function would just look like that. And so if you want the distance, you would find the, the distance traveled, I should say, you would find the integral over the appropriate change in time of the speed function, right over here, which we have graphed. So notice, if we want the distance traveled, so I'll just say, I'll write it out, distance traveled over first five seconds, first five seconds, first five seconds, what would it be? Well, it would be the integral from zero to five of the absolute value of our velocity function, which is, you could just view it as our speed function right over here, dt. And so it would be this area, which we already know to be 12.5 meters. So for the first five seconds, your distance and displacement are consistent. Well, that's because you have, in this case, the velocity function is positive, so the absolute value of it is still going to be positive."}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it would be the integral from zero to five of the absolute value of our velocity function, which is, you could just view it as our speed function right over here, dt. And so it would be this area, which we already know to be 12.5 meters. So for the first five seconds, your distance and displacement are consistent. Well, that's because you have, in this case, the velocity function is positive, so the absolute value of it is still going to be positive. But if you think about over the first 10 seconds, your distance, 10 seconds, what is it going to be? Pause the video and try to think about it. Well, that's going to be the integral from zero to 10 of the absolute value of our velocity function, which is going to be equal to what?"}, {"video_title": "Motion problems with integrals displacement vs. distance   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's because you have, in this case, the velocity function is positive, so the absolute value of it is still going to be positive. But if you think about over the first 10 seconds, your distance, 10 seconds, what is it going to be? Pause the video and try to think about it. Well, that's going to be the integral from zero to 10 of the absolute value of our velocity function, which is going to be equal to what? Well, it's going to be this area plus this area right over here. So plus this area right over here. And so this is going to be five times five times 1 1\u20442 plus five times five times 1 1\u20442, which is going to be 25 meters."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've been given some interesting information here about the functions f, g, and h. For f, they tell us for given values of x what f of x is equal to and what f prime of x is equal to. Then they define g of x for us in terms of this kind of absolute value expression. And then they define h of x for us in terms of both f of x and g of x. And what we're curious about is what is the derivative with respect to x of h of x at x is equal to 9. And so I encourage you to pause this video and think about it on your own before I work through it. So let's think about it a little bit. So another way just to get familiar with the notation of writing this, the derivative of h of x with respect to x at x equals 9."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we're curious about is what is the derivative with respect to x of h of x at x is equal to 9. And so I encourage you to pause this video and think about it on your own before I work through it. So let's think about it a little bit. So another way just to get familiar with the notation of writing this, the derivative of h of x with respect to x at x equals 9. This is equivalent to h. Let me do it in that blue color. It is equivalent to h prime. And the prime signifies that we're taking the derivative."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So another way just to get familiar with the notation of writing this, the derivative of h of x with respect to x at x equals 9. This is equivalent to h. Let me do it in that blue color. It is equivalent to h prime. And the prime signifies that we're taking the derivative. h prime of x when x equals 9. So h prime of 9 is what this really is. Actually, let me do this in a different color."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the prime signifies that we're taking the derivative. h prime of x when x equals 9. So h prime of 9 is what this really is. Actually, let me do this in a different color. So this is h prime of 9. So let's think about what that is. Let's take the derivative of both sides of this expression to figure out what the derivative with respect to x of h is."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me do this in a different color. So this is h prime of 9. So let's think about what that is. Let's take the derivative of both sides of this expression to figure out what the derivative with respect to x of h is. So we get derivative with respect to x of h of x is going to be equal to the derivative with respect to x of all of this business. So I could actually just rewrite it. 3 times f of x plus 2 times g of x."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's take the derivative of both sides of this expression to figure out what the derivative with respect to x of h is. So we get derivative with respect to x of h of x is going to be equal to the derivative with respect to x of all of this business. So I could actually just rewrite it. 3 times f of x plus 2 times g of x. Now, this right over here, the derivative of the sum of two terms, that's going to be the same thing as the sum of the derivatives of each of the terms. So this is going to be the same thing as the derivative with respect to x of 3 times f of x plus the derivative with respect to x of 2 times g of x. Now, the derivative of a number, or I guess you could say a scaling factor, times a function."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "3 times f of x plus 2 times g of x. Now, this right over here, the derivative of the sum of two terms, that's going to be the same thing as the sum of the derivatives of each of the terms. So this is going to be the same thing as the derivative with respect to x of 3 times f of x plus the derivative with respect to x of 2 times g of x. Now, the derivative of a number, or I guess you could say a scaling factor, times a function. The derivative of a scalar times a function is the same thing as the scalar times the derivative of the function. What does that mean? Well, that just means that this first term right over here, that's going to be equivalent to 3 times the derivative with respect to x of f of x plus this part over here is the same thing as 2."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, the derivative of a number, or I guess you could say a scaling factor, times a function. The derivative of a scalar times a function is the same thing as the scalar times the derivative of the function. What does that mean? Well, that just means that this first term right over here, that's going to be equivalent to 3 times the derivative with respect to x of f of x plus this part over here is the same thing as 2. Let me make sure I don't run out of space here. Plus 2 times the derivative with respect to x of g of x. So the derivative of h with respect to x is equal to 3 times the derivative of f with respect to x plus 2 times the derivative of g with respect to x."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that just means that this first term right over here, that's going to be equivalent to 3 times the derivative with respect to x of f of x plus this part over here is the same thing as 2. Let me make sure I don't run out of space here. Plus 2 times the derivative with respect to x of g of x. So the derivative of h with respect to x is equal to 3 times the derivative of f with respect to x plus 2 times the derivative of g with respect to x. And if we wanted to write it in this kind of prime notation here, we could rewrite it as h prime of x is equal to 3 times f prime of x. So this whole part right over here, that is the same thing as f prime of x. So it's 3 times f prime of x plus 2 times g prime of x."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of h with respect to x is equal to 3 times the derivative of f with respect to x plus 2 times the derivative of g with respect to x. And if we wanted to write it in this kind of prime notation here, we could rewrite it as h prime of x is equal to 3 times f prime of x. So this whole part right over here, that is the same thing as f prime of x. So it's 3 times f prime of x plus 2 times g prime of x. And once you kind of are more fluent with this property, the derivative of the sum of two things is the sum of the derivatives. And the derivative of a scalar times something is the same thing as the scalar times the derivative of that something. You really could have gone straight from here to here pretty quickly."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's 3 times f prime of x plus 2 times g prime of x. And once you kind of are more fluent with this property, the derivative of the sum of two things is the sum of the derivatives. And the derivative of a scalar times something is the same thing as the scalar times the derivative of that something. You really could have gone straight from here to here pretty quickly. Now why is this interesting? Well, now we can evaluate this function when x is equal to 9. So h prime of 9 is the same thing as 3 times f prime of 9 plus 2 times g prime of 9."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You really could have gone straight from here to here pretty quickly. Now why is this interesting? Well, now we can evaluate this function when x is equal to 9. So h prime of 9 is the same thing as 3 times f prime of 9 plus 2 times g prime of 9. Now what is f prime of 9, the derivative of our function f when x is equal to 9? Well, they tell us when x is equal to 9, f of 9 is 1. But more importantly, f prime of 9 is 3."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So h prime of 9 is the same thing as 3 times f prime of 9 plus 2 times g prime of 9. Now what is f prime of 9, the derivative of our function f when x is equal to 9? Well, they tell us when x is equal to 9, f of 9 is 1. But more importantly, f prime of 9 is 3. So this part right over here evaluates that part's 3. But what's g prime of 9? So let's look at this function a little bit more closely."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But more importantly, f prime of 9 is 3. So this part right over here evaluates that part's 3. But what's g prime of 9? So let's look at this function a little bit more closely. So there's a couple of ways we could think about it. Actually, let's try to graph it. Now I think that could be interesting just to visualize what's going on here."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at this function a little bit more closely. So there's a couple of ways we could think about it. Actually, let's try to graph it. Now I think that could be interesting just to visualize what's going on here. So let's say that's our y-axis. And this right over here is our x-axis. Now when does an absolute value function like this, when is this going to hit a minimum point?"}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now I think that could be interesting just to visualize what's going on here. So let's say that's our y-axis. And this right over here is our x-axis. Now when does an absolute value function like this, when is this going to hit a minimum point? Well, the absolute value of something is always going to be non-negative. So it hits a minimum point when this thing is equal to 0. Well, when is this thing equal to 0?"}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now when does an absolute value function like this, when is this going to hit a minimum point? Well, the absolute value of something is always going to be non-negative. So it hits a minimum point when this thing is equal to 0. Well, when is this thing equal to 0? When x equals 1, this thing is equal to 0. So we hit a minimum point when x is equal to 1. And when x equals 1, this term is 0, absolute value of 0 is 0. g of 1 is 1."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, when is this thing equal to 0? When x equals 1, this thing is equal to 0. So we hit a minimum point when x is equal to 1. And when x equals 1, this term is 0, absolute value of 0 is 0. g of 1 is 1. So we have this point right over there. Now what happens after that? What happens for x greater than 1?"}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when x equals 1, this term is 0, absolute value of 0 is 0. g of 1 is 1. So we have this point right over there. Now what happens after that? What happens for x greater than 1? Actually, let me write this down. So g of x is equal to. And in general, whenever you have an absolute value or a relatively simple absolute value function like this, you could think of it, you could break it up into two functions."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "What happens for x greater than 1? Actually, let me write this down. So g of x is equal to. And in general, whenever you have an absolute value or a relatively simple absolute value function like this, you could think of it, you could break it up into two functions. Or you could think about this function over different intervals, when the absolute value is non-negative and when the absolute value is negative. So when the absolute value is non-negative, that's when x is greater than or equal to 0. And when the absolute value is non-negative, if you're taking the absolute value of a non-negative number, then it's just going to be itself."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in general, whenever you have an absolute value or a relatively simple absolute value function like this, you could think of it, you could break it up into two functions. Or you could think about this function over different intervals, when the absolute value is non-negative and when the absolute value is negative. So when the absolute value is non-negative, that's when x is greater than or equal to 0. And when the absolute value is non-negative, if you're taking the absolute value of a non-negative number, then it's just going to be itself. The absolute value of 0 is 0. Absolute value of 1 is 1. Absolute value of 100 is 100."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when the absolute value is non-negative, if you're taking the absolute value of a non-negative number, then it's just going to be itself. The absolute value of 0 is 0. Absolute value of 1 is 1. Absolute value of 100 is 100. So then, you can ignore the absolute value for x as greater than or equal to 0. for x is greater than or equal to 1. x is greater than or equal to 1, this thing right over here is non-negative. And so it'll just evaluate to x minus 1. So this is going to be x minus 1 plus 1, which is the same thing as just x."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Absolute value of 100 is 100. So then, you can ignore the absolute value for x as greater than or equal to 0. for x is greater than or equal to 1. x is greater than or equal to 1, this thing right over here is non-negative. And so it'll just evaluate to x minus 1. So this is going to be x minus 1 plus 1, which is the same thing as just x. Minus 1 plus 1, they just cancel out. Now, when this term right over here is negative, and that's going to happen for x is less than 1, well, then the absolute value is going to be the opposite of it. You give me the absolute value of a negative number, it's going to be the opposite."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be x minus 1 plus 1, which is the same thing as just x. Minus 1 plus 1, they just cancel out. Now, when this term right over here is negative, and that's going to happen for x is less than 1, well, then the absolute value is going to be the opposite of it. You give me the absolute value of a negative number, it's going to be the opposite. The absolute value of negative 8 is positive 8. So it's going to be that the negative of x minus 1 is 1 minus x plus 1. Or we could say 2 minus x."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You give me the absolute value of a negative number, it's going to be the opposite. The absolute value of negative 8 is positive 8. So it's going to be that the negative of x minus 1 is 1 minus x plus 1. Or we could say 2 minus x. 2 minus x. Now, so for x is greater than or equal to 1, we would look at this expression. Now, what's the slope of that?"}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or we could say 2 minus x. 2 minus x. Now, so for x is greater than or equal to 1, we would look at this expression. Now, what's the slope of that? Well, the slope of that is 1. So we're going to have a curve that looks like, or a line, I guess we could say, that looks like this. For all x is greater than or equal to 1."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, what's the slope of that? Well, the slope of that is 1. So we're going to have a curve that looks like, or a line, I guess we could say, that looks like this. For all x is greater than or equal to 1. So the important thing, remember, we're going to think about the slope of the tangent line when we think about the derivative of g. So slope is equal to 1. And for x less than 1, well, our slope now, if we look right over here, our slope is negative. Our slope is negative 1."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "For all x is greater than or equal to 1. So the important thing, remember, we're going to think about the slope of the tangent line when we think about the derivative of g. So slope is equal to 1. And for x less than 1, well, our slope now, if we look right over here, our slope is negative. Our slope is negative 1. So it's going to look like this. It's going to look like that. But for the point in question, if we're thinking about g prime of 9, so 9 is someplace out here."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our slope is negative 1. So it's going to look like this. It's going to look like that. But for the point in question, if we're thinking about g prime of 9, so 9 is someplace out here. So what is g prime of 9? So g prime of 9, let me make it clear, this graph right over here. This is the graph of g of x."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But for the point in question, if we're thinking about g prime of 9, so 9 is someplace out here. So what is g prime of 9? So g prime of 9, let me make it clear, this graph right over here. This is the graph of g of x. Or we could say y. This is the graph y equals g of x. y is equal to g of x. So what is g prime of 9?"}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the graph of g of x. Or we could say y. This is the graph y equals g of x. y is equal to g of x. So what is g prime of 9? Well, that's the slope when x is equal to 9. Well, the slope is going to be equal to 1. So g prime of 9 is 1."}, {"video_title": "Basic derivative rules table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is g prime of 9? Well, that's the slope when x is equal to 9. Well, the slope is going to be equal to 1. So g prime of 9 is 1. So what does this evaluate to? This is going to be 3 times 3. So this part right over here is 9 plus 2 times 1, plus 2, which is equal to 11."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause this video and see if you can evaluate this. So you might be saying, oh, what kind of fancy technique could I use? But you will see sometimes the fanciest or maybe the least fancy, but the best technique is to just simplify this algebraically. So in this situation, what happens if we distribute this x squared? Well, then we're going to get a polynomial here within the integral. So this is going to be equal to the integral of x squared times three x is three x to the third and then negative one times x squared is minus x squared and then that times dx. And now this is pretty straightforward to evaluate."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in this situation, what happens if we distribute this x squared? Well, then we're going to get a polynomial here within the integral. So this is going to be equal to the integral of x squared times three x is three x to the third and then negative one times x squared is minus x squared and then that times dx. And now this is pretty straightforward to evaluate. This is going to be equal to the antiderivative of x to the third is x to the fourth over four. So this is going to be three times x to the fourth over four I could write it that way or let me just write it, x to the fourth over four and then the antiderivative of x squared is x to the third over three. So minus x to the third over three and this is an indefinite integral."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now this is pretty straightforward to evaluate. This is going to be equal to the antiderivative of x to the third is x to the fourth over four. So this is going to be three times x to the fourth over four I could write it that way or let me just write it, x to the fourth over four and then the antiderivative of x squared is x to the third over three. So minus x to the third over three and this is an indefinite integral. There might be a constant there. So let me write that down and we're done. The big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "So minus x to the third over three and this is an indefinite integral. There might be a constant there. So let me write that down and we're done. The big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative. Let's do another example. Let's say that we want to take the indefinite integral of, it's going to be a hairy expression, so x to the third plus three x squared minus five, all of that over x squared dx. What would this be?"}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "The big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative. Let's do another example. Let's say that we want to take the indefinite integral of, it's going to be a hairy expression, so x to the third plus three x squared minus five, all of that over x squared dx. What would this be? Pause the video again and see if you can figure it out. So once again, your brain might want to try to do some fancy tricks or whatever else but the main insight here is to realize that you could just simplify it algebraically. What happens if you just divide each of these terms by x squared?"}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "What would this be? Pause the video again and see if you can figure it out. So once again, your brain might want to try to do some fancy tricks or whatever else but the main insight here is to realize that you could just simplify it algebraically. What happens if you just divide each of these terms by x squared? Well then this thing is going to be equal to, put some parentheses here, x to the third divided by x squared is just going to be x. Three x squared divided by x squared is just three and then negative five divided by x squared, you could just write that as negative five times x to the negative two power. And so once again, we just need to use the reverse power rule here to take the antiderivative."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "What happens if you just divide each of these terms by x squared? Well then this thing is going to be equal to, put some parentheses here, x to the third divided by x squared is just going to be x. Three x squared divided by x squared is just three and then negative five divided by x squared, you could just write that as negative five times x to the negative two power. And so once again, we just need to use the reverse power rule here to take the antiderivative. This is going to be, let's see, the antiderivative of x is x squared over two, x squared over two plus the antiderivative of three and the antiderivative of three would just be three x. The antiderivative of negative five x to the negative two. So we would increment the exponent by one, positive one and then divide by that value."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so once again, we just need to use the reverse power rule here to take the antiderivative. This is going to be, let's see, the antiderivative of x is x squared over two, x squared over two plus the antiderivative of three and the antiderivative of three would just be three x. The antiderivative of negative five x to the negative two. So we would increment the exponent by one, positive one and then divide by that value. So it would be negative five x to the negative one. We're adding one to negative one. All of that divided by negative one which is the same, we could write it like that."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would increment the exponent by one, positive one and then divide by that value. So it would be negative five x to the negative one. We're adding one to negative one. All of that divided by negative one which is the same, we could write it like that. Well these two would just, you'd have minus divided by negative one so it's really just going, you can rewrite it like this, plus five x to the negative one and you could take the derivative of this to verify that it would indeed give you that. And of course, we can't forget our plus c. Never forget that if you're taking an indefinite integral. All right, let's just do one more for good measure."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that divided by negative one which is the same, we could write it like that. Well these two would just, you'd have minus divided by negative one so it's really just going, you can rewrite it like this, plus five x to the negative one and you could take the derivative of this to verify that it would indeed give you that. And of course, we can't forget our plus c. Never forget that if you're taking an indefinite integral. All right, let's just do one more for good measure. Let's say we're taking the indefinite integral of the cube root of x to the fifth dx. Cube root of x to the fifth dx. Pause the video and see if you can evaluate this."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's just do one more for good measure. Let's say we're taking the indefinite integral of the cube root of x to the fifth dx. Cube root of x to the fifth dx. Pause the video and see if you can evaluate this. Try to write a little bit neater. X to the fifth dx. Pause the video and try to figure it out."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and see if you can evaluate this. Try to write a little bit neater. X to the fifth dx. Pause the video and try to figure it out. So here the realization is, well if you just rewrite all of this as one exponent, so this is equal to the indefinite integral of x to the fifth to the 1 3rd. I just rewrote the cube root as the 1 3rd power dx which is the same thing as the integral of x to the, if I raise something to a power and then raise that to a power, I can multiply those two exponents. That's just exponent properties."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and try to figure it out. So here the realization is, well if you just rewrite all of this as one exponent, so this is equal to the indefinite integral of x to the fifth to the 1 3rd. I just rewrote the cube root as the 1 3rd power dx which is the same thing as the integral of x to the, if I raise something to a power and then raise that to a power, I can multiply those two exponents. That's just exponent properties. X to the 5 3rd dx. Many of you might have just gone straight to this step right over here. And then once again, we just have to use the reverse power rule."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's just exponent properties. X to the 5 3rd dx. Many of you might have just gone straight to this step right over here. And then once again, we just have to use the reverse power rule. This is going to be x to the, we increment this 5 3rds by one or we can add 3 3rds to it. So it's x to the 8 3rds and then we divide by 8 3rds and multiply by its reciprocal. So we could just say 3 8ths times x to the 8 3rds and of course we have r plus c. And verify this."}, {"video_title": "Rewriting before integrating   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then once again, we just have to use the reverse power rule. This is going to be x to the, we increment this 5 3rds by one or we can add 3 3rds to it. So it's x to the 8 3rds and then we divide by 8 3rds and multiply by its reciprocal. So we could just say 3 8ths times x to the 8 3rds and of course we have r plus c. And verify this. If you use the power rule here, you'd have 8 3rds times the 3 8ths which just gives you a coefficient of one. And then you decrement this by 3 3rds or one, you get to 5 3rds which is exactly what we originally had. So the big takeaway of this video, many times the most powerful integration technique is literally just algebraic simplification first."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that this is a toll booth, right? You're on the turnpike, and this is a toll booth at point A, and you get your toll, you get a, you reach it at exactly 1 p.m., and the highway's computers and stuff register that. Let's say you have some type of a, one of those devices so that when you pay the toll, it just knows who you are, and it takes your money from an account someplace. So it sees that you got there at exactly, you got there at exactly 1 p.m. And then, let's say that there's, and let's say you get off of the toll highway, the turnpike, let's say you get off of it at point B, and you get there at exactly 2 p.m. I'm making these numbers very easy to work with. And let's say that they are 80 miles apart. So this distance right over here is 80 miles."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it sees that you got there at exactly, you got there at exactly 1 p.m. And then, let's say that there's, and let's say you get off of the toll highway, the turnpike, let's say you get off of it at point B, and you get there at exactly 2 p.m. I'm making these numbers very easy to work with. And let's say that they are 80 miles apart. So this distance right over here is 80 miles. 80 miles. And let's say that the speed limit on this stretch of highway is 55 miles per hour. Speed limit."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this distance right over here is 80 miles. 80 miles. And let's say that the speed limit on this stretch of highway is 55 miles per hour. Speed limit. Speed limit is 55 miles per hour. So the question is, is can the authorities prove that you went over the speed limit? Well, let's just graph this."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Speed limit. Speed limit is 55 miles per hour. So the question is, is can the authorities prove that you went over the speed limit? Well, let's just graph this. I think you know where this is going. So let's graph it. So let's say this right over here is our position."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's just graph this. I think you know where this is going. So let's graph it. So let's say this right over here is our position. So I'll call that the S axis, S for position. And that's going to be in miles. And S is, it's, you know, obviously, S doesn't really stand for position, but P, you know, it kind of looks like rho for density, and D we use for differentials, or distance, or displacement."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say this right over here is our position. So I'll call that the S axis, S for position. And that's going to be in miles. And S is, it's, you know, obviously, S doesn't really stand for position, but P, you know, it kind of looks like rho for density, and D we use for differentials, or distance, or displacement. So S is what gets used for position very often. So let's say S is our position. And let's see, this is T for time."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And S is, it's, you know, obviously, S doesn't really stand for position, but P, you know, it kind of looks like rho for density, and D we use for differentials, or distance, or displacement. So S is what gets used for position very often. So let's say S is our position. And let's see, this is T for time. T for time. And let's say this is in hours. And let's see, we care about the interval from time going from time one, time one, to time two, time two."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, this is T for time. T for time. And let's say this is in hours. And let's see, we care about the interval from time going from time one, time one, to time two, time two. I'm not really drawing the axes completely at scale. Let me just assume that there's a gap here, just because I don't want to, I don't want to, I actually want to, don't make you think that I'm drawing it completely at scale, because I really want to focus on this part of the interval. So this is time equals two, two hours."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we care about the interval from time going from time one, time one, to time two, time two. I'm not really drawing the axes completely at scale. Let me just assume that there's a gap here, just because I don't want to, I don't want to, I actually want to, don't make you think that I'm drawing it completely at scale, because I really want to focus on this part of the interval. So this is time equals two, two hours. And so at time equal one, you're right over here. And let's say this position is, we'll just call that S of one. And at time two, at time two, you're at this position right over here."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is time equals two, two hours. And so at time equal one, you're right over here. And let's say this position is, we'll just call that S of one. And at time two, at time two, you're at this position right over here. You're right over there. And so your position is S of two, you're at that coordinate right over there. And that's all we know."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And at time two, at time two, you're at this position right over here. You're right over there. And so your position is S of two, you're at that coordinate right over there. And that's all we know. That's all we know. Well, we know a few other things. We know what our change in time is."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's all we know. That's all we know. Well, we know a few other things. We know what our change in time is. Two minus one. And we know what our change in position is. We know that our change in position, our change in position, which is equal to S of two, S of two minus S of one, minus S of one, is equal to 80 miles."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what our change in time is. Two minus one. And we know what our change in position is. We know that our change in position, our change in position, which is equal to S of two, S of two minus S of one, minus S of one, is equal to 80 miles. The change in position is 80 miles. So let me write that. And we'll just, for simplicity, assume it's a straight highway, so our change in distance is the same as our change in position, same as change in displacement."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that our change in position, our change in position, which is equal to S of two, S of two minus S of one, minus S of one, is equal to 80 miles. The change in position is 80 miles. So let me write that. And we'll just, for simplicity, assume it's a straight highway, so our change in distance is the same as our change in position, same as change in displacement. So this is 80 miles. 80 miles. And then what is our change in time?"}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we'll just, for simplicity, assume it's a straight highway, so our change in distance is the same as our change in position, same as change in displacement. So this is 80 miles. 80 miles. And then what is our change in time? Over our change in time, well, that's going to be two minus one. Two minus one. Which is just going to be one hour."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then what is our change in time? Over our change in time, well, that's going to be two minus one. Two minus one. Which is just going to be one hour. Over one hour. Or we could say that the slope of the line that connects these two points, let me do that in another color, the slope of, that's the same color, the slope of this line right over here, the slope right over here is 80 miles per hour. Slope is equal to 80 miles per hour."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Which is just going to be one hour. Over one hour. Or we could say that the slope of the line that connects these two points, let me do that in another color, the slope of, that's the same color, the slope of this line right over here, the slope right over here is 80 miles per hour. Slope is equal to 80 miles per hour. Or you could say that your average, your average velocity over that hour was 80 miles per hour. And so what the authorities could do in a court of law, and I've never heard a mathematical theorem cited like this, but they could, and I remember reading about this about 10 years ago, and it was very controversial, the authorities literally said, look, over this interval, your average velocity was clearly 80 miles per hour, so at some point in that hour, and they could have cited, they could have said by the mean value theorem, at some point in that hour, you must have been going at exactly 80 miles, at least frankly, 80 miles per hour. And it would have been very hard to disprove because your position as a function of time is definitely continuous and differentiable over that interval, it's continuous, you're not just getting teleported from one place to another, that'd be a pretty amazing car, and it is also differentiable, you always have a well-defined velocity."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Slope is equal to 80 miles per hour. Or you could say that your average, your average velocity over that hour was 80 miles per hour. And so what the authorities could do in a court of law, and I've never heard a mathematical theorem cited like this, but they could, and I remember reading about this about 10 years ago, and it was very controversial, the authorities literally said, look, over this interval, your average velocity was clearly 80 miles per hour, so at some point in that hour, and they could have cited, they could have said by the mean value theorem, at some point in that hour, you must have been going at exactly 80 miles, at least frankly, 80 miles per hour. And it would have been very hard to disprove because your position as a function of time is definitely continuous and differentiable over that interval, it's continuous, you're not just getting teleported from one place to another, that'd be a pretty amazing car, and it is also differentiable, you always have a well-defined velocity. And so I challenge anyone, try to connect these two points with a continuous and differentiable curve where at some point, the instantaneous velocity, the slope of the tangent line is not the same thing as the slope of this line, it's impossible, the mean value theorem tells us it's impossible. So let me just draw, so we can imagine, let's say I had to stop to pay, to get to kind of register where I am on the highway, then I start to accelerate a little bit, so right now, my instantaneous velocity is less than my average velocity, I'm accelerating the slope of the tangent line, but if I wanna get there at that time, and especially because I have to slow down as I approach it, as I approach the toll booth, the only way I could connect these two things, well, let's see, I'm gonna have to, at some point, I'm actually, at this point, I'm actually going faster than the 80 miles per hour. And the mean value theorem just tells us that, look, if this function is continuous and differentiable over this interval, continuous over the closed interval, differentiable over the open interval, that there's at least one point in the open interval, which it calls C, so there's at least one point where your instantaneous rate of change, where the slope of the tangent line is the same as the slope of the secant line."}, {"video_title": "Mean value theorem application   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it would have been very hard to disprove because your position as a function of time is definitely continuous and differentiable over that interval, it's continuous, you're not just getting teleported from one place to another, that'd be a pretty amazing car, and it is also differentiable, you always have a well-defined velocity. And so I challenge anyone, try to connect these two points with a continuous and differentiable curve where at some point, the instantaneous velocity, the slope of the tangent line is not the same thing as the slope of this line, it's impossible, the mean value theorem tells us it's impossible. So let me just draw, so we can imagine, let's say I had to stop to pay, to get to kind of register where I am on the highway, then I start to accelerate a little bit, so right now, my instantaneous velocity is less than my average velocity, I'm accelerating the slope of the tangent line, but if I wanna get there at that time, and especially because I have to slow down as I approach it, as I approach the toll booth, the only way I could connect these two things, well, let's see, I'm gonna have to, at some point, I'm actually, at this point, I'm actually going faster than the 80 miles per hour. And the mean value theorem just tells us that, look, if this function is continuous and differentiable over this interval, continuous over the closed interval, differentiable over the open interval, that there's at least one point in the open interval, which it calls C, so there's at least one point where your instantaneous rate of change, where the slope of the tangent line is the same as the slope of the secant line. So that point right over there, that point looks like that right over there. And so if this is time C, that looks like it's at around 1.15, this, the mean value theorem says that at some point, there exists some time where your, where S prime of C is equal to this average velocity, is equal to 80 miles per hour. It doesn't look like that's the only one, it looks like this one over here, this could also be a candidate for C."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we see that right at x equals two, if we try to evaluate f of two, we get two over one minus cosine of two minus two, which is the same thing as cosine of zero, and cosine of zero is just one, and so one minus one is zero, and so the function is not defined at x equals two, and that's why it might be interesting to find the limit as x approaches two, and especially the one-sided limits. And if the one-sided limits, well, I'll just leave it at that. So let's try to approach this. And there's actually a couple of ways you could do it. There's one way you could do this without a calculator, by just inspecting what's going on here and thinking about the properties of the cosine function. And if that inspires you, pause the video and work it out, and I will do that at the end of this video. The other way, if you have a calculator, is to do it with a little bit of a table, like we've done in other example problems."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there's actually a couple of ways you could do it. There's one way you could do this without a calculator, by just inspecting what's going on here and thinking about the properties of the cosine function. And if that inspires you, pause the video and work it out, and I will do that at the end of this video. The other way, if you have a calculator, is to do it with a little bit of a table, like we've done in other example problems. So if we think about x approaching two from the positive direction, well, then we can make a little table here where you have x and then you have f of x. And so if we're approaching two from values greater than two you could have 2.1, 2.01. Now, the reason why I said calculators, these aren't trivial to evaluate because this would be, what, 2.1 over one minus cosine of 2.1 minus two is 0.1."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The other way, if you have a calculator, is to do it with a little bit of a table, like we've done in other example problems. So if we think about x approaching two from the positive direction, well, then we can make a little table here where you have x and then you have f of x. And so if we're approaching two from values greater than two you could have 2.1, 2.01. Now, the reason why I said calculators, these aren't trivial to evaluate because this would be, what, 2.1 over one minus cosine of 2.1 minus two is 0.1. I do not know what cosine of 0.1 is without a calculator. I do know that cosine of zero is one, so this is very, very close to one without getting to one. And it's going to be less than one."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, the reason why I said calculators, these aren't trivial to evaluate because this would be, what, 2.1 over one minus cosine of 2.1 minus two is 0.1. I do not know what cosine of 0.1 is without a calculator. I do know that cosine of zero is one, so this is very, very close to one without getting to one. And it's going to be less than one. Cosine is never going to be greater than one. The cosine function is bounded between negative one is less than cosine of x. I'll just write the x there, I don't need the parentheses, which is less than one. The cosine function just oscillates between these two values."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's going to be less than one. Cosine is never going to be greater than one. The cosine function is bounded between negative one is less than cosine of x. I'll just write the x there, I don't need the parentheses, which is less than one. The cosine function just oscillates between these two values. So this thing is going to be approaching one, but it's going to be less than one. It definitely cannot be greater than one. And that's actually a good hint for how you can just explore the structure here."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The cosine function just oscillates between these two values. So this thing is going to be approaching one, but it's going to be less than one. It definitely cannot be greater than one. And that's actually a good hint for how you can just explore the structure here. And then you could say, all right, 2.01, well, that's going to be 2.01 over one minus cosine of 0.01. And this is going to be even closer to one without being one. So this could, you know, this is, but it's going to be less than one."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's actually a good hint for how you can just explore the structure here. And then you could say, all right, 2.01, well, that's going to be 2.01 over one minus cosine of 0.01. And this is going to be even closer to one without being one. So this could, you know, this is, but it's going to be less than one. No matter what, cosine of anything is going to be less than, it's going to be between negative one and one, and it could even be including those things. But as we approach two, this thing is going to approach, it's going to approach one, I guess you could say approach one from below. And so you can start to make some intuitions here."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this could, you know, this is, but it's going to be less than one. No matter what, cosine of anything is going to be less than, it's going to be between negative one and one, and it could even be including those things. But as we approach two, this thing is going to approach, it's going to approach one, I guess you could say approach one from below. And so you can start to make some intuitions here. If it's approaching one from below, this thing over here, this whole expression, is going to be positive. And as we approach x equals two, well, the numerator is positive, it's approaching two, the denominator is positive. So this whole thing has to be approaching a positive value, or it could become unbounded in the positive direction."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you can start to make some intuitions here. If it's approaching one from below, this thing over here, this whole expression, is going to be positive. And as we approach x equals two, well, the numerator is positive, it's approaching two, the denominator is positive. So this whole thing has to be approaching a positive value, or it could become unbounded in the positive direction. As we'll see, this is unbounded, because this thing is even closer to one than this thing. And you would see that if you have a calculator. But needless to say, this is going to be unbounded in the positive direction, so we're going to be going towards positive infinity."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this whole thing has to be approaching a positive value, or it could become unbounded in the positive direction. As we'll see, this is unbounded, because this thing is even closer to one than this thing. And you would see that if you have a calculator. But needless to say, this is going to be unbounded in the positive direction, so we're going to be going towards positive infinity. So these two choices have that. And we can make the exact same argument as we go, as we approach x in the negative, or from below, as we approach two from below, I should say. So that's x, and that is f of x."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But needless to say, this is going to be unbounded in the positive direction, so we're going to be going towards positive infinity. So these two choices have that. And we can make the exact same argument as we go, as we approach x in the negative, or from below, as we approach two from below, I should say. So that's x, and that is f of x. And once again, I don't have a calculator in front of me. You could evaluate these things at a calculator, and it'll become very clear that these are positive, and as we get closer to two, they become even larger and larger positive values. And the same thing would happen if you did 1.9, and 1.9, and if you did 1.99."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's x, and that is f of x. And once again, I don't have a calculator in front of me. You could evaluate these things at a calculator, and it'll become very clear that these are positive, and as we get closer to two, they become even larger and larger positive values. And the same thing would happen if you did 1.9, and 1.9, and if you did 1.99. Because here you'd be 1.9 over one minus cosine. Now here you'd have 1.9 minus two, so this would be negative 0.1. Let me scroll over a little bit."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the same thing would happen if you did 1.9, and 1.9, and if you did 1.99. Because here you'd be 1.9 over one minus cosine. Now here you'd have 1.9 minus two, so this would be negative 0.1. Let me scroll over a little bit. This second one would be 1.99 over one minus cosine of negative 0.01, 0.01. And cosine of negative 0.1 is the same thing as cosine of 0.1. Cosine of negative 0.01 is the same thing as cosine of 0.01."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me scroll over a little bit. This second one would be 1.99 over one minus cosine of negative 0.01, 0.01. And cosine of negative 0.1 is the same thing as cosine of 0.1. Cosine of negative 0.01 is the same thing as cosine of 0.01. So these two things, this is going to be equivalent to that, that is going to be equivalent to that. And once again, we're gonna be approaching positive infinity. So the only choice where all of that is true is this first one."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of negative 0.01 is the same thing as cosine of 0.01. So these two things, this is going to be equivalent to that, that is going to be equivalent to that. And once again, we're gonna be approaching positive infinity. So the only choice where all of that is true is this first one. Whether we approach two from the right-hand side or the left-hand side, we are approaching positive infinity. Now the other way you could have deduced that is say, okay, as we approach two, the numerator is going to be positive, because two is positive. And then over here, as you approach two, cosine of anything can never be greater than one."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the only choice where all of that is true is this first one. Whether we approach two from the right-hand side or the left-hand side, we are approaching positive infinity. Now the other way you could have deduced that is say, okay, as we approach two, the numerator is going to be positive, because two is positive. And then over here, as you approach two, cosine of anything can never be greater than one. It's going to approach one, but be less than one. So if this is less than one, as x approaches two, it becomes one when x is equal to two. Well then this right over here, one minus something less than one is going to be positive."}, {"video_title": "Analyzing unbounded limits mixed function   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then over here, as you approach two, cosine of anything can never be greater than one. It's going to approach one, but be less than one. So if this is less than one, as x approaches two, it becomes one when x is equal to two. Well then this right over here, one minus something less than one is going to be positive. So you have a positive divided by a positive, so you're definitely going to get positive values as you approach two. And we know, and they've already told us that these are going to be unbounded based on the choices, so you would also pick that. But you should also feel good about it, that the closer that we get to two, the closer that this value right over here gets to zero, and the closer that this value gets to zero, the closer we get to one, the closer we get to one, the smaller the denominator gets, and then you divide by smaller and smaller denominators, you're going to become unbounded towards infinity, which is exactly what we see in that first choice."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's called the chain rule. And when you're first exposed to it, it can seem a little daunting and a little bit convoluted. But as you see more and more examples, it'll start to make sense. And hopefully, it'll even start to seem a little bit simple and intuitive over time. So let's say that I had a function. Let's say I have a function h of x. And it is equal to, just for example, let's say it's equal to sine of x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And hopefully, it'll even start to seem a little bit simple and intuitive over time. So let's say that I had a function. Let's say I have a function h of x. And it is equal to, just for example, let's say it's equal to sine of x. Let's say it's equal to sine of x squared. Now, I could have written it like this, sine squared of x. But it'll be a little bit clearer using that type of notation."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it is equal to, just for example, let's say it's equal to sine of x. Let's say it's equal to sine of x squared. Now, I could have written it like this, sine squared of x. But it'll be a little bit clearer using that type of notation. So let me make it. So I have h of x. And what I'm curious about is what is h prime of x?"}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it'll be a little bit clearer using that type of notation. So let me make it. So I have h of x. And what I'm curious about is what is h prime of x? So I want to know h prime of x, which another way of writing it is the derivative of h with respect to x. These are just different notations. And to do this, I'm going to use the chain rule."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what I'm curious about is what is h prime of x? So I want to know h prime of x, which another way of writing it is the derivative of h with respect to x. These are just different notations. And to do this, I'm going to use the chain rule. And the chain rule comes into play anytime your function can be used as a composition of more than one function. And that might not seem obvious right now. But it will hopefully maybe by the end of this video or the next one."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to do this, I'm going to use the chain rule. And the chain rule comes into play anytime your function can be used as a composition of more than one function. And that might not seem obvious right now. But it will hopefully maybe by the end of this video or the next one. Now, what I want to do is a little bit of a thought experiment. If I were to ask you, what is the derivative with respect to x, if I were to supply the derivative operator to x squared with respect to x, what do I get? Well, this gives me 2x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it will hopefully maybe by the end of this video or the next one. Now, what I want to do is a little bit of a thought experiment. If I were to ask you, what is the derivative with respect to x, if I were to supply the derivative operator to x squared with respect to x, what do I get? Well, this gives me 2x. We've seen that many, many, many, many times. Now, what if I were to take the derivative with respect to a of a squared? Well, it's the exact same thing."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this gives me 2x. We've seen that many, many, many, many times. Now, what if I were to take the derivative with respect to a of a squared? Well, it's the exact same thing. I just swapped an a for the x's. This is still going to be equal to 2a. Now, I will do something that might be a little bit more bizarre."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's the exact same thing. I just swapped an a for the x's. This is still going to be equal to 2a. Now, I will do something that might be a little bit more bizarre. What if I were to take the derivative with respect to sine of x with respect to sine of x of sine of x squared? Well, wherever I had the x's up here or the a's over here, I just replaced it with a sine of x. So this is just going to be 2 times the thing that I had."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, I will do something that might be a little bit more bizarre. What if I were to take the derivative with respect to sine of x with respect to sine of x of sine of x squared? Well, wherever I had the x's up here or the a's over here, I just replaced it with a sine of x. So this is just going to be 2 times the thing that I had. So whatever I'm taking the derivative with respect to. Here it was with respect to x. Here it was with respect to a."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is just going to be 2 times the thing that I had. So whatever I'm taking the derivative with respect to. Here it was with respect to x. Here it was with respect to a. Here it's with respect to sine of x. So it's going to be 2 times sine of x. Now, so the chain rule tells us that this derivative is going to be the derivative of our whole function with respect or the derivative of this outer function, x squared, the derivative of x squared, the derivative of this outer function with respect to sine of x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here it was with respect to a. Here it's with respect to sine of x. So it's going to be 2 times sine of x. Now, so the chain rule tells us that this derivative is going to be the derivative of our whole function with respect or the derivative of this outer function, x squared, the derivative of x squared, the derivative of this outer function with respect to sine of x. So that's going to be 2 sine of x. So we could view it as the derivative of the outer function with respect to the inner, 2 sine of x. We could just treat sine of x like it's kind of an x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, so the chain rule tells us that this derivative is going to be the derivative of our whole function with respect or the derivative of this outer function, x squared, the derivative of x squared, the derivative of this outer function with respect to sine of x. So that's going to be 2 sine of x. So we could view it as the derivative of the outer function with respect to the inner, 2 sine of x. We could just treat sine of x like it's kind of an x. And it would have been just 2x. But instead, it's a sine of x. So we say 2 sine of x times the derivative of sine of x with respect to x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could just treat sine of x like it's kind of an x. And it would have been just 2x. But instead, it's a sine of x. So we say 2 sine of x times the derivative of sine of x with respect to x. Well, that's more straightforward, a little bit more intuitive. The derivative of sine of x with respect to x, we've seen multiple times, is cosine of x. So times cosine of x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we say 2 sine of x times the derivative of sine of x with respect to x. Well, that's more straightforward, a little bit more intuitive. The derivative of sine of x with respect to x, we've seen multiple times, is cosine of x. So times cosine of x. And so there we've applied the chain rule. It was the derivative of the outer function with respect to the inner. So derivative of sine of x squared with respect to sine of x is 2 sine of x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So times cosine of x. And so there we've applied the chain rule. It was the derivative of the outer function with respect to the inner. So derivative of sine of x squared with respect to sine of x is 2 sine of x. And then we multiply that times the derivative of sine of x with respect to x. Times the derivative of sine of x with respect to x. So let me make it clear."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So derivative of sine of x squared with respect to sine of x is 2 sine of x. And then we multiply that times the derivative of sine of x with respect to x. Times the derivative of sine of x with respect to x. So let me make it clear. This right over here is the derivative. We're taking the derivative of sine of x squared. So let me make it clear."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me make it clear. This right over here is the derivative. We're taking the derivative of sine of x squared. So let me make it clear. That's what we're taking the derivative of with respect to sine of x. With respect to sine of x. And then we're multiplying that times the derivative of sine of x with respect to x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me make it clear. That's what we're taking the derivative of with respect to sine of x. With respect to sine of x. And then we're multiplying that times the derivative of sine of x with respect to x. And this is where it might start making a little bit of intuition. You can't really treat these differentials, this d whatever, this dx, this d sine of x, as a number. And the notation makes it look like a fraction, because intuitively that's what we're doing."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're multiplying that times the derivative of sine of x with respect to x. And this is where it might start making a little bit of intuition. You can't really treat these differentials, this d whatever, this dx, this d sine of x, as a number. And the notation makes it look like a fraction, because intuitively that's what we're doing. But if you were to treat them like fractions, then you could think about canceling that and that. And once again, this isn't a rigorous thing to do, but it can help with the intuition. And then what you're left with is the derivative of this whole sine of x squared with respect to x."}, {"video_title": "Chain rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the notation makes it look like a fraction, because intuitively that's what we're doing. But if you were to treat them like fractions, then you could think about canceling that and that. And once again, this isn't a rigorous thing to do, but it can help with the intuition. And then what you're left with is the derivative of this whole sine of x squared with respect to x. So you're left with the derivative of essentially our original function, sine of x squared with respect to x. Which is exactly what dh dx is. This right over here is our original function h. That's our original function h. So it might seem a little bit daunting now."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Since it involves an integral, we can assume that it's going to involve f prime somehow, especially since they've given us so much information about f prime, including its graph, and the area under or above the curve at different intervals. And then they want us to find f of four and f of negative two. So let's think about how we can connect, how we can connect f prime and integral and f of x. Well, if we took the integral from a to b of f prime of x dx, what is this going to be equal to? Well, this is going to be equal to the antiderivative of f prime of x, which is f of x, and you're going to evaluate that at b and a and subtract the difference. So this is going to be f of b minus f of a. This is straight out of the fundamental theorem of calculus."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if we took the integral from a to b of f prime of x dx, what is this going to be equal to? Well, this is going to be equal to the antiderivative of f prime of x, which is f of x, and you're going to evaluate that at b and a and subtract the difference. So this is going to be f of b minus f of a. This is straight out of the fundamental theorem of calculus. All right, so this is interesting. Well, what is it, instead of having a b, what if we had an x there? And if we're going to put an x as one of our bounds of integration, well, we'll use a different variable for our integration here."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is straight out of the fundamental theorem of calculus. All right, so this is interesting. Well, what is it, instead of having a b, what if we had an x there? And if we're going to put an x as one of our bounds of integration, well, we'll use a different variable for our integration here. So let me write it this way. So if I had the integral from a to x of f prime of u du, well, what is this going to be equal to? Well, this is going to be, by the same logic, the antiderivative of this evaluated at x, so f of x minus f of a."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we're going to put an x as one of our bounds of integration, well, we'll use a different variable for our integration here. So let me write it this way. So if I had the integral from a to x of f prime of u du, well, what is this going to be equal to? Well, this is going to be, by the same logic, the antiderivative of this evaluated at x, so f of x minus f of a. Minus f of a. Or if we wanted to solve for f of x, we could add f of a to both sides, and we would get f of x is equal to the integral from a to x of f prime of u du, and once again, why did I pick u? Well, I just needed some letter other than x, since I already used x as one of my bounds of integration."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is going to be, by the same logic, the antiderivative of this evaluated at x, so f of x minus f of a. Minus f of a. Or if we wanted to solve for f of x, we could add f of a to both sides, and we would get f of x is equal to the integral from a to x of f prime of u du, and once again, why did I pick u? Well, I just needed some letter other than x, since I already used x as one of my bounds of integration. And I'm adding f of a to both sides. I swapped the sides, too. So f of x is going to be equal to this, plus f of a."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I just needed some letter other than x, since I already used x as one of my bounds of integration. And I'm adding f of a to both sides. I swapped the sides, too. So f of x is going to be equal to this, plus f of a. Plus f of a. So this is a general form if a is a lower bound, but they gave us some information. They said that f of one is equal to three."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of x is going to be equal to this, plus f of a. Plus f of a. So this is a general form if a is a lower bound, but they gave us some information. They said that f of one is equal to three. So if we choose a to be equal to one, if we say this is one right over here, this is one, then we know that this is going to be equal to three. So we can write f of x, f of x is equal to the integral, I'm using one as my a, since I know what f of one is, from one to x of f prime of u du plus f of one, they told us what that is, that is going to be equal to three. So this is the first part right here."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "They said that f of one is equal to three. So if we choose a to be equal to one, if we say this is one right over here, this is one, then we know that this is going to be equal to three. So we can write f of x, f of x is equal to the integral, I'm using one as my a, since I know what f of one is, from one to x of f prime of u du plus f of one, they told us what that is, that is going to be equal to three. So this is the first part right here. We just did it. That is the first part of the problem. Now let's try to find f of four and f of negative two."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the first part right here. We just did it. That is the first part of the problem. Now let's try to find f of four and f of negative two. Alright, f of four, well, wherever we see the next, we substitute a four there, it's going to be integral from one to four of f prime of u du plus three. So what is going to be the integral from one to four of f prime of u du? So let's look up here."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's try to find f of four and f of negative two. Alright, f of four, well, wherever we see the next, we substitute a four there, it's going to be integral from one to four of f prime of u du plus three. So what is going to be the integral from one to four of f prime of u du? So let's look up here. So the integral from one to four of f prime, that's that curve right over here. Well, that's going to give this area, but it's going to be the negative of the area because if you just take the integral, it's the area that you could view above the x-axis and below, or if we're thinking in terms of view, above the u-axis and below the function. But this is the other way around."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look up here. So the integral from one to four of f prime, that's that curve right over here. Well, that's going to give this area, but it's going to be the negative of the area because if you just take the integral, it's the area that you could view above the x-axis and below, or if we're thinking in terms of view, above the u-axis and below the function. But this is the other way around. The function is below the horizontal axis here. So this area, which they told us in the problem, the area bounded by the x-axis in the graph of f prime over the interval one, four is 12. So this area over here is 12, but the integral is going to be a negative because once again, our function is below the x-axis."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this is the other way around. The function is below the horizontal axis here. So this area, which they told us in the problem, the area bounded by the x-axis in the graph of f prime over the interval one, four is 12. So this area over here is 12, but the integral is going to be a negative because once again, our function is below the x-axis. So this integral, this right over here is going to be a negative 12. This is negative 12. So negative 12 plus three is negative nine."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this area over here is 12, but the integral is going to be a negative because once again, our function is below the x-axis. So this integral, this right over here is going to be a negative 12. This is negative 12. So negative 12 plus three is negative nine. All right, now let's evaluate f of negative two. f of negative two is equal to the integral from one to negative two of f prime of u du plus three. Well, it's kind of, feels a little strange to have the upper bound being lower than the lower bound, so we can swap the bounds and then add a negative out here."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative 12 plus three is negative nine. All right, now let's evaluate f of negative two. f of negative two is equal to the integral from one to negative two of f prime of u du plus three. Well, it's kind of, feels a little strange to have the upper bound being lower than the lower bound, so we can swap the bounds and then add a negative out here. So this is going to be equal to the negative of, if we swap the bounds here, so from negative two to one, f prime of u du plus three. So what's the integral from negative two to one? Well, they give us that."}, {"video_title": "2015 AP Calculus AB 5d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's kind of, feels a little strange to have the upper bound being lower than the lower bound, so we can swap the bounds and then add a negative out here. So this is going to be equal to the negative of, if we swap the bounds here, so from negative two to one, f prime of u du plus three. So what's the integral from negative two to one? Well, they give us that. From negative two to one, they told us that this area over here is nine, but once again, because it's below the horizontal axis and above the curve, we would say that the, or the integral will evaluate to be negative nine. So the area is nine, but once again, the curve is below the x-axis, so the integral would give us negative nine. So this would evaluate to negative nine."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x. If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. The derivative with respect to x of pi of a constant is just 0. The derivative with respect to x of 1 is just a constant. It's just 0."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of x squared is 2x. The derivative with respect to x of pi of a constant is just 0. The derivative with respect to x of 1 is just a constant. It's just 0. So once again, this is just going to be equal to 2x. In general, the derivative with respect to x of x squared plus any constant is going to be equal to 2x. The derivative of x squared with respect to x is 2x."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's just 0. So once again, this is just going to be equal to 2x. In general, the derivative with respect to x of x squared plus any constant is going to be equal to 2x. The derivative of x squared with respect to x is 2x. The derivative of a constant with respect to x, a constant does not change with respect to x, so it's just equal to 0. You apply the derivative operator to any of these expressions, and you get 2x. Now let's go the other way around."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of x squared with respect to x is 2x. The derivative of a constant with respect to x, a constant does not change with respect to x, so it's just equal to 0. You apply the derivative operator to any of these expressions, and you get 2x. Now let's go the other way around. Let's think about the antiderivative. One way to think about it is we're doing the opposite of the derivative operator. In the derivative operator, you get an expression, and you find its derivative."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's go the other way around. Let's think about the antiderivative. One way to think about it is we're doing the opposite of the derivative operator. In the derivative operator, you get an expression, and you find its derivative. Now what we want to do is, given some expression, we want to find what it could be the derivative of. If someone were to ask you, what is 2x the derivative of? They're essentially asking you for the antiderivative."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "In the derivative operator, you get an expression, and you find its derivative. Now what we want to do is, given some expression, we want to find what it could be the derivative of. If someone were to ask you, what is 2x the derivative of? They're essentially asking you for the antiderivative. You could say, well, 2x is the derivative of x squared. You could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "They're essentially asking you for the antiderivative. You could say, well, 2x is the derivative of x squared. You could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi. I think you get the general idea. If you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. This is what you would consider the antiderivative of 2x."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could also say that 2x is the derivative of x squared plus pi. I think you get the general idea. If you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. This is what you would consider the antiderivative of 2x. That's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. The convention here is to use a strange-looking notation. It's to use a big, elongated, S-looking thing like that, and a dx around the function that you're trying to take the antiderivative of."}, {"video_title": "Antiderivatives and indefinite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is what you would consider the antiderivative of 2x. That's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. The convention here is to use a strange-looking notation. It's to use a big, elongated, S-looking thing like that, and a dx around the function that you're trying to take the antiderivative of. In this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c. Now, you might be saying, why do we use this type of crazy notation? It will become more obvious when we study the definite integral and areas under curves, and taking sums of rectangles in order to approximate the area of the curve."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now let's tackle part B. Find the average velocity of the particle for the time period from zero is less than or equal to t is less than or equal to six. So our average velocity, that's just going to be our change in position, which we could view as our displacement, divided by our change in time. Well, what is our change in position going to be? Well, they don't give us our position function, but they do give us our velocity function. And so to figure out the average velocity, we could figure out our displacement, which is going to be equal to the integral of our velocity, the integral of v of t dt, and we wanna find the average velocity for this time period. So we're gonna go t equals zero to t equals six, and then we're going to divide that by the amount of time that goes by."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what is our change in position going to be? Well, they don't give us our position function, but they do give us our velocity function. And so to figure out the average velocity, we could figure out our displacement, which is going to be equal to the integral of our velocity, the integral of v of t dt, and we wanna find the average velocity for this time period. So we're gonna go t equals zero to t equals six, and then we're going to divide that by the amount of time that goes by. Well, our change in time is going to be equal to six, and so this is going to be equal to the integral from zero to six of two sine of e to the t over four power plus one dt, and then all of that divided by six. Where did I get this from? Well, they tell us what our velocity as a function of time is."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna go t equals zero to t equals six, and then we're going to divide that by the amount of time that goes by. Well, our change in time is going to be equal to six, and so this is going to be equal to the integral from zero to six of two sine of e to the t over four power plus one dt, and then all of that divided by six. Where did I get this from? Well, they tell us what our velocity as a function of time is. It's that right over there. So we can get our calculator out to solve this part. So on our calculator, we would hit Math, and then we would wanna do number nine, which is definite integral, so we'll hit nine, and we are going to go from zero to six of, this is going to be two sine of e to the, so let me do second e to the, and I'll just use the variable x instead of t because it's easier to type in, x divided by four power."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, they tell us what our velocity as a function of time is. It's that right over there. So we can get our calculator out to solve this part. So on our calculator, we would hit Math, and then we would wanna do number nine, which is definite integral, so we'll hit nine, and we are going to go from zero to six of, this is going to be two sine of e to the, so let me do second e to the, and I'll just use the variable x instead of t because it's easier to type in, x divided by four power. So that's my sine, and then let me close the parentheses for the sine, plus one, and then I'm integrating here with respect to x just because it was more convenient, and I am going to get that, and then I divide by six, divide by six, and I get, and when you're taking an AP calculus exam, it's important to round to three decimal places unless they tell you otherwise. That's what they expect. So it's approximately equal to 1.949, approximately 1.949."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on our calculator, we would hit Math, and then we would wanna do number nine, which is definite integral, so we'll hit nine, and we are going to go from zero to six of, this is going to be two sine of e to the, so let me do second e to the, and I'll just use the variable x instead of t because it's easier to type in, x divided by four power. So that's my sine, and then let me close the parentheses for the sine, plus one, and then I'm integrating here with respect to x just because it was more convenient, and I am going to get that, and then I divide by six, divide by six, and I get, and when you're taking an AP calculus exam, it's important to round to three decimal places unless they tell you otherwise. That's what they expect. So it's approximately equal to 1.949, approximately 1.949. Now let's do part c. So part c, find the total distance traveled by the particle from time t equals zero to t equals six. So you might say, hey, didn't we just figure that out? No, this is displacement, and to remember the difference between distance and displacement, if I have something that starts here, it goes over there, and then it comes back to right over there, the distance traveled is the total path length, so it would be that plus this, while the displacement would just be this right over here."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's approximately equal to 1.949, approximately 1.949. Now let's do part c. So part c, find the total distance traveled by the particle from time t equals zero to t equals six. So you might say, hey, didn't we just figure that out? No, this is displacement, and to remember the difference between distance and displacement, if I have something that starts here, it goes over there, and then it comes back to right over there, the distance traveled is the total path length, so it would be that plus this, while the displacement would just be this right over here. So if you wanna figure out distance, what you wanna do is take the integral of the absolute value of velocity. You could think about it as you're taking the integral of the speed function. So this is just going to be the integral from zero to six of the absolute value of v of t dt, and then we can type this, actually let me just write it out since you would wanna do this if you're actually taking the test."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "No, this is displacement, and to remember the difference between distance and displacement, if I have something that starts here, it goes over there, and then it comes back to right over there, the distance traveled is the total path length, so it would be that plus this, while the displacement would just be this right over here. So if you wanna figure out distance, what you wanna do is take the integral of the absolute value of velocity. You could think about it as you're taking the integral of the speed function. So this is just going to be the integral from zero to six of the absolute value of v of t dt, and then we can type this, actually let me just write it out since you would wanna do this if you're actually taking the test. So it's the integral from zero to six of the absolute value of two sine of e to the t over four plus one, close the absolute value, dt, which is going to be approximately equal to, so once again, we hit math, definite integral. We're going from zero to six of, now we'll hit math again, and then we'll go to number. We see absolute value."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is just going to be the integral from zero to six of the absolute value of v of t dt, and then we can type this, actually let me just write it out since you would wanna do this if you're actually taking the test. So it's the integral from zero to six of the absolute value of two sine of e to the t over four plus one, close the absolute value, dt, which is going to be approximately equal to, so once again, we hit math, definite integral. We're going from zero to six of, now we'll hit math again, and then we'll go to number. We see absolute value. The absolute value of two sine of e to the, I'll use x again just because it's easier to use on the calculator, and then I'll close the parentheses on, whoops, let me make sure I do this right. So let me close the parentheses on the sine, and then I have plus one, and then there you go, and then I'm integrating with respect to x approximately 12.573. Now let's do part d. I'll do it right over here, part d. For t between zero and six, the particle changes direction exactly once."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "We see absolute value. The absolute value of two sine of e to the, I'll use x again just because it's easier to use on the calculator, and then I'll close the parentheses on, whoops, let me make sure I do this right. So let me close the parentheses on the sine, and then I have plus one, and then there you go, and then I'm integrating with respect to x approximately 12.573. Now let's do part d. I'll do it right over here, part d. For t between zero and six, the particle changes direction exactly once. Find the position of the particle at that time. So the way that I'm gonna tackle it is first I need to think about at what time does the particle change direction, and then I can figure out its position by taking the integral of the velocity function. So how do I think about where that particle changes direction?"}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's do part d. I'll do it right over here, part d. For t between zero and six, the particle changes direction exactly once. Find the position of the particle at that time. So the way that I'm gonna tackle it is first I need to think about at what time does the particle change direction, and then I can figure out its position by taking the integral of the velocity function. So how do I think about where that particle changes direction? Well, you're going to change direction if your velocity is at zero, and right before that, your velocity was positive, and then right after that, your velocity is negative, or the other way around. Right before that, your velocity was negative, and then your velocity right after that is positive. So we just have to figure out when does v of t equal zero."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how do I think about where that particle changes direction? Well, you're going to change direction if your velocity is at zero, and right before that, your velocity was positive, and then right after that, your velocity is negative, or the other way around. Right before that, your velocity was negative, and then your velocity right after that is positive. So we just have to figure out when does v of t equal zero. So let's just say two sine of e to the t over four plus one is equal to zero, and there's multiple ways you can solve this, but because we're allowed to use a calculator, we might as well do that. So we hit Math, and then we hit B right over there. I'll just hit Enter."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to figure out when does v of t equal zero. So let's just say two sine of e to the t over four plus one is equal to zero, and there's multiple ways you can solve this, but because we're allowed to use a calculator, we might as well do that. So we hit Math, and then we hit B right over there. I'll just hit Enter. Equation Solver. So I could just say, look, the left side of my equation is two sine of, we are going to go e to the, I'll just use x as what I'm solving for because it's easier to type in in the calculator, and close that parentheses, plus one. The right-hand side is equal to zero, and now I just have to give an initial guess, and I'll just go right in between that interval between t equals zero and t equals six."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll just hit Enter. Equation Solver. So I could just say, look, the left side of my equation is two sine of, we are going to go e to the, I'll just use x as what I'm solving for because it's easier to type in in the calculator, and close that parentheses, plus one. The right-hand side is equal to zero, and now I just have to give an initial guess, and I'll just go right in between that interval between t equals zero and t equals six. So I'll just put a three right over there, and now I just need to solve it, and I can do that by hitting Alpha, and then Solve, and then you see it got us to 5.196. So t is approximately equal to 5.196, but remember, they're not asking us for at what time does the particle change direction. They wanna know what's the position of the particle at that time."}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "The right-hand side is equal to zero, and now I just have to give an initial guess, and I'll just go right in between that interval between t equals zero and t equals six. So I'll just put a three right over there, and now I just need to solve it, and I can do that by hitting Alpha, and then Solve, and then you see it got us to 5.196. So t is approximately equal to 5.196, but remember, they're not asking us for at what time does the particle change direction. They wanna know what's the position of the particle at that time. Well, we know how to find our change in position. Our change in position is going to be our displacement from time equals zero to time 5.196, and the displacement, we're integrating the velocity function, so that's going to be two sine of e to the t over four plus one dt. Now, this is going to be our change in position, but where did we start?"}, {"video_title": "2011 Calculus AB Free Response #1 parts b c d   AP Calculus AB   Khan Academy.mp3", "Sentence": "They wanna know what's the position of the particle at that time. Well, we know how to find our change in position. Our change in position is going to be our displacement from time equals zero to time 5.196, and the displacement, we're integrating the velocity function, so that's going to be two sine of e to the t over four plus one dt. Now, this is going to be our change in position, but where did we start? Well, they say at x equals zero, or when x, at time zero, our position was two. So if this is our change, we would start here. This would give us our total position, where we started, plus our change, and so this is going to be approximately equal to, we are going to have two plus, and then I go to Math, definite integral from zero to 5.196, and I've typed this in many times already, two sine of e to the x divided by four power, and then close that parentheses, plus one, and then I'm integrating with respect to x, and so this gives us 14.135, approximately 14.135 is its position, and let me make sure that I put those parentheses there in the right place, and we are done."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I also want to evaluate that at x equals two. So I want to figure that out. And I also want to figure out what does that evaluate to when x is equal to two. So what is the slope of the tangent line to the graph of g when x is equal to two? And like always, pause this video and see if you can work this out on your own before I work through it. With you. And I'll give you some hints."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is the slope of the tangent line to the graph of g when x is equal to two? And like always, pause this video and see if you can work this out on your own before I work through it. With you. And I'll give you some hints. All you really need to do is apply the power rule, a little bit of basic exponent properties, and some basic derivative properties to be able to do this. All right, now let's just do this together. And I'll just rewrite it."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll give you some hints. All you really need to do is apply the power rule, a little bit of basic exponent properties, and some basic derivative properties to be able to do this. All right, now let's just do this together. And I'll just rewrite it. G of x is equal to this first term here, two over x to the third. Well, that could be rewritten as two times x to the negative three. We know that one over x to the n is the same thing as x to the negative n. So I just rewrote it, and now this might be ringing a bell of how the power rule might be useful."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll just rewrite it. G of x is equal to this first term here, two over x to the third. Well, that could be rewritten as two times x to the negative three. We know that one over x to the n is the same thing as x to the negative n. So I just rewrote it, and now this might be ringing a bell of how the power rule might be useful. And then we have minus, well, one over x squared, that is the same thing as x to the negative two. And so this, if we're gonna take the derivative of both sides of this, let's do that. Derivative with respect to x, dx."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that one over x to the n is the same thing as x to the negative n. So I just rewrote it, and now this might be ringing a bell of how the power rule might be useful. And then we have minus, well, one over x squared, that is the same thing as x to the negative two. And so this, if we're gonna take the derivative of both sides of this, let's do that. Derivative with respect to x, dx. We're gonna do that on the left-hand side. We're also gonna do it on the right-hand side. Well, on the left-hand side, the derivative with respect to x of g of x, we can write that as g prime of x is going to be equal to, well, the derivative of this first, what we have right here written in green, this is going to be, we're just going to apply the power rule."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative with respect to x, dx. We're gonna do that on the left-hand side. We're also gonna do it on the right-hand side. Well, on the left-hand side, the derivative with respect to x of g of x, we can write that as g prime of x is going to be equal to, well, the derivative of this first, what we have right here written in green, this is going to be, we're just going to apply the power rule. We're going to take our exponent, multiply it by our coefficient out front. Actually, let me write that out. So that's going to be, let me finish this equal sign."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, on the left-hand side, the derivative with respect to x of g of x, we can write that as g prime of x is going to be equal to, well, the derivative of this first, what we have right here written in green, this is going to be, we're just going to apply the power rule. We're going to take our exponent, multiply it by our coefficient out front. Actually, let me write that out. So that's going to be, let me finish this equal sign. That is going to be two times negative three, times x, and now we're going to decrement this exponent. You have to be very careful here because sometimes your brain might say, okay, one less than three is two, so maybe this is x to the negative two. But remember, you're going down."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's going to be, let me finish this equal sign. That is going to be two times negative three, times x, and now we're going to decrement this exponent. You have to be very careful here because sometimes your brain might say, okay, one less than three is two, so maybe this is x to the negative two. But remember, you're going down. So if you're at negative three, and you subtract one, we're going to go to the negative three minus one power. Well, that's going to take us to negative four. So this is x to the negative four power."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But remember, you're going down. So if you're at negative three, and you subtract one, we're going to go to the negative three minus one power. Well, that's going to take us to negative four. So this is x to the negative four power. So two times negative three x to the negative four, or we could have also written that as negative six x to the negative four power, and then minus, well, we're going to do the same thing again right over here. We take this negative two, multiply it times the coefficient that's implicitly here. You could say there's a one there, so negative two times one."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is x to the negative four power. So two times negative three x to the negative four, or we could have also written that as negative six x to the negative four power, and then minus, well, we're going to do the same thing again right over here. We take this negative two, multiply it times the coefficient that's implicitly here. You could say there's a one there, so negative two times one. So you have the negative two there, and then you have the x to the, well, what's negative two minus one? Well, that's negative three to the negative three power. And so we can rewrite all of this business as the derivative g prime of x is equal to negative six, negative six x to the negative fourth, and now we're subtracting a negative, so we could just write this as plus two x to the negative three."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say there's a one there, so negative two times one. So you have the negative two there, and then you have the x to the, well, what's negative two minus one? Well, that's negative three to the negative three power. And so we can rewrite all of this business as the derivative g prime of x is equal to negative six, negative six x to the negative fourth, and now we're subtracting a negative, so we could just write this as plus two x to the negative three. This negative cancels out with that negative. Subtract a negative, the same thing as adding the positive. So we did the first part."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can rewrite all of this business as the derivative g prime of x is equal to negative six, negative six x to the negative fourth, and now we're subtracting a negative, so we could just write this as plus two x to the negative three. This negative cancels out with that negative. Subtract a negative, the same thing as adding the positive. So we did the first part. We can express g prime of x as a function of x. Now let's just evaluate what g prime of two is. So g prime of two is going to be equal to negative six times two to the negative fourth power plus two times two to the negative third power."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we did the first part. We can express g prime of x as a function of x. Now let's just evaluate what g prime of two is. So g prime of two is going to be equal to negative six times two to the negative fourth power plus two times two to the negative third power. Well, what's this going to be? This is equal to negative six over two to the fourth plus two over two to the third, which is equal to negative six over, two to the fourth is 16, plus two over, two to the third is eight. And so let's see, this is, we could rewrite this."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g prime of two is going to be equal to negative six times two to the negative fourth power plus two times two to the negative third power. Well, what's this going to be? This is equal to negative six over two to the fourth plus two over two to the third, which is equal to negative six over, two to the fourth is 16, plus two over, two to the third is eight. And so let's see, this is, we could rewrite this. Let's write all of this with a common denominator. I could write this as one fourth, but then this one won't work out as cleanly. I could write them both as eights."}, {"video_title": "Negative powers differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's see, this is, we could rewrite this. Let's write all of this with a common denominator. I could write this as one fourth, but then this one won't work out as cleanly. I could write them both as eights. This is negative three eighths, negative three eighths. So you have negative three eighths plus two eighths is equal to negative one eighth. So the slope of the tangent line at x equals two to the graph y equals g of x has a slope, or that slope is negative one eighth."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's start with an example. Let's say we want to evaluate the definite integral going from negative four to negative two of f of x, d of x, plus the definite integral going from negative two to zero of f of x, dx. Pause this video and see if you can evaluate this entire expression. So this first part of our expression, the definite integral from negative four to negative two of f of x, dx, we're going from x equals negative four to x equals negative two. And so this would evaluate as this area between our curve and our x-axis, but it would be the negative of that area because our curve is below the x-axis. And we could try to estimate it based on the information they've given us, but they haven't given us exactly that value. But we also need to figure out this right over here."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this first part of our expression, the definite integral from negative four to negative two of f of x, dx, we're going from x equals negative four to x equals negative two. And so this would evaluate as this area between our curve and our x-axis, but it would be the negative of that area because our curve is below the x-axis. And we could try to estimate it based on the information they've given us, but they haven't given us exactly that value. But we also need to figure out this right over here. And here we're going from x equals negative two to zero of f of x, d of x, so that's gonna be this area. So if you're looking at the sum of these two definite integrals, and notice the upper bound here is the lower bound here, you're really thinking about this is really going to be the same thing as this is equal to the definite integral going from x equals negative four all the way to x equals zero of f of x, dx. And this is indeed one of our integration properties."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we also need to figure out this right over here. And here we're going from x equals negative two to zero of f of x, d of x, so that's gonna be this area. So if you're looking at the sum of these two definite integrals, and notice the upper bound here is the lower bound here, you're really thinking about this is really going to be the same thing as this is equal to the definite integral going from x equals negative four all the way to x equals zero of f of x, dx. And this is indeed one of our integration properties. If our upper bound here is the same as our lower bound here and we are integrating the same thing, well, then you can merge these two definite integrals in this way. And this is just going to be this entire area, but because we are below the x-axis and above our curve here it would be the negative of that area. So this is going to be equal to negative seven."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is indeed one of our integration properties. If our upper bound here is the same as our lower bound here and we are integrating the same thing, well, then you can merge these two definite integrals in this way. And this is just going to be this entire area, but because we are below the x-axis and above our curve here it would be the negative of that area. So this is going to be equal to negative seven. Let's do another example. Let's say someone were to ask you, walk up to you on the street and say, quick, here's a graph, what is the value of the expression that I'm about to write down? The definite integral going from zero to four of f of x, dx, plus the definite integral going from four to six of f of x, dx."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to negative seven. Let's do another example. Let's say someone were to ask you, walk up to you on the street and say, quick, here's a graph, what is the value of the expression that I'm about to write down? The definite integral going from zero to four of f of x, dx, plus the definite integral going from four to six of f of x, dx. Pause this video and see if you can figure that out. Well, once again, this first part right over here going from zero to four, so what would be is would be this area, it would be this five right over here, but then we would need to subtract this area because this area is below our x-axis and above our curve. We don't know exactly what this is, but luckily we also need to take the sum of everything I just showed, but plus this right over here."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "The definite integral going from zero to four of f of x, dx, plus the definite integral going from four to six of f of x, dx. Pause this video and see if you can figure that out. Well, once again, this first part right over here going from zero to four, so what would be is would be this area, it would be this five right over here, but then we would need to subtract this area because this area is below our x-axis and above our curve. We don't know exactly what this is, but luckily we also need to take the sum of everything I just showed, but plus this right over here. And this we're going from four to six, so it's going to be this area. Well, once again, when you look at it this way, you can see that this expression is going to be equivalent to taking the definite integral all the way from zero to six, zero to six of f of x, dx. And once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x, dx, and our upper bound here is our lower bound here."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know exactly what this is, but luckily we also need to take the sum of everything I just showed, but plus this right over here. And this we're going from four to six, so it's going to be this area. Well, once again, when you look at it this way, you can see that this expression is going to be equivalent to taking the definite integral all the way from zero to six, zero to six of f of x, dx. And once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x, dx, and our upper bound here is our lower bound here. So once again, we're able to merge the integrals. And what is this going to be equal to? Well, we have this area here, which is five, and then we have this area, which is six."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x, dx, and our upper bound here is our lower bound here. So once again, we're able to merge the integrals. And what is this going to be equal to? Well, we have this area here, which is five, and then we have this area, which is six. That was given to us. But since it's below the x-axis and above our curve, when we evaluate it as a definite integral, it would evaluate as a negative six. So this is going to be five plus negative six, which is equal to negative one, and we're done."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The number of words that I remember t days after studying is modeled by, so w of t, so this is the number of words I have in my head as a function of time is going to be equal to 80 times the one minus 0.1t squared, for t is between zero and 10, including the two boundaries. That's why we have brackets right over here. What is the rate of change of the number of words, of the number of known words per day two days after studying for the test? And I encourage you to pause this video and try it on your own. So the key here is we come up with this equation for modeling how many words have retained in my brain every day after I first memorized them, after I got the 80 of them into my head. And that's this expression here. And they wanna know the rate of change two days after studying."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and try it on your own. So the key here is we come up with this equation for modeling how many words have retained in my brain every day after I first memorized them, after I got the 80 of them into my head. And that's this expression here. And they wanna know the rate of change two days after studying. Well, the rate of change, I can take the derivative of this with respect to time. So let's do that. So let's take the derivative, the derivative of the number of words I know with respect to time is going to be equal to, well, we have this 80 out front."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they wanna know the rate of change two days after studying. Well, the rate of change, I can take the derivative of this with respect to time. So let's do that. So let's take the derivative, the derivative of the number of words I know with respect to time is going to be equal to, well, we have this 80 out front. That's just a constant. And now I can apply the chain rule right over here. So the derivative of one minus 0.1t whole thing squared with respect to one minus 0.1t is going to be, so I'm essentially taking the derivative of this whole pink thing, with this whole expression squared with respect to the expression."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's take the derivative, the derivative of the number of words I know with respect to time is going to be equal to, well, we have this 80 out front. That's just a constant. And now I can apply the chain rule right over here. So the derivative of one minus 0.1t whole thing squared with respect to one minus 0.1t is going to be, so I'm essentially taking the derivative of this whole pink thing, with this whole expression squared with respect to the expression. So that's going to be two times one minus 0.1t. And now I can find the derivative of this inner expression with respect to t. So derivative of this inner expression with respect to t is just going to be zero minus 0.1. So it's just going to be negative 0.1."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of one minus 0.1t whole thing squared with respect to one minus 0.1t is going to be, so I'm essentially taking the derivative of this whole pink thing, with this whole expression squared with respect to the expression. So that's going to be two times one minus 0.1t. And now I can find the derivative of this inner expression with respect to t. So derivative of this inner expression with respect to t is just going to be zero minus 0.1. So it's just going to be negative 0.1. So it's gonna be negative 0.1. And of course, we can simplify this a little bit. This is going to be equal to, if we take 80 times two is 160, times negative 0.1."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's just going to be negative 0.1. So it's gonna be negative 0.1. And of course, we can simplify this a little bit. This is going to be equal to, if we take 80 times two is 160, times negative 0.1. That's gonna be negative 16. 160 times 0.1 is 16. So negative 16 times one minus 0.1t."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be equal to, if we take 80 times two is 160, times negative 0.1. That's gonna be negative 16. 160 times 0.1 is 16. So negative 16 times one minus 0.1t. And if we want, we could distribute the 16 or we could just leave it like this. But we're ready now to answer our questions. We could write this as the rate of change of the number of words we know with respect to time."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative 16 times one minus 0.1t. And if we want, we could distribute the 16 or we could just leave it like this. But we're ready now to answer our questions. We could write this as the rate of change of the number of words we know with respect to time. Or we could use the alternate notation. We could say this is w prime of t. Either way, it's going to be equal to this thing. It's negative, let me do that same color."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could write this as the rate of change of the number of words we know with respect to time. Or we could use the alternate notation. We could say this is w prime of t. Either way, it's going to be equal to this thing. It's negative, let me do that same color. It's equal to negative 16 times one minus 0.1t. So what's this going to be? What is the rate of change of the number of words known per day two days after studying for the test?"}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's negative, let me do that same color. It's equal to negative 16 times one minus 0.1t. So what's this going to be? What is the rate of change of the number of words known per day two days after studying for the test? Well, we just have to evaluate this when t is equal to two. So w prime of two, do that in magenta. W prime of two is going to be equal to, whoops, I'll just go with the magenta, is equal to negative 16 times one minus 0.1 times two."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the rate of change of the number of words known per day two days after studying for the test? Well, we just have to evaluate this when t is equal to two. So w prime of two, do that in magenta. W prime of two is going to be equal to, whoops, I'll just go with the magenta, is equal to negative 16 times one minus 0.1 times two. Times two, close parentheses. And that's going to be equal to, well let's see, what is this? This is one minus, essentially 0.2, this is going to be 0.8."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "W prime of two is going to be equal to, whoops, I'll just go with the magenta, is equal to negative 16 times one minus 0.1 times two. Times two, close parentheses. And that's going to be equal to, well let's see, what is this? This is one minus, essentially 0.2, this is going to be 0.8. So this is going to be equal to negative 16 times, is that right, one minus 0.2 is, yep, it's going to be times 0.8. And what is that going to be? If I were to multiply 16 times eight, it would be 128."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is one minus, essentially 0.2, this is going to be 0.8. So this is going to be equal to negative 16 times, is that right, one minus 0.2 is, yep, it's going to be times 0.8. And what is that going to be? If I were to multiply 16 times eight, it would be 128. It's two times eight times eight. So two times 64, 128. So this is going to be negative 12.8."}, {"video_title": "Applied rate of change forgetfulness   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I were to multiply 16 times eight, it would be 128. It's two times eight times eight. So two times 64, 128. So this is going to be negative 12.8. And just to really hit the point home of what we're doing, this is, we're going to, it's negative, the number of words, the rate of change is negative 12.8 words per day. So if you believe this model for how many words we know on a given day, this is saying on day two, or right at the day two point, right after two, exactly two days after studying for the test, I'm going to be, right at that moment, I am essentially losing 12.8 words per day. The number of words I know is decreasing by 12.8 words per day."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just remind ourselves what a point of inflection is. A point of inflection is where we go from being, where we change our concavity, or you could say where our second derivative, g prime of x, switches signs. Switches, switches signs. So let's study our second derivative. In order to study our second derivative, let's find it. So we know that g of x is equal to 1 4th x to the 4th minus four x to the 3rd power plus 24 x squared. So given that, let's now find g prime of x. G prime of x is going to be equal to, I'm just gonna apply the power rule multiple times, four times 1 4th is just one, I'm not even gonna write the one down, it's gonna be one times x to the four minus one power."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's study our second derivative. In order to study our second derivative, let's find it. So we know that g of x is equal to 1 4th x to the 4th minus four x to the 3rd power plus 24 x squared. So given that, let's now find g prime of x. G prime of x is going to be equal to, I'm just gonna apply the power rule multiple times, four times 1 4th is just one, I'm not even gonna write the one down, it's gonna be one times x to the four minus one power. So four to the 3rd power minus three times four is 12, x to the three minus one power, or x to the 2nd power, plus two times 24, or 48, x to the two minus one, or x to the 1st power, I could just write that as x. So there you have it, I have our first derivative, now we wanna find our second derivative. So g prime prime of x is just the derivative of the first derivative with respect to x, and so more of the power rule, three x squared minus 24 x to the first, or just 24 x, plus 48."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So given that, let's now find g prime of x. G prime of x is going to be equal to, I'm just gonna apply the power rule multiple times, four times 1 4th is just one, I'm not even gonna write the one down, it's gonna be one times x to the four minus one power. So four to the 3rd power minus three times four is 12, x to the three minus one power, or x to the 2nd power, plus two times 24, or 48, x to the two minus one, or x to the 1st power, I could just write that as x. So there you have it, I have our first derivative, now we wanna find our second derivative. So g prime prime of x is just the derivative of the first derivative with respect to x, and so more of the power rule, three x squared minus 24 x to the first, or just 24 x, plus 48. So let's think about where this switches signs. And this is a continuous function, it's going to be defined for all x's, so the only potential candidates of where it could switch signs are when this thing equals zero. So let's see where it equals zero."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g prime prime of x is just the derivative of the first derivative with respect to x, and so more of the power rule, three x squared minus 24 x to the first, or just 24 x, plus 48. So let's think about where this switches signs. And this is a continuous function, it's going to be defined for all x's, so the only potential candidates of where it could switch signs are when this thing equals zero. So let's see where it equals zero. So let's set it equal to zero. Three x squared minus 24 x plus 48 is equal to zero. Let's see, everything is divisible by three, so let's divide everything by three."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see where it equals zero. So let's set it equal to zero. Three x squared minus 24 x plus 48 is equal to zero. Let's see, everything is divisible by three, so let's divide everything by three. So you get x squared minus eight x plus 16, plus 16 is equal to zero, and let's see, can I factor this? Yeah, this would be x minus four times x minus four. Or I could just do this as x minus four squared is equal to zero, or x minus four is equal to zero, so, or where x equals four."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see, everything is divisible by three, so let's divide everything by three. So you get x squared minus eight x plus 16, plus 16 is equal to zero, and let's see, can I factor this? Yeah, this would be x minus four times x minus four. Or I could just do this as x minus four squared is equal to zero, or x minus four is equal to zero, so, or where x equals four. So g prime prime of four is equal to zero. So let's see what's happening on either side of that. Let's see if g, if we're actually, if we're actually switching signs or not."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or I could just do this as x minus four squared is equal to zero, or x minus four is equal to zero, so, or where x equals four. So g prime prime of four is equal to zero. So let's see what's happening on either side of that. Let's see if g, if we're actually, if we're actually switching signs or not. So let me draw a number line here. And so this is, so this is two, three, four, five, and I could keep going. And so we know that something interesting is happening right over here."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see if g, if we're actually, if we're actually switching signs or not. So let me draw a number line here. And so this is, so this is two, three, four, five, and I could keep going. And so we know that something interesting is happening right over here. G prime prime of four is equal to zero. G prime prime of four is equal to zero. So let's think about what the second derivative is when we are less than four."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we know that something interesting is happening right over here. G prime prime of four is equal to zero. G prime prime of four is equal to zero. So let's think about what the second derivative is when we are less than four. And so, actually, let me just try g prime prime of zero, since that'll be easy to evaluate. G prime prime of zero, well, it's just going to be equal to 48. So when we are less than four, our second derivative, g prime, the second derivative is greater than zero."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what the second derivative is when we are less than four. And so, actually, let me just try g prime prime of zero, since that'll be easy to evaluate. G prime prime of zero, well, it's just going to be equal to 48. So when we are less than four, our second derivative, g prime, the second derivative is greater than zero. So we are actually going to be concave upwards over this interval to the left of four. Now let's think about to the right of four. Two is a different color."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when we are less than four, our second derivative, g prime, the second derivative is greater than zero. So we are actually going to be concave upwards over this interval to the left of four. Now let's think about to the right of four. Two is a different color. So what about to the right of four? And so let me just evaluate, what would be an easy thing to evaluate? Well, I could evaluate g prime of, well, why not do g, or the second derivative, g prime prime, I should say, of 10."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two is a different color. So what about to the right of four? And so let me just evaluate, what would be an easy thing to evaluate? Well, I could evaluate g prime of, well, why not do g, or the second derivative, g prime prime, I should say, of 10. So g, I'll do it right over here. Let me do, well, I'm running a little bit out of space, so I'll just scroll down. So g prime prime of 10 is going to be equal to three times 10 squared, so it's 300 minus 24 times 10, so minus 240 plus 48."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I could evaluate g prime of, well, why not do g, or the second derivative, g prime prime, I should say, of 10. So g, I'll do it right over here. Let me do, well, I'm running a little bit out of space, so I'll just scroll down. So g prime prime of 10 is going to be equal to three times 10 squared, so it's 300 minus 24 times 10, so minus 240 plus 48. So let's see, this is 60. This is, so 300 minus 240 is 60 plus 48, so this is equal to 108, so it's still positive. So on either side of four, g prime prime of x is greater than zero."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g prime prime of 10 is going to be equal to three times 10 squared, so it's 300 minus 24 times 10, so minus 240 plus 48. So let's see, this is 60. This is, so 300 minus 240 is 60 plus 48, so this is equal to 108, so it's still positive. So on either side of four, g prime prime of x is greater than zero. So even though the second derivative at x equals four is equal to zero, on either side, we are concave upwards. On either side, the second derivative is positive, and so, and that was the only potential candidate, so there are no values of x for which g has a point of inflection. X equals four would have been a value of x at which g had a point of inflection if we switch, if the second derivative switched signs here, if it went from positive to negative or negative to positive, but it's just staying from positive to positive."}, {"video_title": "Inflection points (algebraic)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on either side of four, g prime prime of x is greater than zero. So even though the second derivative at x equals four is equal to zero, on either side, we are concave upwards. On either side, the second derivative is positive, and so, and that was the only potential candidate, so there are no values of x for which g has a point of inflection. X equals four would have been a value of x at which g had a point of inflection if we switch, if the second derivative switched signs here, if it went from positive to negative or negative to positive, but it's just staying from positive to positive. So the second derivative is positive. It just touches zero right here, and then it goes positive again. So going back to the question, for what x values does the graph of g have a point of inflection?"}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Part B, using correct units, explain the meaning of the definite integral, so it's the definite integral from zero, from t equals zero to t equals 40, the absolute value v of t, dt, in the context of the problem. Approximate the value of that integral using a right Riemann sum with the four subintervals indicated in the table. All right, so let's first tackle the first part. So what's the meaning of this definite integral? So the velocity, velocity is a vector, it has direction. And you can see, when you look up here, that sometimes the velocity is positive and sometimes it is negative. And a reasonable interpretation of that is, well, this is when she's walking in one direction and then that's whatever direction we assume is the positive direction, and this is when she's walking the other way."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's the meaning of this definite integral? So the velocity, velocity is a vector, it has direction. And you can see, when you look up here, that sometimes the velocity is positive and sometimes it is negative. And a reasonable interpretation of that is, well, this is when she's walking in one direction and then that's whatever direction we assume is the positive direction, and this is when she's walking the other way. And so she has a positive velocity here and then she has a negative velocity, she has some speed in the other direction, and then she has a positive velocity again. So if you were to take the absolute value of this, if you just cared about the magnitude of the velocity, well, we're talking about speed. But we're not just talking about the absolute value of velocity, we're integrating the absolute value of velocity over time, from time equals zero to time equals 40."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And a reasonable interpretation of that is, well, this is when she's walking in one direction and then that's whatever direction we assume is the positive direction, and this is when she's walking the other way. And so she has a positive velocity here and then she has a negative velocity, she has some speed in the other direction, and then she has a positive velocity again. So if you were to take the absolute value of this, if you just cared about the magnitude of the velocity, well, we're talking about speed. But we're not just talking about the absolute value of velocity, we're integrating the absolute value of velocity over time, from time equals zero to time equals 40. So if you integrate speed, if you're saying, okay, for every little amount of time, dt, she has some speed, well, that's gonna give you the distance that she travels. So the distance, remember, is the difference between distance and displacement, is displacement you can think of as the net distance. It takes, if you go back and forth a bunch of times, it's going to add up to the distance, but they're going to net out with respect to each other in displacement."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we're not just talking about the absolute value of velocity, we're integrating the absolute value of velocity over time, from time equals zero to time equals 40. So if you integrate speed, if you're saying, okay, for every little amount of time, dt, she has some speed, well, that's gonna give you the distance that she travels. So the distance, remember, is the difference between distance and displacement, is displacement you can think of as the net distance. It takes, if you go back and forth a bunch of times, it's going to add up to the distance, but they're going to net out with respect to each other in displacement. So this right over here is going to give us the total distance, not displacement, that she travels over those 40 minutes. If there was no absolute value sign here, then we'd be talking about the displacement. And so let me write that."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It takes, if you go back and forth a bunch of times, it's going to add up to the distance, but they're going to net out with respect to each other in displacement. So this right over here is going to give us the total distance, not displacement, that she travels over those 40 minutes. If there was no absolute value sign here, then we'd be talking about the displacement. And so let me write that. So this integral from zero to 40, absolute value v of t, dt, is the total distance, distance she travels, she travels over the 40 minutes. Over the 40, over the 40 minutes. Remember, our unit of time here is in minutes."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let me write that. So this integral from zero to 40, absolute value v of t, dt, is the total distance, distance she travels, she travels over the 40 minutes. Over the 40, over the 40 minutes. Remember, our unit of time here is in minutes. All right, so using correct units, and so this is, and so the units, let me write it this way, is the total distance she travels over the 40 minutes. The units are meters. The units are meters."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, our unit of time here is in minutes. All right, so using correct units, and so this is, and so the units, let me write it this way, is the total distance she travels over the 40 minutes. The units are meters. The units are meters. Are meters. All right. Then they say approximate the value of that integral using a right Riemann sum with the four subintervals indicated in the table."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The units are meters. Are meters. All right. Then they say approximate the value of that integral using a right Riemann sum with the four subintervals indicated in the table. So what do they mean by a right Riemann sum? Well, actually, let me just draw, let me draw the absolute value of v of t, and I already took the trouble in the last part of drawing the axes, and you wouldn't actually have to do this on the test, but I'm doing this to just help you understand what we're actually going to do when we talk about a right Riemann sum. So in orange, I have v of t, but if I were to do the absolute value of v of t, I'll do that in magenta."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then they say approximate the value of that integral using a right Riemann sum with the four subintervals indicated in the table. So what do they mean by a right Riemann sum? Well, actually, let me just draw, let me draw the absolute value of v of t, and I already took the trouble in the last part of drawing the axes, and you wouldn't actually have to do this on the test, but I'm doing this to just help you understand what we're actually going to do when we talk about a right Riemann sum. So in orange, I have v of t, but if I were to do the absolute value of v of t, I'll do that in magenta. So the absolute value of that is that. Absolute value of that is that. Absolute value of that is that."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in orange, I have v of t, but if I were to do the absolute value of v of t, I'll do that in magenta. So the absolute value of that is that. Absolute value of that is that. Absolute value of that is that. Absolute value of negative 220 is going to be positive 220. So we saw here, when t is equal to 24, v of t is negative 220. She's running in the opposite direction, or jogging in the other direction, but we're going to take the absolute value of that."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Absolute value of that is that. Absolute value of negative 220 is going to be positive 220. So we saw here, when t is equal to 24, v of t is negative 220. She's running in the opposite direction, or jogging in the other direction, but we're going to take the absolute value of that. So when t is equal to 24, her speed, we could say, is going to be 220. We're going to take the absolute value of it. So it's going to be right around, right around there."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "She's running in the opposite direction, or jogging in the other direction, but we're going to take the absolute value of that. So when t is equal to 24, her speed, we could say, is going to be 220. We're going to take the absolute value of it. So it's going to be right around, right around there. And then the absolute value of this is just going to be that point again. And so we don't know the actual velocity function. The actual velocity function, I don't know, it might look something like, it might look something like this."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be right around, right around there. And then the absolute value of this is just going to be that point again. And so we don't know the actual velocity function. The actual velocity function, I don't know, it might look something like, it might look something like this. It goes like that. Maybe it dips down. Or I should say the speed function."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The actual velocity function, I don't know, it might look something like, it might look something like this. It goes like that. Maybe it dips down. Or I should say the speed function. We don't know the actual speed function. So this is the absolute value of v of t, which is the speed function, you could say. Speed, speed function."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or I should say the speed function. We don't know the actual speed function. So this is the absolute value of v of t, which is the speed function, you could say. Speed, speed function. And so if we wanted to actually integrate it, so if we wanted to integrate it, well, we don't know how to do it if we don't know, we don't know how to do it exactly without knowing the exact function. But we can approximate what the integral is going to be with a right Riemann sum. And a Riemann sum is just breaking up this area into rectangles and then finding the total area of those rectangles."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Speed, speed function. And so if we wanted to actually integrate it, so if we wanted to integrate it, well, we don't know how to do it if we don't know, we don't know how to do it exactly without knowing the exact function. But we can approximate what the integral is going to be with a right Riemann sum. And a Riemann sum is just breaking up this area into rectangles and then finding the total area of those rectangles. And then there's a couple of ways you could do it. You could use, for each rectangle, you could use the left boundary as the height, or you could use the right boundary as the height. And so they tell us to use a right Riemann sum."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And a Riemann sum is just breaking up this area into rectangles and then finding the total area of those rectangles. And then there's a couple of ways you could do it. You could use, for each rectangle, you could use the left boundary as the height, or you could use the right boundary as the height. And so they tell us to use a right Riemann sum. A right Riemann sum. So for each of the rectangles, for each of the intervals, we're going to use the value of our function on the right side to approximate the area. So let's do that."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so they tell us to use a right Riemann sum. A right Riemann sum. So for each of the rectangles, for each of the intervals, we're going to use the value of our function on the right side to approximate the area. So let's do that. So let me divide this into four rectangles. And they give us the four intervals. So one interval goes from zero to t equals 12."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So let me divide this into four rectangles. And they give us the four intervals. So one interval goes from zero to t equals 12. We see that there. The other interval goes from 12 to 20. The next interval goes from 20 to 24."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one interval goes from zero to t equals 12. We see that there. The other interval goes from 12 to 20. The next interval goes from 20 to 24. The next interval goes from 24 to 40. And if the whole notion of a Riemann sum is completely foreign to you, I encourage you to watch the videos on Khan Academy on Riemann sums. And so here we are."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The next interval goes from 20 to 24. The next interval goes from 24 to 40. And if the whole notion of a Riemann sum is completely foreign to you, I encourage you to watch the videos on Khan Academy on Riemann sums. And so here we are. So the first interval is from zero to 12. And so we want to use the value of our function on the right side to define the height of the rectangle. So let me define."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so here we are. So the first interval is from zero to 12. And so we want to use the value of our function on the right side to define the height of the rectangle. So let me define. So that rectangle is going to look like this. So that's a rectangle right over there. Then the next interval goes from 12 to 20."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me define. So that rectangle is going to look like this. So that's a rectangle right over there. Then the next interval goes from 12 to 20. The next rectangle goes from 12 to 20. We see that there. From 12 to 20."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then the next interval goes from 12 to 20. The next rectangle goes from 12 to 20. We see that there. From 12 to 20. So let's use the value of our function at 20 to define the height of this rectangle. So just like that. Look like."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "From 12 to 20. So let's use the value of our function at 20 to define the height of this rectangle. So just like that. Look like. Let me do it. Let me just hand draw that. It would look like that."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Look like. Let me do it. Let me just hand draw that. It would look like that. And then the next rectangle goes from t equals 20 to t equals 24. And we want to use the right hand side to define the height of that rectangle. And you can see it's a little irregular, and that's okay."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would look like that. And then the next rectangle goes from t equals 20 to t equals 24. And we want to use the right hand side to define the height of that rectangle. And you can see it's a little irregular, and that's okay. They all don't have to have the same width. So the height is something like this. And then the last interval is t equals 24 to t equals 40."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can see it's a little irregular, and that's okay. They all don't have to have the same width. So the height is something like this. And then the last interval is t equals 24 to t equals 40. And so we use the right side as the height. So the interval. So this was the last interval we did."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the last interval is t equals 24 to t equals 40. And so we use the right side as the height. So the interval. So this was the last interval we did. The next interval is going to be that. And then the height, we use the right hand side to define the height. That's why it's a right Riemann sum."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this was the last interval we did. The next interval is going to be that. And then the height, we use the right hand side to define the height. That's why it's a right Riemann sum. And so our approximation of the integral of the absolute value of v of t is going to be the area of these rectangles over here. And so we could write, we could write the integral from zero to 40 of the absolute value of v of t dt is approximately, well it's going to be this first change in time times this height. So this first change in time, this is a change in time of 12 times, so it's going to be 12 times, what's the height?"}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's why it's a right Riemann sum. And so our approximation of the integral of the absolute value of v of t is going to be the area of these rectangles over here. And so we could write, we could write the integral from zero to 40 of the absolute value of v of t dt is approximately, well it's going to be this first change in time times this height. So this first change in time, this is a change in time of 12 times, so it's going to be 12 times, what's the height? Well v of 12 we knew was 200. v of 12 is 200. So it's going to be 12 times, 12 times 200 plus, okay the height here, or the next interval goes from 12 to 20. So it's going to be eight times."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this first change in time, this is a change in time of 12 times, so it's going to be 12 times, what's the height? Well v of 12 we knew was 200. v of 12 is 200. So it's going to be 12 times, 12 times 200 plus, okay the height here, or the next interval goes from 12 to 20. So it's going to be eight times. And then the height here is going to be, what is v of 20? I think that was, v of 20 was 240. v of 20 was 240. And then the next interval goes from 20 to 24."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be eight times. And then the height here is going to be, what is v of 20? I think that was, v of 20 was 240. v of 20 was 240. And then the next interval goes from 20 to 24. So that only has a change in time of four. And then what is the height at 24? The height at 24."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the next interval goes from 20 to 24. So that only has a change in time of four. And then what is the height at 24? The height at 24. And remember we took the absolute value of v of t, so it's 220. So four times 220. And then our last rectangle, our change in time, we go from 24 to 40, so our change in time is 16, times the height at 40, what was that?"}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The height at 24. And remember we took the absolute value of v of t, so it's 220. So four times 220. And then our last rectangle, our change in time, we go from 24 to 40, so our change in time is 16, times the height at 40, what was that? That was 150 I think. Yep, that was 150. So times 150."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then our last rectangle, our change in time, we go from 24 to 40, so our change in time is 16, times the height at 40, what was that? That was 150 I think. Yep, that was 150. So times 150. 150. And so just to make clear what I just did, I'm just adding up the sums of the areas of these rectangles. The area of this rectangle is 12 times 200."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So times 150. 150. And so just to make clear what I just did, I'm just adding up the sums of the areas of these rectangles. The area of this rectangle is 12 times 200. Area of this rectangle is eight times 240. The area of this rectangle is four times 220. Area of this rectangle is 16 times 150."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The area of this rectangle is 12 times 200. Area of this rectangle is eight times 240. The area of this rectangle is four times 220. Area of this rectangle is 16 times 150. And then let's see, we could calculate that. We're not allowed to use our calculator for this part, so we're gonna do it by hand. 12 times 200 is going to be 2,400, plus eight times 240, that's going to be 1,600, 1,600 plus 3,200, that is 19,20."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Area of this rectangle is 16 times 150. And then let's see, we could calculate that. We're not allowed to use our calculator for this part, so we're gonna do it by hand. 12 times 200 is going to be 2,400, plus eight times 240, that's going to be 1,600, 1,600 plus 3,200, that is 19,20. Did I do that right? Eight times 40 is 3,200, or sorry, eight times 40 is 320, plus 1,600, yep, that's right, plus, this is going to be 880, plus, let's see, 15 times 15 is going to be, would be 2,250, that would be 15 times 150, and so we're gonna have another 150, and so this is going to be 2,400. 2,400."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "12 times 200 is going to be 2,400, plus eight times 240, that's going to be 1,600, 1,600 plus 3,200, that is 19,20. Did I do that right? Eight times 40 is 3,200, or sorry, eight times 40 is 320, plus 1,600, yep, that's right, plus, this is going to be 880, plus, let's see, 15 times 15 is going to be, would be 2,250, that would be 15 times 150, and so we're gonna have another 150, and so this is going to be 2,400. 2,400. Now we want to add all these together. Let me just do it the way that, the easiest way. 2,400 plus 19,20, plus 880, plus 2,400 again."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "2,400. Now we want to add all these together. Let me just do it the way that, the easiest way. 2,400 plus 19,20, plus 880, plus 2,400 again. Plus 2,400 again, will give us, will give us, we get a zero, we get a 10. So let's see, we have nine plus nine is 18, plus eight is 26, and then we have seven. So we get 7,600."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "2,400 plus 19,20, plus 880, plus 2,400 again. Plus 2,400 again, will give us, will give us, we get a zero, we get a 10. So let's see, we have nine plus nine is 18, plus eight is 26, and then we have seven. So we get 7,600. So it's equal to 7,600. And just to be clear, this would be in meters. In meters."}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get 7,600. So it's equal to 7,600. And just to be clear, this would be in meters. In meters. In meters. Now you might be saying, wait, wait, wait, didn't we just find an area? Wouldn't areas be in meters squared?"}, {"video_title": "2015 AP Calculus AB BC 3b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "In meters. In meters. Now you might be saying, wait, wait, wait, didn't we just find an area? Wouldn't areas be in meters squared? But remember, we're not, we're finding the area under the velocity, under the velocity graph. And so if you look at our units, say for this rectangle right here, you're multiplying 12 minutes, you're multiplying 12 minutes, times 200, times 200 meters per minute. Times 200 meters per minute."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "The twice differentiable function g and its second derivative g prime prime are graphed. And you can see it right over here. I'm actually working off of the article on Khan Academy called Justifying Using Second Derivatives. So we see our function g, and we see not its first derivative, but its second derivative here in this brown color. So then the article goes on to say, or the problem goes on to say, four students were asked to give an appropriate calculus-based justification for the fact that g has an inflection point at x equals negative two. So let's just feel good that it at least intuitively feels right. So at x equals negative two, remember what an inflection point is."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we see our function g, and we see not its first derivative, but its second derivative here in this brown color. So then the article goes on to say, or the problem goes on to say, four students were asked to give an appropriate calculus-based justification for the fact that g has an inflection point at x equals negative two. So let's just feel good that it at least intuitively feels right. So at x equals negative two, remember what an inflection point is. It's where we're going from concave downwards to concave upwards, or concave upwards to concave downwards. Or another way to think about it, it's a situation where our slope goes from decreasing to increasing, or from increasing to decreasing. And when we look at it over here, it looks like, look, it looks like our slope is decreasing."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "So at x equals negative two, remember what an inflection point is. It's where we're going from concave downwards to concave upwards, or concave upwards to concave downwards. Or another way to think about it, it's a situation where our slope goes from decreasing to increasing, or from increasing to decreasing. And when we look at it over here, it looks like, look, it looks like our slope is decreasing. It's positive, but it's decreasing, goes to zero. Then it keeps decreasing. It becomes, it's negative now, keeps decreasing until we get to about x equals negative two, and then it seems that it's increasing."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when we look at it over here, it looks like, look, it looks like our slope is decreasing. It's positive, but it's decreasing, goes to zero. Then it keeps decreasing. It becomes, it's negative now, keeps decreasing until we get to about x equals negative two, and then it seems that it's increasing. It's getting less and less and less and less negative. It looks like it's zero right over here. Then it just keeps increasing, gets more and more and more positive."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "It becomes, it's negative now, keeps decreasing until we get to about x equals negative two, and then it seems that it's increasing. It's getting less and less and less and less negative. It looks like it's zero right over here. Then it just keeps increasing, gets more and more and more positive. So it does indeed look like at x equals negative two, we go from being concave downwards to concave upwards. Now a calculus-based justification is we could look at the second derivative and see where the second derivative crosses the x-axis. Because where the second derivative is negative, that means our slope is decreasing."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then it just keeps increasing, gets more and more and more positive. So it does indeed look like at x equals negative two, we go from being concave downwards to concave upwards. Now a calculus-based justification is we could look at the second derivative and see where the second derivative crosses the x-axis. Because where the second derivative is negative, that means our slope is decreasing. We are concave downwards. And where the second derivative is positive, it means our first derivative is increasing. Our slope of our original function is increasing, and we are concave upwards."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because where the second derivative is negative, that means our slope is decreasing. We are concave downwards. And where the second derivative is positive, it means our first derivative is increasing. Our slope of our original function is increasing, and we are concave upwards. So notice, we do indeed, the second derivative does indeed cross the x-axis at x equals negative two. It's not enough for it to just be zero or touch the x-axis. It needs to cross the x-axis in order for us to have an inflection point there."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our slope of our original function is increasing, and we are concave upwards. So notice, we do indeed, the second derivative does indeed cross the x-axis at x equals negative two. It's not enough for it to just be zero or touch the x-axis. It needs to cross the x-axis in order for us to have an inflection point there. So given that, let's look at the students' justifications and see what we could, if we kind of put the teacher hat in our mind and say what a teacher would say for the different justifications. So the first student says the second derivative of g changes signs at x equals negative two. Well, that's exactly what we were just talking about."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "It needs to cross the x-axis in order for us to have an inflection point there. So given that, let's look at the students' justifications and see what we could, if we kind of put the teacher hat in our mind and say what a teacher would say for the different justifications. So the first student says the second derivative of g changes signs at x equals negative two. Well, that's exactly what we were just talking about. If the second derivative changes signs, in this case, it goes from negative to positive, that means our first derivative went from decreasing to increasing, which is indeed good for saying this is a calculus-based justification. So at least for now, I'm gonna put, kudos, you are correct there. It crosses the x-axis."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's exactly what we were just talking about. If the second derivative changes signs, in this case, it goes from negative to positive, that means our first derivative went from decreasing to increasing, which is indeed good for saying this is a calculus-based justification. So at least for now, I'm gonna put, kudos, you are correct there. It crosses the x-axis. So this is ambiguous. What is crossing the x-axis? If a student wrote this, I'd say, well, are they talking about the function?"}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "It crosses the x-axis. So this is ambiguous. What is crossing the x-axis? If a student wrote this, I'd say, well, are they talking about the function? Are they talking about the first derivative, the second derivative? And so I would say please use more precise language. This cannot be accepted as a correct justification."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "If a student wrote this, I'd say, well, are they talking about the function? Are they talking about the first derivative, the second derivative? And so I would say please use more precise language. This cannot be accepted as a correct justification. All right, let's read the other ones. The second derivative of g is increasing at x equals negative two. Well, no, that doesn't justify why you have an inflection point there."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "This cannot be accepted as a correct justification. All right, let's read the other ones. The second derivative of g is increasing at x equals negative two. Well, no, that doesn't justify why you have an inflection point there. For example, the second derivative is increasing at x equals negative 2.5. The second derivative is even increasing at x equals negative one, but you don't have inflection points at those places. So I would say this doesn't justify why g has an inflection point."}, {"video_title": "Justification using second derivative inflection point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, no, that doesn't justify why you have an inflection point there. For example, the second derivative is increasing at x equals negative 2.5. The second derivative is even increasing at x equals negative one, but you don't have inflection points at those places. So I would say this doesn't justify why g has an inflection point. And then the last student responds, the graph of g changes concavity at x equals negative two. That is true, but that isn't a calculus-based justification. We'd wanna use our second derivative here."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it seems like we've met that because if you're differentiable over an interval, you're definitely continuous over that interval. I'm not saying that it's just a generally differentiable function f, I guess, over any interval. But the next part is to say, all right, that if that condition is met, then the slope of the secant line between four comma f of four and six comma f of six, that some, at least one point in between four and six, will have a derivative that is equal to the slope of the secant line. And so let's figure out what the slope of the secant line is between four comma f of four and six comma f of six. And if it's equal to five, then we could use the mean value theorem. If it's not equal to five, then the mean value theorem would not apply. And so let's do that."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's figure out what the slope of the secant line is between four comma f of four and six comma f of six. And if it's equal to five, then we could use the mean value theorem. If it's not equal to five, then the mean value theorem would not apply. And so let's do that. F of six minus f of four, all of that over six minus four is equal to seven minus three over two, which is equal to two. So two not equal to five, so mean value theorem doesn't apply. Apply."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's do that. F of six minus f of four, all of that over six minus four is equal to seven minus three over two, which is equal to two. So two not equal to five, so mean value theorem doesn't apply. Apply. All right, let's, I'll put an exclamation mark there for emphasis. All right, let's do the next part. Can we use the mean value theorem to say that the equation f prime of x is equal to negative one has a solution, and now the interval is from zero to two."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Apply. All right, let's, I'll put an exclamation mark there for emphasis. All right, let's do the next part. Can we use the mean value theorem to say that the equation f prime of x is equal to negative one has a solution, and now the interval is from zero to two. If so, write a justification. All right, so let's see this. So if we were to take the slope of the secant line, so f of two minus f of zero, all that over two minus zero, this is equal to negative two minus zero."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can we use the mean value theorem to say that the equation f prime of x is equal to negative one has a solution, and now the interval is from zero to two. If so, write a justification. All right, so let's see this. So if we were to take the slope of the secant line, so f of two minus f of zero, all that over two minus zero, this is equal to negative two minus zero. All of that over two, which is equal to negative two over two, which is equal to negative one. And so, and we also know that we meet the continuity and differentiability conditions. And so we could say, and since f is generally differentiable, generally differentiable, differentiable, it will be differentiable, differentiable, and continuous over the interval from zero to two."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we were to take the slope of the secant line, so f of two minus f of zero, all that over two minus zero, this is equal to negative two minus zero. All of that over two, which is equal to negative two over two, which is equal to negative one. And so, and we also know that we meet the continuity and differentiability conditions. And so we could say, and since f is generally differentiable, generally differentiable, differentiable, it will be differentiable, differentiable, and continuous over the interval from zero to two. And I say the closed interval. You just have to be differentiable over the open interval, but it's even better, I guess, if you're differentiable over the closed interval, because you have to be continuous over the closed interval. And so, and since f is generally differentiable, it will be differentiable and continuous over zero, two."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we could say, and since f is generally differentiable, generally differentiable, differentiable, it will be differentiable, differentiable, and continuous over the interval from zero to two. And I say the closed interval. You just have to be differentiable over the open interval, but it's even better, I guess, if you're differentiable over the closed interval, because you have to be continuous over the closed interval. And so, and since f is generally differentiable, it will be differentiable and continuous over zero, two. So the mean value theorem tells us, tells us, that there is an x in that interval from zero to two such that f prime of x is equal to that secant slope, or you could say that average rate of change is equal to negative one. And so I could write yes, yes, and then this would be my justification. This is the slope of the secant line or the average rate of change."}, {"video_title": "Justification with the mean value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so, and since f is generally differentiable, it will be differentiable and continuous over zero, two. So the mean value theorem tells us, tells us, that there is an x in that interval from zero to two such that f prime of x is equal to that secant slope, or you could say that average rate of change is equal to negative one. And so I could write yes, yes, and then this would be my justification. This is the slope of the secant line or the average rate of change. And since f is generally differentiable, it will be differentiable and continuous over the closed interval. So the mean value theorem tells us that there is an x in this interval such that f prime of x is equal to negative one. And we're done."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then they're asking us, what is h prime of negative 14? And if you're not familiar with how functions and their derivatives relate to their inverses and the derivatives of the inverse, well this will seem like a very hard thing to do. Because if you were tempted to take the inverse of f to figure out what h is, well it's tough to figure out the inverse of a third degree polynomial defined function like this. So the key, I guess, property to realize or the key truth to realize is that if f and h are inverses, then h prime of x, h prime of x, is going to be equal to, it's going to be equal to one over f prime of h of x. One over f prime of h of x. And you could now use this in order to figure out what h prime of negative 14 is. Now I know what some of you are thinking because it's exactly what I would be thinking if someone just sprung this on me, is where does this come from?"}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the key, I guess, property to realize or the key truth to realize is that if f and h are inverses, then h prime of x, h prime of x, is going to be equal to, it's going to be equal to one over f prime of h of x. One over f prime of h of x. And you could now use this in order to figure out what h prime of negative 14 is. Now I know what some of you are thinking because it's exactly what I would be thinking if someone just sprung this on me, is where does this come from? And I would tell you this comes straight out of the chain rule. We know that if a function and its inverse, we know that if we have a function and its inverse, that f of the inverse of our function, so f of h of x, f of h of x, we know that this is going to be equal to x. This literally, this comes out of them being each other's inverses."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now I know what some of you are thinking because it's exactly what I would be thinking if someone just sprung this on me, is where does this come from? And I would tell you this comes straight out of the chain rule. We know that if a function and its inverse, we know that if we have a function and its inverse, that f of the inverse of our function, so f of h of x, f of h of x, we know that this is going to be equal to x. This literally, this comes out of them being each other's inverses. We could have also said h of f of x will also be equal to x. Remember, f is going to map, or h is going to map from some x to h of x, and then f is going to map back to that original x. That's what inverses do."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "This literally, this comes out of them being each other's inverses. We could have also said h of f of x will also be equal to x. Remember, f is going to map, or h is going to map from some x to h of x, and then f is going to map back to that original x. That's what inverses do. So that's because they are inverses. This is by definition, this is what inverses do to each other. But then if you took the derivative of both sides of this, what would you get?"}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's what inverses do. So that's because they are inverses. This is by definition, this is what inverses do to each other. But then if you took the derivative of both sides of this, what would you get? Let me do that. So if we take the derivative of both sides of this, d dx on the left-hand side, d dx on the right-hand side, and I think you see where this is going. You're essentially going to get a version of that."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then if you took the derivative of both sides of this, what would you get? Let me do that. So if we take the derivative of both sides of this, d dx on the left-hand side, d dx on the right-hand side, and I think you see where this is going. You're essentially going to get a version of that. The left-hand side, use the chain rule. You're going to get f prime of h of x, f prime of h of x times h prime of x, comes straight out of the chain rule, is equal to, is equal to the derivative of x, is just going to be equal to one. And then you divide both sides by f prime of h of x, and you get our original property there."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're essentially going to get a version of that. The left-hand side, use the chain rule. You're going to get f prime of h of x, f prime of h of x times h prime of x, comes straight out of the chain rule, is equal to, is equal to the derivative of x, is just going to be equal to one. And then you divide both sides by f prime of h of x, and you get our original property there. So now with that out of the way, let's just actually apply this. So we want to evaluate h prime of 14. Or sorry, h prime of negative 14 is going to be equal to one over f prime of h of negative 14."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you divide both sides by f prime of h of x, and you get our original property there. So now with that out of the way, let's just actually apply this. So we want to evaluate h prime of 14. Or sorry, h prime of negative 14 is going to be equal to one over f prime of h of negative 14. H of negative 14. Now have they given us h of negative 14? Well, they didn't give it to us explicitly, but we have to remember that f and h are inverses of each other."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or sorry, h prime of negative 14 is going to be equal to one over f prime of h of negative 14. H of negative 14. Now have they given us h of negative 14? Well, they didn't give it to us explicitly, but we have to remember that f and h are inverses of each other. So if f of negative two is negative 14, well, h is gonna go from the other way around. If you input negative 14 into h, you're going to get negative two. So h of negative 14, well, this is going to be equal to negative two."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, they didn't give it to us explicitly, but we have to remember that f and h are inverses of each other. So if f of negative two is negative 14, well, h is gonna go from the other way around. If you input negative 14 into h, you're going to get negative two. So h of negative 14, well, this is going to be equal to negative two. Once again, they are inverses of each other. So h of negative 14 is equal to negative negative two. And once again, I just swapped these two around."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So h of negative 14, well, this is going to be equal to negative two. Once again, they are inverses of each other. So h of negative 14 is equal to negative negative two. And once again, I just swapped these two around. That's what the inverse function will do. If you're mapping from, if f goes from negative two to negative 14, h is going to go from negative 14 back to negative two. So now we want to evaluate f prime of negative two."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, I just swapped these two around. That's what the inverse function will do. If you're mapping from, if f goes from negative two to negative 14, h is going to go from negative 14 back to negative two. So now we want to evaluate f prime of negative two. Well, let's figure out what f prime of x is. So f prime of x is equal to, see, we're just gonna leverage the power rule. So three times 1 1\u20442 is 3 1\u20442 times x to the three minus one power, which is just the second power, plus the derivative of three x with respect to x."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we want to evaluate f prime of negative two. Well, let's figure out what f prime of x is. So f prime of x is equal to, see, we're just gonna leverage the power rule. So three times 1 1\u20442 is 3 1\u20442 times x to the three minus one power, which is just the second power, plus the derivative of three x with respect to x. Well, that's just going to be three. And you could view that as just the power rule. If this was x to the first power, one times three x to the zeroth power."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So three times 1 1\u20442 is 3 1\u20442 times x to the three minus one power, which is just the second power, plus the derivative of three x with respect to x. Well, that's just going to be three. And you could view that as just the power rule. If this was x to the first power, one times three x to the zeroth power. Well, x to the zero is just one, so you're just left with three. And the derivative of a constant, that's just gonna be zero. So that's f prime of x."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "If this was x to the first power, one times three x to the zeroth power. Well, x to the zero is just one, so you're just left with three. And the derivative of a constant, that's just gonna be zero. So that's f prime of x. So f prime of negative two is going to be 3 1\u20442 times negative two squared is four, positive four. So plus three. So this is going to be equal to two times three plus three."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's f prime of x. So f prime of negative two is going to be 3 1\u20442 times negative two squared is four, positive four. So plus three. So this is going to be equal to two times three plus three. So six plus three is equal to nine. So this denominator right here is going to be equal to nine. So this whole thing is equal to one over nine."}, {"video_title": "Derivatives of inverse functions from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to two times three plus three. So six plus three is equal to nine. So this denominator right here is going to be equal to nine. So this whole thing is equal to one over nine. So this involved, this isn't something you're gonna see every day. This isn't a typical problem in your calculus class. But it's interesting."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "And let's say that the graph of a particular solution to this the graph of a particular solution passes through the point one comma negative one. So my question to you is, what is y, what is y when x is equal to three for this particular solution? So the particular solution to the differential equation that passes through the point one comma negative one, what is y when x is equal to three? And I encourage you to pause the video and try to work through it on your own. So I'm assuming you had a go at it and the key with a separable differential equation, and that's a big clue, that even calling it a separable differential equation is that you separate the x's from the y's, or all the x's and the dx's from the y's and dy's. So how do you do that here? Well, what I could do, let me just rewrite it."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to work through it on your own. So I'm assuming you had a go at it and the key with a separable differential equation, and that's a big clue, that even calling it a separable differential equation is that you separate the x's from the y's, or all the x's and the dx's from the y's and dy's. So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be dy dx is equal to two y squared. Is equal to two y, equal to two y squared. So let's see, we can multiply both sides by dx."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "Well, what I could do, let me just rewrite it. So it's gonna be dy dx is equal to two y squared. Is equal to two y, equal to two y squared. So let's see, we can multiply both sides by dx. And let's see, so then we're gonna have, that cancels with that if we treat it as just a value or as a variable, we're gonna have dy is equal to two y squared dx. Well, we're not quite done yet. We need to get this two y squared on the left-hand side."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "So let's see, we can multiply both sides by dx. And let's see, so then we're gonna have, that cancels with that if we treat it as just a value or as a variable, we're gonna have dy is equal to two y squared dx. Well, we're not quite done yet. We need to get this two y squared on the left-hand side. So we can divide both sides by two y squared. So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy. Let me, dy is equal to dx."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "We need to get this two y squared on the left-hand side. So we can divide both sides by two y squared. So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy. Let me, dy is equal to dx. And now we can integrate both sides. So we can integrate both sides. Let me get myself a little bit more space."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "Let me, dy is equal to dx. And now we can integrate both sides. So we can integrate both sides. Let me get myself a little bit more space. And so what is, what is this left-hand side going to be? Well, we increment the exponent and then divide by that value. So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "Let me get myself a little bit more space. And so what is, what is this left-hand side going to be? Well, we increment the exponent and then divide by that value. So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power. And we could do a plus c like we did in the previous video, but we're gonna have a plus c on both sides. And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other. So I'm just gonna write the constant only on one side."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power. And we could do a plus c like we did in the previous video, but we're gonna have a plus c on both sides. And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other. So I'm just gonna write the constant only on one side. So you have that is equal to, well, if I integrate just dx, that's just going to give me x. That's just gonna give me x. So this right over here is x, and of course I can have a plus c over there."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "So I'm just gonna write the constant only on one side. So you have that is equal to, well, if I integrate just dx, that's just going to give me x. That's just gonna give me x. So this right over here is x, and of course I can have a plus c over there. And if I want, I can solve for y. If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y. Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "So this right over here is x, and of course I can have a plus c over there. And if I want, I can solve for y. If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y. Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant. It's still going to be negative two times this arbitrary constant, but I could still just call it some arbitrary constant. And then if we want, we can take the reciprocal of both sides. And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant. It's still going to be negative two times this arbitrary constant, but I could still just call it some arbitrary constant. And then if we want, we can take the reciprocal of both sides. And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one. So we get negative one is equal to one over negative two plus c. Or we could say c minus two. We could multiply both sides times c minus two. If then we will get, actually let me just scroll down a little bit."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one. So we get negative one is equal to one over negative two plus c. Or we could say c minus two. We could multiply both sides times c minus two. If then we will get, actually let me just scroll down a little bit. So if you multiply both sides times c minus two, negative one times c minus two is gonna be negative c plus two, or two minus c is equal to one. All I did is I multiplied c minus two times both sides. And then let's see, I can subtract two from both sides."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "If then we will get, actually let me just scroll down a little bit. So if you multiply both sides times c minus two, negative one times c minus two is gonna be negative c plus two, or two minus c is equal to one. All I did is I multiplied c minus two times both sides. And then let's see, I can subtract two from both sides. So negative c is equal to negative one. And then if I multiply both sides by negative one, we get c is equal to one. So our particular solution is y is equal to one over negative two x plus one."}, {"video_title": "Particular solution to differential equation example   Khan Academy.mp3", "Sentence": "And then let's see, I can subtract two from both sides. So negative c is equal to negative one. And then if I multiply both sides by negative one, we get c is equal to one. So our particular solution is y is equal to one over negative two x plus one. And we are almost done. They didn't just ask for, we didn't just ask for the particular solution. We asked what is y when x is equal to three."}, {"video_title": "Limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can find the limit as x approaches negative one of six x squared plus five x minus one. Now the first thing that might jump out at you is this right over here, this expression could be used to define the graph of a parabola. And when you think about this, I'm not doing a rigorous proof here, a parabola would look something like this, and this would be an upward opening parabola, looks something like this. It is a, this graph visually is continuous, you don't see any jumps or gaps in it. And in general, a quadratic like this is going to be defined for all values of x, for all real numbers, and it's going to be continuous for all real numbers. And so if something is continuous for all real numbers, well then the limit as x approaches some real number is going to be the same thing, it's just evaluating the expression at that real number. So what am I saying?"}, {"video_title": "Limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is a, this graph visually is continuous, you don't see any jumps or gaps in it. And in general, a quadratic like this is going to be defined for all values of x, for all real numbers, and it's going to be continuous for all real numbers. And so if something is continuous for all real numbers, well then the limit as x approaches some real number is going to be the same thing, it's just evaluating the expression at that real number. So what am I saying? I'm just gonna say it another way. We know that some function is continuous, is continuous at some x value, at x equals a, if and only if, I'll write that as if, or if, if and only if the limit as x approaches a of f of x is equal to f of, is equal to f of a. So I didn't do a rigorous proof here, but just, it's conceptually not a big jump to say okay, well this is just a standard quadratic right over here, it's defined for all real numbers, and in fact it's continuous for all real numbers, and so we know that this expression, it could define a continuous function, so that means that the limit as x approaches a for this expression is just the same thing as evaluating this expression at a."}, {"video_title": "Limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what am I saying? I'm just gonna say it another way. We know that some function is continuous, is continuous at some x value, at x equals a, if and only if, I'll write that as if, or if, if and only if the limit as x approaches a of f of x is equal to f of, is equal to f of a. So I didn't do a rigorous proof here, but just, it's conceptually not a big jump to say okay, well this is just a standard quadratic right over here, it's defined for all real numbers, and in fact it's continuous for all real numbers, and so we know that this expression, it could define a continuous function, so that means that the limit as x approaches a for this expression is just the same thing as evaluating this expression at a. And in this case our a is negative one. So all I have to do is evaluate this at negative one. This is going to be six times negative one squared plus five times negative one minus one."}, {"video_title": "Limits by direct substitution   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I didn't do a rigorous proof here, but just, it's conceptually not a big jump to say okay, well this is just a standard quadratic right over here, it's defined for all real numbers, and in fact it's continuous for all real numbers, and so we know that this expression, it could define a continuous function, so that means that the limit as x approaches a for this expression is just the same thing as evaluating this expression at a. And in this case our a is negative one. So all I have to do is evaluate this at negative one. This is going to be six times negative one squared plus five times negative one minus one. So that's just one, this is negative five. So it's six minus five minus one, which is equal to zero. And we are done."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "This seems like a hairy integral. Now the key insight here is to realize that you have this expression x to the fourth plus 7 and you also have its derivative up here. The derivative of x to the fourth plus 7 is equal to 4x to the third. The derivative of x to the fourth is 4x to the third. The derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. I'll just write u. I'll write the whole thing."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of x to the fourth is 4x to the third. The derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. I'll just write u. I'll write the whole thing. U-substitution could be the tool of choice. So given that, what would you want to set your u equal to? I'll let you think about that because if you can figure out this part, then the rest will just boil down to a fairly straightforward integral."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll just write u. I'll write the whole thing. U-substitution could be the tool of choice. So given that, what would you want to set your u equal to? I'll let you think about that because if you can figure out this part, then the rest will just boil down to a fairly straightforward integral. You want to set u to be equal to the expression that you have its derivative laying around. So we could set u equal to x to the fourth plus 7. Now what is du going to be equal to?"}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll let you think about that because if you can figure out this part, then the rest will just boil down to a fairly straightforward integral. You want to set u to be equal to the expression that you have its derivative laying around. So we could set u equal to x to the fourth plus 7. Now what is du going to be equal to? I will do it in magenta. du is just going to be the derivative of x to the fourth plus 7 with respect to x, so 4x to the third plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x to the third power."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is du going to be equal to? I will do it in magenta. du is just going to be the derivative of x to the fourth plus 7 with respect to x, so 4x to the third plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x to the third power. When someone writes du over dx like this, this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but oftentimes you can kind of pseudo-manipulate them like fractions. So if you wanted to go from here to there, you could kind of pretend that you're multiplying both sides by dx."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x to the third power. When someone writes du over dx like this, this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but oftentimes you can kind of pseudo-manipulate them like fractions. So if you wanted to go from here to there, you could kind of pretend that you're multiplying both sides by dx. But these are equivalent statements, and we want to get it in differential form in order to do proper u-substitution. The reason why this is useful, and I'll just rewrite it up here so that it becomes pretty obvious, our original integral we can rewrite as 4x to the third dx over x to the fourth plus 7. Then it's pretty clear what's du and what's u. u, which we set to be equal to x to the fourth plus 7, and then du is equal to this."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you wanted to go from here to there, you could kind of pretend that you're multiplying both sides by dx. But these are equivalent statements, and we want to get it in differential form in order to do proper u-substitution. The reason why this is useful, and I'll just rewrite it up here so that it becomes pretty obvious, our original integral we can rewrite as 4x to the third dx over x to the fourth plus 7. Then it's pretty clear what's du and what's u. u, which we set to be equal to x to the fourth plus 7, and then du is equal to this. It's equal to 4x to the third dx. We saw it right over here. So we can rewrite this integral, and I'll try to stay consistent with the colors, as the indefinite integral."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then it's pretty clear what's du and what's u. u, which we set to be equal to x to the fourth plus 7, and then du is equal to this. It's equal to 4x to the third dx. We saw it right over here. So we can rewrite this integral, and I'll try to stay consistent with the colors, as the indefinite integral. What we have in magenta right over here, that's du over x to the fourth plus 7, which is just u. Or we could rewrite this entire thing as the integral of 1 over u du. What is the indefinite integral of 1 over u du?"}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can rewrite this integral, and I'll try to stay consistent with the colors, as the indefinite integral. What we have in magenta right over here, that's du over x to the fourth plus 7, which is just u. Or we could rewrite this entire thing as the integral of 1 over u du. What is the indefinite integral of 1 over u du? That's just going to be equal to the natural log of the absolute value, and we use the absolute value so it'll be defined even for negative u's. It actually does work out, and I'll do another video where I show you that it definitely does. The natural log of the absolute value of u, and then we might have had a constant there that was lost when we took the derivative."}, {"video_title": "_-substitution rational function   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the indefinite integral of 1 over u du? That's just going to be equal to the natural log of the absolute value, and we use the absolute value so it'll be defined even for negative u's. It actually does work out, and I'll do another video where I show you that it definitely does. The natural log of the absolute value of u, and then we might have had a constant there that was lost when we took the derivative. That's essentially our answer in terms of u, but now we need to unsubstitute the u. What happens when we unsubstitute the u? Then we are left with, this is going to be equal to the natural log of the absolute value of, well u is x to the fourth plus 7."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "All right, let's see if we can find the indefinite integral of one over five x squared minus 30x plus 65 dx. Pause this video and see if you can figure it out. All right, so this is going to be an interesting one. It'll be a little bit hairy, but we're gonna work through it together. So immediately, you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "It'll be a little bit hairy, but we're gonna work through it together. So immediately, you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again, pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again, pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I can just factor a five out of the denominator. And if I did that, then this integral will become 1 5th times the integral of one over, so I factored a five out of the denominator. So it is x squared minus six x plus 13 dx."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I can just factor a five out of the denominator. And if I did that, then this integral will become 1 5th times the integral of one over, so I factored a five out of the denominator. So it is x squared minus six x plus 13 dx. And then as I mentioned, I'm gonna complete the square down here. So let me rewrite it. So this is equal to 1 5th times the integral of one over, and so x squared minus six x."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So it is x squared minus six x plus 13 dx. And then as I mentioned, I'm gonna complete the square down here. So let me rewrite it. So this is equal to 1 5th times the integral of one over, and so x squared minus six x. It's clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square?"}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So this is equal to 1 5th times the integral of one over, and so x squared minus six x. It's clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? Well, we've done this before. You take half of your coefficient here, which is negative three, and you square that. So you wanna add a nine here."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? Well, we've done this before. You take half of your coefficient here, which is negative three, and you square that. So you wanna add a nine here. But if you add a nine, then you have to subtract a nine as well. And so this part is going to be x minus three squared. And then this part right over here is going to be equal to a positive four."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So you wanna add a nine here. But if you add a nine, then you have to subtract a nine as well. And so this part is going to be x minus three squared. And then this part right over here is going to be equal to a positive four. And we of course don't wanna forget our dx out here. And so let me write it in this form. So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "And then this part right over here is going to be equal to a positive four. And we of course don't wanna forget our dx out here. And so let me write it in this form. So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. Actually, let me do it that way. Plus two squared dx. Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. Actually, let me do it that way. Plus two squared dx. Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm gonna do is, let's factor a four out of the denominator here. So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm gonna do is, let's factor a four out of the denominator here. So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. And then this is going to be a plus one. And of course, we have our dx. And then we could write this as, and I'm trying to just do every step here."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. And then this is going to be a plus one. And of course, we have our dx. And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as x minus three over two squared plus one. Plus one, and then dx."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as x minus three over two squared plus one. Plus one, and then dx. And now the u substitution is pretty clear. I am just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "Plus one, and then dx. And now the u substitution is pretty clear. I am just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. And du is going to be equal to 1 1 2 dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit, so that we see a 1 1 2 here. So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. And du is going to be equal to 1 1 2 dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit, so that we see a 1 1 2 here. So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. This becomes 1 1. And so doing my u substitution, I get 1 1, that's that 1 1 there, times the integral of, well, I have 1 1 2 dx right over here, which is the same thing as du. So I could put the du either in the numerator, I could put it out here."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. This becomes 1 1. And so doing my u substitution, I get 1 1, that's that 1 1 there, times the integral of, well, I have 1 1 2 dx right over here, which is the same thing as du. So I could put the du either in the numerator, I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arc tan of u? Well, that would be one over u squared plus one."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "So I could put the du either in the numerator, I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arc tan of u? Well, that would be one over u squared plus one. So this is going to be equal to 1 1 0 times the arc tangent of u. And of course, we can't forget our big constant c, because we're taking an indefinite integral. And now we just wanna do the reverse substitution."}, {"video_title": "Integration using completing the square and the derivative of arctan(x)   Khan Academy.mp3", "Sentence": "Well, that would be one over u squared plus one. So this is going to be equal to 1 1 0 times the arc tangent of u. And of course, we can't forget our big constant c, because we're taking an indefinite integral. And now we just wanna do the reverse substitution. We know that u is equal to this business right over here, so we deserve a little bit of a drum roll. This is going to be equal to 1 1 0 times the arc tangent of u. Well, u is just x minus three over two, which could also be written like this."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It has a vertical tangent at the point three comma zero. So three comma zero has a vertical tangent. Let me draw that. So it has a vertical tangent right over there. And a horizontal tangent at the point zero comma negative three, zero comma negative three. So it has a horizontal tangent right over there. And also has a horizontal tangent at six comma three."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it has a vertical tangent right over there. And a horizontal tangent at the point zero comma negative three, zero comma negative three. So it has a horizontal tangent right over there. And also has a horizontal tangent at six comma three. So six comma three, let me draw the horizontal tangent. Just like that. Select all the x values for which f is not differentiable."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And also has a horizontal tangent at six comma three. So six comma three, let me draw the horizontal tangent. Just like that. Select all the x values for which f is not differentiable. Select all that apply. So f prime, f prime, I'll write it in shorthand. So we say no f prime under, it's going to happen under three conditions."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Select all the x values for which f is not differentiable. Select all that apply. So f prime, f prime, I'll write it in shorthand. So we say no f prime under, it's going to happen under three conditions. The first condition, you could say, well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative?"}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we say no f prime under, it's going to happen under three conditions. The first condition, you could say, well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. When you have a vertical tangent, you change your x a very small amount. You have an infinite change in y, either in the positive or the negative direction."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Why is a vertical tangent a place where it's hard to define our derivative? Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. When you have a vertical tangent, you change your x a very small amount. You have an infinite change in y, either in the positive or the negative direction. So that's one situation where you have no derivative. And they tell us where we have a vertical tangent in here, where x is equal to three. So we have no, f is not differentiable at x equals three because of the vertical tangent."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have an infinite change in y, either in the positive or the negative direction. So that's one situation where you have no derivative. And they tell us where we have a vertical tangent in here, where x is equal to three. So we have no, f is not differentiable at x equals three because of the vertical tangent. You might say, what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have no, f is not differentiable at x equals three because of the vertical tangent. You might say, what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. So f prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios?"}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Horizontal tangents are places where the derivative is equal to zero. So f prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not going to have a defined derivative is where the graph is not continuous. Not continuous. And we see right over here at x equals negative three, our graph is not continuous."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "What are other scenarios? Well another scenario where you're not going to have a defined derivative is where the graph is not continuous. Not continuous. And we see right over here at x equals negative three, our graph is not continuous. So x equals negative three, it's not continuous. And those are the only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we see right over here at x equals negative three, our graph is not continuous. So x equals negative three, it's not continuous. And those are the only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These I guess would be interesting cases, but they haven't given us those choices here. And we already said at x equals zero, the derivative is zero. It's defined, it's differentiable there."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know what the graph is doing to the left or the right. These I guess would be interesting cases, but they haven't given us those choices here. And we already said at x equals zero, the derivative is zero. It's defined, it's differentiable there. And at x equals six, the derivative is zero. We have a flat, flat tangent. So once again, it's defined there as well."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's defined, it's differentiable there. And at x equals six, the derivative is zero. We have a flat, flat tangent. So once again, it's defined there as well. Let's do another one of these. Oh, and actually I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, it's defined there as well. Let's do another one of these. Oh, and actually I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. And this isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that. Or like, or like, well no, that doesn't look too sharp."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "A sharp turn. And this isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that. Or like, or like, well no, that doesn't look too sharp. Or like this. And the reason why I think where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that, the reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here where as x increases, y is increasing, while our slope is negative here."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or like, or like, well no, that doesn't look too sharp. Or like this. And the reason why I think where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that, the reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here where as x increases, y is increasing, while our slope is negative here. So as you try to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left-hand side and the right-hand side. So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. Let's do one more examples."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Notice our slope is positive right over here where as x increases, y is increasing, while our slope is negative here. So as you try to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left-hand side and the right-hand side. So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. Let's do one more examples. And actually this one does have some sharp turns, so this could be interesting. The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do one more examples. And actually this one does have some sharp turns, so this could be interesting. The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three. We see that. And horizontal asymptotes at y equals zero. Yep, this end of the curve as x approaches negative infinity, it looks like y is approaching zero."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It has a vertical asymptote at x equals negative three. We see that. And horizontal asymptotes at y equals zero. Yep, this end of the curve as x approaches negative infinity, it looks like y is approaching zero. And it has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yep, this end of the curve as x approaches negative infinity, it looks like y is approaching zero. And it has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable. So first of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Select the x values for which f is not differentiable. So first of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. So we're not continuous at x equals negative three. We're also not continuous at x is equal to one."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. So we're not continuous at x equals negative three. We're also not continuous at x is equal to one. And then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point on our graph. And I see a sharp point right over there. Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1\u20442."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're also not continuous at x is equal to one. And then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point on our graph. And I see a sharp point right over there. Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1\u20442. Well, as we go to the right side of that, it looks like our slope turns negative. And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. So f is also not differentiable at the x value that gives us that little sharp point right over there."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1\u20442. Well, as we go to the right side of that, it looks like our slope turns negative. And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. So f is also not differentiable at the x value that gives us that little sharp point right over there. And if you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. So let me mark that off. And then we could check x equals zero."}, {"video_title": "Differentiability at a point graphical   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f is also not differentiable at the x value that gives us that little sharp point right over there. And if you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. So let me mark that off. And then we could check x equals zero. X equals zero is completely cool. We're at a point that our tangent line is definitely not vertical. We're definitely continuous there."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here it says f prime of five. So this notation, prime, this is another way of saying, well, what's the derivative? Let's estimate the derivative of our function at five. And when we say f prime of five, this is the slope, slope of tangent line, tangent line at five. Or you could view it as the, you could view it as the rate of change of y with respect to x, which is really how we define slope, with respect to x of our function f. So let's think about that a little bit. We see they put the point, the point five comma f of five right over here. And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when we say f prime of five, this is the slope, slope of tangent line, tangent line at five. Or you could view it as the, you could view it as the rate of change of y with respect to x, which is really how we define slope, with respect to x of our function f. So let's think about that a little bit. We see they put the point, the point five comma f of five right over here. And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point. And so let me see if I can do that. So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. Now what makes this an interesting thing in nonlinear is that it's constantly changing."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point. And so let me see if I can do that. So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. Now what makes this an interesting thing in nonlinear is that it's constantly changing. The steepness, it's very low here, and it gets steeper and steeper and steeper as we move to the right for larger and larger x values. But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what makes this an interesting thing in nonlinear is that it's constantly changing. The steepness, it's very low here, and it gets steeper and steeper and steeper as we move to the right for larger and larger x values. But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction. Delta y is equal to two when delta x is equal to one. So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. And they told us to estimate it, but all of these are way off."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction. Delta y is equal to two when delta x is equal to one. So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. And they told us to estimate it, but all of these are way off. Having a negative two derivative would mean that as we increase our x, our y is decreasing. So if our curve looks something like this, we would have a slope of negative two. If having slopes in this, a positive of.1, that would be very flat."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they told us to estimate it, but all of these are way off. Having a negative two derivative would mean that as we increase our x, our y is decreasing. So if our curve looks something like this, we would have a slope of negative two. If having slopes in this, a positive of.1, that would be very flat. Down here we might have a slope closer to.1. Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If having slopes in this, a positive of.1, that would be very flat. Down here we might have a slope closer to.1. Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing. The slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. Let's do one more of these."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing. The slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. Let's do one more of these. All right, so they're telling us to compare the derivative of g at four to the derivative of g at six. And which one of these is greater? And like always, pause the video and see if you could figure this out."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do one more of these. All right, so they're telling us to compare the derivative of g at four to the derivative of g at six. And which one of these is greater? And like always, pause the video and see if you could figure this out. Well, this is just an exercise. Let's see if we were to make a line that indicates the slope there. And you could view this as a tangent line."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And like always, pause the video and see if you could figure this out. Well, this is just an exercise. Let's see if we were to make a line that indicates the slope there. And you could view this as a tangent line. So let me try to do that. So, no, that doesn't do a good job. So right over here at, so that looks like a pretty, I think I can do a better job than that."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could view this as a tangent line. So let me try to do that. So, no, that doesn't do a good job. So right over here at, so that looks like a pretty, I think I can do a better job than that. No, that's too shallow. Let's see, not shallow, that's too flat. So let me try to really, okay, that looks pretty good."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So right over here at, so that looks like a pretty, I think I can do a better job than that. No, that's too shallow. Let's see, not shallow, that's too flat. So let me try to really, okay, that looks pretty good. So that line that I just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line, you could view it as a tangent line so that we can think about what its slope is going to be. And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. So it looks like it is steeper for sure, but it's in the negative direction."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me try to really, okay, that looks pretty good. So that line that I just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line, you could view it as a tangent line so that we can think about what its slope is going to be. And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. So it looks like it is steeper for sure, but it's in the negative direction. As we increase, think of it this way, as we increase x one here, it looks like we are decreasing y by about one. So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks like it is steeper for sure, but it's in the negative direction. As we increase, think of it this way, as we increase x one here, it looks like we are decreasing y by about one. So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three. So g prime of six looks like it's closer to negative three. So which one of these is larger? Well, this one is less negative, so it's gonna be greater than the other one."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three. So g prime of six looks like it's closer to negative three. So which one of these is larger? Well, this one is less negative, so it's gonna be greater than the other one. And you could have done this intuitively. If you just look at the curve, this is some type of a sinusoid here. You have right over here, the curve is flat."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this one is less negative, so it's gonna be greater than the other one. And you could have done this intuitively. If you just look at the curve, this is some type of a sinusoid here. You have right over here, the curve is flat. It's, you have right at that moment, you have no change in y with respect to x. Then it starts to decrease at a, then it decreases at an even faster rate, then it decreases at a faster rate. Then it starts, it's still decreasing, but it's decreasing at slower and slower rates, decreasing at slower rates."}, {"video_title": "Derivative as slope of curve   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have right over here, the curve is flat. It's, you have right at that moment, you have no change in y with respect to x. Then it starts to decrease at a, then it decreases at an even faster rate, then it decreases at a faster rate. Then it starts, it's still decreasing, but it's decreasing at slower and slower rates, decreasing at slower rates. And right at that moment, it's not, you have your slope of your tangent line is zero. Then it starts to increase, increase, so on and so forth. And it just keeps happening over and over again."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "The function d gives the total distance Eddie has driven in kilometers t hours after he left. What is the best interpretation for the following statement? D prime of two is equal to 100. So pause this video and I encourage you to write it out. What do you think this means? And be sure to include the appropriate units. All right, now let's do this together."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video and I encourage you to write it out. What do you think this means? And be sure to include the appropriate units. All right, now let's do this together. If d is equal to the distance driven, then to get d prime, you're taking the derivative with respect to time. So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now?"}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now let's do this together. If d is equal to the distance driven, then to get d prime, you're taking the derivative with respect to time. So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now? Well, that is our t, and that's in hours. So two hours, actually, let me color code it. So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units?"}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now? Well, that is our t, and that's in hours. So two hours, actually, let me color code it. So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units? Well, the distance was given in kilometers, and now we're gonna be thinking about kilometers per unit time, kilometers per hour. So this is 100 kilometers, kilometers per hour. So that's the interpretation there."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units? Well, the distance was given in kilometers, and now we're gonna be thinking about kilometers per unit time, kilometers per hour. So this is 100 kilometers, kilometers per hour. So that's the interpretation there. Let's do another example. Here we are told a tank is being drained of water. The function v gives the volume of liquid in the tank in liters after t minutes."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's the interpretation there. Let's do another example. Here we are told a tank is being drained of water. The function v gives the volume of liquid in the tank in liters after t minutes. What is the best interpretation for the following statement? The slope of the line tangent to the graph of v at t equals seven is equal to negative three. So pause this video again and try to do what we just did with the previous example."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "The function v gives the volume of liquid in the tank in liters after t minutes. What is the best interpretation for the following statement? The slope of the line tangent to the graph of v at t equals seven is equal to negative three. So pause this video again and try to do what we just did with the previous example. Write out that interpretation, and make sure to get the units right. All right, so let's just remind ourselves what's going on. V is going to give us the volume as a function of time."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video again and try to do what we just did with the previous example. Write out that interpretation, and make sure to get the units right. All right, so let's just remind ourselves what's going on. V is going to give us the volume as a function of time. Volume is in liters, and time is in minutes. And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "V is going to give us the volume as a function of time. Volume is in liters, and time is in minutes. And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three. So this, which is the same thing as the slope of tangent line slope of tangent, tangent line, and they tell us that v prime of, at time equals seven minutes, our rate of change of volume with respect to time is equal to negative three. And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. That's why we need that calculus for that instantaneous rate."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three. So this, which is the same thing as the slope of tangent line slope of tangent, tangent line, and they tell us that v prime of, at time equals seven minutes, our rate of change of volume with respect to time is equal to negative three. And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. That's why we need that calculus for that instantaneous rate. At an instantaneous rate of, now, you might be tempted to say it's being drained at an instantaneous rate of negative three liters per minute. But remember, the negative three just shows that the volume is decreasing. So one way to think about it is, this negative is already being accounted for when you're saying it's being drained."}, {"video_title": "Interpreting the meaning of the derivative in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's why we need that calculus for that instantaneous rate. At an instantaneous rate of, now, you might be tempted to say it's being drained at an instantaneous rate of negative three liters per minute. But remember, the negative three just shows that the volume is decreasing. So one way to think about it is, this negative is already being accounted for when you're saying it's being drained. If this was positive, that means it is being filled. So it is being drained at an instantaneous rate of three liters per minute. Three liters per minute."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that y is equal to log base four of x squared plus x. What is the derivative of y with respect to x going to be equal to? Now you might recognize immediately that this is a composite function. We're taking the log base four, not just of x, but we're taking that of another expression that involves x. So we could say, we could say this thing in blue, that's u of x. Let me do that in blue. So this thing in blue, that is u of x. U of x is equal to x squared plus x."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're taking the log base four, not just of x, but we're taking that of another expression that involves x. So we could say, we could say this thing in blue, that's u of x. Let me do that in blue. So this thing in blue, that is u of x. U of x is equal to x squared plus x. And it's going to be useful later on to know what u prime of x is. So that's going to be, just going to use the power rule here. So two x plus one, brought that two out front and decremented the exponent."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this thing in blue, that is u of x. U of x is equal to x squared plus x. And it's going to be useful later on to know what u prime of x is. So that's going to be, just going to use the power rule here. So two x plus one, brought that two out front and decremented the exponent. Derivative with respect to x of x is one. And we could say the log base four of this stuff, well we could call that a function v. We could say v of, well if we said v of x, this would be log base four of x. And then we've shown in other videos that v prime of x is going to be very similar to if this was log base e or natural log, except we're going to scale it."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So two x plus one, brought that two out front and decremented the exponent. Derivative with respect to x of x is one. And we could say the log base four of this stuff, well we could call that a function v. We could say v of, well if we said v of x, this would be log base four of x. And then we've shown in other videos that v prime of x is going to be very similar to if this was log base e or natural log, except we're going to scale it. So it's going to be one over, one over log base four, oh sorry, one over the natural log, the natural log of four times x. If this was v of x, if v of x was just natural log of x, our derivative would be one over x. But since it's log base four, and this comes straight out of the change of base formulas that you might have seen."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we've shown in other videos that v prime of x is going to be very similar to if this was log base e or natural log, except we're going to scale it. So it's going to be one over, one over log base four, oh sorry, one over the natural log, the natural log of four times x. If this was v of x, if v of x was just natural log of x, our derivative would be one over x. But since it's log base four, and this comes straight out of the change of base formulas that you might have seen. And we have a video where we show this. But we just scale it in the denominator with this natural log of four. Or you could think of scaling the whole expression by one over the natural log of four."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But since it's log base four, and this comes straight out of the change of base formulas that you might have seen. And we have a video where we show this. But we just scale it in the denominator with this natural log of four. Or you could think of scaling the whole expression by one over the natural log of four. But we can now use this information because y, this y can be viewed as v of, v of, remember v is the log base four of something. But it's not v of x. We don't have just an x here."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could think of scaling the whole expression by one over the natural log of four. But we can now use this information because y, this y can be viewed as v of, v of, remember v is the log base four of something. But it's not v of x. We don't have just an x here. We have the whole expression that defines u of x. We have u of x right there. And let me draw a little line here so that we don't get those two sides confused."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't have just an x here. We have the whole expression that defines u of x. We have u of x right there. And let me draw a little line here so that we don't get those two sides confused. And so we know from the chain rule, the derivative of y with respect to x, this is going to be, this is going to be the derivative of v with respect to u. Or we could call that v prime, v prime of u of x. v prime of u of x. Let me do the u of x in blue."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me draw a little line here so that we don't get those two sides confused. And so we know from the chain rule, the derivative of y with respect to x, this is going to be, this is going to be the derivative of v with respect to u. Or we could call that v prime, v prime of u of x. v prime of u of x. Let me do the u of x in blue. v prime of u of x times u prime of x. Well, what is v prime of u of x? We know what v prime of x is."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do the u of x in blue. v prime of u of x times u prime of x. Well, what is v prime of u of x? We know what v prime of x is. If we wanted to do v prime of u of x, we would just replace wherever we see an x with the u of x. So this is going to be equal to v prime of u of x. And you just use it, you're taking the derivative of the green function with respect to the blue function."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what v prime of x is. If we wanted to do v prime of u of x, we would just replace wherever we see an x with the u of x. So this is going to be equal to v prime of u of x. And you just use it, you're taking the derivative of the green function with respect to the blue function. So it's going to be one over the natural log of four, natural log of four, times, instead of putting an x there, it would be times u of x. Times u of x. And of course, that whole thing times u prime of x."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you just use it, you're taking the derivative of the green function with respect to the blue function. So it's going to be one over the natural log of four, natural log of four, times, instead of putting an x there, it would be times u of x. Times u of x. And of course, that whole thing times u prime of x. And so, and I'm doing more steps, just hopefully so it's clearer what I'm doing here. So this is one over the natural log of four. u of x is x squared plus x."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And of course, that whole thing times u prime of x. And so, and I'm doing more steps, just hopefully so it's clearer what I'm doing here. So this is one over the natural log of four. u of x is x squared plus x. So x squared plus x. And then we're going to multiply that times u prime of x. So times two x plus one."}, {"video_title": "Logarithmic functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "u of x is x squared plus x. So x squared plus x. And then we're going to multiply that times u prime of x. So times two x plus one. And so we can just rewrite this as two x plus one over, over, over the natural log of four, natural log of four, times x squared plus x. Times x squared plus x. And we're done."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And right now, right at this moment, it is 0.8 miles from the intersection. Now we have this truck over here. It's approaching the same intersection on a street that is perpendicular to the street that the car is on. And right now it is 0.6 miles. So that is 0.6 miles from the intersection and is approaching the intersection at 30 miles per hour. Now my question to you is, what is the rate at which the distance between the car and the truck is changing? Well, to think about that, let's first just think about what we're asking."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And right now it is 0.6 miles. So that is 0.6 miles from the intersection and is approaching the intersection at 30 miles per hour. Now my question to you is, what is the rate at which the distance between the car and the truck is changing? Well, to think about that, let's first just think about what we're asking. So we're asking about the distance between the car and the truck. So right at this moment when the car is 0.8 miles from the intersection, the truck is 0.6 miles from the intersection, the truck is traveling at 30 miles per hour towards the intersection, the car is traveling 60 miles per hour towards the intersection, right at this moment, what is the rate at which this distance right over here is changing? And just so that we have some variables in place, let's call this distance s. So what we really are trying to figure out is right at this moment, what is ds dt going to be equal to?"}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to think about that, let's first just think about what we're asking. So we're asking about the distance between the car and the truck. So right at this moment when the car is 0.8 miles from the intersection, the truck is 0.6 miles from the intersection, the truck is traveling at 30 miles per hour towards the intersection, the car is traveling 60 miles per hour towards the intersection, right at this moment, what is the rate at which this distance right over here is changing? And just so that we have some variables in place, let's call this distance s. So what we really are trying to figure out is right at this moment, what is ds dt going to be equal to? Well, let's think about what we know that we could use to somehow come to terms or figure out what ds dt is. Well, we know the distance of the car and the intersection. And let's just call that distance y."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just so that we have some variables in place, let's call this distance s. So what we really are trying to figure out is right at this moment, what is ds dt going to be equal to? Well, let's think about what we know that we could use to somehow come to terms or figure out what ds dt is. Well, we know the distance of the car and the intersection. And let's just call that distance y. So y is equal to 0.8 miles. We also know that d, so let me write this. We know that y is 0.8 miles right now."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's just call that distance y. So y is equal to 0.8 miles. We also know that d, so let me write this. We know that y is 0.8 miles right now. 0.8 miles. We also know that dy dt, the rate at which y is changing with respect to time, is what? Well, y is decreasing by 60 miles per hour."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that y is 0.8 miles right now. 0.8 miles. We also know that dy dt, the rate at which y is changing with respect to time, is what? Well, y is decreasing by 60 miles per hour. So let me write it as negative 60 miles per hour. Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment. So we know that x is equal to 0.6 miles."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, y is decreasing by 60 miles per hour. So let me write it as negative 60 miles per hour. Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment. So we know that x is equal to 0.6 miles. What is the rate at which x is changing with respect to time? Well, we know it's 30 miles per hour is how fast we're approaching the intersection. But x is decreasing by 30 miles every hour."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that x is equal to 0.6 miles. What is the rate at which x is changing with respect to time? Well, we know it's 30 miles per hour is how fast we're approaching the intersection. But x is decreasing by 30 miles every hour. So we should say it's negative 30 miles per hour. So we know what y is. We know what x is."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But x is decreasing by 30 miles every hour. So we should say it's negative 30 miles per hour. So we know what y is. We know what x is. We know how fast y is changing, how fast x is changing with respect to time. So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time. And it seems like we have pretty much everything we need to solve for this."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what x is. We know how fast y is changing, how fast x is changing with respect to time. So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time. And it seems like we have pretty much everything we need to solve for this. So what's the relationship between x, y, and s? Well, we know that this is a right triangle. The streets are perpendicular to each other."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it seems like we have pretty much everything we need to solve for this. So what's the relationship between x, y, and s? Well, we know that this is a right triangle. The streets are perpendicular to each other. So we can use the Pythagorean theorem. We know that x squared plus y squared is going to be equal to s squared. And then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The streets are perpendicular to each other. So we can use the Pythagorean theorem. We know that x squared plus y squared is going to be equal to s squared. And then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about. So what's the derivative of x squared with respect to time? Well, it's going to be the derivative of x squared with respect to x, which is just 2x, times the derivative of x with respect to time, times dx dt. Once again, just a chain rule."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about. So what's the derivative of x squared with respect to time? Well, it's going to be the derivative of x squared with respect to x, which is just 2x, times the derivative of x with respect to time, times dx dt. Once again, just a chain rule. Derivative of something squared with respect to the something times the derivative of the something with respect to time. And we use similar logic right over here when we want to take the derivative of y squared with respect to time. Derivative of y squared with respect to y times the derivative of y with respect to time."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, just a chain rule. Derivative of something squared with respect to the something times the derivative of the something with respect to time. And we use similar logic right over here when we want to take the derivative of y squared with respect to time. Derivative of y squared with respect to y times the derivative of y with respect to time. Now on the right hand side of this equation, we once again take the derivative with respect to time. So it's the derivative of s squared with respect to s, which is just 2s, times the derivative of s with respect to time. Once again, this is all just an application of the chain rule."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative of y squared with respect to y times the derivative of y with respect to time. Now on the right hand side of this equation, we once again take the derivative with respect to time. So it's the derivative of s squared with respect to s, which is just 2s, times the derivative of s with respect to time. Once again, this is all just an application of the chain rule. So now it looks like we know what x is. We know what dx dt is. We know what y is."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, this is all just an application of the chain rule. So now it looks like we know what x is. We know what dx dt is. We know what y is. We know what dy dt is. All we need to figure out is what s and then what ds dt is, the rate at which this distance is changing with respect to time. Well, what's s right now?"}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what y is. We know what dy dt is. All we need to figure out is what s and then what ds dt is, the rate at which this distance is changing with respect to time. Well, what's s right now? Well, we can actually use the Pythagorean theorem at this exact moment. We know that x squared, so x is 0.6, we know 0.6 squared plus y squared, 0.8 squared, is equal to s squared. Well, this is 0.36 plus 0.64 is equal to s squared."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's s right now? Well, we can actually use the Pythagorean theorem at this exact moment. We know that x squared, so x is 0.6, we know 0.6 squared plus y squared, 0.8 squared, is equal to s squared. Well, this is 0.36 plus 0.64 is equal to s squared. This is 1 is equal to s squared. And we only care about positive distances, so we have s is equal to 1 right now. So we also know what s is."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is 0.36 plus 0.64 is equal to s squared. This is 1 is equal to s squared. And we only care about positive distances, so we have s is equal to 1 right now. So we also know what s is. So let's substitute all of these numbers in and then try to solve for what we came here to do, solve for ds dt. So the rate at which 2 times x, maybe I'll do that in yellow, 2 times x, x is 0.6, is going to be 1.2 times dx dt. So that's negative 30 miles per hour."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we also know what s is. So let's substitute all of these numbers in and then try to solve for what we came here to do, solve for ds dt. So the rate at which 2 times x, maybe I'll do that in yellow, 2 times x, x is 0.6, is going to be 1.2 times dx dt. So that's negative 30 miles per hour. So times negative 30 miles per hour plus 2 times y is 1.6, times dy dt is negative 60 miles per hour. And I'm not writing the units here, but if you were to write the units, you will see that all of our distances are in miles and all of our time is within hours. So we're going to get an answer when we solve for ds dt."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's negative 30 miles per hour. So times negative 30 miles per hour plus 2 times y is 1.6, times dy dt is negative 60 miles per hour. And I'm not writing the units here, but if you were to write the units, you will see that all of our distances are in miles and all of our time is within hours. So we're going to get an answer when we solve for ds dt. That's miles per hour. But I encourage you, if you want to, to actually write out the units and see how they work out. And so this is going to be equal to 2 times s. Well, s is 1 mile, so it's just going to be 2 times ds dt, which is what we're trying to solve for."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to get an answer when we solve for ds dt. That's miles per hour. But I encourage you, if you want to, to actually write out the units and see how they work out. And so this is going to be equal to 2 times s. Well, s is 1 mile, so it's just going to be 2 times ds dt, which is what we're trying to solve for. So what do we get here on the left-hand side? So 1.2 times negative 30, that's negative 36. 1 over 5 of 30 is 6."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to 2 times s. Well, s is 1 mile, so it's just going to be 2 times ds dt, which is what we're trying to solve for. So what do we get here on the left-hand side? So 1.2 times negative 30, that's negative 36. 1 over 5 of 30 is 6. Yep, that's right. And then 1.6 times negative 60, that's going to be negative 96, is equal to 2 times ds dt, is equal to 2 times the rate at which our distance is changing with respect to time. On the left-hand side right over here, this is negative 132."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "1 over 5 of 30 is 6. Yep, that's right. And then 1.6 times negative 60, that's going to be negative 96, is equal to 2 times ds dt, is equal to 2 times the rate at which our distance is changing with respect to time. On the left-hand side right over here, this is negative 132. Negative 132 is equal to 2 times ds dt. Divide both sides by 2. We get negative 66."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "On the left-hand side right over here, this is negative 132. Negative 132 is equal to 2 times ds dt. Divide both sides by 2. We get negative 66. And now we can put our units if we want. Miles per hour is the rate at which our distance is changing with respect to time. So ds dt is negative 66 miles per hour."}, {"video_title": "Related rates Approaching cars   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get negative 66. And now we can put our units if we want. Miles per hour is the rate at which our distance is changing with respect to time. So ds dt is negative 66 miles per hour. Does it make sense that we got a negative number here? Well, sure. This distance is decreasing right at this moment as they approach the intersection."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know from our limit properties that this is going to be the same thing as the limit as x approaches zero of f of x times, times the limit limit as x approaches zero of h of x. And let's think about what each of these are. So let's first think about f of x right over here. So on f of x, as x approaches zero, notice the function itself isn't defined there, but we see when we approach from the left, we are approaching, the function seems to be approaching the value of negative one right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on f of x, as x approaches zero, notice the function itself isn't defined there, but we see when we approach from the left, we are approaching, the function seems to be approaching the value of negative one right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one. As we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "As we approach from the left, we're approaching negative one. As we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one. We can see that as we approach it from the left, we are approaching one. As we approach from the right, we are approaching one. As we approach x equals zero from the left, we approach, the function approaches one."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the limit is also equal to one. We can see that as we approach it from the left, we are approaching one. As we approach from the right, we are approaching one. As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, and is defined at x equals zero. And the limit as x approaches zero is equal to the same as the, is equal to the value of the function at that point, because this is a continuous function."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "As we approach x equals zero from the left, we approach, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, and is defined at x equals zero. And the limit as x approaches zero is equal to the same as the, is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the limit as x approaches zero is equal to the same as the, is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more. All right, so these are both, looks like continuous functions. So we have the limit as x approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do one more. All right, so these are both, looks like continuous functions. So we have the limit as x approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four, and as we approach x equals zero from the right, our function seems to be approaching four And that's also what the value of the function is at x equals zero. That makes sense because this is a continuous function."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, using our limit properties, this is going to be the same thing as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four, and as we approach x equals zero from the right, our function seems to be approaching four And that's also what the value of the function is at x equals zero. That makes sense because this is a continuous function. So the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four. And let's think about the limit of g of x as x approaches zero."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That makes sense because this is a continuous function. So the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four. And let's think about the limit of g of x as x approaches zero. So from the left, it looks like as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be g of zero. G of zero is also zero."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's think about the limit of g of x as x approaches zero. So from the left, it looks like as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be g of zero. G of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same because it's continuous. So this also is zero. But now we're in a strange situation."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same because it's continuous. So this also is zero. But now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist because we can't take four and divide it by zero. So even though the limit of h of x is x equals, as x approaches zero exists, and the limit of g of x as x approaches zero exists, we can't divide four by zero."}, {"video_title": "Limits of combined functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist because we can't take four and divide it by zero. So even though the limit of h of x is x equals, as x approaches zero exists, and the limit of g of x as x approaches zero exists, we can't divide four by zero. So this whole entire limit does not exist. Does not exist. And actually, if you were to plot h of x over g of x, if you were to plot that graph, you would see it even clearer that that limit does not exist."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But to do that, let's refresh our memory about continuity at a point. So we say that f is continuous when x is equal to c if and only if, so I'm gonna make these two-way arrows right over here, the limit of f of x as x approaches c is equal to f of c. And when we first introduced this, we said, hey, this looks a little bit technical, but it's actually pretty intuitive. Think about what's happening. The limit as x approaches c of f of x, so let's say that f of x as x approaches c is approaching some value. So if we approach from the left, we're getting to this value. If we approach from the right, we're getting this value. Well, in order for the function to be continuous, if I had to draw this function without picking up my pen, well, the value of the function at that point should be the same as the limit."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "The limit as x approaches c of f of x, so let's say that f of x as x approaches c is approaching some value. So if we approach from the left, we're getting to this value. If we approach from the right, we're getting this value. Well, in order for the function to be continuous, if I had to draw this function without picking up my pen, well, the value of the function at that point should be the same as the limit. This is really just a more rigorous way of describing this notion of not having to pick up your pencil, this notion of connectedness, that you don't have any jumps or any discontinuities of any kind. So with that out of the way, let's discuss continuity over intervals. Let me delete this really fast so I have space to work with."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, in order for the function to be continuous, if I had to draw this function without picking up my pen, well, the value of the function at that point should be the same as the limit. This is really just a more rigorous way of describing this notion of not having to pick up your pencil, this notion of connectedness, that you don't have any jumps or any discontinuities of any kind. So with that out of the way, let's discuss continuity over intervals. Let me delete this really fast so I have space to work with. So we say, so I'm gonna first talk about an open interval, and then we're gonna talk about a closed interval, because a closed interval gets a little bit more involved. So we say f is continuous over an open interval from a to b. So the parentheses instead of brackets, this shows that we're not including the endpoints."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me delete this really fast so I have space to work with. So we say, so I'm gonna first talk about an open interval, and then we're gonna talk about a closed interval, because a closed interval gets a little bit more involved. So we say f is continuous over an open interval from a to b. So the parentheses instead of brackets, this shows that we're not including the endpoints. So this would be all of the points between x equals a and x equals b, but not equaling x equals a and x equals b. So f is continuous over this open interval if and only if, if and only if, f is continuous, f is continuous over every point in, over every point in the interval. So let's do a couple of examples of that."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the parentheses instead of brackets, this shows that we're not including the endpoints. So this would be all of the points between x equals a and x equals b, but not equaling x equals a and x equals b. So f is continuous over this open interval if and only if, if and only if, f is continuous, f is continuous over every point in, over every point in the interval. So let's do a couple of examples of that. So let's say we're talking about the open interval from negative seven to negative five. Is f continuous over that interval? Let's see, we're going from negative seven to negative five, and there's a couple of ways you could do it."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do a couple of examples of that. So let's say we're talking about the open interval from negative seven to negative five. Is f continuous over that interval? Let's see, we're going from negative seven to negative five, and there's a couple of ways you could do it. There's the not-so-mathematically-rigorous way, where you could say, hey, look, if I start here, I can get all the way to negative five without having to pick up my pencil. If you wanted to do more rigorously, and you actually had the definition of the function, you might be able to do a proof that for any of these points over the interval, that the limit as x approaches any one of these points of f of x is equal to the value of the function at that point. It's harder to do when you only have a graph."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see, we're going from negative seven to negative five, and there's a couple of ways you could do it. There's the not-so-mathematically-rigorous way, where you could say, hey, look, if I start here, I can get all the way to negative five without having to pick up my pencil. If you wanted to do more rigorously, and you actually had the definition of the function, you might be able to do a proof that for any of these points over the interval, that the limit as x approaches any one of these points of f of x is equal to the value of the function at that point. It's harder to do when you only have a graph. When you only have a graph, you can only just do it by inspection and say, okay, I can go from that point to that point without picking up my pencil, so I feel pretty good about it. Now let's do another interval. Let's say the, so let me put a check mark here."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's harder to do when you only have a graph. When you only have a graph, you can only just do it by inspection and say, okay, I can go from that point to that point without picking up my pencil, so I feel pretty good about it. Now let's do another interval. Let's say the, so let me put a check mark here. That is continuous. Let's think about the interval from negative two to positive one, the open interval. So this is interesting, because the function at negative two is up here, and so if you really wanted to start at negative two, you would have to start here, and then jump immediately down as soon as you get slightly larger than negative two, and then keep going, but this is an open interval, so we're not actually concerned with what exactly happens at negative two."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say the, so let me put a check mark here. That is continuous. Let's think about the interval from negative two to positive one, the open interval. So this is interesting, because the function at negative two is up here, and so if you really wanted to start at negative two, you would have to start here, and then jump immediately down as soon as you get slightly larger than negative two, and then keep going, but this is an open interval, so we're not actually concerned with what exactly happens at negative two. We're concerned what happens when we are all the numbers larger than negative two, so we would actually start right over here, and then we would go to one, and once again, based on the intuitive, I didn't have to pick up my pen idea, this function would be continuous over this interval. So what's an example of an interval where the function would not be continuous? Well, think about the interval from, well, this is a pretty straightforward one, the open interval from three to five."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is interesting, because the function at negative two is up here, and so if you really wanted to start at negative two, you would have to start here, and then jump immediately down as soon as you get slightly larger than negative two, and then keep going, but this is an open interval, so we're not actually concerned with what exactly happens at negative two. We're concerned what happens when we are all the numbers larger than negative two, so we would actually start right over here, and then we would go to one, and once again, based on the intuitive, I didn't have to pick up my pen idea, this function would be continuous over this interval. So what's an example of an interval where the function would not be continuous? Well, think about the interval from, well, this is a pretty straightforward one, the open interval from three to five. The function is here when x is equal to three, but if we wanted to get to five, it looks like we're asymptoting, it looks like we're asymptoting up towards infinity, and we just keep on going for a very long time, and then we would have to pick up our pencil and jump over, and then we would come back down right over here, and so here, we are not continuous over that interval. So now let's think about the more, the slightly more involved interval, the slightly more involved case is when you have a closed interval. F is continuous over the closed interval from a to b, so this includes not just the points between a and b, but the endpoints as well, if and only if f is continuous over the open interval and the one-sided limits, let me write this, and the limit as x approaches a from the right of f of x is equal to f of a, and the limit as x approaches b from the left of f of x is equal to f of b."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, think about the interval from, well, this is a pretty straightforward one, the open interval from three to five. The function is here when x is equal to three, but if we wanted to get to five, it looks like we're asymptoting, it looks like we're asymptoting up towards infinity, and we just keep on going for a very long time, and then we would have to pick up our pencil and jump over, and then we would come back down right over here, and so here, we are not continuous over that interval. So now let's think about the more, the slightly more involved interval, the slightly more involved case is when you have a closed interval. F is continuous over the closed interval from a to b, so this includes not just the points between a and b, but the endpoints as well, if and only if f is continuous over the open interval and the one-sided limits, let me write this, and the limit as x approaches a from the right of f of x is equal to f of a, and the limit as x approaches b from the left of f of x is equal to f of b. Now what's going on here? Well, it's just saying that the one-sided limit when you're operating within the interval has to approach the same value as the function. So for example, if we said the closed interval from negative seven to negative five, well, this one is still reasonable, just based on the picking up your pencil thing, you don't have to pick up your pencil, and what you would do is at the endpoint, and at negative seven, this function is just plain old continuous, but if it wasn't defined over here, it could still be continuous because you would do the right-handed limit towards it, and you'd say, okay, the right-handed limit is equal to the value of the function, and then at this endpoint, at the second endpoint, you'd say, okay, the left-handed limit is equal to the function, even if it wasn't defined here, even if the two-sided limit were not defined."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "F is continuous over the closed interval from a to b, so this includes not just the points between a and b, but the endpoints as well, if and only if f is continuous over the open interval and the one-sided limits, let me write this, and the limit as x approaches a from the right of f of x is equal to f of a, and the limit as x approaches b from the left of f of x is equal to f of b. Now what's going on here? Well, it's just saying that the one-sided limit when you're operating within the interval has to approach the same value as the function. So for example, if we said the closed interval from negative seven to negative five, well, this one is still reasonable, just based on the picking up your pencil thing, you don't have to pick up your pencil, and what you would do is at the endpoint, and at negative seven, this function is just plain old continuous, but if it wasn't defined over here, it could still be continuous because you would do the right-handed limit towards it, and you'd say, okay, the right-handed limit is equal to the value of the function, and then at this endpoint, at the second endpoint, you'd say, okay, the left-handed limit is equal to the function, even if it wasn't defined here, even if the two-sided limit were not defined. And so we could actually look at an example of that. If we were looking at the interval from the closed, and you could have one side open, one side closed, but let's just do the closed interval from negative three to negative two. So notice I did not have to pick up my pencil."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, if we said the closed interval from negative seven to negative five, well, this one is still reasonable, just based on the picking up your pencil thing, you don't have to pick up your pencil, and what you would do is at the endpoint, and at negative seven, this function is just plain old continuous, but if it wasn't defined over here, it could still be continuous because you would do the right-handed limit towards it, and you'd say, okay, the right-handed limit is equal to the value of the function, and then at this endpoint, at the second endpoint, you'd say, okay, the left-handed limit is equal to the function, even if it wasn't defined here, even if the two-sided limit were not defined. And so we could actually look at an example of that. If we were looking at the interval from the closed, and you could have one side open, one side closed, but let's just do the closed interval from negative three to negative two. So notice I did not have to pick up my pencil. I'm including negative three, and I'm getting all the way to negative two. If you knew the analytic definition of this function, you could prove that, hey, the limit at any of these points inside between negative three and negative two is equal to the value of the function. Negative three, the function is clearly, at negative three, the function is just plain old continuous."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So notice I did not have to pick up my pencil. I'm including negative three, and I'm getting all the way to negative two. If you knew the analytic definition of this function, you could prove that, hey, the limit at any of these points inside between negative three and negative two is equal to the value of the function. Negative three, the function is clearly, at negative three, the function is just plain old continuous. The two-sided limit approaches the value of the function, but at negative two, the two-sided limit does not exist. When you approach from the left, looks like you're approaching zero, f of x is equal to zero. When you approach from the right, it looks like f of x is approaching negative three."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative three, the function is clearly, at negative three, the function is just plain old continuous. The two-sided limit approaches the value of the function, but at negative two, the two-sided limit does not exist. When you approach from the left, looks like you're approaching zero, f of x is equal to zero. When you approach from the right, it looks like f of x is approaching negative three. So even though the two-sided limit does not exist, we can still be good because the left-handed limit does exist, and the left-handed limit is approaching the value of the function. So we actually are continuous over that interval. But then if we did the interval, if we did the closed interval from negative two to negative two to one, pause the video and think about, based on what we just talked about, are we continuous over this interval?"}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you approach from the right, it looks like f of x is approaching negative three. So even though the two-sided limit does not exist, we can still be good because the left-handed limit does exist, and the left-handed limit is approaching the value of the function. So we actually are continuous over that interval. But then if we did the interval, if we did the closed interval from negative two to negative two to one, pause the video and think about, based on what we just talked about, are we continuous over this interval? Well, we're going from negative two to one, and negative two is the lower bound. So is this right over here, is this right over here true? Is the limit as we approach negative two from the right, is that the same thing as f of negative two?"}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then if we did the interval, if we did the closed interval from negative two to negative two to one, pause the video and think about, based on what we just talked about, are we continuous over this interval? Well, we're going from negative two to one, and negative two is the lower bound. So is this right over here, is this right over here true? Is the limit as we approach negative two from the right, is that the same thing as f of negative two? Well, the limit as we approach from the right seems to be approaching negative three, and f of negative two is zero. So this limit does not, these two things, the limit as we approach from the right and the value of the function are not the same, and so we do not have that, I guess you could say that one-sided continuity at negative two. And that also makes sense."}, {"video_title": "Continuity over an interval   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is the limit as we approach negative two from the right, is that the same thing as f of negative two? Well, the limit as we approach from the right seems to be approaching negative three, and f of negative two is zero. So this limit does not, these two things, the limit as we approach from the right and the value of the function are not the same, and so we do not have that, I guess you could say that one-sided continuity at negative two. And that also makes sense. If I start at negative two, let me do this in a color you can see. If I start at negative two and I wanna go the rest of the interval to one, I have to pick up my pencil, pick up my pencil, go here, and then keep on going. So this is, we are not continuous over that interval."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to figure it out on your own. All right, so in order to evaluate this, we need to remember the fundamental theorem of calculus, which connects the notion of a definite integral and an antiderivative. So the fundamental theorem of calculus tells us that our definite integral from a to b of f of x dx is going to be equal to the antiderivative of our function f, which we denote with a capital F, evaluated at the upper bound minus our antiderivative evaluated at the lower bound. So we just have to do that right over here. So this is going to be equal to, well, what is the antiderivative of four? Well, you might immediately say, well, that's just going to be four x. You could even think of it in terms of reverse power rule."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to do that right over here. So this is going to be equal to, well, what is the antiderivative of four? Well, you might immediately say, well, that's just going to be four x. You could even think of it in terms of reverse power rule. Four is the same thing as four x to the zero. So you increase zero by one, so it's going to be four x to the first, and then you divide by that new exponent. Four x to the first divided by one, well, that's just going to be four x."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could even think of it in terms of reverse power rule. Four is the same thing as four x to the zero. So you increase zero by one, so it's going to be four x to the first, and then you divide by that new exponent. Four x to the first divided by one, well, that's just going to be four x. So the antiderivative is four x. This is, you could say, our capital F of x. And we're going to evaluate that at five and at negative three and we're gonna find the difference between these two."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Four x to the first divided by one, well, that's just going to be four x. So the antiderivative is four x. This is, you could say, our capital F of x. And we're going to evaluate that at five and at negative three and we're gonna find the difference between these two. So what we have right over here, evaluating the antiderivative at our upper bound, that is going to be four x five. And then from that, we're going to subtract evaluating our antiderivative at the lower bound. So that's four x negative three."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to evaluate that at five and at negative three and we're gonna find the difference between these two. So what we have right over here, evaluating the antiderivative at our upper bound, that is going to be four x five. And then from that, we're going to subtract evaluating our antiderivative at the lower bound. So that's four x negative three. Four x negative three. And what is that going to be equal to? So this is 20 and then minus negative 12."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's four x negative three. Four x negative three. And what is that going to be equal to? So this is 20 and then minus negative 12. So this is going to be plus 12, which is going to be equal to 32. Let's do another example where we're going to do the reverse power rule. So let's say that we want to find the indefinite, or we want to find the definite integral going from negative one to three of seven x squared dx."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is 20 and then minus negative 12. So this is going to be plus 12, which is going to be equal to 32. Let's do another example where we're going to do the reverse power rule. So let's say that we want to find the indefinite, or we want to find the definite integral going from negative one to three of seven x squared dx. What is this going to be equal to? Well, what we want to do is evaluate what is the antiderivative of this? Or you could say, if this is lowercase f of x, what is capital F of x?"}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that we want to find the indefinite, or we want to find the definite integral going from negative one to three of seven x squared dx. What is this going to be equal to? Well, what we want to do is evaluate what is the antiderivative of this? Or you could say, if this is lowercase f of x, what is capital F of x? Well, the reverse power rule, we increase this exponent by one. So we're going to have seven times x to the third and then we divide by that increased exponent. So seven x to the third divided by three."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could say, if this is lowercase f of x, what is capital F of x? Well, the reverse power rule, we increase this exponent by one. So we're going to have seven times x to the third and then we divide by that increased exponent. So seven x to the third divided by three. And we want to evaluate that at our upper bound and then subtract from that and evaluate it at our lower bound. So this is going to be equal to, so evaluating it at our upper bound, it's going to be seven times three to the third. I'll just write that, three to the third over three."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So seven x to the third divided by three. And we want to evaluate that at our upper bound and then subtract from that and evaluate it at our lower bound. So this is going to be equal to, so evaluating it at our upper bound, it's going to be seven times three to the third. I'll just write that, three to the third over three. And then from that, we are going to subtract this capital F of x, the antiderivative, evaluated at the lower bound. So that is going to be seven times negative one to the third, all of that over three. And so this first expression, let's see, this is going to be seven times three to the third over three this is 27 over three, this is going to be the same thing as seven times nine."}, {"video_title": "Definite integrals reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll just write that, three to the third over three. And then from that, we are going to subtract this capital F of x, the antiderivative, evaluated at the lower bound. So that is going to be seven times negative one to the third, all of that over three. And so this first expression, let's see, this is going to be seven times three to the third over three this is 27 over three, this is going to be the same thing as seven times nine. So this is going to be 63. And this over here, negative one to the third power is negative one, but then we're subtracting a negative, so this is just going to be adding. And so this is just going to be plus seven over three."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "I thought I would do a few more examples of taking antiderivatives, just so we feel comfortable taking antiderivatives of all of the functions, the basic functions that we know how to take the derivatives of. And on top of that, I just want to make it clear that it doesn't always have to be functions of x. Here we have a function of t, and we're taking the antiderivative with respect to t. And so you would not write a dx here. That is not the notation. You'll see why when we focus on definite integrals. So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t. It's going to be the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative of cosine of t. Plus the antiderivative of cosine of t. So let's think about what these antiderivatives are."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is not the notation. You'll see why when we focus on definite integrals. So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t. It's going to be the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative of cosine of t. Plus the antiderivative of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. So if we want a sine of t here, we would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. Derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t?"}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's going to be the same thing as the antiderivative of sine of t. It's going to be the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative of cosine of t. Plus the antiderivative of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. So if we want a sine of t here, we would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. Derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t. So plus sine of t. And we're done. We found the antiderivative. Now let's tackle this."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t. So plus sine of t. And we're done. We found the antiderivative. Now let's tackle this. Now we don't have a t. We're taking the indefinite integral with respect to, actually, this is a mistake. This should be with respect to a. Let me clean this up."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's tackle this. Now we don't have a t. We're taking the indefinite integral with respect to, actually, this is a mistake. This should be with respect to a. Let me clean this up. This should be a dA. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me clean this up. This should be a dA. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be dA. That's what we are integrating."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "But I don't want to confuse you right now. Let me make it clear. This is going to be dA. That's what we are integrating. We're taking the antiderivative with respect to. So what is this going to be equal to? Well, once again, we can rewrite it as the sum of integrals."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's what we are integrating. We're taking the antiderivative with respect to. So what is this going to be equal to? Well, once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a dA. So this one right over here. I'll do it in green."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a dA. So this one right over here. I'll do it in green. Plus the indefinite integral, or the antiderivative, of 1 over a dA. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll do it in green. Plus the indefinite integral, or the antiderivative, of 1 over a dA. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e and the exponential function in general is so amazing. And if we just replaced a with x, or x with a, you get the derivative with respect to a of e to the a is equal to e to the a."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e and the exponential function in general is so amazing. And if we just replaced a with x, or x with a, you get the derivative with respect to a of e to the a is equal to e to the a. So the antiderivative here, the derivative of e to the a is e to the a. The antiderivative is going to be e to the a. And maybe you could shift it by some type of a constant."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we just replaced a with x, or x with a, you get the derivative with respect to a of e to the a is equal to e to the a. So the antiderivative here, the derivative of e to the a is e to the a. The antiderivative is going to be e to the a. And maybe you could shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor. So let me, always important, remember the constant."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "And maybe you could shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor. So let me, always important, remember the constant. So you have a constant factor right over here. Never forget that. I almost did."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me, always important, remember the constant. So you have a constant factor right over here. Never forget that. I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a. What's the antiderivative of 1 over a?"}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a. What's the antiderivative of 1 over a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a. And then we want to have the most general antiderivative."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's the antiderivative of 1 over a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a. And then we want to have the most general antiderivative. So there could be a constant factor out here as well. And we are done. We found the antiderivative of both of these expressions."}, {"video_title": "Second derivatives   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you're wondering where this notation comes from for a second derivative, imagine if you started with your y and you first take a derivative, and we've seen this notation before. So that would be the first derivative. Then we want to take the derivative of that. So we then want to take the derivative of that to get us our second derivative. And so that's where that notation comes from. It looks like you have a d squared, d times d, although you're not really multiplying them, you're applying the derivative operator twice. It looks like you have a dx squared."}, {"video_title": "Second derivatives   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we then want to take the derivative of that to get us our second derivative. And so that's where that notation comes from. It looks like you have a d squared, d times d, although you're not really multiplying them, you're applying the derivative operator twice. It looks like you have a dx squared. Once again, you're not multiplying them, you're just applying the operator twice. But that's where that notation actually comes from. Well, let's first take the first derivative of y with respect to x."}, {"video_title": "Second derivatives   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like you have a dx squared. Once again, you're not multiplying them, you're just applying the operator twice. But that's where that notation actually comes from. Well, let's first take the first derivative of y with respect to x. And to do that, let's just remind ourselves that we just have to apply the power rule here. And we can just remind ourselves, based on the fact that y is equal to six x to the negative two. So let's take the derivative of both sides of this with respect to x."}, {"video_title": "Second derivatives   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's first take the first derivative of y with respect to x. And to do that, let's just remind ourselves that we just have to apply the power rule here. And we can just remind ourselves, based on the fact that y is equal to six x to the negative two. So let's take the derivative of both sides of this with respect to x. So with respect to x, I'm gonna do that. And so on the left-hand side, I'm gonna have dy dx is equal to, now on the right-hand side, take our negative two, multiply it times the six, it's gonna get negative 12x to the, negative two minus one is x to the negative three. And actually, let me give myself a little bit more space here so this is negative 12x to the negative three."}, {"video_title": "Second derivatives   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's take the derivative of both sides of this with respect to x. So with respect to x, I'm gonna do that. And so on the left-hand side, I'm gonna have dy dx is equal to, now on the right-hand side, take our negative two, multiply it times the six, it's gonna get negative 12x to the, negative two minus one is x to the negative three. And actually, let me give myself a little bit more space here so this is negative 12x to the negative three. And now let's take the derivative of that with respect to x. So I'm gonna apply the derivative operator again. So the derivative with respect to x."}, {"video_title": "One-sided limits from graphs asymptote   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here we have the graph of y is equal to g of x. And what I want to do is I want to figure out the limit of g of x as x approaches positive six from values that are less than positive six, or you could say from the left, from the, you could say the negative direction. So what is this going to be equal to? And if you have a sense of it, pause the video and give a go at it. Well, to think about this, let's just take different x values that approach six from the left and look at what the values of the function are. So g of two looks like it's a little bit more than one. G of three, it's a little bit more than that."}, {"video_title": "One-sided limits from graphs asymptote   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you have a sense of it, pause the video and give a go at it. Well, to think about this, let's just take different x values that approach six from the left and look at what the values of the function are. So g of two looks like it's a little bit more than one. G of three, it's a little bit more than that. G of four looks like it's a little under two. G of five, it looks like it's around three. G of 5.5 looks like it's around five."}, {"video_title": "One-sided limits from graphs asymptote   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of three, it's a little bit more than that. G of four looks like it's a little under two. G of five, it looks like it's around three. G of 5.5 looks like it's around five. G of, let's say 5.75, looks like it's like nine. And so as x gets closer and closer to six from the left, it looks like the value of our function just becomes unbounded. It's just getting infinitely large."}, {"video_title": "One-sided limits from graphs asymptote   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "G of 5.5 looks like it's around five. G of, let's say 5.75, looks like it's like nine. And so as x gets closer and closer to six from the left, it looks like the value of our function just becomes unbounded. It's just getting infinitely large. And so in some context, you might see someone write that. Maybe this is equal to infinity. But infinity isn't, we're not talking about a specific number."}, {"video_title": "One-sided limits from graphs asymptote   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's just getting infinitely large. And so in some context, you might see someone write that. Maybe this is equal to infinity. But infinity isn't, we're not talking about a specific number. And if we're talking technically about limits, the way that we've looked at it, what is, you'll sometimes see this in some classes, but in this context, especially on the exercises on Khan Academy, we'll say that this does not exist. Not exist. This thing right over here is unbounded."}, {"video_title": "One-sided limits from graphs asymptote   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But infinity isn't, we're not talking about a specific number. And if we're talking technically about limits, the way that we've looked at it, what is, you'll sometimes see this in some classes, but in this context, especially on the exercises on Khan Academy, we'll say that this does not exist. Not exist. This thing right over here is unbounded. And this is interesting because the left-handed limit here doesn't exist, but the right-handed limit does. If I were to say the limit of g of x as x approaches six from the right-hand side, well, let's see. We have g of eight is there, g of seven is there, g of 6.5, looks like it's a little less than negative three, g of 6.01, a little even closer to negative three, g of 6.0000001 is very close to negative three."}, {"video_title": "Derivative of __   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we're going to know by the end of this video is one of the most fascinating ideas in calculus. And once again, it reinforces the idea that e is really this somewhat magical number. So we're gonna do a little bit of an exploration. Let's just pick some points on this curve of y is equal to e to the x, and think about what the slope of the tangent line is, or what the derivative looks like. And so let's say when y is equal to one, or when e to the x is equal to one, this is the case when x is equal to zero, well, the slope of the tangent line looks like it is one, which is curious, because that's exactly the value of the function at that point. What about when e to the x is equal to two, right over here? Well, here, let me do it in another color, the slope of the tangent line sure looks pretty close, sure looks pretty close to two."}, {"video_title": "Derivative of __   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just pick some points on this curve of y is equal to e to the x, and think about what the slope of the tangent line is, or what the derivative looks like. And so let's say when y is equal to one, or when e to the x is equal to one, this is the case when x is equal to zero, well, the slope of the tangent line looks like it is one, which is curious, because that's exactly the value of the function at that point. What about when e to the x is equal to two, right over here? Well, here, let me do it in another color, the slope of the tangent line sure looks pretty close, sure looks pretty close to two. What about, what about when e to the x is equal to 1 1\u20442? So that's happening right about here. Well, it sure looks like the slope of the tangent line is about 1 1\u20442."}, {"video_title": "Derivative of __   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, here, let me do it in another color, the slope of the tangent line sure looks pretty close, sure looks pretty close to two. What about, what about when e to the x is equal to 1 1\u20442? So that's happening right about here. Well, it sure looks like the slope of the tangent line is about 1 1\u20442. We could try, what happens when e to the x is equal to five? Well, the slope of the tangent line here sure does look pretty close, sure does look pretty close to five. And so, just eyeballing it, is it the case that the slope of the tangent line of e to the x is the same thing, is e to the x?"}, {"video_title": "Derivative of __   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it sure looks like the slope of the tangent line is about 1 1\u20442. We could try, what happens when e to the x is equal to five? Well, the slope of the tangent line here sure does look pretty close, sure does look pretty close to five. And so, just eyeballing it, is it the case that the slope of the tangent line of e to the x is the same thing, is e to the x? And I will tell you, and this is an amazing thing, that that is indeed true. That if I have some function, f of x, that is equal to e to the x, and if I were to take the derivative of this, this is going to be equal to e to the x as well. Or another way of saying it, the derivative with respect to x of e to the x is equal to e to the x."}, {"video_title": "Derivative of __   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so, just eyeballing it, is it the case that the slope of the tangent line of e to the x is the same thing, is e to the x? And I will tell you, and this is an amazing thing, that that is indeed true. That if I have some function, f of x, that is equal to e to the x, and if I were to take the derivative of this, this is going to be equal to e to the x as well. Or another way of saying it, the derivative with respect to x of e to the x is equal to e to the x. And that is an amazing thing. In previous lessons or courses, you've learned about ways to define e. And this could be a new one. E is the number that where if you take that number to the power x, if you define a function or expression as e to the x, it's that number where if you take the derivative of that, it's still going to be e to the x."}, {"video_title": "Derivative of __   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or another way of saying it, the derivative with respect to x of e to the x is equal to e to the x. And that is an amazing thing. In previous lessons or courses, you've learned about ways to define e. And this could be a new one. E is the number that where if you take that number to the power x, if you define a function or expression as e to the x, it's that number where if you take the derivative of that, it's still going to be e to the x. And what you're looking here, this curve, it's a curve where the value, that's y value, at any point is the same as the slope of the tangent line. If that doesn't strike you as mysterious and magical and amazing just yet, it will. Maybe tonight, you'll wake up in the middle of the night and you'll realize just what's going on."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do another example. And this time we're going to rotate our function. We're going to rotate it around a vertical line that is not the y-axis. And if we do that, so we're going to rotate y is equal to x squared minus 1, or at least this part of it, we're going to rotate it around the vertical line x is equal to negative 2. And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disk method. So what I want to do is construct some disks."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we do that, so we're going to rotate y is equal to x squared minus 1, or at least this part of it, we're going to rotate it around the vertical line x is equal to negative 2. And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disk method. So what I want to do is construct some disks. So construct some disks. So let's place one of the disks right over here. It's going to have some depth."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what I want to do is construct some disks. So construct some disks. So let's place one of the disks right over here. It's going to have some depth. And that depth is going to be dy. It's going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to have some depth. And that depth is going to be dy. It's going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disk is going to be the area as a function of y times the depth of the disk times dy. And then we just have to integrate it over the interval that we care about. And we're doing it all in terms of y."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disk is going to be the area as a function of y times the depth of the disk times dy. And then we just have to integrate it over the interval that we care about. And we're doing it all in terms of y. And in this case, we're going to integrate from y is equal to well, this is going to hit this y intercept right over here. So y is equal to negative 1. And let's go all the way to y is equal to 3. y is equal to 3 right over here."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're doing it all in terms of y. And in this case, we're going to integrate from y is equal to well, this is going to hit this y intercept right over here. So y is equal to negative 1. And let's go all the way to y is equal to 3. y is equal to 3 right over here. So from y equals negative 1 to y equals 3. So y equals negative 1 to y is equal to 3. And that's going to give us the volume of our upside down gum drop type looking thing."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's go all the way to y is equal to 3. y is equal to 3 right over here. So from y equals negative 1 to y equals 3. So y equals negative 1 to y is equal to 3. And that's going to give us the volume of our upside down gum drop type looking thing. So the key here is, so that we can start evaluating the double integral, is to just figure out what the area of each of these disks are as a function of y. And we know that area is a function of y. It's just going to be pi times radius as a function of y squared."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's going to give us the volume of our upside down gum drop type looking thing. So the key here is, so that we can start evaluating the double integral, is to just figure out what the area of each of these disks are as a function of y. And we know that area is a function of y. It's just going to be pi times radius as a function of y squared. So the real key is what is the radius as a function of y for any one of these y's. So what is the radius as a function of y? So let's think about that a little bit."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's just going to be pi times radius as a function of y squared. So the real key is what is the radius as a function of y for any one of these y's. So what is the radius as a function of y? So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. If you add 1 to both sides, then I'm going to swap sides."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. If you add 1 to both sides, then I'm going to swap sides. So you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And then you get x is equal to the principal root of the square root of y plus 1."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you add 1 to both sides, then I'm going to swap sides. So you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And then you get x is equal to the principal root of the square root of y plus 1. So this we can write as x, or we can even write it as f of y if we want. f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you get x is equal to the principal root of the square root of y plus 1. So this we can write as x, or we can even write it as f of y if we want. f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance here at any point? Well, this distance, let me make it very clear. So it's going to be our total distance in the horizontal direction."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance here at any point? Well, this distance, let me make it very clear. So it's going to be our total distance in the horizontal direction. So this first part, as we're doing it in a different color, you can't see. So this part right over here is just going to be the value of the function. It's going to give you an x value."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be our total distance in the horizontal direction. So this first part, as we're doing it in a different color, you can't see. So this part right over here is just going to be the value of the function. It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius is a function of y. Your radius as a function of y is going to be equal to the square root of y plus 1."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius is a function of y. Your radius as a function of y is going to be equal to the square root of y plus 1. This essentially will give you one of these x values when you're sitting on this curve. So x as a function of y will give you one of these x values. And then from that, you add another 2."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Your radius as a function of y is going to be equal to the square root of y plus 1. This essentially will give you one of these x values when you're sitting on this curve. So x as a function of y will give you one of these x values. And then from that, you add another 2. So plus 2. Another way of thinking about it, you get an x value here. And from that x value, you subtract out x is equal to negative 2."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then from that, you add another 2. So plus 2. Another way of thinking about it, you get an x value here. And from that x value, you subtract out x is equal to negative 2. And when you subtract x is equal to negative 2, you're adding 2 here. But hopefully this makes intuitive sense. This is the x value."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And from that x value, you subtract out x is equal to negative 2. And when you subtract x is equal to negative 2, you're adding 2 here. But hopefully this makes intuitive sense. This is the x value. Let me do this in a better color. This right over here, this distance right over here, is the x value you get when you just evaluate the function of y. But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the x value. Let me do this in a better color. This right over here, this distance right over here, is the x value you get when you just evaluate the function of y. But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation. Once again, if you just take a given y right over there, you evaluate the y, you get an x value. That x value will just give you this distance. If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation. Once again, if you just take a given y right over there, you evaluate the y, you get an x value. That x value will just give you this distance. If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius. So our radius as a function of y is this thing right over here. Substituting back into this, we can now write our definite integral for our volume. The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius. So our radius as a function of y is this thing right over here. Substituting back into this, we can now write our definite integral for our volume. The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy. I can write the pi out here. We've done this multiple times. So it's going to be square root of y plus 1 plus 2 squared."}, {"video_title": "Disc method rotating around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy. I can write the pi out here. We've done this multiple times. So it's going to be square root of y plus 1 plus 2 squared. That's our radius. Times dy. So we've set up the definite integral, and now we just have to evaluate this thing."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that we're interested in approximating what the square root of 4.36 is equal to. So we want to figure out an approximation of this, and we don't have a calculator at hand. Well, one way to think about it is we know what the square root of four is. We know that this is positive two. The principal root of four is positive two. So you say, okay, this is going to be a little bit more than two. But let's say that we want to get a little bit more accurate."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We know that this is positive two. The principal root of four is positive two. So you say, okay, this is going to be a little bit more than two. But let's say that we want to get a little bit more accurate. And so what I'm going to show you in this video is a method for doing that, for approximating the value of a function near a value where we already know the value. So what am I talking about? So let's just imagine that we had the function."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But let's say that we want to get a little bit more accurate. And so what I'm going to show you in this video is a method for doing that, for approximating the value of a function near a value where we already know the value. So what am I talking about? So let's just imagine that we had the function. We have the function f of x is equal to the square root of x, which is, of course, the same thing as x to the 1 1\u20442 power. So we know what f of two is. We know that f of two, oh, sorry, we know that f of four is."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's just imagine that we had the function. We have the function f of x is equal to the square root of x, which is, of course, the same thing as x to the 1 1\u20442 power. So we know what f of two is. We know that f of two, oh, sorry, we know that f of four is. We know that f of four is the square root of four, which is going to be equal to two, or the principal root of four, which is equal to positive two. And what we want to approximate, we want to figure out what f of, we want to figure out what f of 4.36 is equal to. This is just another way of framing the exact same question that we started off this video."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We know that f of two, oh, sorry, we know that f of four is. We know that f of four is the square root of four, which is going to be equal to two, or the principal root of four, which is equal to positive two. And what we want to approximate, we want to figure out what f of, we want to figure out what f of 4.36 is equal to. This is just another way of framing the exact same question that we started off this video. So let's just imagine our function. Let's just imagine it for a second. So let me draw some axes."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is just another way of framing the exact same question that we started off this video. So let's just imagine our function. Let's just imagine it for a second. So let me draw some axes. This is my y-axis. This is my x-axis. And let's graph y is equal to f of x."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me draw some axes. This is my y-axis. This is my x-axis. And let's graph y is equal to f of x. So let's say it looks something like this. Y equals f of x looks something like that. So that's pretty decent."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And let's graph y is equal to f of x. So let's say it looks something like this. Y equals f of x looks something like that. So that's pretty decent. All right, so that right there is y is equal to f of x. And we know f of four is equal to two. F of four is equal to two, so this is when x is equal to four."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's pretty decent. All right, so that right there is y is equal to f of x. And we know f of four is equal to two. F of four is equal to two, so this is when x is equal to four. I haven't drawn it really to scale, but hopefully this is clear enough. So that right over here is going to be two. That's f of four."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "F of four is equal to two, so this is when x is equal to four. I haven't drawn it really to scale, but hopefully this is clear enough. So that right over here is going to be two. That's f of four. And what we want to approximate is f of 4.36. So 4.36 might be right around there. And so we want to approximate this y value right over here."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That's f of four. And what we want to approximate is f of 4.36. So 4.36 might be right around there. And so we want to approximate this y value right over here. We want to approximate that right over here is f of 4.36. And once again, we're assuming we don't have a calculator at hand. So how can we do that using what we know about derivatives?"}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And so we want to approximate this y value right over here. We want to approximate that right over here is f of 4.36. And once again, we're assuming we don't have a calculator at hand. So how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to four. And then we use that linearization, that linearization to find, to approximate values local to it."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to four. And then we use that linearization, that linearization to find, to approximate values local to it. And this technique is called local linearization. So what I'm saying is let's figure out what this equation of this line is. Let's call that L of x."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we use that linearization, that linearization to find, to approximate values local to it. And this technique is called local linearization. So what I'm saying is let's figure out what this equation of this line is. Let's call that L of x. And then we can use that to, then we can evaluate that at 4.36. And hopefully that'll be a little bit easier to do than to try to figure out this right over here. So how would we do that?"}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's call that L of x. And then we can use that to, then we can evaluate that at 4.36. And hopefully that'll be a little bit easier to do than to try to figure out this right over here. So how would we do that? Well, one way to think about it, obviously there's many ways to express a line, but one way to think about it is, okay, it's going to, L of x is going to be, it's going to be, it's going to be f of four. It's going to be f of four, which is two. It's going to be f of four plus the slope, the slope at x equals four, which is of course the derivative, f prime of four."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So how would we do that? Well, one way to think about it, obviously there's many ways to express a line, but one way to think about it is, okay, it's going to, L of x is going to be, it's going to be, it's going to be f of four. It's going to be f of four, which is two. It's going to be f of four plus the slope, the slope at x equals four, which is of course the derivative, f prime of four. So that's going to be the slope of this line, of L of x is f prime of four. Let me make that clear. So this right over here is the slope, the slope when x is, at x equals four."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It's going to be f of four plus the slope, the slope at x equals four, which is of course the derivative, f prime of four. So that's going to be the slope of this line, of L of x is f prime of four. Let me make that clear. So this right over here is the slope, the slope when x is, at x equals four. So it's the slope of this entire line. And so any other point on this, it's going to be f of four plus the slope times how far you are away from x equals four. So it's going to be times x minus four."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this right over here is the slope, the slope when x is, at x equals four. So it's the slope of this entire line. And so any other point on this, it's going to be f of four plus the slope times how far you are away from x equals four. So it's going to be times x minus four. Let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So if this is, so I'm going to do a zoom in, I'm going to do a zoom in, I'm going to try to zoom in into this region right over here."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be times x minus four. Let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So if this is, so I'm going to do a zoom in, I'm going to do a zoom in, I'm going to try to zoom in into this region right over here. So this is the point, this is the point four comma f of four. And we are going to graph L of x. So let me do that."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So if this is, so I'm going to do a zoom in, I'm going to do a zoom in, I'm going to try to zoom in into this region right over here. So this is the point, this is the point four comma f of four. And we are going to graph L of x. So let me do that. So this right over here is L of x. That's L of x. And let's say this right over here, this right over here is the point 4.36 comma f of 4.36."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me do that. So this right over here is L of x. That's L of x. And let's say this right over here, this right over here is the point 4.36 comma f of 4.36. And the way we're going to approximate this value is to figure out what, to figure out what this value is right over here. And what is this one going to be? This right over here is going to be, this is going to be 4.36 comma L of 4.36."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And let's say this right over here, this right over here is the point 4.36 comma f of 4.36. And the way we're going to approximate this value is to figure out what, to figure out what this value is right over here. And what is this one going to be? This right over here is going to be, this is going to be 4.36 comma L of 4.36. This line evaluated when x is equal to 4.36. And what is that going to be? What is that going to be equal to?"}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This right over here is going to be, this is going to be 4.36 comma L of 4.36. This line evaluated when x is equal to 4.36. And what is that going to be? What is that going to be equal to? Well let's see, let's just evaluate it. L of 4.36 is going to be f of four. So it's going to be two plus the derivative, so the slope of this line, plus f prime of four times x minus four."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "What is that going to be equal to? Well let's see, let's just evaluate it. L of 4.36 is going to be f of four. So it's going to be two plus the derivative, so the slope of this line, plus f prime of four times x minus four. So 4.36 minus four is going to be times 0.36. And that makes sense, you're starting at two, and you're saying okay, my change in x, my change in x is 4.36, so my change in y is going to be my slope times that change in x to get me that value, to get me that value right over there. So let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be two plus the derivative, so the slope of this line, plus f prime of four times x minus four. So 4.36 minus four is going to be times 0.36. And that makes sense, you're starting at two, and you're saying okay, my change in x, my change in x is 4.36, so my change in y is going to be my slope times that change in x to get me that value, to get me that value right over there. So let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is. So to do that we need to figure out f prime of four. So let's go back up here. I'll try to leave, actually, I'll leave this little visualization here."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is. So to do that we need to figure out f prime of four. So let's go back up here. I'll try to leave, actually, I'll leave this little visualization here. So let's see, f prime, f prime of x is going to be 1.5 x to the negative 1.5, just using the power rule over here. So f prime of four, f prime of four is equal to 1.5 times four to the negative 1.5, which is of course equal to 1.5 times 1.5. Four to the 1.5 would be two, four to the negative 1.5 is going to be 1.5."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I'll try to leave, actually, I'll leave this little visualization here. So let's see, f prime, f prime of x is going to be 1.5 x to the negative 1.5, just using the power rule over here. So f prime of four, f prime of four is equal to 1.5 times four to the negative 1.5, which is of course equal to 1.5 times 1.5. Four to the 1.5 would be two, four to the negative 1.5 is going to be 1.5. So this is equal to 1.4. So L of, we deserve a little bit of a drum roll now, L of 4.36 is equal to f of four, it's equal to f of four, which is, let me just rewrite it, it's f of four plus f prime of four plus, gee, why am I switching to that color? Let me do the yellow."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Four to the 1.5 would be two, four to the negative 1.5 is going to be 1.5. So this is equal to 1.4. So L of, we deserve a little bit of a drum roll now, L of 4.36 is equal to f of four, it's equal to f of four, which is, let me just rewrite it, it's f of four plus f prime of four plus, gee, why am I switching to that color? Let me do the yellow. Plus f prime of four times, times, times 4.36, 4.36, let me make this actually in a new color just so we see it. So 4.36, so times 4.36 minus four, minus four. Actually, let me make all the fours one color too so you see it's the same."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do the yellow. Plus f prime of four times, times, times 4.36, 4.36, let me make this actually in a new color just so we see it. So 4.36, so times 4.36 minus four, minus four. Actually, let me make all the fours one color too so you see it's the same. So just like that. So what is this going to be? Well, this we already established is positive two."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Actually, let me make all the fours one color too so you see it's the same. So just like that. So what is this going to be? Well, this we already established is positive two. This we already established, let me do this in the yellow color. This we already established is 1.4, and this part right over here is 0.36. So this is going to be equal to two plus 1.4 times 0.36 is 0.9, or 0.09."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, this we already established is positive two. This we already established, let me do this in the yellow color. This we already established is 1.4, and this part right over here is 0.36. So this is going to be equal to two plus 1.4 times 0.36 is 0.9, or 0.09. So this is going to be equal to, this is going to be equal to 2.09. So that is our approximation, and it should be, at least based on how I graphed it, a little bit higher than the actual value of the square root of 4.36. But we could write that up here."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to two plus 1.4 times 0.36 is 0.9, or 0.09. So this is going to be equal to, this is going to be equal to 2.09. So that is our approximation, and it should be, at least based on how I graphed it, a little bit higher than the actual value of the square root of 4.36. But we could write that up here. This is going to be approximately, let me just write it this way. The square root, I'll just write it down here. So we could say the square root of 4.36, which is the same thing as f of 4.36, this is approximately equal to 2.09."}, {"video_title": "Local linearization   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But we could write that up here. This is going to be approximately, let me just write it this way. The square root, I'll just write it down here. So we could say the square root of 4.36, which is the same thing as f of 4.36, this is approximately equal to 2.09. Now, let's just say we happen to find a calculator, and just out of curiosity, let's see how good of an approximation this actually is. Let's get a calculator out. And so we want to do the square root of 4.36, and we get 2.088."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "In the last video, we established that if we say the rate of change of a population with respect to time is going to be proportional to the population, we were able to solve that differential equation and find a general solution which involves an exponential, that the population is going to be equal to some constant times e to some other constant times time. In the last video, we assumed time was days. So let's just apply this just to feel good that we can truly model population in this way. So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K?"}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right?"}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50."}, {"video_title": "Particular solution given initial conditions for population   Khan Academy.mp3", "Sentence": "Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50. And we're done. We can now write the particular solution that meets these conditions. So we can now write that our population, and I can even write our population as a function of time, is going to be equal to, is going to be equal to 100, 100 times E to the, now K is natural log of two over 50, so I'll write that."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we were to make the general statement that the limit of some function f of x as x approaches infinity is equal to three, what I wanna do in this video is show some examples of that and to show that we can keep creating more and more examples, really an infinite number of examples where that is going to be true. So for example, we could look at this graph over here. And in other videos, we'll think about why this is the case, but just think about what happens when you have very, very large x's. When you have very, very large x's, the plus five doesn't matter as much, and so it gets closer and closer to three x squared over x squared, which is equal to three. And you could see that right over here. It's graphed in this green color. And you can see, even when x is equal to 10, we're getting awfully close to three right over there."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you have very, very large x's, the plus five doesn't matter as much, and so it gets closer and closer to three x squared over x squared, which is equal to three. And you could see that right over here. It's graphed in this green color. And you can see, even when x is equal to 10, we're getting awfully close to three right over there. Let me zoom out a little bit so you see our axis. So that is three. Actually, let me draw a dotted line at the asymptote."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can see, even when x is equal to 10, we're getting awfully close to three right over there. Let me zoom out a little bit so you see our axis. So that is three. Actually, let me draw a dotted line at the asymptote. That is y is equal to three, and so you see the function's getting closer and closer as x approaches infinity. But that's not the only function that could do that. As I keep saying, there's an infinite number of functions that could do that."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me draw a dotted line at the asymptote. That is y is equal to three, and so you see the function's getting closer and closer as x approaches infinity. But that's not the only function that could do that. As I keep saying, there's an infinite number of functions that could do that. You could have this somewhat wild function that involves natural logs. That too, as x approaches infinity, it is getting closer and closer to three. It might be getting closer to three at a slightly slower rate than the one in green, but we're talking about infinity."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "As I keep saying, there's an infinite number of functions that could do that. You could have this somewhat wild function that involves natural logs. That too, as x approaches infinity, it is getting closer and closer to three. It might be getting closer to three at a slightly slower rate than the one in green, but we're talking about infinity. As x approaches infinity, this thing is approaching three. And as we've talked about in other videos, you could even have things that keep oscillating around the asymptote as long as they're getting closer and closer and closer to it as x gets larger and larger and larger. So, for example, that function right over there."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It might be getting closer to three at a slightly slower rate than the one in green, but we're talking about infinity. As x approaches infinity, this thing is approaching three. And as we've talked about in other videos, you could even have things that keep oscillating around the asymptote as long as they're getting closer and closer and closer to it as x gets larger and larger and larger. So, for example, that function right over there. Let me zoom in. So let's zoom in. Let's say when x is equal to 14, we can see that they're all approaching three."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, for example, that function right over there. Let me zoom in. So let's zoom in. Let's say when x is equal to 14, we can see that they're all approaching three. The purple one is oscillating around it. The other two are approaching three from below. But as we get much larger, let me actually zoom out a ways, and then I'll zoom in."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say when x is equal to 14, we can see that they're all approaching three. The purple one is oscillating around it. The other two are approaching three from below. But as we get much larger, let me actually zoom out a ways, and then I'll zoom in. So let's get to really large values. So, actually, even 100 isn't even that large if we're thinking about infinity. Even a trillion wouldn't be that large if we're thinking about infinity."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But as we get much larger, let me actually zoom out a ways, and then I'll zoom in. So let's get to really large values. So, actually, even 100 isn't even that large if we're thinking about infinity. Even a trillion wouldn't be that large if we're thinking about infinity. But let's go to 200. 200 is much larger than the numbers we've been looking at. And let me zoom in when x is equal to 200."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Even a trillion wouldn't be that large if we're thinking about infinity. But let's go to 200. 200 is much larger than the numbers we've been looking at. And let me zoom in when x is equal to 200. And you can see we have to zoom in an awfully lot, an awful lot, just to even see that the graphs still aren't quite stabilized around the asymptote, that they are a little bit different than the asymptote. I am really zoomed in. I mean, look at the scale."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me zoom in when x is equal to 200. And you can see we have to zoom in an awfully lot, an awful lot, just to even see that the graphs still aren't quite stabilized around the asymptote, that they are a little bit different than the asymptote. I am really zoomed in. I mean, look at the scale. This is, each of these are now a hundredth, each square. And so we've gotten much, much, much closer to the asymptote. In fact, the green function, we still can't tell the difference."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "I mean, look at the scale. This is, each of these are now a hundredth, each square. And so we've gotten much, much, much closer to the asymptote. In fact, the green function, we still can't tell the difference. You can see the calculation. This is up to three or four decimal places. We're getting awfully close to three now, but we aren't there."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "In fact, the green function, we still can't tell the difference. You can see the calculation. This is up to three or four decimal places. We're getting awfully close to three now, but we aren't there. So the green function's got there the fastest, is an argument. But the whole point of this is to emphasize the fact that there's an infinite number of functions for which you could make the statement that we made, that the limit of the function as x approaches infinity, in this case, we said that limit is going to be equal to three and I just picked three arbitrarily. This could be true for any, for any function."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're getting awfully close to three now, but we aren't there. So the green function's got there the fastest, is an argument. But the whole point of this is to emphasize the fact that there's an infinite number of functions for which you could make the statement that we made, that the limit of the function as x approaches infinity, in this case, we said that limit is going to be equal to three and I just picked three arbitrarily. This could be true for any, for any function. And I'm trying to, I didn't realize how much I had zoomed in. So let me now go back to the origin where we had our original expression. So there we have it."}, {"video_title": "Functions with same limit at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This could be true for any, for any function. And I'm trying to, I didn't realize how much I had zoomed in. So let me now go back to the origin where we had our original expression. So there we have it. Maybe I could zoom in this way. So there you have it. Limit of any of these as x approaches infinity is equal to three."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what I want to do is figure out the slope at x is equal to 1. So when x is equal to 1, as you can imagine, once we implicitly take the derivative of this, we're going to have that as a function of x and y. So it will be useful to know what y value we get to when our x is equal to 1. So let's figure that out right now. So when x is equal to 1, our relationship right over here becomes 1 squared, which is just 1, plus y minus 1 to the third power is equal to 28. Subtract 1 from both sides, you get y minus 1 to the third power is equal to 27. It looks like the numbers work out quite neatly for us."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's figure that out right now. So when x is equal to 1, our relationship right over here becomes 1 squared, which is just 1, plus y minus 1 to the third power is equal to 28. Subtract 1 from both sides, you get y minus 1 to the third power is equal to 27. It looks like the numbers work out quite neatly for us. Take the cube root of both sides, you get y minus 1 is equal to 3. Add 1 to both sides, you get y is equal to 4. So we really want to figure out the slope at the point 1, 1, 4, which is right over here."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like the numbers work out quite neatly for us. Take the cube root of both sides, you get y minus 1 is equal to 3. Add 1 to both sides, you get y is equal to 4. So we really want to figure out the slope at the point 1, 1, 4, which is right over here. When x is 1, y is 4. So we want to figure out the slope of the tangent line right over there. So let's start doing some implicit differentiation."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we really want to figure out the slope at the point 1, 1, 4, which is right over here. When x is 1, y is 4. So we want to figure out the slope of the tangent line right over there. So let's start doing some implicit differentiation. So we're going to take the derivative of both sides of this relationship or this equation, depending on how you want to view it. And so let's skip down here, past the orange. So the derivative with respect to x of x squared is going to be 2x."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's start doing some implicit differentiation. So we're going to take the derivative of both sides of this relationship or this equation, depending on how you want to view it. And so let's skip down here, past the orange. So the derivative with respect to x of x squared is going to be 2x. And the derivative with respect to x of something to the third power is going to be 3 times that something squared times the derivative of that something with respect to x. So what's the derivative of this with respect to x? Well, the derivative of y with respect to x is just dy dx."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative with respect to x of x squared is going to be 2x. And the derivative with respect to x of something to the third power is going to be 3 times that something squared times the derivative of that something with respect to x. So what's the derivative of this with respect to x? Well, the derivative of y with respect to x is just dy dx. And then the derivative of x with respect to x is just 1. So we have minus 1. And on the right-hand side, we just get 0."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the derivative of y with respect to x is just dy dx. And then the derivative of x with respect to x is just 1. So we have minus 1. And on the right-hand side, we just get 0. Derivative of a constant is just equal to 0. Now we need to solve for dy dx. So we get 2x."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And on the right-hand side, we just get 0. Derivative of a constant is just equal to 0. Now we need to solve for dy dx. So we get 2x. And so if we distribute this business times the dy dx and times the negative 1, when we multiply it times the dy dx, we get, and actually I'm going to write it over here, so we get plus 3 times y minus x squared times dy dx. And then when we multiply it times the negative 1, we get negative 3 times y minus x squared. And then, of course, all of that is going to be equal to 0."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get 2x. And so if we distribute this business times the dy dx and times the negative 1, when we multiply it times the dy dx, we get, and actually I'm going to write it over here, so we get plus 3 times y minus x squared times dy dx. And then when we multiply it times the negative 1, we get negative 3 times y minus x squared. And then, of course, all of that is going to be equal to 0. Now all we have to do is take this and put it on the right-hand side. So we'll subtract it from both sides of this equation. So on the left-hand side, and actually all the stuff that's not a dy dx I'm going to write in green."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then, of course, all of that is going to be equal to 0. Now all we have to do is take this and put it on the right-hand side. So we'll subtract it from both sides of this equation. So on the left-hand side, and actually all the stuff that's not a dy dx I'm going to write in green. So on the left-hand side, we're just left with 3 times y minus x squared times dy dx. dy, the derivative of y with respect to x, is equal to, I'm just going to subtract this from both sides, is equal to negative 2x plus this. So I could write it as 3 times y minus x squared plus, sorry, minus 2x."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So on the left-hand side, and actually all the stuff that's not a dy dx I'm going to write in green. So on the left-hand side, we're just left with 3 times y minus x squared times dy dx. dy, the derivative of y with respect to x, is equal to, I'm just going to subtract this from both sides, is equal to negative 2x plus this. So I could write it as 3 times y minus x squared plus, sorry, minus 2x. So we're adding this to both sides, and we're subtracting this from both sides, minus 2x. And then to solve for dy dx, we've done this multiple times already, to solve for the derivative of y with respect to x. The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x, all of that over this stuff, 3 times y minus x squared."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I could write it as 3 times y minus x squared plus, sorry, minus 2x. So we're adding this to both sides, and we're subtracting this from both sides, minus 2x. And then to solve for dy dx, we've done this multiple times already, to solve for the derivative of y with respect to x. The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x, all of that over this stuff, 3 times y minus x squared. And we can leave it just like that for now. So what is the derivative of y with respect to x? What is the slope of the tangent line when x is 1 and y is equal to 4?"}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x, all of that over this stuff, 3 times y minus x squared. And we can leave it just like that for now. So what is the derivative of y with respect to x? What is the slope of the tangent line when x is 1 and y is equal to 4? Well, we just have to substitute x is equal to 1 and y equals 4 into this expression. So it's going to be equal to 3 times 4 minus 1 squared minus 2 times 1, all of that over 3 times 4 minus 1 squared, which is equal to 4 minus 1 is 3. You square it, you get 9."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the slope of the tangent line when x is 1 and y is equal to 4? Well, we just have to substitute x is equal to 1 and y equals 4 into this expression. So it's going to be equal to 3 times 4 minus 1 squared minus 2 times 1, all of that over 3 times 4 minus 1 squared, which is equal to 4 minus 1 is 3. You square it, you get 9. 9 times 3 is 27. You get 27 minus 2 in the numerator, which is going to be equal to 25. And then in the denominator, you get 3 times 9, which is 27."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "You square it, you get 9. 9 times 3 is 27. You get 27 minus 2 in the numerator, which is going to be equal to 25. And then in the denominator, you get 3 times 9, which is 27. So the slope is 25 27ths. So it's almost 1, but not quite. And that's actually what it looks like on this graph."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then in the denominator, you get 3 times 9, which is 27. So the slope is 25 27ths. So it's almost 1, but not quite. And that's actually what it looks like on this graph. And actually, just to make sure you know where I got this graph, this was from Wolfram Alpha. Wolfram Alpha. I should have told you that from the beginning."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we have here, we have the graph of f of x is equal to one plus 0.1 x squared. That's this curve right over here. And then we have these rectangles that are trying to approximate the area under the curve, the area under the function f between x equals 0 and x equals 8. And the way that this diagram, or the way that we're attempting to do it, is by splitting it into four rectangles. And so we could call this rectangle one, this is rectangle two, rectangle three, and rectangle four. And each of their heights, let's see, the interval looks like they each have a width of two, so they're equally spaced. So we go from zero to eight, and we split into four sections, so each has a width of two."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the way that this diagram, or the way that we're attempting to do it, is by splitting it into four rectangles. And so we could call this rectangle one, this is rectangle two, rectangle three, and rectangle four. And each of their heights, let's see, the interval looks like they each have a width of two, so they're equally spaced. So we go from zero to eight, and we split into four sections, so each has a width of two. So they're each going to be two wide. So that's two, that's two, that's two, that's two. And their height seems to be based on, their height seems to be based on the midpoint."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we go from zero to eight, and we split into four sections, so each has a width of two. So they're each going to be two wide. So that's two, that's two, that's two, that's two. And their height seems to be based on, their height seems to be based on the midpoint. So between the start, between the left side and the right side of the rectangle, you take the value of the function at at the middle value right over here. So for example, this height right over here looks like f of one. This height right over here looks like f of three."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And their height seems to be based on, their height seems to be based on the midpoint. So between the start, between the left side and the right side of the rectangle, you take the value of the function at at the middle value right over here. So for example, this height right over here looks like f of one. This height right over here looks like f of three. This height of this rectangle is f of five. This height right over here is f of seven. So given the way that this has been constructed, and we want to take the sum of the areas of these rectangles as an approximation under, as the area under this curve, how would we write that as sigma notation?"}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "This height right over here looks like f of three. This height of this rectangle is f of five. This height right over here is f of seven. So given the way that this has been constructed, and we want to take the sum of the areas of these rectangles as an approximation under, as the area under this curve, how would we write that as sigma notation? And I'll get us started, and then I encourage you to pause the video and try to finish it. So the sum of these, the sum of these rectangles, we could say it's the sum of, so we'll have n equals one to four, because we have four rectangles. And I encourage you to, and I encourage you to, to finish this up."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So given the way that this has been constructed, and we want to take the sum of the areas of these rectangles as an approximation under, as the area under this curve, how would we write that as sigma notation? And I'll get us started, and then I encourage you to pause the video and try to finish it. So the sum of these, the sum of these rectangles, we could say it's the sum of, so we'll have n equals one to four, because we have four rectangles. And I encourage you to, and I encourage you to, to finish this up. Once, actually just write it in terms of the function, use function notation. You don't have to write it out as one plus point oh one times something squared. So I'm assuming you've had a go at it."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to, and I encourage you to, to finish this up. Once, actually just write it in terms of the function, use function notation. You don't have to write it out as one plus point oh one times something squared. So I'm assuming you've had a go at it. So for each of these, so for the first, so for the first rectangle over here, we're going to multiply two times the height. So the height right over here is one, and this is the first rectangle, so you might be tempted to say times f of n. But then that breaks down as we go into the second rectangle. The second rectangle, the two still applies."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it. So for each of these, so for the first, so for the first rectangle over here, we're going to multiply two times the height. So the height right over here is one, and this is the first rectangle, so you might be tempted to say times f of n. But then that breaks down as we go into the second rectangle. The second rectangle, the two still applies. This two is the width of the rectangle. But now we want to multiply it times f of three, not f of two. So this f of n isn't going to, this f of n isn't going to, isn't going to pass muster."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "The second rectangle, the two still applies. This two is the width of the rectangle. But now we want to multiply it times f of three, not f of two. So this f of n isn't going to, this f of n isn't going to, isn't going to pass muster. And so let's see, let's see how we want to think about it. So when n is, so when n is one, two, three, four, we're gonna take f of, so f of n, or I should say f of n, we're gonna take f of something. So here, this first one, we're gonna take f of one."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this f of n isn't going to, this f of n isn't going to, isn't going to pass muster. And so let's see, let's see how we want to think about it. So when n is, so when n is one, two, three, four, we're gonna take f of, so f of n, or I should say f of n, we're gonna take f of something. So here, this first one, we're gonna take f of one. Then here, for the second rectangle, we're taking f of three. f of three for the height. For the third rectangle, we're taking f of five."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here, this first one, we're gonna take f of one. Then here, for the second rectangle, we're taking f of three. f of three for the height. For the third rectangle, we're taking f of five. f of five. And then, so for the fourth rectangle, we're taking f of seven. f of seven."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "For the third rectangle, we're taking f of five. f of five. And then, so for the fourth rectangle, we're taking f of seven. f of seven. So what's the relationship over here? Let's see, it looks like if you multiply by two and subtract one, so two times one minus one is one, two times two minus one is three. Two times three minus five, two times three minus one is five."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "f of seven. So what's the relationship over here? Let's see, it looks like if you multiply by two and subtract one, so two times one minus one is one, two times two minus one is three. Two times three minus five, two times three minus one is five. Two times four minus one is seven. So this is two n minus one. So the area of each of these rectangles, the base is two, and the height is f of two n minus, f of two n minus one."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two times three minus five, two times three minus one is five. Two times four minus one is seven. So this is two n minus one. So the area of each of these rectangles, the base is two, and the height is f of two n minus, f of two n minus one. So that's, that hopefully makes a little bit clearer, kind of mapping between the sigma notation and what we're actually trying to do. And now let's just, just for fun, let's actually try to evaluate this thing. What is this thing going to evaluate to?"}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area of each of these rectangles, the base is two, and the height is f of two n minus, f of two n minus one. So that's, that hopefully makes a little bit clearer, kind of mapping between the sigma notation and what we're actually trying to do. And now let's just, just for fun, let's actually try to evaluate this thing. What is this thing going to evaluate to? Well, this is going to evaluate to two times f of, when n is equal to one, this is one, f of one, plus two times, when n is two, this is going to be f of two times two minus one is three, f of three. When n is three, this is going to be two times f of five. When n is four, this is going to be two times, two times f of seven."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is this thing going to evaluate to? Well, this is going to evaluate to two times f of, when n is equal to one, this is one, f of one, plus two times, when n is two, this is going to be f of two times two minus one is three, f of three. When n is three, this is going to be two times f of five. When n is four, this is going to be two times, two times f of seven. Four times two minus one is seven. f of seven. And so that is going to be, we're going to have to evaluate a bunch of these things over here."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "When n is four, this is going to be two times, two times f of seven. Four times two minus one is seven. f of seven. And so that is going to be, we're going to have to evaluate a bunch of these things over here. So let me actually, let me clear this out so I have a little bit more real estate. I'm feeling this might get a little bit messy now. So this is going to be, actually we could factor out a two."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that is going to be, we're going to have to evaluate a bunch of these things over here. So let me actually, let me clear this out so I have a little bit more real estate. I'm feeling this might get a little bit messy now. So this is going to be, actually we could factor out a two. So this is going to be equal to two times, f of one is one plus 0.1 times one squared. So it's one plus 0.1, so it's one point, let me color code a little bit so we can keep track of things. So this right over here is one point one."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be, actually we could factor out a two. So this is going to be equal to two times, f of one is one plus 0.1 times one squared. So it's one plus 0.1, so it's one point, let me color code a little bit so we can keep track of things. So this right over here is one point one. So one plus 0.1 is 1.1. This right over here, f of three, so it's one plus 0.1 times three squared, nine. So one plus 0.9."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is one point one. So one plus 0.1 is 1.1. This right over here, f of three, so it's one plus 0.1 times three squared, nine. So one plus 0.9. So one plus, so it's 1.9. And then let's see, this one right over here, f of five, this is going to be one plus, see five squared is 25 times 0.1 is 2.5. So one plus 2.5 is going to be 3.5."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one plus 0.9. So one plus, so it's 1.9. And then let's see, this one right over here, f of five, this is going to be one plus, see five squared is 25 times 0.1 is 2.5. So one plus 2.5 is going to be 3.5. And then finally f of seven, f of seven is going to be one plus 0.1 times seven squared. So this is 49 times 0.1. 49 times 0.1 is 4.9 plus one, so plus 5.9."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one plus 2.5 is going to be 3.5. And then finally f of seven, f of seven is going to be one plus 0.1 times seven squared. So this is 49 times 0.1. 49 times 0.1 is 4.9 plus one, so plus 5.9. And so what is this going to be equal to? So let's see, 1.1 plus 1.9, these two are going to sum up to be equal to three. And then these two are going to sum up to be, let's see, if we add the five, we get to 8.5, and then we add the 0.9, we get to 9.4."}, {"video_title": "Worked example Riemann sums in summation notation   AP Calculus AB   Khan Academy.mp3", "Sentence": "49 times 0.1 is 4.9 plus one, so plus 5.9. And so what is this going to be equal to? So let's see, 1.1 plus 1.9, these two are going to sum up to be equal to three. And then these two are going to sum up to be, let's see, if we add the five, we get to 8.5, and then we add the 0.9, we get to 9.4. So plus 9.4, did I do that right? Three plus five is 8.5 plus 0.9 is 1.4, yep. And so this is going to be, so once again we have the two times it all, so this is going to be equal to two times 12.4, which is equal to 24.8, which is our approximation."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we want to evaluate the definite integral from negative four to zero of f of x dx. And like always, pause this video and see if you could work through this. Now when you first do this, you might stumble around a little bit because how do you take the antiderivative of an absolute value function? And the key here is to, one way to approach it, is to rewrite f of x without the absolute value and we can do that by rewriting it as a piecewise function. And the way I'm gonna do it, I'm gonna think about intervals where whatever we take inside the absolute value is going to be positive, and other intervals where everything that we take inside the absolute value is going to be negative. And the point at which we change is where x plus two is equal to zero or x is equal to negative two. So let's just think about the intervals x is less than negative two and x is greater than or equal to negative two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the key here is to, one way to approach it, is to rewrite f of x without the absolute value and we can do that by rewriting it as a piecewise function. And the way I'm gonna do it, I'm gonna think about intervals where whatever we take inside the absolute value is going to be positive, and other intervals where everything that we take inside the absolute value is going to be negative. And the point at which we change is where x plus two is equal to zero or x is equal to negative two. So let's just think about the intervals x is less than negative two and x is greater than or equal to negative two. And this could have been less than or equal, in which case this would have been greater than. Either way, it would have been equal to this absolute value. This is a continuous function here."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just think about the intervals x is less than negative two and x is greater than or equal to negative two. And this could have been less than or equal, in which case this would have been greater than. Either way, it would have been equal to this absolute value. This is a continuous function here. And so when, well let's do the easier case. When x is greater than or equal to negative two, then x plus two is going to be positive or it's going to be greater than or equal to zero and so the absolute value of it is just going to be x plus two. So it's going to be x plus two when x is greater than or equal to negative two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is a continuous function here. And so when, well let's do the easier case. When x is greater than or equal to negative two, then x plus two is going to be positive or it's going to be greater than or equal to zero and so the absolute value of it is just going to be x plus two. So it's going to be x plus two when x is greater than or equal to negative two. And what about when x is less than negative two? Well when x is less than negative two, x plus two is going to be negative and then if you take the absolute value of a negative number, you're going to take the opposite of it. So this is going to be negative x plus two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be x plus two when x is greater than or equal to negative two. And what about when x is less than negative two? Well when x is less than negative two, x plus two is going to be negative and then if you take the absolute value of a negative number, you're going to take the opposite of it. So this is going to be negative x plus two. And to really help grok this, because frankly this is the hardest part of what we're doing and really this is more algebra than calculus, let me draw the absolute value function to make this clear. So that is my x-axis. That is my y-axis."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be negative x plus two. And to really help grok this, because frankly this is the hardest part of what we're doing and really this is more algebra than calculus, let me draw the absolute value function to make this clear. So that is my x-axis. That is my y-axis. And let's say we're here at negative two. And so when we are less than negative two, when x is less than negative two, my graph is going to look like this. It is going to, it is going to look, look something, let's see what is, it's going to look like that."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is my y-axis. And let's say we're here at negative two. And so when we are less than negative two, when x is less than negative two, my graph is going to look like this. It is going to, it is going to look, look something, let's see what is, it's going to look like that. And when we are greater than negative two, use that in a different color, when we are greater than negative two, it's going to look like this. It's going to look like that. And so notice, this is in blue, we have, this is the graph x plus two, we can say this is a graph of y equals x plus two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is going to, it is going to look, look something, let's see what is, it's going to look like that. And when we are greater than negative two, use that in a different color, when we are greater than negative two, it's going to look like this. It's going to look like that. And so notice, this is in blue, we have, this is the graph x plus two, we can say this is a graph of y equals x plus two. And what we have in magenta right over here, this is the graph of negative x minus two. It has a negative slope and we intercept the y-axis at negative two. So it makes sense."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so notice, this is in blue, we have, this is the graph x plus two, we can say this is a graph of y equals x plus two. And what we have in magenta right over here, this is the graph of negative x minus two. It has a negative slope and we intercept the y-axis at negative two. So it makes sense. There's multiple ways that you could reason through this. Now once we break it up, then we can break up the integral. We could say that what we wrote here, this is equal to the integral from negative four to two, sorry, negative four to negative two of f of x, which is in that case, it's going to be negative x minus two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it makes sense. There's multiple ways that you could reason through this. Now once we break it up, then we can break up the integral. We could say that what we wrote here, this is equal to the integral from negative four to two, sorry, negative four to negative two of f of x, which is in that case, it's going to be negative x minus two. I just distributed the negative sign there. Dx, and then plus, plus the definite integral going from negative two to zero of x plus two dx. And just to make sure we know what we're doing here, this, if this is negative four right over here, this is zero, that first integral is going to give us, is going to give us this area right over here."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could say that what we wrote here, this is equal to the integral from negative four to two, sorry, negative four to negative two of f of x, which is in that case, it's going to be negative x minus two. I just distributed the negative sign there. Dx, and then plus, plus the definite integral going from negative two to zero of x plus two dx. And just to make sure we know what we're doing here, this, if this is negative four right over here, this is zero, that first integral is going to give us, is going to give us this area right over here. What's the area under the curve negative x minus two, under that curve or under that line and above the x-axis? And the second integral is gonna give us this area right over here between x plus two and the x-axis going from negative two to zero. And so let's evaluate each of these."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just to make sure we know what we're doing here, this, if this is negative four right over here, this is zero, that first integral is going to give us, is going to give us this area right over here. What's the area under the curve negative x minus two, under that curve or under that line and above the x-axis? And the second integral is gonna give us this area right over here between x plus two and the x-axis going from negative two to zero. And so let's evaluate each of these. And you might even be able to just evaluate these with a little bit of triangle areas, but let's just do this analytically or algebraically. And so what's the antiderivative of negative x? Well, that's negative x squared over two, and then we have the negative two, or so the antiderivative is negative two x."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's evaluate each of these. And you might even be able to just evaluate these with a little bit of triangle areas, but let's just do this analytically or algebraically. And so what's the antiderivative of negative x? Well, that's negative x squared over two, and then we have the negative two, or so the antiderivative is negative two x. We're gonna evaluate that at negative, at negative two and negative four. And so that part is going to be what? Negative two squared, so it's the negative of negative two squared."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's negative x squared over two, and then we have the negative two, or so the antiderivative is negative two x. We're gonna evaluate that at negative, at negative two and negative four. And so that part is going to be what? Negative two squared, so it's the negative of negative two squared. So it's negative four over two minus two times negative two, so plus four. So that's it evaluated at negative two. And then minus, if we evaluated it at negative four, so minus, so we're gonna have minus, negative four squared is 16 over two minus two times negative four."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative two squared, so it's the negative of negative two squared. So it's negative four over two minus two times negative two, so plus four. So that's it evaluated at negative two. And then minus, if we evaluated it at negative four, so minus, so we're gonna have minus, negative four squared is 16 over two minus two times negative four. So that is plus eight. So what is that? What is that going to give us?"}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then minus, if we evaluated it at negative four, so minus, so we're gonna have minus, negative four squared is 16 over two minus two times negative four. So that is plus eight. So what is that? What is that going to give us? So this is negative two. This right over here is negative eight. So the second term right over here is just going to be equal to zero."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is that going to give us? So this is negative two. This right over here is negative eight. So the second term right over here is just going to be equal to zero. Did I do that right? Yeah, 16 over two, it's negative, and it's plus, okay, so this is just going to be zero. And this is negative two plus four, which is going to be equal to two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the second term right over here is just going to be equal to zero. Did I do that right? Yeah, 16 over two, it's negative, and it's plus, okay, so this is just going to be zero. And this is negative two plus four, which is going to be equal to two. So what we have here in magenta is equal to two. And what we have here in blue, well, let's see, this is the antiderivative of x squared over two plus two x. We're gonna evaluate it at zero and negative two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is negative two plus four, which is going to be equal to two. So what we have here in magenta is equal to two. And what we have here in blue, well, let's see, this is the antiderivative of x squared over two plus two x. We're gonna evaluate it at zero and negative two. You evaluate this thing at zero, it's just gonna be zero. And from that, you're going to subtract negative two squared over two. That is positive four over two, which is positive two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna evaluate it at zero and negative two. You evaluate this thing at zero, it's just gonna be zero. And from that, you're going to subtract negative two squared over two. That is positive four over two, which is positive two. And then plus two times negative two. So minus, minus four. And so this is going to be, this is going to be the negative of negative two, or positive two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is positive four over two, which is positive two. And then plus two times negative two. So minus, minus four. And so this is going to be, this is going to be the negative of negative two, or positive two. So it's two plus two, and that makes sense. That what we have in magenta here is two, and what we have over here is two. There's a symmetry here."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be, this is going to be the negative of negative two, or positive two. So it's two plus two, and that makes sense. That what we have in magenta here is two, and what we have over here is two. There's a symmetry here. There is a symmetry here. And so you add them all together, and you get our integral is going to be equal to four. And once again, just as a reality check, you could say, look, the height here is two, the width, the base here is two."}, {"video_title": "Definite integral of absolute value function   AP Calculus AB   Khan Academy.mp3", "Sentence": "There's a symmetry here. There is a symmetry here. And so you add them all together, and you get our integral is going to be equal to four. And once again, just as a reality check, you could say, look, the height here is two, the width, the base here is two. Two times two times 1 1\u20442 is indeed equal to two. Same thing over here. So that's the more geometric argument for why that area is two, that area is two."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to rotate it around the vertical line x is equal to 2. We're going to rotate it right around like that. And we will get, if we rotate it like that, we will get this shape, so this strange-looking shape. It's hollowed out in the middle. You can kind of use the y equals x squared part kind of hollows out the middle, and it's really the stuff in between that forms the wall of this kind of rotated shape. So let's think about how we can figure out the volume. And we're going to do it using the disk method, sometimes called the ring method."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's hollowed out in the middle. You can kind of use the y equals x squared part kind of hollows out the middle, and it's really the stuff in between that forms the wall of this kind of rotated shape. So let's think about how we can figure out the volume. And we're going to do it using the disk method, sometimes called the ring method. Actually, it's going to be more of the washer method to do this one right over here. So we're rotating around a vertical line. We want to use our disk or ring or washer method."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to do it using the disk method, sometimes called the ring method. Actually, it's going to be more of the washer method to do this one right over here. So we're rotating around a vertical line. We want to use our disk or ring or washer method. And so it will be really helpful to have a bunch of rings stacked up in the y direction. So we're probably going to want to integrate with respect to y. Let me make it clear what I'm talking about."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "We want to use our disk or ring or washer method. And so it will be really helpful to have a bunch of rings stacked up in the y direction. So we're probably going to want to integrate with respect to y. Let me make it clear what I'm talking about. Each of our rings are going to look something like this. So let me do my best attempt to draw. So that's the inner radius of the ring defined by y is equal to x squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make it clear what I'm talking about. Each of our rings are going to look something like this. So let me do my best attempt to draw. So that's the inner radius of the ring defined by y is equal to x squared. That's the inner radius. And then the outside radius of the ring might look something like this. So the outside radius, my best attempt to draw it reasonably, the outside radius of the ring might look something like that."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's the inner radius of the ring defined by y is equal to x squared. That's the inner radius. And then the outside radius of the ring might look something like this. So the outside radius, my best attempt to draw it reasonably, the outside radius of the ring might look something like that. And the depth of the ring would be a little dy, just like that. So let me draw the depth. So let me see, right over here."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the outside radius, my best attempt to draw it reasonably, the outside radius of the ring might look something like that. And the depth of the ring would be a little dy, just like that. So let me draw the depth. So let me see, right over here. You can kind of view it as the height of the ring because we are thinking in a vertical direction. So we have a ring for a given y, for say this y right over here. It's essentially the ring generated if you were to take this rectangle, this rectangle of height dy and rotate it around the line x equals 2."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me see, right over here. You can kind of view it as the height of the ring because we are thinking in a vertical direction. So we have a ring for a given y, for say this y right over here. It's essentially the ring generated if you were to take this rectangle, this rectangle of height dy and rotate it around the line x equals 2. And we're going to construct a ring like that for each of the y's in our interval. So you can imagine stacking up a whole set of rings just like this and then taking the sum of the volumes of all of those rings and the limit of that sum as you have an infinite number of rings with, or close to infinite, or really infinite number of rings with infinitesimally small height or depth or dy. So let's figure out how we would do this."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's essentially the ring generated if you were to take this rectangle, this rectangle of height dy and rotate it around the line x equals 2. And we're going to construct a ring like that for each of the y's in our interval. So you can imagine stacking up a whole set of rings just like this and then taking the sum of the volumes of all of those rings and the limit of that sum as you have an infinite number of rings with, or close to infinite, or really infinite number of rings with infinitesimally small height or depth or dy. So let's figure out how we would do this. Well, we know that all we have to do is figure out what the volume of one of these rings are for a given y as a function of y and then integrate along all of them, sum them all up. So let's figure out what the volume of one of these rings are. And to do it, we're going to express these functions as functions of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's figure out how we would do this. Well, we know that all we have to do is figure out what the volume of one of these rings are for a given y as a function of y and then integrate along all of them, sum them all up. So let's figure out what the volume of one of these rings are. And to do it, we're going to express these functions as functions of y. So our purple function, y is equal to square root of x. If we square both sides, we would get y squared is equal to x. Let me swap the sides so it makes it clear that now x is a function of y. x is equal to y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to do it, we're going to express these functions as functions of y. So our purple function, y is equal to square root of x. If we square both sides, we would get y squared is equal to x. Let me swap the sides so it makes it clear that now x is a function of y. x is equal to y squared. That's our top function, the way we've drawn it, the kind of outer shell for our figure. And then y is equal to x squared. If you take the principal root of both sides of that, it all works out because we're operating in the first quadrant here."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me swap the sides so it makes it clear that now x is a function of y. x is equal to y squared. That's our top function, the way we've drawn it, the kind of outer shell for our figure. And then y is equal to x squared. If you take the principal root of both sides of that, it all works out because we're operating in the first quadrant here. That's the part of it that we care about. So you're going to get x is equal to the square root of y. x is equal to square root of y. That is our yellow function right over there."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you take the principal root of both sides of that, it all works out because we're operating in the first quadrant here. That's the part of it that we care about. So you're going to get x is equal to the square root of y. x is equal to square root of y. That is our yellow function right over there. Now, how do we figure out the area of the surface of one of these rings or one of these washers? Well, the area, let me do this in orange because I drew that ring in orange. So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is our yellow function right over there. Now, how do we figure out the area of the surface of one of these rings or one of these washers? Well, the area, let me do this in orange because I drew that ring in orange. So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius. And then I subtract out the area of the circle constructed by the inner radius. Just kind of subtract it out. So the outer circle radius, so it's going to be pi times outer radius squared minus pi times inner radius squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius. And then I subtract out the area of the circle constructed by the inner radius. Just kind of subtract it out. So the outer circle radius, so it's going to be pi times outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the outer circle radius, so it's going to be pi times outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize. Actually, I'll try it both places. So it's this entire distance right over here. Essentially, the distance between the vertical line, the horizontal distance between our vertical line and our outer function, the horizontal distance between our vertical line and our outer function."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it might be easier to visualize. Actually, I'll try it both places. So it's this entire distance right over here. Essentially, the distance between the vertical line, the horizontal distance between our vertical line and our outer function, the horizontal distance between our vertical line and our outer function. So if you think about it in terms of x, it's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this as a function of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Essentially, the distance between the vertical line, the horizontal distance between our vertical line and our outer function, the horizontal distance between our vertical line and our outer function. So if you think about it in terms of x, it's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this as a function of y. So our outer radius, this whole distance is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, we want this as a function of y. So our outer radius, this whole distance is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus y squared. We want it as a function of y. 2 minus y squared. And then our inner radius is going to be equal to what?"}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the outer radius is 2 minus y squared. We want it as a function of y. 2 minus y squared. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between 2 and whatever x value this is. But this x value as a function of y is just the square root of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between 2 and whatever x value this is. But this x value as a function of y is just the square root of y. So it's going to be 2 minus the square root of y. And so now we can come up with an expression for area. It's going to be, and I'll just factor out, I'll leave the pi there."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this x value as a function of y is just the square root of y. So it's going to be 2 minus the square root of y. And so now we can come up with an expression for area. It's going to be, and I'll just factor out, I'll leave the pi there. So it's going to be pi, I'll write it over here. It's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be, and I'll just factor out, I'll leave the pi there. So it's going to be pi, I'll write it over here. It's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared. And let me just, well, I'll just write it, 2 minus y squared. And we're going to square that. Squared minus pi times the inner radius squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the outer radius is 2 minus y squared. And let me just, well, I'll just write it, 2 minus y squared. And we're going to square that. Squared minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one too."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Squared minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one too. So this gives us the area of one of our rings as a function of y, the top of the ring where I shaded in orange. And now if we want the volume of one of those rings, we have to multiply it by its depth or its height, the way we've drawn it right over here. And its height, we've done this multiple times already, right over here, is an infinitesimal change in y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to square that one too. So this gives us the area of one of our rings as a function of y, the top of the ring where I shaded in orange. And now if we want the volume of one of those rings, we have to multiply it by its depth or its height, the way we've drawn it right over here. And its height, we've done this multiple times already, right over here, is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings. And then we want to sum up all of the rings over our interval."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And its height, we've done this multiple times already, right over here, is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings. And then we want to sum up all of the rings over our interval. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval?"}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we want to sum up all of the rings over our interval. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval? So we've looked at this multiple times. These two graphs, you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at, remember, we care about our y interval."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what's our interval? So we've looked at this multiple times. These two graphs, you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at, remember, we care about our y interval. They intersect at y is equal to 0 and y is equal to 1. Intersect at y equals 0 and y equals 1. And there you have it."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "The alternate form of the derivative of the function f at a number a denoted by f prime of a is given by this stuff. Now, this might look a little strange to you, but if you really think about what it's saying, it's really just taking the slope of the tangent line between a comma f of a. So let's imagine some arbitrary function like this. Let's say that that is, well, I'll just write, that's our function f. That's our function f. And so you could have the point when x is equal to a. This is our x-axis. When x is equal to a, this is the point a f of a. You notice a f of a."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that that is, well, I'll just write, that's our function f. That's our function f. And so you could have the point when x is equal to a. This is our x-axis. When x is equal to a, this is the point a f of a. You notice a f of a. And then we could take the slope between that and some arbitrary point. Let's call that x. So this is the point x f of x."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "You notice a f of a. And then we could take the slope between that and some arbitrary point. Let's call that x. So this is the point x f of x. And notice, this top, the numerator right here, this is just our change in the value of our function. Or you could view that as a change in the vertical axis. So that would give you this distance right over here."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is the point x f of x. And notice, this top, the numerator right here, this is just our change in the value of our function. Or you could view that as a change in the vertical axis. So that would give you this distance right over here. That's what we're doing up here in the numerator. And then in the denominator, we're finding the change in our horizontal values, horizontal coordinates. Let me do that in a different color."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "So that would give you this distance right over here. That's what we're doing up here in the numerator. And then in the denominator, we're finding the change in our horizontal values, horizontal coordinates. Let me do that in a different color. So the change in the horizontal, that's this right over here. And then they're trying to find the limit as x approaches a. So as x gets closer and closer and closer and closer to a, what's going to happen is that when x is out here, we have this secant line."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do that in a different color. So the change in the horizontal, that's this right over here. And then they're trying to find the limit as x approaches a. So as x gets closer and closer and closer and closer to a, what's going to happen is that when x is out here, we have this secant line. We're finding the slope of this secant line. But as x gets closer and closer, the secant lines better and better and better approximate the slope of the tangent line. Where the limit as x approaches a, but doesn't quite equal a, is going to be, this is actually our definition of our derivative, or I guess the alternate form of the derivative definition."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "So as x gets closer and closer and closer and closer to a, what's going to happen is that when x is out here, we have this secant line. We're finding the slope of this secant line. But as x gets closer and closer, the secant lines better and better and better approximate the slope of the tangent line. Where the limit as x approaches a, but doesn't quite equal a, is going to be, this is actually our definition of our derivative, or I guess the alternate form of the derivative definition. And this would be the slope of the tangent line, if it exists. So with all that out of the way, let's try to answer their question. With the alternative form of the derivative as an aid, make sense of the following limit expression by identifying the function f and the number a."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Where the limit as x approaches a, but doesn't quite equal a, is going to be, this is actually our definition of our derivative, or I guess the alternate form of the derivative definition. And this would be the slope of the tangent line, if it exists. So with all that out of the way, let's try to answer their question. With the alternative form of the derivative as an aid, make sense of the following limit expression by identifying the function f and the number a. So right here, they want to find the slope of the tangent line at 5. Here, they wanted to find the slope of the tangent line at a. So it's pretty clear that a is equal to 5 and that f of a is equal to 125."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "With the alternative form of the derivative as an aid, make sense of the following limit expression by identifying the function f and the number a. So right here, they want to find the slope of the tangent line at 5. Here, they wanted to find the slope of the tangent line at a. So it's pretty clear that a is equal to 5 and that f of a is equal to 125. Now, what about f of x? Well, here, it's the limit of f of x minus f of a. Well, here's the limit as x to the third minus 125."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's pretty clear that a is equal to 5 and that f of a is equal to 125. Now, what about f of x? Well, here, it's the limit of f of x minus f of a. Well, here's the limit as x to the third minus 125. And this makes sense. If f of x is equal to x to the third, then it makes sense that f of 5 is going to be 5 to the third is going to be 125. And we're also taking up here, the limit as x approaches a."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, here's the limit as x to the third minus 125. And this makes sense. If f of x is equal to x to the third, then it makes sense that f of 5 is going to be 5 to the third is going to be 125. And we're also taking up here, the limit as x approaches a. Here, we're taking the limit as x approaches 5. So this is the derivative of the function f of x is equal to x to the third. Let me write that down in the green color."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "And we're also taking up here, the limit as x approaches a. Here, we're taking the limit as x approaches 5. So this is the derivative of the function f of x is equal to x to the third. Let me write that down in the green color. x to the third at the number a is equal to 5. And so we can imagine this. Let's try to actually graph it just so we can imagine it."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me write that down in the green color. x to the third at the number a is equal to 5. And so we can imagine this. Let's try to actually graph it just so we can imagine it. Actually, I'll do it out here where I have a little bit better contrast with the colors. So let's say that is my y-axis. Let's say that this is my x-axis."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's try to actually graph it just so we can imagine it. Actually, I'll do it out here where I have a little bit better contrast with the colors. So let's say that is my y-axis. Let's say that this is my x-axis. I'm not going to quite draw it to scale. Let's say this right over here is a point 125 or y. This is when y equals 125."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that this is my x-axis. I'm not going to quite draw it to scale. Let's say this right over here is a point 125 or y. This is when y equals 125. This is when x is equal to 5. So they're clearly not at the same scale. But the function is going to look something like this."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "This is when y equals 125. This is when x is equal to 5. So they're clearly not at the same scale. But the function is going to look something like this. We know what x to the third looks like. It looks something like this. We know it's going to look something like this."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "But the function is going to look something like this. We know what x to the third looks like. It looks something like this. We know it's going to look something like this. So here, our a is equal to 5. This point right over here is 5, 125. And then we're taking the slope between that point and an arbitrary x value, or I should say an arbitrary other point on the curve."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "We know it's going to look something like this. So here, our a is equal to 5. This point right over here is 5, 125. And then we're taking the slope between that point and an arbitrary x value, or I should say an arbitrary other point on the curve. So this right over here would be the point. We could call that x comma x to the third. We know that f of x is equal to x to the third."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we're taking the slope between that point and an arbitrary x value, or I should say an arbitrary other point on the curve. So this right over here would be the point. We could call that x comma x to the third. We know that f of x is equal to x to the third. And let me make it clear. This is a graph of y is equal to x to the third. And so this expression right over here, all of this, this is the slope between these two points."}, {"video_title": "Formal and alternate form of the derivative example 1   Differential Calculus   Khan Academy.mp3", "Sentence": "We know that f of x is equal to x to the third. And let me make it clear. This is a graph of y is equal to x to the third. And so this expression right over here, all of this, this is the slope between these two points. And as we take the limit as x approaches 5, so right now this is our x. As x gets closer and closer and closer to 5, these secant lines are going to better and better approximate the slope of the tangent line at x equals 5. So the slope of the tangent line would look something like that."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Maybe nothing crazy happens, so let's just try it out. If we try to do x equals 0, what happens? We get 2 sine of 0, which is 0, minus sine of 2 times 0. That's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0, and then we have another sine of 0 there, so all 0s. And our denominator, we're going to have a 0 minus sine of 0."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0, and then we have another sine of 0 there, so all 0s. And our denominator, we're going to have a 0 minus sine of 0. Well, that's also going to be 0, but we have that indeterminate form, we have that undefined 0 over 0 that we talked about in the last video. So maybe we can use L'Hopital's Rule here. In order to use L'Hopital's Rule, then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And our denominator, we're going to have a 0 minus sine of 0. Well, that's also going to be 0, but we have that indeterminate form, we have that undefined 0 over 0 that we talked about in the last video. So maybe we can use L'Hopital's Rule here. In order to use L'Hopital's Rule, then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's Rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "In order to use L'Hopital's Rule, then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's Rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then minus, well, the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then minus, well, the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x. Just use the chain rule there. Derivative of the inside is just 2. That's the 2 out there."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So minus 2 cosine of 2x. Just use the chain rule there. Derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x. And we had that negative number out there. So that's the derivative of our numerator."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That's the 2 out there. Derivative of the outside is cosine of 2x. And we had that negative number out there. So that's the derivative of our numerator. And what is the derivative of our denominator? Well, derivative of x is just 1. And derivative of sine of x is just cosine of x."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's the derivative of our numerator. And what is the derivative of our denominator? Well, derivative of x is just 1. And derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get?"}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here, we're going to get 2 times cosine of 0, which is 2. Let me write it like this. So this is 2 times cosine of 0, which is 1."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "What do we get? If we put a 0 up here, we're going to get 2 times cosine of 0, which is 2. Let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Cosine of, let me write it this way. Actually, let me just do it this way."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Cosine of, let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and denominator, what are we going to get? We're going to get 2 cosine of 0, which is 2, minus 2 times cosine of, well, this 2 times 0 is still going to be 0. So minus 2 times cosine of 0, which is 2."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and denominator, what are we going to get? We're going to get 2 cosine of 0, which is 2, minus 2 times cosine of, well, this 2 times 0 is still going to be 0. So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0 over 0. So does this mean that the limit doesn't exist?"}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0 over 0. So does this mean that the limit doesn't exist? No, it still might exist. We might just want to do L'Hopital's Rule again. Let me take the derivative of that and put it over the derivative of that and then take the limit."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So does this mean that the limit doesn't exist? No, it still might exist. We might just want to do L'Hopital's Rule again. Let me take the derivative of that and put it over the derivative of that and then take the limit. And maybe L'Hopital's Rule will help us on the next stage. So let's see if it gets us anywhere. So this should be equal to the limit, if L'Hopital's Rule applies here, we're not 100% sure yet."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me take the derivative of that and put it over the derivative of that and then take the limit. And maybe L'Hopital's Rule will help us on the next stage. So let's see if it gets us anywhere. So this should be equal to the limit, if L'Hopital's Rule applies here, we're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, the derivative of cosine of x is negative sine of x."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this should be equal to the limit, if L'Hopital's Rule applies here, we're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, the derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. And then the derivative of cosine of 2x is negative 2 sine of 2x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times a 2."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, the derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. And then the derivative of cosine of 2x is negative 2 sine of 2x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times a 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times a 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times the negative right there is a plus."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times the negative right there is a plus. The positive sign, so it's sine of 2x. So that's the numerator when you take the derivative. And the denominator, this is just an exercise in taking derivatives."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The negative sine of x times the negative right there is a plus. The positive sign, so it's sine of 2x. So that's the numerator when you take the derivative. And the denominator, this is just an exercise in taking derivatives. What's the derivative of the denominator? Derivative of 1 is 0. And derivative of negative cosine of x is just sine of x."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And the denominator, this is just an exercise in taking derivatives. What's the derivative of the denominator? Derivative of 1 is 0. And derivative of negative cosine of x is just sine of x. So let's take this limit. So this is going to be equal to, well immediately if I take x is equal to 0 in the denominator, I know that sine of 0 is just 0. Let's see what happens in the numerator."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And derivative of negative cosine of x is just sine of x. So let's take this limit. So this is going to be equal to, well immediately if I take x is equal to 0 in the denominator, I know that sine of 0 is just 0. Let's see what happens in the numerator. Negative 2 times sine of 0, that's going to be 0. And then plus 4 times sine of 2 times 0, that's still sine of 0. So that's still going to be 0."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's see what happens in the numerator. Negative 2 times sine of 0, that's going to be 0. And then plus 4 times sine of 2 times 0, that's still sine of 0. So that's still going to be 0. So once again we've got indeterminate form again. Are we done? Do we give up?"}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's still going to be 0. So once again we've got indeterminate form again. Are we done? Do we give up? Do we say that L'Hopital's rule didn't work? No. Because this could have been our first limit problem."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Do we give up? Do we say that L'Hopital's rule didn't work? No. Because this could have been our first limit problem. And if this is our first limit problem, we say hey, maybe we could use L'Hopital's rule here because we've got an indeterminate form where both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to, if the limit exists, the limit as x approaches 0."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Because this could have been our first limit problem. And if this is our first limit problem, we say hey, maybe we could use L'Hopital's rule here because we've got an indeterminate form where both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to, if the limit exists, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then plus the derivative of 4 sine of 2x, well it's 2 times 4, which is 8 times cosine of 2x."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This will be equal to, if the limit exists, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then plus the derivative of 4 sine of 2x, well it's 2 times 4, which is 8 times cosine of 2x. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then plus the derivative of 4 sine of 2x, well it's 2 times 4, which is 8 times cosine of 2x. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. It looks like we've made some headway, or maybe L'Hopital's rule will stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. It looks like we've made some headway, or maybe L'Hopital's rule will stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0 over 0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That is 1. So we're definitely not going to get that indeterminate form, that 0 over 0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well, that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, x is 0, so it's going to be cosine of 0, which is 1."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We get negative 2 times cosine of 0. Well, that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well, this thing right here, negative 2 plus 8 is 6."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well, this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule, it applies to this last step."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule, it applies to this last step. If this was the problem we were given, we said, hey, when we tried to apply the limit, we get the limit as the numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. The limit as the derivative of the numerator over the derivative of the denominator, that exists and it equals 6."}, {"video_title": "L'H\u00f4pital's rule example 1   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So L'Hopital's rule, it applies to this last step. If this was the problem we were given, we said, hey, when we tried to apply the limit, we get the limit as the numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. The limit as the derivative of the numerator over the derivative of the denominator, that exists and it equals 6. So this limit must be equal to 6. Well, if this limit is equal to 6, by the same argument, this limit is also going to be equal to 6. And by the same argument, this limit has got to also be equal to 6."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Give a reason for your answer. Alright, so f needs to be both concave down and decreasing. So f concave down, concave down, that's going to be true if and only if the second derivative of f is less than zero, which is true if and only if f prime is decreasing. f prime is decreasing. So that's how we can tell when f is concave down. And I want to see what do I need to observe on f prime because that's the graph they give us. Alright, now for f to be decreasing, f is going to be decreasing if and only if, well you're decreasing when you have a negative slope, if your rate of change is negative."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "f prime is decreasing. So that's how we can tell when f is concave down. And I want to see what do I need to observe on f prime because that's the graph they give us. Alright, now for f to be decreasing, f is going to be decreasing if and only if, well you're decreasing when you have a negative slope, if your rate of change is negative. So that's true if and only if f prime of x is less than zero. So both true, both true when f prime of x is less than zero and decreasing. And let me write, and f prime is decreasing."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, now for f to be decreasing, f is going to be decreasing if and only if, well you're decreasing when you have a negative slope, if your rate of change is negative. So that's true if and only if f prime of x is less than zero. So both true, both true when f prime of x is less than zero and decreasing. And let me write, and f prime is decreasing. And f prime decreasing. So let's see if we can identify that. Where, let's go back to our graph and see when is f prime of x less than zero and decreasing."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me write, and f prime is decreasing. And f prime decreasing. So let's see if we can identify that. Where, let's go back to our graph and see when is f prime of x less than zero and decreasing. So if we look at less than zero, let me highlight this enough. So less than zero, I'm highlighting it in magenta right over here. This is all the places where we are less than zero in magenta."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Where, let's go back to our graph and see when is f prime of x less than zero and decreasing. So if we look at less than zero, let me highlight this enough. So less than zero, I'm highlighting it in magenta right over here. This is all the places where we are less than zero in magenta. And now of that, where are we decreasing? Well we are decreasing over this interval, right over here. So that would be negative two is less than x, which is less than negative one."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is all the places where we are less than zero in magenta. And now of that, where are we decreasing? Well we are decreasing over this interval, right over here. So that would be negative two is less than x, which is less than negative one. So that's, we're decreasing. Of the places where we are negative, I'm taking the subset of where we are negative to think about, well where are we decreasing? We're decreasing right there."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would be negative two is less than x, which is less than negative one. So that's, we're decreasing. Of the places where we are negative, I'm taking the subset of where we are negative to think about, well where are we decreasing? We're decreasing right there. And we're also decreasing, we're also decreasing right over there. So that is, one is less than x, which is less than three. Now we are decreasing before x equals negative two over here, but we're not less than zero there."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're decreasing right there. And we're also decreasing, we're also decreasing right over there. So that is, one is less than x, which is less than three. Now we are decreasing before x equals negative two over here, but we're not less than zero there. So I'm not gonna include that. And so we have our intervals. So our open intervals."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we are decreasing before x equals negative two over here, but we're not less than zero there. So I'm not gonna include that. And so we have our intervals. So our open intervals. So we have both true when these two things are true. And so this happens. This happens when negative two is less than x, is less than negative one."}, {"video_title": "2015 AP Calculus AB 5b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our open intervals. So we have both true when these two things are true. And so this happens. This happens when negative two is less than x, is less than negative one. Or one is less than x, is less than three. So those are our two open intervals. This one and that one."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So before I even look at this, what do we know about the intermediate value theorem? Well, it applies here. It's a continuous function on this closed interval. We know what the value of the function is at negative two. It's three, so let me write that. F of negative two is equal to three and f of one, they tell us right over here, is equal to six. And all the intermediate value theorem tells us, and if this is completely unfamiliar to you, I encourage you to watch the video on the intermediate value theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the end points of the interval."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know what the value of the function is at negative two. It's three, so let me write that. F of negative two is equal to three and f of one, they tell us right over here, is equal to six. And all the intermediate value theorem tells us, and if this is completely unfamiliar to you, I encourage you to watch the video on the intermediate value theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the end points of the interval. Or another way to say it is, for any L between three and six, there is at least one C, there is at least one C, one C between, or I could say one C in the interval from negative two to one, the closed interval, such that f of C is equal to L. This comes straight out of the intermediate value theorem. And just saying it in everyday language is, look, this is a continuous function. Actually, I'll draw it visually in a few seconds, but it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well, then it makes sense that I would have to take on every value between three and six, or there's at least one point in this interval where I take on any given value between three and six."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And all the intermediate value theorem tells us, and if this is completely unfamiliar to you, I encourage you to watch the video on the intermediate value theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the end points of the interval. Or another way to say it is, for any L between three and six, there is at least one C, there is at least one C, one C between, or I could say one C in the interval from negative two to one, the closed interval, such that f of C is equal to L. This comes straight out of the intermediate value theorem. And just saying it in everyday language is, look, this is a continuous function. Actually, I'll draw it visually in a few seconds, but it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well, then it makes sense that I would have to take on every value between three and six, or there's at least one point in this interval where I take on any given value between three and six. So let's see which of these answers are consistent with that. We only pick one. So f of C equals four, so that would be a case where L is equal to four."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, I'll draw it visually in a few seconds, but it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well, then it makes sense that I would have to take on every value between three and six, or there's at least one point in this interval where I take on any given value between three and six. So let's see which of these answers are consistent with that. We only pick one. So f of C equals four, so that would be a case where L is equal to four. So there's at least one C in this interval such that f of C is equal to four. We could say that, but that's not exactly what they're saying here. F of C could be four for at least one C, not in this interval."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of C equals four, so that would be a case where L is equal to four. So there's at least one C in this interval such that f of C is equal to four. We could say that, but that's not exactly what they're saying here. F of C could be four for at least one C, not in this interval. Remember, the C is our X. This is our X right over here. So the C is going to be in this interval."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "F of C could be four for at least one C, not in this interval. Remember, the C is our X. This is our X right over here. So the C is going to be in this interval. I'll take a look at it visually in a second so that we can validate that. We're not saying for at least one C between three and six, f of C is equal to four. We're saying for at least one C in this interval, f of C is going to be equal to four."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the C is going to be in this interval. I'll take a look at it visually in a second so that we can validate that. We're not saying for at least one C between three and six, f of C is equal to four. We're saying for at least one C in this interval, f of C is going to be equal to four. It's important that four is between three and six because that's the value of our function, and the C needs to be in our closed interval along the X axis. So I'm gonna rule this out. They're trying to confuse us."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're saying for at least one C in this interval, f of C is going to be equal to four. It's important that four is between three and six because that's the value of our function, and the C needs to be in our closed interval along the X axis. So I'm gonna rule this out. They're trying to confuse us. All right, f of C equals zero for at least one C between negative two and one. Well, here they got the interval along the X axis right. That's where the C would be between, but it's not guaranteed by the intermediate value theorem that f of C is going to be equal to zero because zero is not between three and six."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "They're trying to confuse us. All right, f of C equals zero for at least one C between negative two and one. Well, here they got the interval along the X axis right. That's where the C would be between, but it's not guaranteed by the intermediate value theorem that f of C is going to be equal to zero because zero is not between three and six. So I'm gonna rule that one out. I'm going to rule this one out, saying f of C equals zero. And let's see, we're only left with this one, so I hope it works."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's where the C would be between, but it's not guaranteed by the intermediate value theorem that f of C is going to be equal to zero because zero is not between three and six. So I'm gonna rule that one out. I'm going to rule this one out, saying f of C equals zero. And let's see, we're only left with this one, so I hope it works. So f of C is equal to four. Well, that seems reasonable because four is between three and six for at least one C between negative two and one. Well, yeah, because that's in this interval right over here."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we're only left with this one, so I hope it works. So f of C is equal to four. Well, that seems reasonable because four is between three and six for at least one C between negative two and one. Well, yeah, because that's in this interval right over here. So I am feeling good about that. And we could think about this visually as well. The intermediate value theorem, when you think about it visually, makes a lot of sense."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, yeah, because that's in this interval right over here. So I am feeling good about that. And we could think about this visually as well. The intermediate value theorem, when you think about it visually, makes a lot of sense. So let me draw the X axis first, actually, and then let me draw my Y axis, and I'm gonna draw them at different scales because my Y axis, well, let's see, if this is six, this is three, that's my Y axis, this is one, this is negative one, this is negative two. And so we're continuous on the closed interval from negative two to one, and f of negative two is equal to three. So let me plot that."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "The intermediate value theorem, when you think about it visually, makes a lot of sense. So let me draw the X axis first, actually, and then let me draw my Y axis, and I'm gonna draw them at different scales because my Y axis, well, let's see, if this is six, this is three, that's my Y axis, this is one, this is negative one, this is negative two. And so we're continuous on the closed interval from negative two to one, and f of negative two is equal to three. So let me plot that. F of negative two is equal to three. So that's right over there. And f of one is equal to six."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me plot that. F of negative two is equal to three. So that's right over there. And f of one is equal to six. So that's right over there. And so let's try to draw a continuous function. So a continuous function includes these points, and it's continuous, so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And f of one is equal to six. So that's right over there. And so let's try to draw a continuous function. So a continuous function includes these points, and it's continuous, so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points. So I can't pick up my, I can't do that. That would be picking up my pencil. So it is a continuous function."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So a continuous function includes these points, and it's continuous, so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points. So I can't pick up my, I can't do that. That would be picking up my pencil. So it is a continuous function. So it takes on every value, as we can see, it definitely does that. It takes on every value between three and six. It might take on other values, but we know for sure it has to take on every value between three and six."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it is a continuous function. So it takes on every value, as we can see, it definitely does that. It takes on every value between three and six. It might take on other values, but we know for sure it has to take on every value between three and six. And so if we think about four, four is right over here. The way I drew it, it actually looks like it's almost taking on that value right at the Y axis. I forgot to label my X axis here."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "It might take on other values, but we know for sure it has to take on every value between three and six. And so if we think about four, four is right over here. The way I drew it, it actually looks like it's almost taking on that value right at the Y axis. I forgot to label my X axis here. But you can see, it took on that value in for a C, in this case, between negative two and one. And I could have drawn that graph multiple different ways. I could have drawn it something like, I could have done it, and actually it takes on, there's multiple times it takes on the value four here."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "I forgot to label my X axis here. But you can see, it took on that value in for a C, in this case, between negative two and one. And I could have drawn that graph multiple different ways. I could have drawn it something like, I could have done it, and actually it takes on, there's multiple times it takes on the value four here. So this could be our C, but once again, it's between the interval negative two and one. This could be our C, once again, in the interval between negative two and one. Or this could be our C, in between the interval of negative two and one."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could have drawn it something like, I could have done it, and actually it takes on, there's multiple times it takes on the value four here. So this could be our C, but once again, it's between the interval negative two and one. This could be our C, once again, in the interval between negative two and one. Or this could be our C, in between the interval of negative two and one. That's just the way I happened to draw it. I could have drawn this thing as just a straight line. I could have drawn it like this."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or this could be our C, in between the interval of negative two and one. That's just the way I happened to draw it. I could have drawn this thing as just a straight line. I could have drawn it like this. And then it looks like it's taking on four, only once, and it's doing it right around there. This isn't necessarily true, that you take on, you take on, that you become four for at least one C between three and six. Three and six aren't even on our graph here."}, {"video_title": "Intermediate value theorem example   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could have drawn it like this. And then it looks like it's taking on four, only once, and it's doing it right around there. This isn't necessarily true, that you take on, you take on, that you become four for at least one C between three and six. Three and six aren't even on our graph here. I would have to go all the way to, so two, three. There's no guarantee that our function, that our function takes on four for one C between three and six. We don't even know what the function does when X is between three and six."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "It is 100 meters. And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire, I'm going to obviously cut it in two. With the left section, I am going to construct an equilateral triangle. And with the right section, I am going to construct a square. I am going to construct a square."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "With the left section of wire, I'm going to obviously cut it in two. With the left section, I am going to construct an equilateral triangle. And with the right section, I am going to construct a square. I am going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out, let's define a variable that we're trying to minimize or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "I am going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out, let's define a variable that we're trying to minimize or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be x meters. And the length for the square would be, well, if we use x up for the left-hand side, we're going to have 100 minus x for the right-hand side. And so what would the dimensions of the triangle and the square be?"}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be x meters. And the length for the square would be, well, if we use x up for the left-hand side, we're going to have 100 minus x for the right-hand side. And so what would the dimensions of the triangle and the square be? Well, the triangle sides are going to be x over 3, x over 3, and x over 3 is an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now, it's easy to figure out an expression for the area of this square in terms of x."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "And so what would the dimensions of the triangle and the square be? Well, the triangle sides are going to be x over 3, x over 3, and x over 3 is an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now, it's easy to figure out an expression for the area of this square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle just like that."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, it's easy to figure out an expression for the area of this square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle just like that. And its sides are length s, s, and s. Now, we know that the area of a triangle is 1 half times the base times the height. So in this case, the height we could consider to be an altitude. If we were to drop an altitude just like this, this length right over here, this is the height."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say we have an equilateral triangle just like that. And its sides are length s, s, and s. Now, we know that the area of a triangle is 1 half times the base times the height. So in this case, the height we could consider to be an altitude. If we were to drop an altitude just like this, this length right over here, this is the height. And this would be perpendicular just like that. So our area is going to be equal to 1 half times our base is s, 1 half times s times whatever our height is. Now, how can we express h as a function of s?"}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "If we were to drop an altitude just like this, this length right over here, this is the height. And this would be perpendicular just like that. So our area is going to be equal to 1 half times our base is s, 1 half times s times whatever our height is. Now, how can we express h as a function of s? Well, to do that, we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, how can we express h as a function of s? Well, to do that, we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two. So this right over here has length s over 2. So to figure out what h is, we can just use the Pythagorean theorem."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two. So this right over here has length s over 2. So to figure out what h is, we can just use the Pythagorean theorem. We would have h squared plus s over 2 squared is going to be equal to the hypotenuse squared. It is going to be equal to s squared. So you would get h squared plus s squared over 4 is equal to s squared."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So to figure out what h is, we can just use the Pythagorean theorem. We would have h squared plus s over 2 squared is going to be equal to the hypotenuse squared. It is going to be equal to s squared. So you would get h squared plus s squared over 4 is equal to s squared. Subtract s squared over 4 from both sides, and you get h squared is equal to s squared minus s squared over 4. Now, to do this, I could call s squared. I could call this 4s squared over 4, just to be able to have a common denominator."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So you would get h squared plus s squared over 4 is equal to s squared. Subtract s squared over 4 from both sides, and you get h squared is equal to s squared minus s squared over 4. Now, to do this, I could call s squared. I could call this 4s squared over 4, just to be able to have a common denominator. And 4s squared minus s squared over 4 is going to be equal to 3s squared over 4. So we get h squared is equal to 3s squared over 4. Now we can take the principal root of both sides, and we get h is equal to the square root of 3 times s over 2."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "I could call this 4s squared over 4, just to be able to have a common denominator. And 4s squared minus s squared over 4 is going to be equal to 3s squared over 4. So we get h squared is equal to 3s squared over 4. Now we can take the principal root of both sides, and we get h is equal to the square root of 3 times s over 2. So now we can just substitute back right over here, and we get our area is equal to 1 half s times h. Well, h is this business, so it's s times this. So it's 1 half times s times the square root of 3s over 2, which is going to be equal to s times s is s squared. So it's going to be square root of 3s squared over 2 times 2 over 4."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "Now we can take the principal root of both sides, and we get h is equal to the square root of 3 times s over 2. So now we can just substitute back right over here, and we get our area is equal to 1 half s times h. Well, h is this business, so it's s times this. So it's 1 half times s times the square root of 3s over 2, which is going to be equal to s times s is s squared. So it's going to be square root of 3s squared over 2 times 2 over 4. So this is the area of an equilateral triangle as the function of the length of its sides. So what's the area of this business going to be? So the area of our little equilateral triangle, let me write combined area."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be square root of 3s squared over 2 times 2 over 4. So this is the area of an equilateral triangle as the function of the length of its sides. So what's the area of this business going to be? So the area of our little equilateral triangle, let me write combined area. Let me do that in a neutral color. So let me do that in white. So the combined area, I'll write it a sub c, is going to be equal to the area of my triangle, a sub t, plus the area of my square."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So the area of our little equilateral triangle, let me write combined area. Let me do that in a neutral color. So let me do that in white. So the combined area, I'll write it a sub c, is going to be equal to the area of my triangle, a sub t, plus the area of my square. Well, the area of my triangle, we know what it's going to be. It's going to be square root of 3 times the length of a side squared divided by 4. So it's going to be square root of 3."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So the combined area, I'll write it a sub c, is going to be equal to the area of my triangle, a sub t, plus the area of my square. Well, the area of my triangle, we know what it's going to be. It's going to be square root of 3 times the length of a side squared divided by 4. So it's going to be square root of 3. Let me do that in that same yellow color. It's going to be, make sure I switch colors, it's going to be square root of 3 over 4 times the side squared, times x over 3 squared. All I did is the length of a side is x over 3."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be square root of 3. Let me do that in that same yellow color. It's going to be, make sure I switch colors, it's going to be square root of 3 over 4 times the side squared, times x over 3 squared. All I did is the length of a side is x over 3. We already know what the area is. It's square root of 3 over 4 times the length of a side squared. And then the area of this square right over here, the area of the square is just going to be 100 minus x over 4 squared."}, {"video_title": "Expression for combined area of triangle and square   Differential Calculus   Khan Academy.mp3", "Sentence": "All I did is the length of a side is x over 3. We already know what the area is. It's square root of 3 over 4 times the length of a side squared. And then the area of this square right over here, the area of the square is just going to be 100 minus x over 4 squared. So our area, our combined area, maybe I could write like this. Our combined area as a function of where we make the cut is all of this business right over here. And this is what we need to minimize."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I have two curves graphed here, and we're used to seeing things where y is a function of x, but here we have x as a function of y. In fact, we can write this top expression as being a function of y, and this second one, just to make it different, we could view this as g of y. Once again, it is a function of y. And what we're concerning ourselves with in this video is how do we find this area in this light blue color between these two curves? And I encourage you to pause the video and try to work through it. All right, so a huge hint here is we're going to want to integrate with respect to y, a definite integral where our bounds are in terms of y. So for example, this is this lower point of intersection right over here."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we're concerning ourselves with in this video is how do we find this area in this light blue color between these two curves? And I encourage you to pause the video and try to work through it. All right, so a huge hint here is we're going to want to integrate with respect to y, a definite integral where our bounds are in terms of y. So for example, this is this lower point of intersection right over here. This would be our lower bound in terms of y. Let's call that y one. And then this up here, this would be our upper y bound."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, this is this lower point of intersection right over here. This would be our lower bound in terms of y. Let's call that y one. And then this up here, this would be our upper y bound. So if we think about where do these two curves intersect, and we look at the y coordinates of those intersections, well, that gives us two nice bounds for our integral. So we're gonna take our integral from y one to y two, from y one to y two, y two. And we're going to integrate with respect to y, dy."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this up here, this would be our upper y bound. So if we think about where do these two curves intersect, and we look at the y coordinates of those intersections, well, that gives us two nice bounds for our integral. So we're gonna take our integral from y one to y two, from y one to y two, y two. And we're going to integrate with respect to y, dy. And so what are we going to sum up? Well, when we integrate, we can think about taking the sum of infinitely thin rectangles. And in this case, it would be infinitely flat rectangles since we're thinking about dy."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to integrate with respect to y, dy. And so what are we going to sum up? Well, when we integrate, we can think about taking the sum of infinitely thin rectangles. And in this case, it would be infinitely flat rectangles since we're thinking about dy. So dy would be the height of each of these rectangles. And what would be, in this case, the width or the length of this rectangle right over here? Well, over this interval from y one to y two, our blue function, f of y, takes on larger x values than g of y."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in this case, it would be infinitely flat rectangles since we're thinking about dy. So dy would be the height of each of these rectangles. And what would be, in this case, the width or the length of this rectangle right over here? Well, over this interval from y one to y two, our blue function, f of y, takes on larger x values than g of y. So this length right over here, this would be f of y, f of y, this x value, minus this x value, minus g of y. So this is going to be f of y minus g of y, g of y. Well, we know what f of y and g of y are."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, over this interval from y one to y two, our blue function, f of y, takes on larger x values than g of y. So this length right over here, this would be f of y, f of y, this x value, minus this x value, minus g of y. So this is going to be f of y minus g of y, g of y. Well, we know what f of y and g of y are. Really, the trickiest part is is figuring out these points of intersection. So let's think about where these two curves intersect. They are both equal to x, so we can set these two y expressions equal to each other."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we know what f of y and g of y are. Really, the trickiest part is is figuring out these points of intersection. So let's think about where these two curves intersect. They are both equal to x, so we can set these two y expressions equal to each other. So we know that negative, let me do it in that other color. So we know that negative y squared plus three y plus 11 is going to be equal to this, is going to be equal to y squared plus y minus one. So let's just subtract all of this from both sides so that on the right side we have a zero and on the left side we just have a quadratic."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "They are both equal to x, so we can set these two y expressions equal to each other. So we know that negative, let me do it in that other color. So we know that negative y squared plus three y plus 11 is going to be equal to this, is going to be equal to y squared plus y minus one. So let's just subtract all of this from both sides so that on the right side we have a zero and on the left side we just have a quadratic. So let's subtract y squared, let's subtract y, and then subtract negative one, which is just adding one, is, and over here we're going to do the same thing, minus y plus one. And what we are left with is going to be hopefully a straightforward quadratic. So let's see, this is going to be negative two y squared plus two y, am I doing that right?"}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just subtract all of this from both sides so that on the right side we have a zero and on the left side we just have a quadratic. So let's subtract y squared, let's subtract y, and then subtract negative one, which is just adding one, is, and over here we're going to do the same thing, minus y plus one. And what we are left with is going to be hopefully a straightforward quadratic. So let's see, this is going to be negative two y squared plus two y, am I doing that right? Yeah, plus two y, plus 12 is equal to zero. And then this over here I can factor out a negative two, and I get negative two times y squared minus y minus six is equal to zero. This we can factor from inspection."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, this is going to be negative two y squared plus two y, am I doing that right? Yeah, plus two y, plus 12 is equal to zero. And then this over here I can factor out a negative two, and I get negative two times y squared minus y minus six is equal to zero. This we can factor from inspection. What two numbers, when we add, equal negative one? When we take their product we get negative six. Well that would be negative three and two."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "This we can factor from inspection. What two numbers, when we add, equal negative one? When we take their product we get negative six. Well that would be negative three and two. So this is going to be negative two times y minus three times y plus two, that's just straightforward factoring a polynomial, a quadratic. Did I do that right? Yep, that looks right, is equal to zero."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well that would be negative three and two. So this is going to be negative two times y minus three times y plus two, that's just straightforward factoring a polynomial, a quadratic. Did I do that right? Yep, that looks right, is equal to zero. So what are the points of intersection? The points of intersection are going to be y is equal to three and y is equal to negative two. So this right over here is y is equal to negative two, and then the upper bound is y is equal to three."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yep, that looks right, is equal to zero. So what are the points of intersection? The points of intersection are going to be y is equal to three and y is equal to negative two. So this right over here is y is equal to negative two, and then the upper bound is y is equal to three. So now we just have to evaluate this from negative two all the way until three. So let's do that, I'm going to clear this out so I get a little bit of real estate. So this is equal to the integral from negative two to three of negative y squared plus three y plus 11 minus all of this stuff."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is y is equal to negative two, and then the upper bound is y is equal to three. So now we just have to evaluate this from negative two all the way until three. So let's do that, I'm going to clear this out so I get a little bit of real estate. So this is equal to the integral from negative two to three of negative y squared plus three y plus 11 minus all of this stuff. So if we just distribute a negative sign here, it's minus y squared minus y plus one and then we have a dy, dy. This is equal to the definite integral from negative two to positive three of, let's see, negative y squared minus y squared, negative two y squared and then three y minus y is going to be plus two y and then 11 plus one plus 12. We saw this just now when we were trying to solve for y, dy."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to the integral from negative two to three of negative y squared plus three y plus 11 minus all of this stuff. So if we just distribute a negative sign here, it's minus y squared minus y plus one and then we have a dy, dy. This is equal to the definite integral from negative two to positive three of, let's see, negative y squared minus y squared, negative two y squared and then three y minus y is going to be plus two y and then 11 plus one plus 12. We saw this just now when we were trying to solve for y, dy. And so what is that going to be equal to? Well, we just take the antiderivative here. This is going to be, let's see, negative two, let's increment the exponent, y to the third, divide by that exponent, reverse power rule, plus two y squared divided by two, which is just y squared, just the reverse power rule, and then plus 12 y and we're going to evaluate that at three and at negative two."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We saw this just now when we were trying to solve for y, dy. And so what is that going to be equal to? Well, we just take the antiderivative here. This is going to be, let's see, negative two, let's increment the exponent, y to the third, divide by that exponent, reverse power rule, plus two y squared divided by two, which is just y squared, just the reverse power rule, and then plus 12 y and we're going to evaluate that at three and at negative two. So if we evaluate that at three, we are going to get, let's see, negative two times 27 over three plus nine plus 36 and then we are going to want to subtract minus, all of this evaluated at negative two. So it's going to be negative two times negative eight over three plus four minus 24. So we just have a little bit of mathematics ahead of us."}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be, let's see, negative two, let's increment the exponent, y to the third, divide by that exponent, reverse power rule, plus two y squared divided by two, which is just y squared, just the reverse power rule, and then plus 12 y and we're going to evaluate that at three and at negative two. So if we evaluate that at three, we are going to get, let's see, negative two times 27 over three plus nine plus 36 and then we are going to want to subtract minus, all of this evaluated at negative two. So it's going to be negative two times negative eight over three plus four minus 24. So we just have a little bit of mathematics ahead of us. So let's see, this is going to be 27 divided by three is nine so this is negative 18, negative 18 plus nine is going to be negative nine plus 36. All of that is going to be equal to, so the stuff in blue is equal to 27, right? Did I do that right?"}, {"video_title": "Horizontal area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have a little bit of mathematics ahead of us. So let's see, this is going to be 27 divided by three is nine so this is negative 18, negative 18 plus nine is going to be negative nine plus 36. All of that is going to be equal to, so the stuff in blue is equal to 27, right? Did I do that right? You got negative 18 plus nine, yep, 27. And then all the stuff in red over here, we have, this is going to be negative times a negative, so it's 16 over three plus four minus 24. So that is going to be 16 over three and then minus 20, but then we have this negative out here so if we distribute that, we'll get plus 20 minus, we could say instead of 16 over three, we could rewrite that as five and 1 3rd, minus five and 1 3rd and so what is that going to get us?"}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that y is equal to seven to the x squared minus x power. What is the derivative of y, derivative of y, with respect to x? And like always, pause this video and see if you can figure it out. Well, based on how this has been color-coded ahead of time, you might immediately recognize that this is a composite function, or it could be viewed as a composite function. If you had a v of x, which if you had a function v of x, which is equal to seven to the x power, and you had another function u of x, u of x, which is equal to x squared minus x, then what we have right over here, y, y is equal to seven to something. So it's equal to v of, and it's not just v of x, it's v of u of x. Instead of an x here, you have the whole function u of x, x squared minus x."}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, based on how this has been color-coded ahead of time, you might immediately recognize that this is a composite function, or it could be viewed as a composite function. If you had a v of x, which if you had a function v of x, which is equal to seven to the x power, and you had another function u of x, u of x, which is equal to x squared minus x, then what we have right over here, y, y is equal to seven to something. So it's equal to v of, and it's not just v of x, it's v of u of x. Instead of an x here, you have the whole function u of x, x squared minus x. So it's v of u of x, and the chain rule tells us that the derivative of y with respect to x, and you'll see different notations here, sometimes you'll see it written as the derivative of v with respect to u, so v prime of u of x, times the derivative of u with respect to x, so that's one way you could do it, or you could say that this is equal to, this is equal to the derivative, the derivative of v with respect to x, derivative, or sorry, derivative of v with respect to u, dv du, times the derivative of u with respect to x, derivative of u with respect to x. And so either way, we can apply that right over here. So what's the derivative of v with respect to u?"}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of an x here, you have the whole function u of x, x squared minus x. So it's v of u of x, and the chain rule tells us that the derivative of y with respect to x, and you'll see different notations here, sometimes you'll see it written as the derivative of v with respect to u, so v prime of u of x, times the derivative of u with respect to x, so that's one way you could do it, or you could say that this is equal to, this is equal to the derivative, the derivative of v with respect to x, derivative, or sorry, derivative of v with respect to u, dv du, times the derivative of u with respect to x, derivative of u with respect to x. And so either way, we can apply that right over here. So what's the derivative of v with respect to u? What is v prime of u of x? Well, we know, we know, let me actually write it right over here. If v of x is equal to seven to the x power, v prime of x would be equal to, and we proved this in other videos where we take derivatives, exponentials of bases other than e, this is going to be the natural log of seven, times seven to the x power."}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's the derivative of v with respect to u? What is v prime of u of x? Well, we know, we know, let me actually write it right over here. If v of x is equal to seven to the x power, v prime of x would be equal to, and we proved this in other videos where we take derivatives, exponentials of bases other than e, this is going to be the natural log of seven, times seven to the x power. So if we are taking v prime of u of x, then notice, instead of an x everywhere, we're going to have a u of x everywhere. So this right over here, this is going to be natural log of seven times seven to the, instead of saying seven to the x power, remember, we're taking v prime of u of x. So it's going to be seven to the x squared minus x power."}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If v of x is equal to seven to the x power, v prime of x would be equal to, and we proved this in other videos where we take derivatives, exponentials of bases other than e, this is going to be the natural log of seven, times seven to the x power. So if we are taking v prime of u of x, then notice, instead of an x everywhere, we're going to have a u of x everywhere. So this right over here, this is going to be natural log of seven times seven to the, instead of saying seven to the x power, remember, we're taking v prime of u of x. So it's going to be seven to the x squared minus x power. X squared, x squared minus x power. And then we want to multiply that times the derivative of u with respect to x. So u prime of x, well that's going to be two x to the first, which is just two x, minus one."}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be seven to the x squared minus x power. X squared, x squared minus x power. And then we want to multiply that times the derivative of u with respect to x. So u prime of x, well that's going to be two x to the first, which is just two x, minus one. So we're going to multiply this times two x, two x minus one. So there you have it. That is the derivative of y with respect to x."}, {"video_title": "Exponential functions differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So u prime of x, well that's going to be two x to the first, which is just two x, minus one. So we're going to multiply this times two x, two x minus one. So there you have it. That is the derivative of y with respect to x. You could, we could try to simplify this or I guess re-express it in different ways, but the main thing to realize is, look, we're just going to take the derivative of the seven to the, this to the u of x power with respect to u of x. So we treat the u of x the way that we would have treated an x right over here. So it's going to be natural log of seven times seven to the u of x power."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So right over here we have the graph of y is equal to x squared, or at least part of the graph of y is equal to x squared, and the first thing I'd like to tackle is think about the average rate of change of y with respect to x over the interval from x equaling one to x equaling three. So let me write that down. We want to know the average rate of change of y with respect to x over the interval from x going from one to three, and it's a closed interval where x could be one and x could be equal to three. Well we could do this even without looking at the graph. If I were to just make a table here where if this is x and this is y is equal to x squared, when x is equal to one, y is equal to one squared, which is just one. You see that right over there. And when x is equal to three, y is equal to three squared, which is equal to nine."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well we could do this even without looking at the graph. If I were to just make a table here where if this is x and this is y is equal to x squared, when x is equal to one, y is equal to one squared, which is just one. You see that right over there. And when x is equal to three, y is equal to three squared, which is equal to nine. And so you can see when x is equal to three, y is equal to nine. And to figure out the average rate of change of y with respect to x, you say okay, well what's my change in x? Well we can see very clearly that our change in x over this interval is equal to positive two."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And when x is equal to three, y is equal to three squared, which is equal to nine. And so you can see when x is equal to three, y is equal to nine. And to figure out the average rate of change of y with respect to x, you say okay, well what's my change in x? Well we can see very clearly that our change in x over this interval is equal to positive two. Well what's our change in y over the same interval? Our change in y is equal to, when x went increased by two from one to three, y increases by eight. So it's gonna be a positive eight."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well we can see very clearly that our change in x over this interval is equal to positive two. Well what's our change in y over the same interval? Our change in y is equal to, when x went increased by two from one to three, y increases by eight. So it's gonna be a positive eight. So what is our average rate of change? Well it's gonna be our change in y over our change in x, which is equal to eight over two, which is equal to four. So that would be our average rate of change."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's gonna be a positive eight. So what is our average rate of change? Well it's gonna be our change in y over our change in x, which is equal to eight over two, which is equal to four. So that would be our average rate of change. Over that interval, on average, every time x increases by one, y is increasing by four. And how did we calculate that? We looked at our change in x."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would be our average rate of change. Over that interval, on average, every time x increases by one, y is increasing by four. And how did we calculate that? We looked at our change in x. Let me draw that here. We looked at our change in x. And we looked at our change in y, which would be this right over here."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We looked at our change in x. Let me draw that here. We looked at our change in x. And we looked at our change in y, which would be this right over here. And we calculated change in y over change of x for average rate of change. Now this might be looking fairly familiar to you, because you're used to thinking about change in y over change in x as the slope of a line connecting two points. And that's indeed what we did calculate."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we looked at our change in y, which would be this right over here. And we calculated change in y over change of x for average rate of change. Now this might be looking fairly familiar to you, because you're used to thinking about change in y over change in x as the slope of a line connecting two points. And that's indeed what we did calculate. If you were to draw a secant line between these two points, we essentially just calculated the slope of that secant line. And so the average rate of change between two points, that is the same thing as the slope of the secant line. And by looking at the secant line in comparison to the curve over that interval, it hopefully gives you a visual intuition for what even average rate of change means."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's indeed what we did calculate. If you were to draw a secant line between these two points, we essentially just calculated the slope of that secant line. And so the average rate of change between two points, that is the same thing as the slope of the secant line. And by looking at the secant line in comparison to the curve over that interval, it hopefully gives you a visual intuition for what even average rate of change means. Because in the beginning part of the interval, you see that the secant line is actually increasing at a faster rate. But then as we get closer to three, it looks like our yellow curve is increasing at a faster rate than the secant line, and then they eventually catch up. And so that's why the slope of the secant line is the average rate of change."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And by looking at the secant line in comparison to the curve over that interval, it hopefully gives you a visual intuition for what even average rate of change means. Because in the beginning part of the interval, you see that the secant line is actually increasing at a faster rate. But then as we get closer to three, it looks like our yellow curve is increasing at a faster rate than the secant line, and then they eventually catch up. And so that's why the slope of the secant line is the average rate of change. Is it the exact rate of change at every point? Absolutely not. The curve's rate of change is constantly changing."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that's why the slope of the secant line is the average rate of change. Is it the exact rate of change at every point? Absolutely not. The curve's rate of change is constantly changing. It's at a slower rate of change in the beginning part of this interval, and then it's actually increasing at a higher rate as we get closer and closer to three. So over the interval, their change in y over the change in x is exactly the same. Now one question you might be wondering is, why are you learning this in a calculus class?"}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "The curve's rate of change is constantly changing. It's at a slower rate of change in the beginning part of this interval, and then it's actually increasing at a higher rate as we get closer and closer to three. So over the interval, their change in y over the change in x is exactly the same. Now one question you might be wondering is, why are you learning this in a calculus class? Couldn't you have learned this in an algebra class? The answer is yes. But what's going to be interesting, and it's really one of the foundational ideas of calculus, is, well, what happens as these points get closer and closer together?"}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now one question you might be wondering is, why are you learning this in a calculus class? Couldn't you have learned this in an algebra class? The answer is yes. But what's going to be interesting, and it's really one of the foundational ideas of calculus, is, well, what happens as these points get closer and closer together? We found the average rate of change between one and three, or the slope of the secant line from one comma one to three comma nine. But what instead if you found the slope of the secant line between two comma four and three comma nine? So what if you found this slope?"}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what's going to be interesting, and it's really one of the foundational ideas of calculus, is, well, what happens as these points get closer and closer together? We found the average rate of change between one and three, or the slope of the secant line from one comma one to three comma nine. But what instead if you found the slope of the secant line between two comma four and three comma nine? So what if you found this slope? But what if you wanted to get even closer? Let's say you wanted to find the slope of the secant line between the point 2.5, 6.25, and three comma nine. And what if you just kept getting closer and closer and closer?"}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what if you found this slope? But what if you wanted to get even closer? Let's say you wanted to find the slope of the secant line between the point 2.5, 6.25, and three comma nine. And what if you just kept getting closer and closer and closer? Well then, the slopes of these secant lines are gonna get closer and closer to the slope of the tangent line at x equals three. And if we can figure out the slope of the tangent line, well then we're in business. Because then we're not talking about average rate of change, we're gonna be talking about instantaneous rate of change, which is one of the central ideas."}, {"video_title": "Secant lines & average rate of change   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what if you just kept getting closer and closer and closer? Well then, the slopes of these secant lines are gonna get closer and closer to the slope of the tangent line at x equals three. And if we can figure out the slope of the tangent line, well then we're in business. Because then we're not talking about average rate of change, we're gonna be talking about instantaneous rate of change, which is one of the central ideas. That is the derivative, and we're going to get there soon. But it's really important to appreciate that average rate of change between two points is the same thing as the slope of the secant line. And as those points get closer and closer together, and as the secant line is connecting two closer and closer points together, as that distance between the points, between the x values of the points approach zero, very interesting things are going to happen."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, this three is right at the interface between these two clauses or these two cases. We go to this first case when x is between zero and three, when it's greater than zero and less than three, and then at three, we hit this case. So in order to find the limit, we want to find the limit from the left-hand side, which will have us dealing with this situation, because if we're less than three, we're in this clause, and we also want to find the limit from the right-hand side, which would put us in this clause right over here. And then if both of those limits exist and if they are the same, then that is going to be the limit of this. So let's do that. So let me first go from the left-hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left, of g of x."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then if both of those limits exist and if they are the same, then that is going to be the limit of this. So let's do that. So let me first go from the left-hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left, of g of x. Well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the limit as x approaches three from values less than three, so we're gonna approach from the left, of g of x. Well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of three x. And since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all x's greater than zero, well, we can just substitute three in here to see what it would be approaching."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of three x. And since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all x's greater than zero, well, we can just substitute three in here to see what it would be approaching. So this would be, this would be equal log of, log of three times three, or logarithm of nine. And once again, when people just write log here without writing the base, it's implied that we're dealing, that it is 10 right over here. So this is log base 10."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all x's greater than zero, well, we can just substitute three in here to see what it would be approaching. So this would be, this would be equal log of, log of three times three, or logarithm of nine. And once again, when people just write log here without writing the base, it's implied that we're dealing, that it is 10 right over here. So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right-hand side, from values greater than three."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right-hand side, from values greater than three. Well, we are now going to be in this scenario right over there. So this is going to be equal to the limit as x approaches three from the positive direction, from the right-hand side, of, well, g of x is in this clause when we are greater than three. So four minus x times log of nine."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's think about the situation where we are approaching three from the right-hand side, from values greater than three. Well, we are now going to be in this scenario right over there. So this is going to be equal to the limit as x approaches three from the positive direction, from the right-hand side, of, well, g of x is in this clause when we are greater than three. So four minus x times log of nine. Four times log of nine. And this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant. Log base 10 of nine, it's gonna be some number close to one."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So four minus x times log of nine. Four times log of nine. And this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant. Log base 10 of nine, it's gonna be some number close to one. This is just, this expression would actually define a line. For x greater than or equal to three, g of x is just a line, even though it looks a little bit complicated. And so this is actually defined for all real numbers."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Log base 10 of nine, it's gonna be some number close to one. This is just, this expression would actually define a line. For x greater than or equal to three, g of x is just a line, even though it looks a little bit complicated. And so this is actually defined for all real numbers. And so, and it's also, it's continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well, we can just evaluate it at three. So it's going to be four minus three times log of nine."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is actually defined for all real numbers. And so, and it's also, it's continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well, we can just evaluate it at three. So it's going to be four minus three times log of nine. Well, that's just one. So that's equal to log base 10 of nine. So the limit from the left equals the limit from the right."}, {"video_title": "Analyzing functions for discontinuities (continuous example)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be four minus three times log of nine. Well, that's just one. So that's equal to log base 10 of nine. So the limit from the left equals the limit from the right. They're both log nine. So the answer here is log, log of nine. Log, log of nine, and we are done."}, {"video_title": "Finding specific antiderivatives exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is f of zero? So to evaluate f of zero, let's take the antiderivative of f prime of x, and then we're going to have a constant of integration there, so we can use the information that they gave us up here, that f of seven is equal to this. This might look like an expression, well it is an expression, but it's really just a number. There's no variables in this. And so we can use that to solve for our constant of integration, and then we will have fully known what f of x is, and we can use that to evaluate f of zero. So let's just do it. So if f prime of x is equal to 5e to the x, then f of x is going to be equal to the antiderivative of f prime of x, so the antiderivative of 5e to the x, dx."}, {"video_title": "Finding specific antiderivatives exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "There's no variables in this. And so we can use that to solve for our constant of integration, and then we will have fully known what f of x is, and we can use that to evaluate f of zero. So let's just do it. So if f prime of x is equal to 5e to the x, then f of x is going to be equal to the antiderivative of f prime of x, so the antiderivative of 5e to the x, dx. And this is the thing that I always find amazing about exponentials, and actually let me just take a step. I'll take that five out of the integral so it becomes a little bit more obvious. And so the antiderivative of e to the x, well that's just e to the x, because the derivative of e to the x is e to the x, which I find amazing every time I have to manipulate or take the derivative or antiderivative of e to the x."}, {"video_title": "Finding specific antiderivatives exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if f prime of x is equal to 5e to the x, then f of x is going to be equal to the antiderivative of f prime of x, so the antiderivative of 5e to the x, dx. And this is the thing that I always find amazing about exponentials, and actually let me just take a step. I'll take that five out of the integral so it becomes a little bit more obvious. And so the antiderivative of e to the x, well that's just e to the x, because the derivative of e to the x is e to the x, which I find amazing every time I have to manipulate or take the derivative or antiderivative of e to the x. So this is going to be 5e to the x plus c. And you can verify, take the derivative of 5e to the x plus c. The derivative of 5e to the x, well that's 5e to the x, so that works out, and the derivative of c is zero, so you wouldn't see it over here. So now let's use this information to figure out what c is so that we know exactly what f of x is, and then we can evaluate f of zero. So we know that f of seven, so when x is equal to seven, we're going to, this expression is going to evaluate to this thing, 40 plus 5e to the seven."}, {"video_title": "Finding specific antiderivatives exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the antiderivative of e to the x, well that's just e to the x, because the derivative of e to the x is e to the x, which I find amazing every time I have to manipulate or take the derivative or antiderivative of e to the x. So this is going to be 5e to the x plus c. And you can verify, take the derivative of 5e to the x plus c. The derivative of 5e to the x, well that's 5e to the x, so that works out, and the derivative of c is zero, so you wouldn't see it over here. So now let's use this information to figure out what c is so that we know exactly what f of x is, and then we can evaluate f of zero. So we know that f of seven, so when x is equal to seven, we're going to, this expression is going to evaluate to this thing, 40 plus 5e to the seven. So five times e to the seventh power plus c is equal to 40, is equal to 40 plus 5e to the seventh power. And notice all I did is said okay, f of seven. Well if this is f of x, f of, let me write this down."}, {"video_title": "Finding specific antiderivatives exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that f of seven, so when x is equal to seven, we're going to, this expression is going to evaluate to this thing, 40 plus 5e to the seven. So five times e to the seventh power plus c is equal to 40, is equal to 40 plus 5e to the seventh power. And notice all I did is said okay, f of seven. Well if this is f of x, f of, let me write this down. So if this is f of seven, if this is f of x, I just replaced the x with a seven to find f of seven, and we know that f of seven is also going to be equal to that, they gave us that information. But when you just look at this, it's pretty easy to figure out what c is going to be. You can subtract 5e to the seventh from both sides, and you see that c is equal to 40."}, {"video_title": "Finding specific antiderivatives exponential function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well if this is f of x, f of, let me write this down. So if this is f of seven, if this is f of x, I just replaced the x with a seven to find f of seven, and we know that f of seven is also going to be equal to that, they gave us that information. But when you just look at this, it's pretty easy to figure out what c is going to be. You can subtract 5e to the seventh from both sides, and you see that c is equal to 40. And so we can rewrite f of x, we can say that f of x is equal to 5e to the x plus c, which is 40. And so now from that, we can evaluate f of zero. F of zero is going to be five times e to the zero power plus 40, e to the zero is one, so it's going to be five times one, which is just five, plus 40, which is equal to 45, and we're done."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I encourage you to pause this video and see if you can figure this out on your own. All right, now let's work through it together. So some of you might have immediately said, hey, this is the form of a differential equation where the solution is going to be an exponential and you just got right to it. But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equations."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I could add a constant, but I'm gonna add a constant on the right hand side so there's no reason to add two arbitrary constants on both sides. I could just add one on one side."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I could add a constant, but I'm gonna add a constant on the right hand side so there's no reason to add two arbitrary constants on both sides. I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the three x. So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the three x. So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in general terms."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative. So let's do that."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative. So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in. And our particular solution is going to be y is equal to c. C is two e to the negative third power times e to the three x. Now I have, I'm taking the product of two things with the same base."}, {"video_title": "Worked example exponential solution to differential equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in. And our particular solution is going to be y is equal to c. C is two e to the negative third power times e to the three x. Now I have, I'm taking the product of two things with the same base. I can add the exponents. So I could say y is equal to two times e to the three x, e to the three x, and then I'll add the exponents to three x minus three. And there you go."}, {"video_title": "2015 AP Calculus AB BC 1d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "For t greater than 8, water continues to flow into and out of the pipe at the given rate until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the pipe will begin to overflow. Alright, so the pipe is going to overflow, so we want to figure out that gives the time w when the pipe begins to overflow. So the pipe will begin to overflow when it crosses 50 cubic feet of water. You could say right when it hits 50 cubic feet of water, then it will begin to overflow. So we can figure out at what time does the pipe have 50 cubic feet of water in it. And so we could just say, well, w of this time, so I used uppercase w as my function for how much water is in the pipe, so capital w of lowercase w is going to be equal to 50."}, {"video_title": "2015 AP Calculus AB BC 1d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the pipe will begin to overflow when it crosses 50 cubic feet of water. You could say right when it hits 50 cubic feet of water, then it will begin to overflow. So we can figure out at what time does the pipe have 50 cubic feet of water in it. And so we could just say, well, w of this time, so I used uppercase w as my function for how much water is in the pipe, so capital w of lowercase w is going to be equal to 50. And so you would just solve for the w. And they say write, but do not solve, an equation. Well, just to make this a little bit clearer, if uppercase w of lowercase w is going to be 30 plus the integral from 0 to w, and actually now since I don't have t as one of my bounds, I could just say r of t minus d of t dt, so let me just do that. r of t minus d of t dt, so this is the amount of total water in the pipe at time w. Well, this is going to be equal to 50."}, {"video_title": "2015 AP Calculus AB BC 1d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we could just say, well, w of this time, so I used uppercase w as my function for how much water is in the pipe, so capital w of lowercase w is going to be equal to 50. And so you would just solve for the w. And they say write, but do not solve, an equation. Well, just to make this a little bit clearer, if uppercase w of lowercase w is going to be 30 plus the integral from 0 to w, and actually now since I don't have t as one of my bounds, I could just say r of t minus d of t dt, so let me just do that. r of t minus d of t dt, so this is the amount of total water in the pipe at time w. Well, this is going to be equal to 50. So we have just written an equation involving one or more integrals that gives the time w when the pipe will begin to overflow. So if you could solve for w, that's the time that the pipe begins to overflow. And we are assuming that it doesn't just right get to 50 and then somehow come back down, that it will cross 50 at this time right here."}, {"video_title": "2015 AP Calculus AB BC 1d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "r of t minus d of t dt, so this is the amount of total water in the pipe at time w. Well, this is going to be equal to 50. So we have just written an equation involving one or more integrals that gives the time w when the pipe will begin to overflow. So if you could solve for w, that's the time that the pipe begins to overflow. And we are assuming that it doesn't just right get to 50 and then somehow come back down, that it will cross 50 at this time right here. And you could test that a little bit more if you want. You could try the slightly larger w, or you could see that the rate that you have more flowing in than flowing out at that time, so this r of w is going to be greater than d of w, so it means you're only going to be increasing, so you're going to cross over right at that time. So if you wanted that, want the w though, you'd solve this."}, {"video_title": "2015 AP Calculus AB BC 1d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we are assuming that it doesn't just right get to 50 and then somehow come back down, that it will cross 50 at this time right here. And you could test that a little bit more if you want. You could try the slightly larger w, or you could see that the rate that you have more flowing in than flowing out at that time, so this r of w is going to be greater than d of w, so it means you're only going to be increasing, so you're going to cross over right at that time. So if you wanted that, want the w though, you'd solve this. Now another option, you could say, okay, we know w is going to be greater than 8, so you could say, okay, how much water do we have right at time equals 8? We figured that out in the last problem. And so you could say we, at time equals 8, we have that much, 48.544."}, {"video_title": "2015 AP Calculus AB BC 1d   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you wanted that, want the w though, you'd solve this. Now another option, you could say, okay, we know w is going to be greater than 8, so you could say, okay, how much water do we have right at time equals 8? We figured that out in the last problem. And so you could say we, at time equals 8, we have that much, 48.544. This is an approximation, but it's pretty close. Plus the amount of water we accumulate between time 8 and time w of r of t minus d of t, dt, is equal to 50. Either one of these would get you to the same place."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're told that the tangent line to the graph of function at the point two comma three passes through the point seven comma six. Find f prime of two. So whenever you see something like this, it doesn't hurt to try to visualize it. You might want to draw it out or just visualize it in your head, but since you can't get in my head, I will draw it out. So let me draw the information that they are giving us. So that's x-axis, that is the y-axis. Let's see, the relevant points here are two comma three and seven comma six."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You might want to draw it out or just visualize it in your head, but since you can't get in my head, I will draw it out. So let me draw the information that they are giving us. So that's x-axis, that is the y-axis. Let's see, the relevant points here are two comma three and seven comma six. So let me go one, two, three, four, five, six, seven along the x-axis. And I'm gonna go one, two, three, four, five, and six along the y-axis. And now this point, so we have the point two comma three."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's see, the relevant points here are two comma three and seven comma six. So let me go one, two, three, four, five, six, seven along the x-axis. And I'm gonna go one, two, three, four, five, and six along the y-axis. And now this point, so we have the point two comma three. So let me mark that. So two comma three is right over there. So that's two comma three."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now this point, so we have the point two comma three. So let me mark that. So two comma three is right over there. So that's two comma three. And we also have the point seven comma six. Seven comma six is going to be right over there. Seven comma six."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's two comma three. And we also have the point seven comma six. Seven comma six is going to be right over there. Seven comma six. Now let's remind ourselves what they're saying. They're saying the tangent line to the graph of function f at this point passes through the point seven comma six. So if it's the tangent line to the graph at that point, it must go through two comma three."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Seven comma six. Now let's remind ourselves what they're saying. They're saying the tangent line to the graph of function f at this point passes through the point seven comma six. So if it's the tangent line to the graph at that point, it must go through two comma three. That's the only place where it intersects our graph, and it goes through seven comma six. So you only need two points to define a line. And so the tangent line is going to look like, it's going to look like, let me see if I can, no, that's not right."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if it's the tangent line to the graph at that point, it must go through two comma three. That's the only place where it intersects our graph, and it goes through seven comma six. So you only need two points to define a line. And so the tangent line is going to look like, it's going to look like, let me see if I can, no, that's not right. Let me draw it. Like it's going to look, oh, that's not exactly right. Let me try one more time."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the tangent line is going to look like, it's going to look like, let me see if I can, no, that's not right. Let me draw it. Like it's going to look, oh, that's not exactly right. Let me try one more time. Okay, there you go. So the tangent line is going to look like that. It goes, it's tangent to f right at two comma three, and it goes through the point seven comma six."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me try one more time. Okay, there you go. So the tangent line is going to look like that. It goes, it's tangent to f right at two comma three, and it goes through the point seven comma six. And so we don't know anything other than f, but we can imagine what f looks like. Our function f could, so our function f, it could look something like this. It just has to be tangent."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It goes, it's tangent to f right at two comma three, and it goes through the point seven comma six. And so we don't know anything other than f, but we can imagine what f looks like. Our function f could, so our function f, it could look something like this. It just has to be tangent. So that line has to be tangent to our function right at that point. So our function f could look something like that. So when they say find f prime of two, they're really saying what is the slope of the tangent line when x is equal to two?"}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "It just has to be tangent. So that line has to be tangent to our function right at that point. So our function f could look something like that. So when they say find f prime of two, they're really saying what is the slope of the tangent line when x is equal to two? So when x is equal to two, well, the slope of the tangent line is the slope of this line. They gave us, they gave us the two points that sit on the tangent line. So we just have to figure out its slope, because that is going to be the rate of change of that function right over there."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when they say find f prime of two, they're really saying what is the slope of the tangent line when x is equal to two? So when x is equal to two, well, the slope of the tangent line is the slope of this line. They gave us, they gave us the two points that sit on the tangent line. So we just have to figure out its slope, because that is going to be the rate of change of that function right over there. It's derivative. It's going to be the slope of the tangent line, because this is the tangent line. So let's do that."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to figure out its slope, because that is going to be the rate of change of that function right over there. It's derivative. It's going to be the slope of the tangent line, because this is the tangent line. So let's do that. So as we know, slope is change in y over change in x. So if we change our, to go from two comma three to seven comma six, our change in x, change in x, we go from x equals two to x equals seven, so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six, so our change in y is equal to three."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So as we know, slope is change in y over change in x. So if we change our, to go from two comma three to seven comma six, our change in x, change in x, we go from x equals two to x equals seven, so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six, so our change in y is equal to three. So our change in y over change in x is going to be three over five, which is the slope of this line, which is the derivative of the function at two, because this is the tangent line at x equals two. Let's do another one of these. So, for a function g, we are given that g of negative one equals three, and g prime of negative one is equal to negative two."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And our change in y, our change in y, we go from y equals three to y equals six, so our change in y is equal to three. So our change in y over change in x is going to be three over five, which is the slope of this line, which is the derivative of the function at two, because this is the tangent line at x equals two. Let's do another one of these. So, for a function g, we are given that g of negative one equals three, and g prime of negative one is equal to negative two. What is the equation of the tangent line to the graph of g at x equals negative one? All right, so once again, I think it will be helpful to graph this. So, we have our y-axis."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, for a function g, we are given that g of negative one equals three, and g prime of negative one is equal to negative two. What is the equation of the tangent line to the graph of g at x equals negative one? All right, so once again, I think it will be helpful to graph this. So, we have our y-axis. We have our x-axis. And let's see. We say for a function g, we are given that g of negative one is equal to three."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, we have our y-axis. We have our x-axis. And let's see. We say for a function g, we are given that g of negative one is equal to three. So the point negative one comma three is on our function. So this is negative one, and then we have one, two, and three. So that's that right over there."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "We say for a function g, we are given that g of negative one is equal to three. So the point negative one comma three is on our function. So this is negative one, and then we have one, two, and three. So that's that right over there. That is the point, that is the point negative one comma three. It's going to be on our function. And we also know that g prime of negative one is equal to negative two."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's that right over there. That is the point, that is the point negative one comma three. It's going to be on our function. And we also know that g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us. The slope of the tangent line when x is equal to negative one is equal to negative two."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we also know that g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us. The slope of the tangent line when x is equal to negative one is equal to negative two. So I could use that information to actually draw the tangent line. So let me see if I can, let me see if I can do this. So it will look, so I think it will, let me just draw it like this."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope of the tangent line when x is equal to negative one is equal to negative two. So I could use that information to actually draw the tangent line. So let me see if I can, let me see if I can do this. So it will look, so I think it will, let me just draw it like this. So it's gonna go, so it's a slope of negative two is going to look something like that. So as we can see, if we move positive one in the x direction, we go down two in the y direction. So that has a slope of negative two."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it will look, so I think it will, let me just draw it like this. So it's gonna go, so it's a slope of negative two is going to look something like that. So as we can see, if we move positive one in the x direction, we go down two in the y direction. So that has a slope of negative two. And so you might say, well where is g? Well we could draw what g could look like. G might look something like this."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that has a slope of negative two. And so you might say, well where is g? Well we could draw what g could look like. G might look something like this. Might look something like that right over there where that is the tangent line. We could make g do all sorts of crazy things after that. But all we really care about is the equation for this green line."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "G might look something like this. Might look something like that right over there where that is the tangent line. We could make g do all sorts of crazy things after that. But all we really care about is the equation for this green line. And there's a couple of ways that you could do this. You could say, well look, a line is generally, there's a bunch of different ways where you can define the equation for a line. You could say a line has a form y is equal to mx plus b where m is the slope and b is the y intercept."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "But all we really care about is the equation for this green line. And there's a couple of ways that you could do this. You could say, well look, a line is generally, there's a bunch of different ways where you can define the equation for a line. You could say a line has a form y is equal to mx plus b where m is the slope and b is the y intercept. Well we already know what the slope of this line is. It is negative two. So we could say y is equal to negative two, negative two times x, times x plus b."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say a line has a form y is equal to mx plus b where m is the slope and b is the y intercept. Well we already know what the slope of this line is. It is negative two. So we could say y is equal to negative two, negative two times x, times x plus b. And then to solve for b, we know that the point negative one comma three is on this line. And this goes back to some of your algebra one that you might have learned a few years ago. So let's substitute negative one and three for x and y."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say y is equal to negative two, negative two times x, times x plus b. And then to solve for b, we know that the point negative one comma three is on this line. And this goes back to some of your algebra one that you might have learned a few years ago. So let's substitute negative one and three for x and y. So when y is equal to three, so three, three is equal to, is equal to negative two, negative two times x times negative one, times negative one plus b, plus b. And so let's see, this is negative two times negative one is positive two. And so if you subtract two from both sides, you get one is equal to b."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's substitute negative one and three for x and y. So when y is equal to three, so three, three is equal to, is equal to negative two, negative two times x times negative one, times negative one plus b, plus b. And so let's see, this is negative two times negative one is positive two. And so if you subtract two from both sides, you get one is equal to b. And there you have it. That is the equation of our line. Y is equal to negative two x plus one."}, {"video_title": "The derivative & tangent line equations   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if you subtract two from both sides, you get one is equal to b. And there you have it. That is the equation of our line. Y is equal to negative two x plus one. And there's other ways that you could have done this. You could have written the line in point slope form or you could have done it this way. You could have written it in standard form."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So let's say that I have the graph y is equal to square root of x. So let's do it so it looks something like this. So that right over there is y is equal to the square root of x. And let's say I also have the graph of y equals x. So let's say y equals x looks something like this. Looks just like that. y equals x."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "And let's say I also have the graph of y equals x. So let's say y equals x looks something like this. Looks just like that. y equals x. And what I care about now is the solid I get if I were to rotate the area between these two things around the x-axis. So let's try to visualize it. So the outside is going to be kind of a truffle shape."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "y equals x. And what I care about now is the solid I get if I were to rotate the area between these two things around the x-axis. So let's try to visualize it. So the outside is going to be kind of a truffle shape. It's going to look like a truffle shape. And then we hollow out a cone inside of it. So my best attempt to draw this shape."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So the outside is going to be kind of a truffle shape. It's going to look like a truffle shape. And then we hollow out a cone inside of it. So my best attempt to draw this shape. So it's going to look something like... So the outside is going to look something like this. It's going to look something like that."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So my best attempt to draw this shape. So it's going to look something like... So the outside is going to look something like this. It's going to look something like that. And we care about the interval. We care about the interval between the points that they intersect. So between this point and this point here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "It's going to look something like that. And we care about the interval. We care about the interval between the points that they intersect. So between this point and this point here. So the outside is going to look something like this. So this is the base of the truffle. That is the base of the truffle."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So between this point and this point here. So the outside is going to look something like this. So this is the base of the truffle. That is the base of the truffle. It's going to have this kind of truffle shape on the outside. But I guess maybe we are on some type of a diet. We don't want to eat the entire truffle."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "That is the base of the truffle. It's going to have this kind of truffle shape on the outside. But I guess maybe we are on some type of a diet. We don't want to eat the entire truffle. So we carve out a cone on the inside. So the inside of it is essentially hollow except for this kind of shell part. So we carve out a cone in the center."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "We don't want to eat the entire truffle. So we carve out a cone on the inside. So the inside of it is essentially hollow except for this kind of shell part. So we carve out a cone in the center. So we rotate it around the x-axis. Truffle on the outside. Carved out a cone on the inside."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So we carve out a cone in the center. So we rotate it around the x-axis. Truffle on the outside. Carved out a cone on the inside. So what's going to be the volume of that thing? So it's essentially, if we take a slice of our figure, this is going to be the wall. And we're essentially going to take the volume of this entire wall that we're rotating around the x-axis."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "Carved out a cone on the inside. So what's going to be the volume of that thing? So it's essentially, if we take a slice of our figure, this is going to be the wall. And we're essentially going to take the volume of this entire wall that we're rotating around the x-axis. So how do we do that? Well, it might dawn on you that if we found the volume of the truffle if it was not carved out, and then subtract from that the volume of the cone, we would essentially find out the volume of the space in between the outside of the truffle and the cone part of the truffle. So how would we do that?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "And we're essentially going to take the volume of this entire wall that we're rotating around the x-axis. So how do we do that? Well, it might dawn on you that if we found the volume of the truffle if it was not carved out, and then subtract from that the volume of the cone, we would essentially find out the volume of the space in between the outside of the truffle and the cone part of the truffle. So how would we do that? Well, so to find the volume of the outer shape, so let me draw it over here. So actually, let me just draw it over here. If we think about the volume of the outer shape, once again, we can use the disk method."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So how would we do that? Well, so to find the volume of the outer shape, so let me draw it over here. So actually, let me just draw it over here. If we think about the volume of the outer shape, once again, we can use the disk method. So at any given point in time, our radius for one of our disks is going to be equal to the function. Let's rotate that disk around. Actually, let me do it in a different color."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "If we think about the volume of the outer shape, once again, we can use the disk method. So at any given point in time, our radius for one of our disks is going to be equal to the function. Let's rotate that disk around. Actually, let me do it in a different color. It's hard to see that disk, since it's in the same magenta. So this is our radius, and let's rotate it around. So I'm rotating the disk around."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "Actually, let me do it in a different color. It's hard to see that disk, since it's in the same magenta. So this is our radius, and let's rotate it around. So I'm rotating the disk around. This is our face of the disk. That's our face of the disk. It's going to have a depth of dx."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So I'm rotating the disk around. This is our face of the disk. That's our face of the disk. It's going to have a depth of dx. We've seen this multiple times. It's going to have a depth of dx. So the volume of this disk is going to be our depth, dx, times the area of the face."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "It's going to have a depth of dx. We've seen this multiple times. It's going to have a depth of dx. So the volume of this disk is going to be our depth, dx, times the area of the face. The area of the face is going to be pi times the radius squared. The radius is going to be equal to the value of the outer function. In this case, it's square root of x."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So the volume of this disk is going to be our depth, dx, times the area of the face. The area of the face is going to be pi times the radius squared. The radius is going to be equal to the value of the outer function. In this case, it's square root of x. So it's going to be pi times our radius squared, which is pi times square root of x squared. And so if we want to find the volume of the entire outer thimble or truffle or whatever we want to call it, before we even carve out the center, we just take a sum of a bunch of these disks that we've created. So that's one disk."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "In this case, it's square root of x. So it's going to be pi times our radius squared, which is pi times square root of x squared. And so if we want to find the volume of the entire outer thimble or truffle or whatever we want to call it, before we even carve out the center, we just take a sum of a bunch of these disks that we've created. So that's one disk. We would have another disk over here, another disk over here. For each x, we have another disk. And as we go, as x's get larger and larger, the disks have a larger and larger radius."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So that's one disk. We would have another disk over here, another disk over here. For each x, we have another disk. And as we go, as x's get larger and larger, the disks have a larger and larger radius. So we're going to sum up all of those disks, and we're going to take the limit as each of those disks get infinitely thin, and we have an infinite number of them. But we have to figure out our boundaries of integration. So what are our boundaries of integration?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "And as we go, as x's get larger and larger, the disks have a larger and larger radius. So we're going to sum up all of those disks, and we're going to take the limit as each of those disks get infinitely thin, and we have an infinite number of them. But we have to figure out our boundaries of integration. So what are our boundaries of integration? What are the two points right over here where they intersect? Well, we could just set these two things to be equal to each other. If you just said x is equal to square root of x, when does x equal square root of x?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So what are our boundaries of integration? What are the two points right over here where they intersect? Well, we could just set these two things to be equal to each other. If you just said x is equal to square root of x, when does x equal square root of x? I mean, you could square both sides of this. You could say, when does x squared equal x? You could, well, we could keep it there."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "If you just said x is equal to square root of x, when does x equal square root of x? I mean, you could square both sides of this. You could say, when does x squared equal x? You could, well, we could keep it there. You could kind of solve this. There's multiple ways you could do it, but you could solve this kind of just thinking about it. If x is equal to 0, x squared is equal to x, and you see that on the graph right over here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "You could, well, we could keep it there. You could kind of solve this. There's multiple ways you could do it, but you could solve this kind of just thinking about it. If x is equal to 0, x squared is equal to x, and you see that on the graph right over here. x is equal to 0. And also, 1 squared is equal to 1. 1 is equal to the square root of 1."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "If x is equal to 0, x squared is equal to x, and you see that on the graph right over here. x is equal to 0. And also, 1 squared is equal to 1. 1 is equal to the square root of 1. You could have done other things. You could say, okay, x squared minus x is equal to 0. You could factor out an x."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "1 is equal to the square root of 1. You could have done other things. You could say, okay, x squared minus x is equal to 0. You could factor out an x. You get x times x minus 1 is equal to 0. And so either one of these could be equal to 0. So x is equal to 0, or x minus 1 is equal to 0."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "You could factor out an x. You get x times x minus 1 is equal to 0. And so either one of these could be equal to 0. So x is equal to 0, or x minus 1 is equal to 0. And then you get x equals 0, or x is equal to 1. x is equal to 0, or x equals 1, which gives us our boundaries of integration. So this goes from x equals 0 to x equals 1. And so for the outside of our shape, we can now figure out the volume."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So x is equal to 0, or x minus 1 is equal to 0. And then you get x equals 0, or x is equal to 1. x is equal to 0, or x equals 1, which gives us our boundaries of integration. So this goes from x equals 0 to x equals 1. And so for the outside of our shape, we can now figure out the volume. But we're not done. We also need to figure out the volume of the inside of our shape that we're going to take out. So we're going to subtract out that volume."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "And so for the outside of our shape, we can now figure out the volume. But we're not done. We also need to figure out the volume of the inside of our shape that we're going to take out. So we're going to subtract out that volume. So we're going to subtract out a volume. Our x values, once again, are going between 0 and 1. And so let's think about those disks."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So we're going to subtract out that volume. So we're going to subtract out a volume. Our x values, once again, are going between 0 and 1. And so let's think about those disks. So let's construct a disk on the inside right over here. So if I construct a disk on the inside, so now I'm carving out the cone part of it. What is the area of the face of one of those disks?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "And so let's think about those disks. So let's construct a disk on the inside right over here. So if I construct a disk on the inside, so now I'm carving out the cone part of it. What is the area of the face of one of those disks? Well, it's going to be pi times the radius squared. In this case, the radius is going to be equal to the value of this inner function, which is just x. So pi times, and so this is just y is equal to x."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "What is the area of the face of one of those disks? Well, it's going to be pi times the radius squared. In this case, the radius is going to be equal to the value of this inner function, which is just x. So pi times, and so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely thin depth right over here, so it's going to be dx."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "So pi times, and so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely thin depth right over here, so it's going to be dx. And so the volume of our truffle with the cone carved out is going to be this integral minus this integral right over here. And we could evaluate it just like that, or we could even say, OK, we could factor out a pi out of both of them. Actually, there's multiple ways that we could write it, but let's just evaluate it like this, and then I'll generalize it in the next video."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "If you imagine a quarter that has an infinitely thin depth right over here, so it's going to be dx. And so the volume of our truffle with the cone carved out is going to be this integral minus this integral right over here. And we could evaluate it just like that, or we could even say, OK, we could factor out a pi out of both of them. Actually, there's multiple ways that we could write it, but let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. We take the pi outside. The square root of x squared is going to be x dx minus the integral."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "Actually, there's multiple ways that we could write it, but let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. We take the pi outside. The square root of x squared is going to be x dx minus the integral. We can factor the pi out from 0 to 1 of x squared dx. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to, and I'm going to arbitrarily switch colors just because the green is getting monotonous, pi times 1 squared over 2 minus 0 squared over 2."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "The square root of x squared is going to be x dx minus the integral. We can factor the pi out from 0 to 1 of x squared dx. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to, and I'm going to arbitrarily switch colors just because the green is getting monotonous, pi times 1 squared over 2 minus 0 squared over 2. I could write it squared. 1 squared over 2 minus 0 squared over 2 minus pi times 1 to the third over 3 minus 0 to the third over 3. And so we get this is equal to, let me do it in that same blue color, it's equal to, so this is just simplified, this is just 0 right over here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "This expression is equal to, and I'm going to arbitrarily switch colors just because the green is getting monotonous, pi times 1 squared over 2 minus 0 squared over 2. I could write it squared. 1 squared over 2 minus 0 squared over 2 minus pi times 1 to the third over 3 minus 0 to the third over 3. And so we get this is equal to, let me do it in that same blue color, it's equal to, so this is just simplified, this is just 0 right over here. This is 1 squared over 2, which is just 1 half. So it's just pi over 2, 1 half times pi minus, well this is just 0, this is 1 third minus pi over 3. Minus pi over 3 and then to simplify this it's just really subtracting fractions."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "And so we get this is equal to, let me do it in that same blue color, it's equal to, so this is just simplified, this is just 0 right over here. This is 1 squared over 2, which is just 1 half. So it's just pi over 2, 1 half times pi minus, well this is just 0, this is 1 third minus pi over 3. Minus pi over 3 and then to simplify this it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "Minus pi over 3 and then to simplify this it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. Pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with 3 of something minus 2 of something, you end up with 1 of something."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method)   Khan Academy.mp3", "Sentence": "This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. Pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with 3 of something minus 2 of something, you end up with 1 of something. We end up with 1 pi over 6. And we are done. We were able to find the volume of that wacky, kind of gutted out truffle."}, {"video_title": "Interpreting direction of motion from velocity-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "For each point on the graph, is the object moving forward, backward, or neither? So pause this video and see if you can figure that out. All right, now let's do this together. And so we can see these different points on this velocity versus time graph. And the important thing to realize is, is if the velocity is positive, we're moving forward. If the velocity is negative, we're moving backward. And if the velocity is zero, we're not moving either forward nor backwards, or neither forward nor backwards."}, {"video_title": "Interpreting direction of motion from velocity-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can see these different points on this velocity versus time graph. And the important thing to realize is, is if the velocity is positive, we're moving forward. If the velocity is negative, we're moving backward. And if the velocity is zero, we're not moving either forward nor backwards, or neither forward nor backwards. So right over here, we see that our velocity is positive. It's a positive two meters per second. So that means that we are moving forward."}, {"video_title": "Interpreting direction of motion from velocity-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if the velocity is zero, we're not moving either forward nor backwards, or neither forward nor backwards. So right over here, we see that our velocity is positive. It's a positive two meters per second. So that means that we are moving forward. Now over here, our velocity is zero meters per second. So this is neither. Now over here, our velocity is negative four meters per second."}, {"video_title": "Interpreting direction of motion from velocity-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that means that we are moving forward. Now over here, our velocity is zero meters per second. So this is neither. Now over here, our velocity is negative four meters per second. So one way to think about it is we're moving four meters per second backward. So I'll write backward. Now this is interesting, this last point."}, {"video_title": "Interpreting direction of motion from velocity-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now over here, our velocity is negative four meters per second. So one way to think about it is we're moving four meters per second backward. So I'll write backward. Now this is interesting, this last point. Because you might be tempted to say, all right, I'm oscillating, I'm going up, then I'm going down, then I'm going back up. Maybe I'm moving forward here. But remember, what we're thinking about here, this isn't position versus time, this is velocity versus time."}, {"video_title": "Interpreting direction of motion from velocity-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this is interesting, this last point. Because you might be tempted to say, all right, I'm oscillating, I'm going up, then I'm going down, then I'm going back up. Maybe I'm moving forward here. But remember, what we're thinking about here, this isn't position versus time, this is velocity versus time. So if our velocity is negative, we're moving backwards. And here, our velocity is still negative. It's becoming less negative, but it's still negative."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that f of x is equal to x times e to the negative 2x squared. And we want to find any critical numbers for f. So I encourage you to pause this video and think about, can you find any critical numbers of f? So assuming you've given a go at it, so let's just remind ourselves what a critical number is. So we would say c is a critical number of f if and only if, I'll write if with two f's, short for if and only if, f prime of c is equal to 0 or f prime of c is undefined. So if we look for the critical numbers for f, we want to figure out all the places where the derivative of this with respect to x is either equal to 0 or it is undefined. So let's think about how we can find the derivative of this. So let's see, f prime of x is going to be, well let's see, we're going to have to apply some combination of the product rule and the chain rule."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would say c is a critical number of f if and only if, I'll write if with two f's, short for if and only if, f prime of c is equal to 0 or f prime of c is undefined. So if we look for the critical numbers for f, we want to figure out all the places where the derivative of this with respect to x is either equal to 0 or it is undefined. So let's think about how we can find the derivative of this. So let's see, f prime of x is going to be, well let's see, we're going to have to apply some combination of the product rule and the chain rule. So it's going to be the derivative with respect to x of x of x, so it's going to be that, times e to the negative 2x squared plus the derivative with respect to x of e to the negative 2x squared of e to the negative 2x squared times x. So this is just the product rule right over here, derivative of this, of the x, times e to the negative 2x squared plus the derivative of e to the negative 2x squared times x right over here. So what is this going to be?"}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, f prime of x is going to be, well let's see, we're going to have to apply some combination of the product rule and the chain rule. So it's going to be the derivative with respect to x of x of x, so it's going to be that, times e to the negative 2x squared plus the derivative with respect to x of e to the negative 2x squared of e to the negative 2x squared times x. So this is just the product rule right over here, derivative of this, of the x, times e to the negative 2x squared plus the derivative of e to the negative 2x squared times x right over here. So what is this going to be? Well, all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to 1. So this first part is going to be equal to e to the negative 2x squared. And now the derivative of e to the negative 2x squared over here, I'll do this in this pink color, so this part right over here, that is going to be equal to, or just apply the chain rule, derivative of e to the negative 2x squared with respect to negative 2x squared, well that's just going to be e to the negative 2x squared, and we're going to multiply that times the derivative of negative 2x squared with respect to x."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is this going to be? Well, all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to 1. So this first part is going to be equal to e to the negative 2x squared. And now the derivative of e to the negative 2x squared over here, I'll do this in this pink color, so this part right over here, that is going to be equal to, or just apply the chain rule, derivative of e to the negative 2x squared with respect to negative 2x squared, well that's just going to be e to the negative 2x squared, and we're going to multiply that times the derivative of negative 2x squared with respect to x. And so that's going to be what? Negative 4x. So times negative 4x, and of course we have this x over here."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now the derivative of e to the negative 2x squared over here, I'll do this in this pink color, so this part right over here, that is going to be equal to, or just apply the chain rule, derivative of e to the negative 2x squared with respect to negative 2x squared, well that's just going to be e to the negative 2x squared, and we're going to multiply that times the derivative of negative 2x squared with respect to x. And so that's going to be what? Negative 4x. So times negative 4x, and of course we have this x over here. We have that x over there, and let's see, can we simplify it? Can we simplify it at all? Well obviously both of these terms have an e to the negative 2x squared."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So times negative 4x, and of course we have this x over here. We have that x over there, and let's see, can we simplify it? Can we simplify it at all? Well obviously both of these terms have an e to the negative 2x squared. So I'm going to try to figure out where this is either undefined or where this is equal to 0. So let's think about this a little bit. So let's see, if we factor in e to the negative 2x squared, I'll do that in green, we're going to have, this is equal to e to the negative 2x squared times, we have here 1 minus 4x squared."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well obviously both of these terms have an e to the negative 2x squared. So I'm going to try to figure out where this is either undefined or where this is equal to 0. So let's think about this a little bit. So let's see, if we factor in e to the negative 2x squared, I'll do that in green, we're going to have, this is equal to e to the negative 2x squared times, we have here 1 minus 4x squared. So this is the derivative of f. Now, where would this be undefined or equal to 0? Well, let's see, e to the negative 2x squared, this is going to be defined for any value of x. This part is going to be defined, and this part is also going to be defined for any value of x."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, if we factor in e to the negative 2x squared, I'll do that in green, we're going to have, this is equal to e to the negative 2x squared times, we have here 1 minus 4x squared. So this is the derivative of f. Now, where would this be undefined or equal to 0? Well, let's see, e to the negative 2x squared, this is going to be defined for any value of x. This part is going to be defined, and this part is also going to be defined for any value of x. So there's no point where this is undefined, but let's think about when this is going to be equal to 0. So you have this product of these two expressions equaling 0. e to the negative 2x squared, that'll never be equal to 0. If you get this exponent to be a really, I guess you could say very negative number, you will approach 0, but you'll never get it to be, you'll never get it to be 0."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "This part is going to be defined, and this part is also going to be defined for any value of x. So there's no point where this is undefined, but let's think about when this is going to be equal to 0. So you have this product of these two expressions equaling 0. e to the negative 2x squared, that'll never be equal to 0. If you get this exponent to be a really, I guess you could say very negative number, you will approach 0, but you'll never get it to be, you'll never get it to be 0. So this part here can't be 0, but if the product of two things are 0, at least one of them has to be 0. So the only way that we can get f prime of x to be equal to 0 is when 1 minus 4x squared is equal to 0. So 1 minus 4x squared is equal to 0."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you get this exponent to be a really, I guess you could say very negative number, you will approach 0, but you'll never get it to be, you'll never get it to be 0. So this part here can't be 0, but if the product of two things are 0, at least one of them has to be 0. So the only way that we can get f prime of x to be equal to 0 is when 1 minus 4x squared is equal to 0. So 1 minus 4x squared is equal to 0. Let me rewrite that. 1 minus 4x squared is equal to 0. When does that happen?"}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 1 minus 4x squared is equal to 0. Let me rewrite that. 1 minus 4x squared is equal to 0. When does that happen? And this one we can just solve. Add 4x squared to both sides, you get 1 is equal to 4x squared. Divide both sides by 4, you get 1 fourth is equal to x squared."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "When does that happen? And this one we can just solve. Add 4x squared to both sides, you get 1 is equal to 4x squared. Divide both sides by 4, you get 1 fourth is equal to x squared. And then what x values is this true at? Well, we just take the plus or minus square root of both sides and you get x is equal to plus or minus 1 half. Negative 1 half squared is 1 fourth, positive 1 half squared is 1 fourth."}, {"video_title": "Finding critical points   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Divide both sides by 4, you get 1 fourth is equal to x squared. And then what x values is this true at? Well, we just take the plus or minus square root of both sides and you get x is equal to plus or minus 1 half. Negative 1 half squared is 1 fourth, positive 1 half squared is 1 fourth. So at x equals plus or minus 1 half, f prime or the derivative is equal to 0. So let me write it this way. f prime of 1 half is equal to 0, and you can verify that right over here."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're told to consider the curve given by the equation, they give this equation. It can be shown that the derivative of y with respect to x is equal to this expression, and you could figure that out with just some implicit differentiation and then solving for the derivative of y with respect to x. We've done that in other videos. Write the equation of the horizontal line that is tangent to the curve and is above the x-axis. Pause this video and see if you can have a go at it. So let's just make sure we're visualizing this right. Let me just draw a quick and dirty diagram."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "Write the equation of the horizontal line that is tangent to the curve and is above the x-axis. Pause this video and see if you can have a go at it. So let's just make sure we're visualizing this right. Let me just draw a quick and dirty diagram. If that's my y-axis, this is my x-axis. I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this. Well, there would be two tangent lines that are horizontal based on how I've drawn it."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just draw a quick and dirty diagram. If that's my y-axis, this is my x-axis. I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this. Well, there would be two tangent lines that are horizontal based on how I've drawn it. One might be right over there, so it might be like there, and then another one might be maybe right over here. And they want the equation of the horizontal line that is tangent to the curve and is above the x-axis. So what do we know, what is true if this tangent line is horizontal?"}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, there would be two tangent lines that are horizontal based on how I've drawn it. One might be right over there, so it might be like there, and then another one might be maybe right over here. And they want the equation of the horizontal line that is tangent to the curve and is above the x-axis. So what do we know, what is true if this tangent line is horizontal? Well, that tells us that at this point, dy dx is equal to zero. In fact, that would be true at both of these points. And we know what dy dx is."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what do we know, what is true if this tangent line is horizontal? Well, that tells us that at this point, dy dx is equal to zero. In fact, that would be true at both of these points. And we know what dy dx is. We know that the derivative of y with respect to x is equal to negative two times x plus three over four y to the third power for any x and y. And so when will this equal zero? Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know what dy dx is. We know that the derivative of y with respect to x is equal to negative two times x plus three over four y to the third power for any x and y. And so when will this equal zero? Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't. So when is our numerator going to be zero? When x is equal to negative three. So when x is equal to negative three, the derivative is equal to zero."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't. So when is our numerator going to be zero? When x is equal to negative three. So when x is equal to negative three, the derivative is equal to zero. So what is going to be the corresponding y value when x is equal to negative three? And if we know that, well, this equation is just going to be y is equal to something. It's going to be that y value."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x is equal to negative three, the derivative is equal to zero. So what is going to be the corresponding y value when x is equal to negative three? And if we know that, well, this equation is just going to be y is equal to something. It's going to be that y value. Well, to figure that out, we just take this x equals negative three, substitute it back into our original equation, and then solve for y. So let's do that. So it's going to be negative three squared plus y to the fourth plus six times negative three is equal to seven."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be that y value. Well, to figure that out, we just take this x equals negative three, substitute it back into our original equation, and then solve for y. So let's do that. So it's going to be negative three squared plus y to the fourth plus six times negative three is equal to seven. This is nine, this is negative 18. And so we're gonna get y to the fourth minus nine is equal to seven. Or adding nine to both sides, we get y to the fourth power is equal to 16."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be negative three squared plus y to the fourth plus six times negative three is equal to seven. This is nine, this is negative 18. And so we're gonna get y to the fourth minus nine is equal to seven. Or adding nine to both sides, we get y to the fourth power is equal to 16. And this would tell us that y is going to be equal to plus or minus two. Well, there would be then two horizontal lines. One would be y is equal to two."}, {"video_title": "Horizontal tangent to implicit curve   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or adding nine to both sides, we get y to the fourth power is equal to 16. And this would tell us that y is going to be equal to plus or minus two. Well, there would be then two horizontal lines. One would be y is equal to two. The other is y is equal to negative two. But they want us the equation of the horizontal line that is tangent to the curve and is above the x-axis. So only this one is going to be above the x-axis."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that we're given the equation that y squared minus x squared is equal to four. And our goal is to find the second derivative of y with respect to x. And we wanna find an expression for it in terms of x's and y's. So pause this video and see if you can work through this. All right, now let's do it together. Now some of you might have wanted to solve for y and then use some traditional techniques. But here we have a y squared, and so it might involve a plus or minus square root."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video and see if you can work through this. All right, now let's do it together. Now some of you might have wanted to solve for y and then use some traditional techniques. But here we have a y squared, and so it might involve a plus or minus square root. And so some of y'all might have realized, hey, we can do a little bit of implicit differentiation, which is really just an application of the chain rule. So let's do that. Let's first find the first derivative of y with respect to x."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "But here we have a y squared, and so it might involve a plus or minus square root. And so some of y'all might have realized, hey, we can do a little bit of implicit differentiation, which is really just an application of the chain rule. So let's do that. Let's first find the first derivative of y with respect to x. And to do that, I'll just take the derivative with respect to x of both sides of this equation. And then what do we get? Well, the derivative with respect to x of y squared, we're gonna use the chain rule here."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's first find the first derivative of y with respect to x. And to do that, I'll just take the derivative with respect to x of both sides of this equation. And then what do we get? Well, the derivative with respect to x of y squared, we're gonna use the chain rule here. First, we can take the derivative of y squared with respect to y, which is going to be equal to two y. And then that times the derivative of y with respect to x. Once again, this comes straight out of the chain rule."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the derivative with respect to x of y squared, we're gonna use the chain rule here. First, we can take the derivative of y squared with respect to y, which is going to be equal to two y. And then that times the derivative of y with respect to x. Once again, this comes straight out of the chain rule. And then from that, we will subtract, what's the derivative of x squared with respect to x? Well, that's just going to be two x. And then last but not least, what is the derivative of a constant with respect to x?"}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, this comes straight out of the chain rule. And then from that, we will subtract, what's the derivative of x squared with respect to x? Well, that's just going to be two x. And then last but not least, what is the derivative of a constant with respect to x? Well, it doesn't change, so it's just going to be equal to zero. All right, now we can solve for our first derivative of y with respect to x. Let's do that."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then last but not least, what is the derivative of a constant with respect to x? Well, it doesn't change, so it's just going to be equal to zero. All right, now we can solve for our first derivative of y with respect to x. Let's do that. We can add two x to both sides. And we would get two y times the derivative of y with respect to x is equal to two x. And now I can divide both sides by two y."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do that. We can add two x to both sides. And we would get two y times the derivative of y with respect to x is equal to two x. And now I can divide both sides by two y. And I am going to get that the derivative of y with respect to x is equal to x, x over y. Now, the next step is let's take the derivative of both sides of this with respect to x. And then we can hopefully find our second derivative of y with respect to x."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now I can divide both sides by two y. And I am going to get that the derivative of y with respect to x is equal to x, x over y. Now, the next step is let's take the derivative of both sides of this with respect to x. And then we can hopefully find our second derivative of y with respect to x. And to help us there, actually, let me rewrite this. And I always forget the quotient rule, although it might be a useful thing for you to remember. But I could rewrite this as a product, which will help me at least."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can hopefully find our second derivative of y with respect to x. And to help us there, actually, let me rewrite this. And I always forget the quotient rule, although it might be a useful thing for you to remember. But I could rewrite this as a product, which will help me at least. So I'm gonna rewrite this as the derivative of y with respect to x is equal to x times y to the negative one power, y to the negative one power. And now, if we wanna find the second derivative, we apply the derivative operator on both sides of this equation, derivative with respect to x. And our left-hand side is exactly what we eventually wanted to get."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "But I could rewrite this as a product, which will help me at least. So I'm gonna rewrite this as the derivative of y with respect to x is equal to x times y to the negative one power, y to the negative one power. And now, if we wanna find the second derivative, we apply the derivative operator on both sides of this equation, derivative with respect to x. And our left-hand side is exactly what we eventually wanted to get. So the second derivative of y with respect to x. And what do we get here on the right-hand side? Well, we can apply the product rule."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "And our left-hand side is exactly what we eventually wanted to get. So the second derivative of y with respect to x. And what do we get here on the right-hand side? Well, we can apply the product rule. So first, we can say the derivative of x with respect to x, well, that is just going to be one times the other thing. So times y to the negative one power, y to the negative one power. And then we have plus x times the derivative of y to the negative one."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can apply the product rule. So first, we can say the derivative of x with respect to x, well, that is just going to be one times the other thing. So times y to the negative one power, y to the negative one power. And then we have plus x times the derivative of y to the negative one. So plus x, what's the derivative of y to the negative one power? Well, first, we can find the derivative of y to the negative one power with respect to y. We'll just leverage the power rule there."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we have plus x times the derivative of y to the negative one. So plus x, what's the derivative of y to the negative one power? Well, first, we can find the derivative of y to the negative one power with respect to y. We'll just leverage the power rule there. So that's going to be negative one times y to the negative two power. And then we would multiply that times the derivative of y with respect to x, just an application of the chain rule, times dy dx. And remember, we know what the derivative of y with respect to x is."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "We'll just leverage the power rule there. So that's going to be negative one times y to the negative two power. And then we would multiply that times the derivative of y with respect to x, just an application of the chain rule, times dy dx. And remember, we know what the derivative of y with respect to x is. We already solved for that. It is x over y. So this over here is going to be x over y."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "And remember, we know what the derivative of y with respect to x is. We already solved for that. It is x over y. So this over here is going to be x over y. And so now we just have to simplify this expression. This is going to be equal to, and I'll try to do it part by part, that part right over there is just going to be a one over y. And then all of this business, let's see if I can simplify that, this negative is going to go out front, so minus."}, {"video_title": "Second derivatives (implicit equations) find expression   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this over here is going to be x over y. And so now we just have to simplify this expression. This is going to be equal to, and I'll try to do it part by part, that part right over there is just going to be a one over y. And then all of this business, let's see if I can simplify that, this negative is going to go out front, so minus. And then I'm gonna have x times x in the numerator, and then it's going to be divided by y squared and then divided by another y. So it's going to be minus x squared over y to the third, over y to the third. Or another way to think about it, x squared times y to the negative three."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "In the last video, we found the slope at a particular point of the curve y is equal to x squared. But let's see if we can generalize this and come up with a formula that finds us the slope at any point of the curve y is equal to x squared. So let me redraw my function here. Never hurts to have a nice drawing. So that is my y-axis. That is my x-axis right there. My x-axis, let me draw my curve."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Never hurts to have a nice drawing. So that is my y-axis. That is my x-axis right there. My x-axis, let me draw my curve. It looks something like that. You've seen that multiple times. This is y is equal to x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "My x-axis, let me draw my curve. It looks something like that. You've seen that multiple times. This is y is equal to x squared. So let's be very general right now. Remember, if we want to find, let me just write the definition of our derivative. So if we have some point right here, let's call that x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is y is equal to x squared. So let's be very general right now. Remember, if we want to find, let me just write the definition of our derivative. So if we have some point right here, let's call that x. So we want to be very general. We want to find the slope at the point x. We want to find a function where you give me an x, and I'll tell you the slope at that point."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So if we have some point right here, let's call that x. So we want to be very general. We want to find the slope at the point x. We want to find a function where you give me an x, and I'll tell you the slope at that point. We're going to call that f prime of x. That's going to be the derivative of f of x. But all it does is, look, when you f of x, it's a function that you give it an x, and it tells you the value of that."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We want to find a function where you give me an x, and I'll tell you the slope at that point. We're going to call that f prime of x. That's going to be the derivative of f of x. But all it does is, look, when you f of x, it's a function that you give it an x, and it tells you the value of that. We draw the curve here. With f of x, you give that same x, but it's not going to tell you the value of the curve. It's not going to say, oh, this is your f of x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But all it does is, look, when you f of x, it's a function that you give it an x, and it tells you the value of that. We draw the curve here. With f of x, you give that same x, but it's not going to tell you the value of the curve. It's not going to say, oh, this is your f of x. It's going to give you the value of the slope of the curve at that point. So f of x, if you put it into that function, it should tell you, oh, the slope at that point is equal to, if you put 3 there, you'll say, oh, the slope there is equal to 6. We saw that in the last example."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's not going to say, oh, this is your f of x. It's going to give you the value of the slope of the curve at that point. So f of x, if you put it into that function, it should tell you, oh, the slope at that point is equal to, if you put 3 there, you'll say, oh, the slope there is equal to 6. We saw that in the last example. So that's what we want to do. And we saw in the last, I think it was two videos ago, that we defined f prime of x to be equal to just the, well, I'll write it this way. It's the slope of the secant line between x and some point that's a little bit further away from x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We saw that in the last example. So that's what we want to do. And we saw in the last, I think it was two videos ago, that we defined f prime of x to be equal to just the, well, I'll write it this way. It's the slope of the secant line between x and some point that's a little bit further away from x. So the slope of the secant line is change in y. So it's the y value of the point that's a little bit further away from x. So f of x plus h minus the y value at x, right?"}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's the slope of the secant line between x and some point that's a little bit further away from x. So the slope of the secant line is change in y. So it's the y value of the point that's a little bit further away from x. So f of x plus h minus the y value at x, right? Because this is right here. This is f of x. So minus f of x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So f of x plus h minus the y value at x, right? Because this is right here. This is f of x. So minus f of x. All of that over the change in x. So if this is x plus h here, the change in x is x plus h minus x, or this distance right here is just h. The change in x is going to be equal to h. So that's your slope of the secant line between any two points like that. And we said, hey, we could find the slope of the tangent line if we just take the limit of this as it approaches, as h approaches 0."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So minus f of x. All of that over the change in x. So if this is x plus h here, the change in x is x plus h minus x, or this distance right here is just h. The change in x is going to be equal to h. So that's your slope of the secant line between any two points like that. And we said, hey, we could find the slope of the tangent line if we just take the limit of this as it approaches, as h approaches 0. Limit as h approaches 0. And then we'll be finding the slope of the tangent line. Now let's apply this idea to a particular function, y, or f of x is equal to x squared, or y is equal to x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And we said, hey, we could find the slope of the tangent line if we just take the limit of this as it approaches, as h approaches 0. Limit as h approaches 0. And then we'll be finding the slope of the tangent line. Now let's apply this idea to a particular function, y, or f of x is equal to x squared, or y is equal to x squared. So here we could consider this to be the point x squared. So f of x is just equal to x squared. And then this would be the point."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Now let's apply this idea to a particular function, y, or f of x is equal to x squared, or y is equal to x squared. So here we could consider this to be the point x squared. So f of x is just equal to x squared. And then this would be the point. Let me do it in a more vibrant color. This is the point x plus h. That's this point right here. It's a little bit further down."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then this would be the point. Let me do it in a more vibrant color. This is the point x plus h. That's this point right here. It's a little bit further down. And then x plus h squared. And in the last video, we did this for a particular x. We did it for 3."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's a little bit further down. And then x plus h squared. And in the last video, we did this for a particular x. We did it for 3. But now I want a general formula. You give me any x, and I won't have to do what I did in the last video for any particular number. I'll have a general function."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We did it for 3. But now I want a general formula. You give me any x, and I won't have to do what I did in the last video for any particular number. I'll have a general function. You give me 7, I'll tell you what the slope is at 7. You give me negative 3, I'll tell you what the slope is at negative 3. You give me 100,000, I'll tell you what the slope is at 100,000."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I'll have a general function. You give me 7, I'll tell you what the slope is at 7. You give me negative 3, I'll tell you what the slope is at negative 3. You give me 100,000, I'll tell you what the slope is at 100,000. So let's apply it here. So we want to find the change in y over the change in x. So first of all, the change in y is this guy's y value, which is x plus h squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You give me 100,000, I'll tell you what the slope is at 100,000. So let's apply it here. So we want to find the change in y over the change in x. So first of all, the change in y is this guy's y value, which is x plus h squared. That's this guy's y value right here. That's this right here. That's x plus h squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So first of all, the change in y is this guy's y value, which is x plus h squared. That's this guy's y value right here. That's this right here. That's x plus h squared. I just took x plus h, evaluated, I squared it, and that's its point on the curve. So it's x plus h squared. So that's there right there."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's x plus h squared. I just took x plus h, evaluated, I squared it, and that's its point on the curve. So it's x plus h squared. So that's there right there. And then what's this value? f of x right here is equal to, I know it's getting messy, is equal to x squared. If you take your x, you evaluate the function at that point."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's there right there. And then what's this value? f of x right here is equal to, I know it's getting messy, is equal to x squared. If you take your x, you evaluate the function at that point. You're going to get x squared. So it's equal to minus x squared. This is your change in y."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If you take your x, you evaluate the function at that point. You're going to get x squared. So it's equal to minus x squared. This is your change in y. That's this distance right there. Change in y. And just to relate it to our definition of a derivative, this blue thing right here is equivalent to this thing right here."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is your change in y. That's this distance right there. Change in y. And just to relate it to our definition of a derivative, this blue thing right here is equivalent to this thing right here. We just evaluated our function. Our function is f of x is equal to x squared. We just evaluated when x is equal to x plus h. So if you have to square it, if I put an a there, it would be a squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And just to relate it to our definition of a derivative, this blue thing right here is equivalent to this thing right here. We just evaluated our function. Our function is f of x is equal to x squared. We just evaluated when x is equal to x plus h. So if you have to square it, if I put an a there, it would be a squared. If I put an apple there, it would be apple squared. If I put an x plus h in there, it's going to be x plus h squared. So this is that thing."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We just evaluated when x is equal to x plus h. So if you have to square it, if I put an a there, it would be a squared. If I put an apple there, it would be apple squared. If I put an x plus h in there, it's going to be x plus h squared. So this is that thing. And then this thing right here is just the function evaluated at the point in question, right there. So this is our change in y. And let's divide that by our change in x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is that thing. And then this thing right here is just the function evaluated at the point in question, right there. So this is our change in y. And let's divide that by our change in x. If this is x plus h and this is just x, our change in x is just going to be h. So that's where we get that term from. So this is just a slope between these two points. But of course, we want to find the limit as this point gets closer and closer to this point, as this point gets closer and closer to that point."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And let's divide that by our change in x. If this is x plus h and this is just x, our change in x is just going to be h. So that's where we get that term from. So this is just a slope between these two points. But of course, we want to find the limit as this point gets closer and closer to this point, as this point gets closer and closer to that point. So this becomes a tangent line. So we're going to take the limit as h approaches 0. And this is our f prime of x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But of course, we want to find the limit as this point gets closer and closer to this point, as this point gets closer and closer to that point. So this becomes a tangent line. So we're going to take the limit as h approaches 0. And this is our f prime of x. And this is the exact same definition of this. Instead of being general and saying for any function, we know what the function was. It was f of x is equal to x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And this is our f prime of x. And this is the exact same definition of this. Instead of being general and saying for any function, we know what the function was. It was f of x is equal to x squared. So we actually applied it. Instead of f of x, we wrote x squared. Instead of f of x plus h, we wrote x plus h squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It was f of x is equal to x squared. So we actually applied it. Instead of f of x, we wrote x squared. Instead of f of x plus h, we wrote x plus h squared. So let's see if we can evaluate this limit. So this is going to be equal to the limit. Let me write a little neater than that."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Instead of f of x plus h, we wrote x plus h squared. So let's see if we can evaluate this limit. So this is going to be equal to the limit. Let me write a little neater than that. The limit as h approaches 0, to square this out. I'll do it in the same color. That's x squared plus 2xh plus h squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me write a little neater than that. The limit as h approaches 0, to square this out. I'll do it in the same color. That's x squared plus 2xh plus h squared. And then we have this minus x squared over here. I just multiplied this guy out over here. And then all of that is divided by h. Now let's see if we can simplify this a little bit."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's x squared plus 2xh plus h squared. And then we have this minus x squared over here. I just multiplied this guy out over here. And then all of that is divided by h. Now let's see if we can simplify this a little bit. Well, you immediately see you have an x squared and you have a minus x squared. So those cancel out. And then we could divide the numerator and the denominator by h. So this simplifies too."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then all of that is divided by h. Now let's see if we can simplify this a little bit. Well, you immediately see you have an x squared and you have a minus x squared. So those cancel out. And then we could divide the numerator and the denominator by h. So this simplifies too. So we get f prime of x is equal to, if we divide the numerator and denominator by h, we get 2x plus h. Oh, sorry. I forgot my limit. It equals the limit."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we could divide the numerator and the denominator by h. So this simplifies too. So we get f prime of x is equal to, if we divide the numerator and denominator by h, we get 2x plus h. Oh, sorry. I forgot my limit. It equals the limit. Very important. Limit as h approaches 0, divide everything by h, and you get 2x plus h squared divided by h is h. If you remember the last video when we did it with a particular x, when we said x is equal to 3, we got 6 plus delta x here, or 6 plus h here. So it's very similar."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It equals the limit. Very important. Limit as h approaches 0, divide everything by h, and you get 2x plus h squared divided by h is h. If you remember the last video when we did it with a particular x, when we said x is equal to 3, we got 6 plus delta x here, or 6 plus h here. So it's very similar. So if you take the limit as h approaches 0 here, that's just going to disappear. So this is just going to be equal to 2x. So we just figured out that if f of x, this is a big result, this is exciting, that if f of x is equal to x squared, f prime of x is equal to 2x."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's very similar. So if you take the limit as h approaches 0 here, that's just going to disappear. So this is just going to be equal to 2x. So we just figured out that if f of x, this is a big result, this is exciting, that if f of x is equal to x squared, f prime of x is equal to 2x. That's what we just figured out. And I want to make sure you understand how to interpret this. f of x, if you give me a value, is going to tell you the value of the function at that point."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we just figured out that if f of x, this is a big result, this is exciting, that if f of x is equal to x squared, f prime of x is equal to 2x. That's what we just figured out. And I want to make sure you understand how to interpret this. f of x, if you give me a value, is going to tell you the value of the function at that point. f prime of x is going to tell you the slope at that point. Let me draw that. Because this is a key realization, and it's kind of maybe initially unintuitive to think of a function that gives us the slope at any point of another function."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "f of x, if you give me a value, is going to tell you the value of the function at that point. f prime of x is going to tell you the slope at that point. Let me draw that. Because this is a key realization, and it's kind of maybe initially unintuitive to think of a function that gives us the slope at any point of another function. So it looks like this. Let me draw it a little neater than that. That's still not that neat."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Because this is a key realization, and it's kind of maybe initially unintuitive to think of a function that gives us the slope at any point of another function. So it looks like this. Let me draw it a little neater than that. That's still not that neat. That's satisfactory. Let me just draw it in the positive coordinate. Well, I'll just draw the whole thing."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's still not that neat. That's satisfactory. Let me just draw it in the positive coordinate. Well, I'll just draw the whole thing. The curve looks like something like that. Now, this is the curve of f of x. This is the curve of f of x is equal to x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, I'll just draw the whole thing. The curve looks like something like that. Now, this is the curve of f of x. This is the curve of f of x is equal to x squared. Just like that. So if you give me a point, you give me the point 7. You apply, you put it in here, you square it, and it is mapped to the number 49."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the curve of f of x is equal to x squared. Just like that. So if you give me a point, you give me the point 7. You apply, you put it in here, you square it, and it is mapped to the number 49. So you get the number 49 right there. This is number 749. You're used to dealing with functions right there."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You apply, you put it in here, you square it, and it is mapped to the number 49. So you get the number 49 right there. This is number 749. You're used to dealing with functions right there. But what is f prime of 7? f prime of 7, you say 2 times 7 is equal to 14. What is this 14 number here?"}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You're used to dealing with functions right there. But what is f prime of 7? f prime of 7, you say 2 times 7 is equal to 14. What is this 14 number here? What is this thing? Well, this is the slope of the tangent line at x is equal to 7. So if I were to take that point and draw a tangent line, a point that just grazes our curve, if I were to just draw a tangent line, that wasn't tangent enough for me."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What is this 14 number here? What is this thing? Well, this is the slope of the tangent line at x is equal to 7. So if I were to take that point and draw a tangent line, a point that just grazes our curve, if I were to just draw a tangent line, that wasn't tangent enough for me. So that's my tangent line right there. Did you get the idea? The slope of this guy, you do your change in y over your change in x, is going to be equal to 14."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So if I were to take that point and draw a tangent line, a point that just grazes our curve, if I were to just draw a tangent line, that wasn't tangent enough for me. So that's my tangent line right there. Did you get the idea? The slope of this guy, you do your change in y over your change in x, is going to be equal to 14. The slope of the curve at y is equal to 7 is a pretty steep curve. If you wanted to find the slope, let's say that this is y, let's say x is equal to 2. I said at x is equal to 7, the slope is 14."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope of this guy, you do your change in y over your change in x, is going to be equal to 14. The slope of the curve at y is equal to 7 is a pretty steep curve. If you wanted to find the slope, let's say that this is y, let's say x is equal to 2. I said at x is equal to 7, the slope is 14. At x is equal to 2, what is the slope? Well, you go, you figure out f prime of 2, which is equal to 2 times 2, which is equal to 4. So the slope here is 4."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I said at x is equal to 7, the slope is 14. At x is equal to 2, what is the slope? Well, you go, you figure out f prime of 2, which is equal to 2 times 2, which is equal to 4. So the slope here is 4. The slope is 4. You could say m is equal to 4, m for slope. What is f prime of 0?"}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So the slope here is 4. The slope is 4. You could say m is equal to 4, m for slope. What is f prime of 0? We know that f of 0 is 0, right? 0 squared is 0. What is f prime of 0?"}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What is f prime of 0? We know that f of 0 is 0, right? 0 squared is 0. What is f prime of 0? Well, 2 times 0 is 0. That's also equal to 0. But what does that mean?"}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What is f prime of 0? Well, 2 times 0 is 0. That's also equal to 0. But what does that mean? What's the interpretation? It means the slope of the tangent line is 0. So a 0-sloped line looks like this."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But what does that mean? What's the interpretation? It means the slope of the tangent line is 0. So a 0-sloped line looks like this. Looks just like a horizontal line. And that looks about right. That a horizontal line would be tangent to the curve at y equals 0."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So a 0-sloped line looks like this. Looks just like a horizontal line. And that looks about right. That a horizontal line would be tangent to the curve at y equals 0. Let's try another one. Let's try the point minus 1. So let's say we're right there."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That a horizontal line would be tangent to the curve at y equals 0. Let's try another one. Let's try the point minus 1. So let's say we're right there. x is equal to minus 1. So f of minus 1, you just square it, because we're dealing with x squared. So it's equal to 1."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say we're right there. x is equal to minus 1. So f of minus 1, you just square it, because we're dealing with x squared. So it's equal to 1. That's that point right there. What is f prime of minus 1? f prime of minus 1 is 2 times minus 1."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's equal to 1. That's that point right there. What is f prime of minus 1? f prime of minus 1 is 2 times minus 1. 2 times minus is minus 2. What does that mean? It means that the slope of the tangent line at x is equal to 1 to this curve, to the function, is minus 2."}, {"video_title": "The derivative of f(x)=x^2 for any x   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "f prime of minus 1 is 2 times minus 1. 2 times minus is minus 2. What does that mean? It means that the slope of the tangent line at x is equal to 1 to this curve, to the function, is minus 2. So if I were to draw the tangent line here, the tangent line looks like that. And look, it is a downward sloping line. It makes sense."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So plus f of x times the derivative of the second function. Times the derivative of the second function. So two terms. In each term, we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "In each term, we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. And I frankly always forget the quotient rule, and I just re-derive it from the product rule."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. And I frankly always forget the quotient rule, and I just re-derive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I frankly always forget the quotient rule, and I just re-derive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business. The derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of, instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. g of x to the negative 1 power."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we want to take the derivative of this business. The derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of, instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule. It's the derivative of the first function right over here. So it's going to be f prime of x times just the second function, which is just g of x to the negative 1 power, plus the first function, which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we just use the product rule. It's the derivative of the first function right over here. So it's going to be f prime of x times just the second function, which is just g of x to the negative 1 power, plus the first function, which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of use something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x, to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of use something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x, to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your bath book. So let's see if we can simplify this a little bit."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your bath book. So let's see if we can simplify this a little bit. All of this is going to be equal to, we can write this term right over here, as f prime of x, as f prime of x over g of x. And we can write all of this as, we can put this negative sign out front, we have negative f of x times g prime of x, and then all of that over g of x squared. All of that over g of x squared."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can simplify this a little bit. All of this is going to be equal to, we can write this term right over here, as f prime of x, as f prime of x over g of x. And we can write all of this as, we can put this negative sign out front, we have negative f of x times g prime of x, and then all of that over g of x squared. All of that over g of x squared. And it still isn't the form that you typically see in your calculus book. To do that, we just have to add these two fractions. So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that over g of x squared. And it still isn't the form that you typically see in your calculus book. To do that, we just have to add these two fractions. So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator. So if we multiply the numerator by g of x, we'll get g of x right over here. And then the denominator will be g of x squared. And now we're ready to add."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator. So if we multiply the numerator by g of x, we'll get g of x right over here. And then the denominator will be g of x squared. And now we're ready to add. And so we get the derivative of f of x over g of x is equal to the derivative of f of x times g of x minus, not plus anymore, minus, let me write it in white, minus f of x times g prime of x. Times g prime of x, all of that over g of x squared. So once again, you can always derive this from the product rule and the chain rule."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we're ready to add. And so we get the derivative of f of x over g of x is equal to the derivative of f of x times g of x minus, not plus anymore, minus, let me write it in white, minus f of x times g prime of x. Times g prime of x, all of that over g of x squared. So once again, you can always derive this from the product rule and the chain rule. Sometimes this might be convenient to remember in order to work through some problems of this form a little bit faster. And if you wanted to kind of see the pattern between the product rule and the quotient rule, the derivative of one function just times the other function, and instead of adding the derivative of the second function times the first function, we now subtract it. And all of that is over the second function squared."}, {"video_title": "Quotient rule from product & chain rules   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, you can always derive this from the product rule and the chain rule. Sometimes this might be convenient to remember in order to work through some problems of this form a little bit faster. And if you wanted to kind of see the pattern between the product rule and the quotient rule, the derivative of one function just times the other function, and instead of adding the derivative of the second function times the first function, we now subtract it. And all of that is over the second function squared. Whatever was in the denominator, it's all of that squared. So when we're taking the derivative of the function in the denominator, up here there's a subtraction. And then we are also putting everything over the second function squared."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The rate at which the water drained from the hot tub in gallons per minute is shown. How many gallons of water were in the hot tub before it started to leak? So let's see what they have over here. So they've plotted time, or they've plotted gallons per minute versus time within minutes. And so we see here, this blue line, this is the rate at which water drained from the tub, rate at which water drained from the hot tub. So at minute zero, the water wasn't really draining from the hot tub, and then more, not just more drained, but the rate at which the water was draining increased. So 10 minutes into his bath, the water was draining at a gallon per minute, and then 20 minutes into his bath, water was draining at two gallons per minute."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they've plotted time, or they've plotted gallons per minute versus time within minutes. And so we see here, this blue line, this is the rate at which water drained from the tub, rate at which water drained from the hot tub. So at minute zero, the water wasn't really draining from the hot tub, and then more, not just more drained, but the rate at which the water was draining increased. So 10 minutes into his bath, the water was draining at a gallon per minute, and then 20 minutes into his bath, water was draining at two gallons per minute. And then he noticed it, and he opens the drain. I guess he wants to accelerate the end of his bath. So he opens the drain, and then all of a sudden, water starts draining out at a much higher rate, at 20 gallons per minute, but then that decreases."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 10 minutes into his bath, the water was draining at a gallon per minute, and then 20 minutes into his bath, water was draining at two gallons per minute. And then he noticed it, and he opens the drain. I guess he wants to accelerate the end of his bath. So he opens the drain, and then all of a sudden, water starts draining out at a much higher rate, at 20 gallons per minute, but then that decreases. And so we could think about physically why that might decrease. Maybe there was just less pressure or whatever. We're not gonna go into the physics of it, but we're just gonna take this chart as fact."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So he opens the drain, and then all of a sudden, water starts draining out at a much higher rate, at 20 gallons per minute, but then that decreases. And so we could think about physically why that might decrease. Maybe there was just less pressure or whatever. We're not gonna go into the physics of it, but we're just gonna take this chart as fact. But the rate at which the water drains decreases all the way to the 60th minute, which is 40 minutes after he opened the drain. 60th minute, all of the water was drained. All of the water was actually drained out."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're not gonna go into the physics of it, but we're just gonna take this chart as fact. But the rate at which the water drains decreases all the way to the 60th minute, which is 40 minutes after he opened the drain. 60th minute, all of the water was drained. All of the water was actually drained out. So given that, how do we think about how many gallons of water were in the hot tub before it started to leak? And I encourage you to pause the video and try to see if you can figure it out yourself before I work through it. Well, to answer this question, how many gallons of water were in the hot tub before it started to leak, that's answering the same question as how much total hot water drained out."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of the water was actually drained out. So given that, how do we think about how many gallons of water were in the hot tub before it started to leak? And I encourage you to pause the video and try to see if you can figure it out yourself before I work through it. Well, to answer this question, how many gallons of water were in the hot tub before it started to leak, that's answering the same question as how much total hot water drained out. So how much drained out? And to think about that, we can just kind of go back to what we knew before we learned about calculus. If I have something happening at a fixed rate, so let's say that this is gallons per minute."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to answer this question, how many gallons of water were in the hot tub before it started to leak, that's answering the same question as how much total hot water drained out. So how much drained out? And to think about that, we can just kind of go back to what we knew before we learned about calculus. If I have something happening at a fixed rate, so let's say that this is gallons per minute. So this is still the same context, I guess you could say. This is time in minutes. Now let's say things are draining out at a constant rate."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I have something happening at a fixed rate, so let's say that this is gallons per minute. So this is still the same context, I guess you could say. This is time in minutes. Now let's say things are draining out at a constant rate. If you wanted to figure out how much drains out over a certain interval of time, let's say this interval of time right over here, let's call that delta t over delta t, you would just multiply the rate over that interval of time, the rate over that interval of time, which we could represent by this orange height right over here, times the amount of time that passed by, which essentially would give you the area under the curve over that interval. The area under the curve over that interval. This area would tell you the gallons that drained over that delta t. And this doesn't just apply when you have a constant, when you have a constant rate."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's say things are draining out at a constant rate. If you wanted to figure out how much drains out over a certain interval of time, let's say this interval of time right over here, let's call that delta t over delta t, you would just multiply the rate over that interval of time, the rate over that interval of time, which we could represent by this orange height right over here, times the amount of time that passed by, which essentially would give you the area under the curve over that interval. The area under the curve over that interval. This area would tell you the gallons that drained over that delta t. And this doesn't just apply when you have a constant, when you have a constant rate. If your rate looked something like this, as we've seen in other videos, as we've seen in other videos, you could figure out the amount that has drained, the amount that has drained in a specific interval, let's say this interval right over here, by essentially figuring out the area under the curve over that interval. And the units work out. If you multiply gallons per minute times minute, this area is going to be in terms of gallons."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This area would tell you the gallons that drained over that delta t. And this doesn't just apply when you have a constant, when you have a constant rate. If your rate looked something like this, as we've seen in other videos, as we've seen in other videos, you could figure out the amount that has drained, the amount that has drained in a specific interval, let's say this interval right over here, by essentially figuring out the area under the curve over that interval. And the units work out. If you multiply gallons per minute times minute, this area is going to be in terms of gallons. This area is going to be in terms of gallons. Another way you could think about it, and this goes back to, well, how do you figure out the area of a trapezoid right over here? Well, the area of a trapezoid, you could find the average height of the trapezoid, which would be the average of the beginning and the end period."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you multiply gallons per minute times minute, this area is going to be in terms of gallons. This area is going to be in terms of gallons. Another way you could think about it, and this goes back to, well, how do you figure out the area of a trapezoid right over here? Well, the area of a trapezoid, you could find the average height of the trapezoid, which would be the average of the beginning and the end period. You take that average over there, and this would work for a line like this. If you take that average height and multiply by the change in time, you are going to actually, you are going to figure out, you are going to figure out that area. That's another way of thinking about it, is you're taking the average rate over that interval times the interval, and that's going to give you the total number of gallons."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the area of a trapezoid, you could find the average height of the trapezoid, which would be the average of the beginning and the end period. You take that average over there, and this would work for a line like this. If you take that average height and multiply by the change in time, you are going to actually, you are going to figure out, you are going to figure out that area. That's another way of thinking about it, is you're taking the average rate over that interval times the interval, and that's going to give you the total number of gallons. And so we just have to apply that idea over here. We just literally figure out the area under the curve over the entire interval when the water was actually leaking or draining. So essentially the area under the curve between zero minutes and 60 minutes."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's another way of thinking about it, is you're taking the average rate over that interval times the interval, and that's going to give you the total number of gallons. And so we just have to apply that idea over here. We just literally figure out the area under the curve over the entire interval when the water was actually leaking or draining. So essentially the area under the curve between zero minutes and 60 minutes. And so it's going to be this area, let's put that in magenta, plus all of this area under this part as well. And to help us think about that, I'm just going to split that up into some sections. So I'll have this triangular section up here."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So essentially the area under the curve between zero minutes and 60 minutes. And so it's going to be this area, let's put that in magenta, plus all of this area under this part as well. And to help us think about that, I'm just going to split that up into some sections. So I'll have this triangular section up here. So I have this triangular section. I could just think about this as a trapezoid, but I'm just going to split it up into a triangle and a rectangle. And then I have this section right over here in green."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll have this triangular section up here. So I have this triangular section. I could just think about this as a trapezoid, but I'm just going to split it up into a triangle and a rectangle. And then I have this section right over here in green. So what is the area of this entire thing? Well, the area right over here, we have 20 gallons per minute. So 20, oh sorry, not 20 gallons per minute, 20 minutes times two gallons per minute, two gallons per minute times 1 1\u20442."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I have this section right over here in green. So what is the area of this entire thing? Well, the area right over here, we have 20 gallons per minute. So 20, oh sorry, not 20 gallons per minute, 20 minutes times two gallons per minute, two gallons per minute times 1 1\u20442. Times 1 1\u20442 gives us the area under this triangle. So that is going to be 20 gallons. 20 gallons, we see that the units work out nicely."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 20, oh sorry, not 20 gallons per minute, 20 minutes times two gallons per minute, two gallons per minute times 1 1\u20442. Times 1 1\u20442 gives us the area under this triangle. So that is going to be 20 gallons. 20 gallons, we see that the units work out nicely. And so that's essentially how much has drained out in the first 20 minutes. And then this green area is going to be 40 minutes, 40 minutes times 10 gallons per minute, times 10, and actually maybe the units, since I'm breaking it up in this strange way, I'll just multiply, I'll just figure out this area in a unitless way. So 40 times 10, which is equal to 400."}, {"video_title": "Worked example problem involving definite integral (graphical)   AP Calculus AB   Khan Academy.mp3", "Sentence": "20 gallons, we see that the units work out nicely. And so that's essentially how much has drained out in the first 20 minutes. And then this green area is going to be 40 minutes, 40 minutes times 10 gallons per minute, times 10, and actually maybe the units, since I'm breaking it up in this strange way, I'll just multiply, I'll just figure out this area in a unitless way. So 40 times 10, which is equal to 400. And then finally in blue, I have 40 times this height right over here between 10 and 20 is another 10, but then I'm gonna multiply that times 1 1\u20442. So it's gonna be 40 times 10 times 1 1\u20442, which is going to be 200. And so you add all of these together, these areas together, you're going to get 400 plus 200 is 600 plus 20, you're going to have 620 gallons."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is use the online graphing calculator Desmos and explore the relationship between vertical and horizontal asymptotes and think about how they relate to what we know about limits. So let's first graph two over x minus one. So let me get that one graphed. And so you can immediately see that something interesting happens at x is equal to one. If you were to just substitute x equals one into this expression, you're going to get two over zero. And whenever you get a non-zero thing over zero, that's a good sign that you might be dealing with a vertical asymptote. In fact, we can draw that vertical asymptote right over here at x equals one."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you can immediately see that something interesting happens at x is equal to one. If you were to just substitute x equals one into this expression, you're going to get two over zero. And whenever you get a non-zero thing over zero, that's a good sign that you might be dealing with a vertical asymptote. In fact, we can draw that vertical asymptote right over here at x equals one. But let's think about how that relates to limits. What if we were to explore the limit as x approaches one of f of x is equal to two over x minus one? And we could think about it from the left and from the right."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "In fact, we can draw that vertical asymptote right over here at x equals one. But let's think about how that relates to limits. What if we were to explore the limit as x approaches one of f of x is equal to two over x minus one? And we could think about it from the left and from the right. So if we approach one from the left, let me zoom in a little bit over here. So we can see as we approach from the left when x is equal to zero, the f of x would be equal to negative two. When x is equal to.5, f of x is equal to negative four."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could think about it from the left and from the right. So if we approach one from the left, let me zoom in a little bit over here. So we can see as we approach from the left when x is equal to zero, the f of x would be equal to negative two. When x is equal to.5, f of x is equal to negative four. And then it just gets more and more negative the closer we get to one from the left. I could really, so I'm not even that close yet. If I get to, let's say.91, I'm still 900, so less than one."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is equal to.5, f of x is equal to negative four. And then it just gets more and more negative the closer we get to one from the left. I could really, so I'm not even that close yet. If I get to, let's say.91, I'm still 900, so less than one. I'm at negative 22.222 already. And so the limit as we approach one from the left is unbounded. Some people would say it goes to negative infinity, but it's really an undefined limit."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I get to, let's say.91, I'm still 900, so less than one. I'm at negative 22.222 already. And so the limit as we approach one from the left is unbounded. Some people would say it goes to negative infinity, but it's really an undefined limit. It is unbounded in the negative direction. And likewise, as we approach from the right, we get unbounded in the positive infinity direction, and technically we would say that that limit does not exist. And this would be the case when we're dealing with a vertical asymptote like we see over here."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Some people would say it goes to negative infinity, but it's really an undefined limit. It is unbounded in the negative direction. And likewise, as we approach from the right, we get unbounded in the positive infinity direction, and technically we would say that that limit does not exist. And this would be the case when we're dealing with a vertical asymptote like we see over here. Now let's compare that to a horizontal asymptote where it turns out that the limit actually can exist. So let me delete these or just erase them for now. And so let's look at this function, which is a pretty neat function."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this would be the case when we're dealing with a vertical asymptote like we see over here. Now let's compare that to a horizontal asymptote where it turns out that the limit actually can exist. So let me delete these or just erase them for now. And so let's look at this function, which is a pretty neat function. I made it up right before this video started, but it's kind of cool looking. So let's think about the behavior as x approaches infinity. So as x approaches infinity, it looks like our y value or the value of the expression, if we said y is equal to that expression, it looks like it's getting closer and closer and closer to three."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's look at this function, which is a pretty neat function. I made it up right before this video started, but it's kind of cool looking. So let's think about the behavior as x approaches infinity. So as x approaches infinity, it looks like our y value or the value of the expression, if we said y is equal to that expression, it looks like it's getting closer and closer and closer to three. And so we could say that we have a horizontal asymptote at y is equal to three. And we could also, and there's a more rigorous way of defining it, say that our limit as x approaches infinity is equal of the expression or of the function is equal to three. Notice my mouse is covering it a little bit."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So as x approaches infinity, it looks like our y value or the value of the expression, if we said y is equal to that expression, it looks like it's getting closer and closer and closer to three. And so we could say that we have a horizontal asymptote at y is equal to three. And we could also, and there's a more rigorous way of defining it, say that our limit as x approaches infinity is equal of the expression or of the function is equal to three. Notice my mouse is covering it a little bit. As we get larger and larger, we're getting closer and closer to three. In fact, we're getting so close now that, well, here, you can see it. We're getting closer and closer and closer to three."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Notice my mouse is covering it a little bit. As we get larger and larger, we're getting closer and closer to three. In fact, we're getting so close now that, well, here, you can see it. We're getting closer and closer and closer to three. And you could also think about what happens as x approaches negative infinity. And here you're getting closer and closer and closer to three from below. Now one thing that's interesting about horizontal asymptotes is you might see that the function actually can cross a horizontal asymptote."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're getting closer and closer and closer to three. And you could also think about what happens as x approaches negative infinity. And here you're getting closer and closer and closer to three from below. Now one thing that's interesting about horizontal asymptotes is you might see that the function actually can cross a horizontal asymptote. It's crossing this horizontal asymptote in this area in between. And even as we approach infinity or negative infinity, you can oscillate around that horizontal asymptote. Let me set this up."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now one thing that's interesting about horizontal asymptotes is you might see that the function actually can cross a horizontal asymptote. It's crossing this horizontal asymptote in this area in between. And even as we approach infinity or negative infinity, you can oscillate around that horizontal asymptote. Let me set this up. Let me multiply this times sine of x. And so there you have it. We are now oscillating around the horizontal asymptote."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me set this up. Let me multiply this times sine of x. And so there you have it. We are now oscillating around the horizontal asymptote. And once again, this limit can exist. Even though we keep crossing the horizontal asymptote, we're getting closer and closer and closer to it the larger x gets. And that's actually a key difference between a horizontal and a vertical asymptote."}, {"video_title": "Infinite limits and asymptotes   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are now oscillating around the horizontal asymptote. And once again, this limit can exist. Even though we keep crossing the horizontal asymptote, we're getting closer and closer and closer to it the larger x gets. And that's actually a key difference between a horizontal and a vertical asymptote. Vertical asymptotes, if you're dealing with a function, you're not going to cross it. While with a horizontal asymptote, you could. And you are just getting closer and closer and closer to it as x goes to positive infinity or as x goes to negative infinity."}, {"video_title": "Inflection points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we talked about the interval over which the function is concave upwards. But we see here that there's a point at which we transition from being concave downwards to concave upwards. Before that point, the slope was decreasing, and then the slope starts increasing. The slope was decreasing, and then the slope started increasing. So that's one way to look at it. Right here in our function, we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing."}, {"video_title": "Inflection points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope was decreasing, and then the slope started increasing. So that's one way to look at it. Right here in our function, we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name, you're probably thinking, and you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has an extremum point, or the point at which our second derivative switches signs like this, we call it an inflection point."}, {"video_title": "Inflection points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name, you're probably thinking, and you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has an extremum point, or the point at which our second derivative switches signs like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u, or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive, but we could have also switched from being positive to negative."}, {"video_title": "Inflection points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has an extremum point, or the point at which our second derivative switches signs like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u, or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive, but we could have also switched from being positive to negative. So inflection point, your second derivative, f prime prime of x, switches signs, goes from being positive to negative or negative to positive, switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards like that, then at this inflection point, up until that point, the slope was increasing."}, {"video_title": "Inflection points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in this case, we went from negative to positive, but we could have also switched from being positive to negative. So inflection point, your second derivative, f prime prime of x, switches signs, goes from being positive to negative or negative to positive, switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards like that, then at this inflection point, up until that point, the slope was increasing. So the second derivative would be positive. And then the slope is decreasing. The derivative is decreasing."}, {"video_title": "Inflection points introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we went from concave upwards to concave downwards like that, then at this inflection point, up until that point, the slope was increasing. So the second derivative would be positive. And then the slope is decreasing. The derivative is decreasing. So your second derivative would be negative. So here, your second derivative is going from positive to negative. Here, your second derivative is going from negative to positive."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "A particle moves in a straight line with velocity v of t is equal to negative t squared plus eight meters per second where t is time in seconds. At t is equal to two, the particle's distance from the starting point was five meters. What is the total distance the particle has traveled between t equals two and t equals six seconds? Which expression should Alexei use to solve the problem? So we don't actually have to figure the actual answer out. We just have to figure out what is the appropriate expression. So like always, pause this video and see if you can work through it on your own."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "Which expression should Alexei use to solve the problem? So we don't actually have to figure the actual answer out. We just have to figure out what is the appropriate expression. So like always, pause this video and see if you can work through it on your own. So now let's tackle this together. So the key question is what is the total distance the particle has traveled between t equals two and t equals six? So we just care what happens between those points."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "So like always, pause this video and see if you can work through it on your own. So now let's tackle this together. So the key question is what is the total distance the particle has traveled between t equals two and t equals six? So we just care what happens between those points. We don't care that the particle's distance from the starting point was five meters at t equals two. So this right over here is actually unnecessary information. So the first thing that you might want to think about is well maybe distance is just the integral of the velocity function."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just care what happens between those points. We don't care that the particle's distance from the starting point was five meters at t equals two. So this right over here is actually unnecessary information. So the first thing that you might want to think about is well maybe distance is just the integral of the velocity function. We see that multiple times. If you want to find the change in a quantity, you just say the starting time and the ending time, and then you integrate the rate function. So wouldn't it just be that?"}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the first thing that you might want to think about is well maybe distance is just the integral of the velocity function. We see that multiple times. If you want to find the change in a quantity, you just say the starting time and the ending time, and then you integrate the rate function. So wouldn't it just be that? Now we have to be very, very careful. If the question was what is the displacement for the particle between time equals two and times equals six, this would have been the correct answer. So this would be displacement, displacement from t equals two to t is equal to six."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "So wouldn't it just be that? Now we have to be very, very careful. If the question was what is the displacement for the particle between time equals two and times equals six, this would have been the correct answer. So this would be displacement, displacement from t equals two to t is equal to six. But they're not saying displacement. They're saying total distance the particle has traveled. So this is the total path length for the particle."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this would be displacement, displacement from t equals two to t is equal to six. But they're not saying displacement. They're saying total distance the particle has traveled. So this is the total path length for the particle. So one way to think about it, this is you would integrate not the velocity function. If you integrate velocity, you get displacement. Instead, you would integrate the speed function."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the total path length for the particle. So one way to think about it, this is you would integrate not the velocity function. If you integrate velocity, you get displacement. Instead, you would integrate the speed function. Now what is speed? It is the magnitude of velocity. In one dimension, it would just be the absolute value of your velocity function."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead, you would integrate the speed function. Now what is speed? It is the magnitude of velocity. In one dimension, it would just be the absolute value of your velocity function. And so the absolute value of the velocity function, this would give you, integrating the speed, this would give you the distance. Distance from t equals two to t is equal to six. And let's see, we have that choice right over here."}, {"video_title": "Analyzing motion problems total distance traveled   AP Calculus AB   Khan Academy.mp3", "Sentence": "In one dimension, it would just be the absolute value of your velocity function. And so the absolute value of the velocity function, this would give you, integrating the speed, this would give you the distance. Distance from t equals two to t is equal to six. And let's see, we have that choice right over here. The displacement one here, this is an interesting distractor, but that is not going to be the choice. This one right over here, v prime of six, that gives you the acceleration. If you're taking the derivative of the velocity function, the acceleration at six seconds, that's not what we're interested in."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let R be the region enclosed by Y is equal to four times the square root of nine minus X and the axes in the first quadrant. And we can see that region R in gray right over here. Region R is the base of a solid. For each Y value, the cross section of the solid taken perpendicular to the Y axis is a rectangle whose base lies in R and whose height is Y. Express the volume of the solid with a definite integral. So pause this video and see if you can do that. All right, now let's do this together."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "For each Y value, the cross section of the solid taken perpendicular to the Y axis is a rectangle whose base lies in R and whose height is Y. Express the volume of the solid with a definite integral. So pause this video and see if you can do that. All right, now let's do this together. And first let's just try to visualize the solid. And I'll try to do it by drawing this with a little bit of perspective. So if that's our Y axis, and then this is our X axis right over here."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now let's do this together. And first let's just try to visualize the solid. And I'll try to do it by drawing this with a little bit of perspective. So if that's our Y axis, and then this is our X axis right over here. And I can redraw region R. It looks something like this. And now let's just imagine a cross section of our solid. So it says the cross section of the solid taken perpendicular to the Y axis."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if that's our Y axis, and then this is our X axis right over here. And I can redraw region R. It looks something like this. And now let's just imagine a cross section of our solid. So it says the cross section of the solid taken perpendicular to the Y axis. So let's pick a Y value right over here. We're gonna go perpendicular to the Y axis. It says whose base lies in R. So the base would look like that."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it says the cross section of the solid taken perpendicular to the Y axis. So let's pick a Y value right over here. We're gonna go perpendicular to the Y axis. It says whose base lies in R. So the base would look like that. It would actually be the X value that corresponds to that particular Y value. So I'll just write X right over here. And then the height is Y."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "It says whose base lies in R. So the base would look like that. It would actually be the X value that corresponds to that particular Y value. So I'll just write X right over here. And then the height is Y. So the height is going to be whatever our Y value is. And then if we wanted to calculate the volume of just a little bit, a slice that has an infinitesimal depth, we could think about that infinitesimal depth in terms of Y. So we could say its depth right over here is DY, DY."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the height is Y. So the height is going to be whatever our Y value is. And then if we wanted to calculate the volume of just a little bit, a slice that has an infinitesimal depth, we could think about that infinitesimal depth in terms of Y. So we could say its depth right over here is DY, DY. And we could draw other cross sections. For example, right over here, our Y is much lower. It might look some, so our height will be like that."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say its depth right over here is DY, DY. And we could draw other cross sections. For example, right over here, our Y is much lower. It might look some, so our height will be like that. But then our base is the corresponding X value that sits on the curve right over that XY pair that would sit on that curve. And so this cross section would look like this. And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's an infinitesimal volume and it would have depth DY."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "It might look some, so our height will be like that. But then our base is the corresponding X value that sits on the curve right over that XY pair that would sit on that curve. And so this cross section would look like this. And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's an infinitesimal volume and it would have depth DY. And so as we've learned many times in integration, what we wanna do is think about the volume of one of these, I guess you could say slices, and then integrate across all of them. Now there's a couple of ways to approach it. You could try to integrate with respect to X or you could integrate with respect to Y. I'm gonna argue it's much easier to integrate with respect to Y here because we already have things in terms of DY."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's an infinitesimal volume and it would have depth DY. And so as we've learned many times in integration, what we wanna do is think about the volume of one of these, I guess you could say slices, and then integrate across all of them. Now there's a couple of ways to approach it. You could try to integrate with respect to X or you could integrate with respect to Y. I'm gonna argue it's much easier to integrate with respect to Y here because we already have things in terms of DY. The volume of this little slice is going to be Y times X times DY. Now if we wanna integrate with respect to Y, we want everything in terms of Y. And so what we need to do is express X in terms of Y."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could try to integrate with respect to X or you could integrate with respect to Y. I'm gonna argue it's much easier to integrate with respect to Y here because we already have things in terms of DY. The volume of this little slice is going to be Y times X times DY. Now if we wanna integrate with respect to Y, we want everything in terms of Y. And so what we need to do is express X in terms of Y. So here we just have to solve for X. So one way to do this is, let's see, we can square both sides of, well actually let's divide both sides by four. So you get Y over four is equal to the square root of nine minus X."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what we need to do is express X in terms of Y. So here we just have to solve for X. So one way to do this is, let's see, we can square both sides of, well actually let's divide both sides by four. So you get Y over four is equal to the square root of nine minus X. Now we can square both sides. Y squared over 16 is equal to nine minus X. And then let's see, we could multiply both sides by negative one."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you get Y over four is equal to the square root of nine minus X. Now we can square both sides. Y squared over 16 is equal to nine minus X. And then let's see, we could multiply both sides by negative one. So negative Y squared over 16 is equal to X minus nine. And now we could add nine to both sides and we get nine minus Y squared over 16 is equal to X. And so we could substitute that right over there."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let's see, we could multiply both sides by negative one. So negative Y squared over 16 is equal to X minus nine. And now we could add nine to both sides and we get nine minus Y squared over 16 is equal to X. And so we could substitute that right over there. So another way to express the volume of this little slice right over here of infinitesimal depth, DY depth, is going to be Y times nine minus Y squared over 16 DY. Y squared over 16 DY. And if we want to find the volume of the whole figure, that's gonna look something like that, we're just going to integrate from Y equals zero to Y is equal to 12."}, {"video_title": "Volume with cross sections perpendicular to y-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we could substitute that right over there. So another way to express the volume of this little slice right over here of infinitesimal depth, DY depth, is going to be Y times nine minus Y squared over 16 DY. Y squared over 16 DY. And if we want to find the volume of the whole figure, that's gonna look something like that, we're just going to integrate from Y equals zero to Y is equal to 12. So integrate from Y is equal to zero to Y is equal to 12. And that's all they asked us to do to express the volume as a definite integral. But this is actually a definite integral that you could solve without a calculator."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "And what might seem challenging is throughout this region I have the same lower function, or I guess the lower boundary is y is equal to x squared over 4 minus 1, but I have a different upper boundary. And the way that we can tackle this is by dividing this area into two sections, or dividing this region into two regions, the region on the left and the region on the right, where for this first region, which I'll color even more in yellow, for this first region, over that entire interval in x, and it looks like x is going between 0 and 1, y equals, when x is equal to 1, this function is equal to 1, when x is equal to 1, this function is also equal to 1. So this is the point 1, 1, that's where they intersect. So for this section, this sub-region right over here, y equals square root of x is the upper function the entire time. And then we can set up a different, we can separately tackle figuring out the area of this region, from x is equal to 1 to x is equal to 2, where y equals 2 minus x is the upper function. So let's do it. So let's first think about this first region."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "So for this section, this sub-region right over here, y equals square root of x is the upper function the entire time. And then we can set up a different, we can separately tackle figuring out the area of this region, from x is equal to 1 to x is equal to 2, where y equals 2 minus x is the upper function. So let's do it. So let's first think about this first region. Well, that's going to be the definite integral from x is equal to 0 to x is equal to 1, and our upper function is square root of x, so square root of x, and then from that we want to subtract our lower function, square root of x minus x squared over 4 minus 1. So minus x squared over 4 minus 1. And then, of course, we have our dx."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "So let's first think about this first region. Well, that's going to be the definite integral from x is equal to 0 to x is equal to 1, and our upper function is square root of x, so square root of x, and then from that we want to subtract our lower function, square root of x minus x squared over 4 minus 1. So minus x squared over 4 minus 1. And then, of course, we have our dx. So this right over here, this is describing the area in yellow, and you can imagine it that this part right over here, the difference between these two functions, is essentially this height, we'll do it in a different color, is essentially this height, this height, and then you multiply it times dx, you get a little rectangle with width dx, and then you do that for each x, each x you get a different rectangle, and then you sum them all up, and you take the limit as your change in x approaches 0, so you get ultra, ultra thin rectangles, and you have an infinite number of them. And that's our definition, the Riemann definition of what a definite integral is. And so this is the area of the left region, and by the exact same logic, we can figure out the area of the right region."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "And then, of course, we have our dx. So this right over here, this is describing the area in yellow, and you can imagine it that this part right over here, the difference between these two functions, is essentially this height, we'll do it in a different color, is essentially this height, this height, and then you multiply it times dx, you get a little rectangle with width dx, and then you do that for each x, each x you get a different rectangle, and then you sum them all up, and you take the limit as your change in x approaches 0, so you get ultra, ultra thin rectangles, and you have an infinite number of them. And that's our definition, the Riemann definition of what a definite integral is. And so this is the area of the left region, and by the exact same logic, we can figure out the area of the right region. The right region, and then we can just sum the two things together. The right region, we're going from x is equal to 0 to x, sorry, x is equal to 1, x is equal to 2, 1 to 2. The upper function is 2 minus x, and from that we're going to subtract, from that we're going to subtract the lower function, which is x squared over 4 minus 1. x squared over 4 minus 1."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "And so this is the area of the left region, and by the exact same logic, we can figure out the area of the right region. The right region, and then we can just sum the two things together. The right region, we're going from x is equal to 0 to x, sorry, x is equal to 1, x is equal to 2, 1 to 2. The upper function is 2 minus x, and from that we're going to subtract, from that we're going to subtract the lower function, which is x squared over 4 minus 1. x squared over 4 minus 1. And now we just have to evaluate. So let's first simplify this right over here. This is equal to the definite integral from 0 to 1 of square root of x minus x squared over 4 plus 1 dx, I'm going to write it all in one color now, plus the definite integral from 1 to 2 of 2 minus x minus x squared over 4, then subtracting a negative 1 is a positive 3, or a positive 1, we could just add it to this 2, and so this 2 just becomes a 3."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "The upper function is 2 minus x, and from that we're going to subtract, from that we're going to subtract the lower function, which is x squared over 4 minus 1. x squared over 4 minus 1. And now we just have to evaluate. So let's first simplify this right over here. This is equal to the definite integral from 0 to 1 of square root of x minus x squared over 4 plus 1 dx, I'm going to write it all in one color now, plus the definite integral from 1 to 2 of 2 minus x minus x squared over 4, then subtracting a negative 1 is a positive 3, or a positive 1, we could just add it to this 2, and so this 2 just becomes a 3. I said 2 minus negative 1 is 3 dx, and now we just have to take the antiderivative and evaluate it at 1 and 0. So the antiderivative of this is, well, this is x to the 1 half, increment it by 1, increment the power by 1, you get x to the 3 halves, and then multiply by the reciprocal of the new exponent, so it's 2 thirds x to the 3 halves, minus the antiderivative of x squared over 4 is x to the 3rd divided by 3 divided by 4, so divided by 12, plus x, that's the antiderivative of 1, we're going to evaluate it at 1 and 0, and then here the antiderivative is going to be 3x minus x squared over 2, minus x to the 3rd over 12, once again evaluated at, well not once again, now we're going to evaluate at 2 and 1. So over here you evaluate all of this stuff at 1, you get 2 thirds minus 1 12th plus 1, and then from that you subtract this evaluated at 0, but this is just all 0, so you get nothing, so this is what the yellow stuff simplified to, and then this purple stuff, or this magenta stuff, or mauve, or whatever color this is, first you evaluate it at 2, you get 6 minus, let's see, 2 squared over 2 is 2, minus 8 over 12, and then from that you're going to subtract this evaluated at 1, so it's going to be 3 times 1, minus 1 half, minus 1 over 12, and now what we're essentially left with is adding a bunch of fractions, so let's see if we can do that, it looks like 12 would be the most obvious common denominator, so here you have 8 over 12 minus 1 over 12, plus 12 over 12, so this simplifies to, this is 19 over 12, the part that we have in yellow, then this business, let me do it in this color, so 6 minus 2, this is just going to be 4, so we can write this as 48 over 12, that's 4, minus 8 over 12, and then we're going to have to subtract a 3, which is 36 over 12, then we're going to add 1 half, which is just plus 6 over 12, and then we're going to add a 1 twelfth, so this is all going to simplify to, let's see, 48 minus 8 is 40, minus 36 is 4, plus 6 is 10, plus 1 is 11, so this becomes plus 11 over 12, let me make sure I did that right, 48 minus 8 is 40, minus 36 is 4, 10, 11, so that looks right, and then we're ready to add these two, 19 plus 11 is equal to 30 over 12, or if we want to simplify this a little bit, we can divide the numerator and the denominator by 6, this is equal to 5 halves, or 2 and a half, and we're done, we figured out the area of this entire region, it is 2 and a half."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is an appropriate calculus-based justification for the fact that f is decreasing when x is greater than three? So we can see that that actually is indeed the case. When x is greater than three, we see that our function is indeed decreasing. As x increases, the y value, the value of our function, decreases. So a calculus-based justification, without even looking at the choices, well, I could look at the derivative, and we're going to be decreasing if the slope of the tangent line is negative, which means that the derivative is negative. And we can see that for x is greater than three, the derivative is less than zero. So my justification, I haven't even looked at these choices yet, I would say for x is greater than three, f prime of x is less than zero."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "As x increases, the y value, the value of our function, decreases. So a calculus-based justification, without even looking at the choices, well, I could look at the derivative, and we're going to be decreasing if the slope of the tangent line is negative, which means that the derivative is negative. And we can see that for x is greater than three, the derivative is less than zero. So my justification, I haven't even looked at these choices yet, I would say for x is greater than three, f prime of x is less than zero. That would be my justification, not even looking at these choices. Now let's look at the choices. F prime is decreasing when x is greater than three."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So my justification, I haven't even looked at these choices yet, I would say for x is greater than three, f prime of x is less than zero. That would be my justification, not even looking at these choices. Now let's look at the choices. F prime is decreasing when x is greater than three. Now this isn't right. What we care about is whether f prime is positive or negative. If f prime is negative, if it's less than zero, then the function itself is decreasing."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "F prime is decreasing when x is greater than three. Now this isn't right. What we care about is whether f prime is positive or negative. If f prime is negative, if it's less than zero, then the function itself is decreasing. The slope of the tangent line will be negative. You could, f prime could be positive while decreasing. For example, f prime could be doing something like this."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "If f prime is negative, if it's less than zero, then the function itself is decreasing. The slope of the tangent line will be negative. You could, f prime could be positive while decreasing. For example, f prime could be doing something like this. And even though f prime would be decreasing in this situation, the actual value of the derivative would be positive, which means the function would be increasing in that scenario. So I would rule this one out. For values of x larger than three, as x values increase, the values of f of x decrease."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "For example, f prime could be doing something like this. And even though f prime would be decreasing in this situation, the actual value of the derivative would be positive, which means the function would be increasing in that scenario. So I would rule this one out. For values of x larger than three, as x values increase, the values of f of x decrease. Now that is actually true. This is actually the definition that f is decreasing. As x values increase, the values of f of x decrease."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "For values of x larger than three, as x values increase, the values of f of x decrease. Now that is actually true. This is actually the definition that f is decreasing. As x values increase, the values of f of x decrease. But this is not a calculus-based justification. So I am going to rule this one out as well. F prime is negative when x is greater than three."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "As x values increase, the values of f of x decrease. But this is not a calculus-based justification. So I am going to rule this one out as well. F prime is negative when x is greater than three. Well, that's exactly what I wrote up here. If f prime is negative, then that means that our slope of the tangent line of our original function f is going to be downward sloping, or that our function is decreasing. So this one is looking good."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "F prime is negative when x is greater than three. Well, that's exactly what I wrote up here. If f prime is negative, then that means that our slope of the tangent line of our original function f is going to be downward sloping, or that our function is decreasing. So this one is looking good. And this one right over here says f prime of zero is equal to negative three. So they're just pointing out this point. This isn't even relative to the interval that we care about, or this isn't even relative when x is greater than three."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this one is looking good. And this one right over here says f prime of zero is equal to negative three. So they're just pointing out this point. This isn't even relative to the interval that we care about, or this isn't even relative when x is greater than three. So we definitely want to rule that one out. Let's do one more of these. So here we're told the differentiable function g and its derivative g prime are graphed."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "This isn't even relative to the interval that we care about, or this isn't even relative when x is greater than three. So we definitely want to rule that one out. Let's do one more of these. So here we're told the differentiable function g and its derivative g prime are graphed. So once again, g is in this bluish color, and then g prime, its derivative, is in this orange color. What is an appropriate calculus-based justification for the fact that g has a relative minimum point at x is equal to negative three? And we could see here, when x is equal to negative three, it looks like g is equal to negative six, and it looks like a relative minimum point there."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here we're told the differentiable function g and its derivative g prime are graphed. So once again, g is in this bluish color, and then g prime, its derivative, is in this orange color. What is an appropriate calculus-based justification for the fact that g has a relative minimum point at x is equal to negative three? And we could see here, when x is equal to negative three, it looks like g is equal to negative six, and it looks like a relative minimum point there. So what's the best justification? So once again, without even looking at the choices, I would say a good justification is before we get to x equals negative three, before we get to x equals negative three, our derivative, and this is a calculus-based justification, before we go to x equals negative three, our derivative is negative, and after x equals negative three, our derivative is positive. That would be my justification, because if our derivative is positive before that value, if our derivative is negative before that value, that means that we are downward sloping before that value, and if it's positive after that value, that means we're upward sloping after that, which is a good justification that we are at a relative minimum point right over there."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could see here, when x is equal to negative three, it looks like g is equal to negative six, and it looks like a relative minimum point there. So what's the best justification? So once again, without even looking at the choices, I would say a good justification is before we get to x equals negative three, before we get to x equals negative three, our derivative, and this is a calculus-based justification, before we go to x equals negative three, our derivative is negative, and after x equals negative three, our derivative is positive. That would be my justification, because if our derivative is positive before that value, if our derivative is negative before that value, that means that we are downward sloping before that value, and if it's positive after that value, that means we're upward sloping after that, which is a good justification that we are at a relative minimum point right over there. So let's see, the point where x equals negative three is the lowest point on the graph of g in its surrounding interval. That is true, but that's not a calculus-based justification. You wouldn't even have to look at the derivative to make that statement, so let's rule that one out."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "That would be my justification, because if our derivative is positive before that value, if our derivative is negative before that value, that means that we are downward sloping before that value, and if it's positive after that value, that means we're upward sloping after that, which is a good justification that we are at a relative minimum point right over there. So let's see, the point where x equals negative three is the lowest point on the graph of g in its surrounding interval. That is true, but that's not a calculus-based justification. You wouldn't even have to look at the derivative to make that statement, so let's rule that one out. G prime has a relative maximum at zero comma three. At zero, zero comma three, it actually does not. Oh, g prime, g prime, yes, g prime actually does have a relative maximum at zero comma three, but that doesn't tell us anything about whether we're at a relative minimum point at x equals negative three, so I would rule that out."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "You wouldn't even have to look at the derivative to make that statement, so let's rule that one out. G prime has a relative maximum at zero comma three. At zero, zero comma three, it actually does not. Oh, g prime, g prime, yes, g prime actually does have a relative maximum at zero comma three, but that doesn't tell us anything about whether we're at a relative minimum point at x equals negative three, so I would rule that out. G prime of negative three is equal to zero. So g prime of negative three is equal to zero. So that tells us that the slope of the tangent line of our function is going to be zero right over there, but that by itself is not enough to say that we are at a relative minimum point."}, {"video_title": "Justification using first derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Oh, g prime, g prime, yes, g prime actually does have a relative maximum at zero comma three, but that doesn't tell us anything about whether we're at a relative minimum point at x equals negative three, so I would rule that out. G prime of negative three is equal to zero. So g prime of negative three is equal to zero. So that tells us that the slope of the tangent line of our function is going to be zero right over there, but that by itself is not enough to say that we are at a relative minimum point. For example, I could be at a point that does something like this, where the slope of the tangent line is zero, and then it keeps increasing again, or it does something like this, and it keeps decreasing. So even though you're at a point where the slope of your tangent line is zero, it doesn't mean you're at a relative minimum point, so I would rule that out. G prime crosses the x-axis from below it to above it at x equals negative three."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Let's now introduce ourselves to the idea of a differential equation. And as we'll see, differential equations are super useful for modeling and simulating phenomena and understanding how they operate. But we'll get into that later. For now, let's just think about, or at least look at, what a differential equation actually is. So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could write this in function notation."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "For now, let's just think about, or at least look at, what a differential equation actually is. So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first derivative of y with respect to x is equal to three times y. All three of these, I guess, equations are really representing the same thing."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Another way we could write it, if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first derivative of y with respect to x is equal to three times y. All three of these, I guess, equations are really representing the same thing. They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "All three of these, I guess, equations are really representing the same thing. They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions. It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions. It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation. So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "And it's important to contrast this relative to a traditional equation. So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set of numbers."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set of numbers. We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "The solutions to this algebraic equation are going to be numbers or a set of numbers. We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like?"}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like? And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like? And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x. One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this together."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x. One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this together. So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it, so let's work through this together. So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x. And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x. And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there. So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So let me, so we first have the second derivative of y, so that's that term right over there. So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x. Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Well three times y is three times e to the negative 3x. Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x. This is absolutely going to be true. E to the x plus two to the x is three e to the x. So y two is also a solution to this differential equation."}, {"video_title": "Differential equation introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x. This is absolutely going to be true. E to the x plus two to the x is three e to the x. So y two is also a solution to this differential equation. So that's a start. In the next few videos, we'll explore this more. We'll start to see what the solutions look like, what classes of solutions are, techniques for solving them, visualizing solutions to differential equations, and a whole toolkit for kind of digging in deeper."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We see that right over there. Use the data in the table to estimate the value of not V of 16, V prime of 16. What we see here, they don't even give us V of 16, so how do we think about V prime of 16? Well, just remind ourselves, this is the rate of change of V with respect to time when time is equal to 16. So we can estimate it, and that's the key word here, estimate, we can estimate V prime of 16 by thinking about what is our change in velocity over our change in time around 16. When we look at it over here, they don't give us V of 16, but they give us V of 12, they give us V of 20. So let's figure out our rate of change between 12, T equals 12, and T equals 20."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, just remind ourselves, this is the rate of change of V with respect to time when time is equal to 16. So we can estimate it, and that's the key word here, estimate, we can estimate V prime of 16 by thinking about what is our change in velocity over our change in time around 16. When we look at it over here, they don't give us V of 16, but they give us V of 12, they give us V of 20. So let's figure out our rate of change between 12, T equals 12, and T equals 20. So our change in velocity, that's going to be V of 20 minus V of 12, and then our change in time is going to be 20 minus 12. And so this is going to be equal to, V of 20 is 240, 240, V of 12, we see right there, is 200. And so this is going to be 40 over eight, which is equal to five."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's figure out our rate of change between 12, T equals 12, and T equals 20. So our change in velocity, that's going to be V of 20 minus V of 12, and then our change in time is going to be 20 minus 12. And so this is going to be equal to, V of 20 is 240, 240, V of 12, we see right there, is 200. And so this is going to be 40 over eight, which is equal to five. And we would be done. For good measure, it's good to put the units there. So this is our rate, this is how fast the velocity is changing with respect to time."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be 40 over eight, which is equal to five. And we would be done. For good measure, it's good to put the units there. So this is our rate, this is how fast the velocity is changing with respect to time. So the units are going to be meters per minute per minute. So we could write this as meters per minute squared, per minute, meters per minute squared. And we're done."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is our rate, this is how fast the velocity is changing with respect to time. So the units are going to be meters per minute per minute. So we could write this as meters per minute squared, per minute, meters per minute squared. And we're done. Now if you want to get a little bit more of a visual understanding of this, and what I'm about to do, you would not actually have to do on the actual exam. Well let's just try to graph, let's graph these points here. Let me give myself some space to do it."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're done. Now if you want to get a little bit more of a visual understanding of this, and what I'm about to do, you would not actually have to do on the actual exam. Well let's just try to graph, let's graph these points here. Let me give myself some space to do it. So if we were, if we tried to graph it, so I'll just do a very rough graph here. So let's say this is Y is equal to V of T, and we see that V of T goes as low as negative 220, goes as high as 240. So let me give, so I want to draw the horizontal axis someplace around here."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me give myself some space to do it. So if we were, if we tried to graph it, so I'll just do a very rough graph here. So let's say this is Y is equal to V of T, and we see that V of T goes as low as negative 220, goes as high as 240. So let me give, so I want to draw the horizontal axis someplace around here. T is positive, it looks something like that. And so let's just make, let's make this, let's make that 200 and, let's make that 300. And so then this would be 200 and 100."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me give, so I want to draw the horizontal axis someplace around here. T is positive, it looks something like that. And so let's just make, let's make this, let's make that 200 and, let's make that 300. And so then this would be 200 and 100. 200 and 100 there. And then that would be negative 100, and then that would be negative 200, and then that would be negative 300. And we see on the T axis, our highest value is 40, we go between zero and 40."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so then this would be 200 and 100. 200 and 100 there. And then that would be negative 100, and then that would be negative 200, and then that would be negative 300. And we see on the T axis, our highest value is 40, we go between zero and 40. And so these obviously aren't at the same scale, but this is gonna be zero if we put 40 here, and then if we put 20 in between, and so this would be 10. Let me do it a little bit to the right. That would be 10, and then that would be 30."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we see on the T axis, our highest value is 40, we go between zero and 40. And so these obviously aren't at the same scale, but this is gonna be zero if we put 40 here, and then if we put 20 in between, and so this would be 10. Let me do it a little bit to the right. That would be 10, and then that would be 30. And so what points do they give us? So they give us, I'll do these in orange, zero comma zero, so that is right over there. They give us, when time is 12, our velocity is 200."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That would be 10, and then that would be 30. And so what points do they give us? So they give us, I'll do these in orange, zero comma zero, so that is right over there. They give us, when time is 12, our velocity is 200. So when the time is 12, which is right over there, our velocity is going to be 200. So that's that point. When our time is 20, our velocity is going to be 240."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "They give us, when time is 12, our velocity is 200. So when the time is 12, which is right over there, our velocity is going to be 200. So that's that point. When our time is 20, our velocity is going to be 240. So when our time is 20, our velocity is 240, which is gonna be right over there. And then when our time is 24, our velocity is negative 220, so she switched directions. So when it's 24, so 24 is gonna be roughly over here."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "When our time is 20, our velocity is going to be 240. So when our time is 20, our velocity is 240, which is gonna be right over there. And then when our time is 24, our velocity is negative 220, so she switched directions. So when it's 24, so 24 is gonna be roughly over here. It's gonna be negative 220. So negative 220 might be right over there. And then finally, when time is 40, her velocity is 150, positive 150."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when it's 24, so 24 is gonna be roughly over here. It's gonna be negative 220. So negative 220 might be right over there. And then finally, when time is 40, her velocity is 150, positive 150. So at 40, it's positive 150. And so these are just sample points from her velocity function, but what we wanted to do is we wanted to find, in this problem, we wanted to say, okay, when t is equal to 16, what is the rate of change? And we don't know much about, we don't know what v of 16 is, but what we could do is, and this is essentially what we did in this problem, we could say, all right, well, we can approximate what the function might do by roughly drawing a line here."}, {"video_title": "2015 AP Calculus AB BC 3a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, when time is 40, her velocity is 150, positive 150. So at 40, it's positive 150. And so these are just sample points from her velocity function, but what we wanted to do is we wanted to find, in this problem, we wanted to say, okay, when t is equal to 16, what is the rate of change? And we don't know much about, we don't know what v of 16 is, but what we could do is, and this is essentially what we did in this problem, we could say, all right, well, we can approximate what the function might do by roughly drawing a line here. So if you draw a line there, and you say, all right, well, v of 16, or v prime of 16, I should say, so v prime of 16 is going to be approximately the slope, is going to be approximately the slope of this line. That's going to be our best job, based on the data that they've given us, of estimating the value of v prime of 16. So we literally just did change in v, which is that one, delta v, over change in t, over, over delta t, to get the slope of this line, which is our best approximation for the derivative, when t is equal to 16."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to figure that out, I have to first figure out what are the critical points for my function f, and then which of those critical points do we achieve a minimum or maximum value. And to determine the critical points, we have to find the derivative of our function, because our critical points are just the points at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you could call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x, is either undefined or 0."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x, is either undefined or 0. Well, this thing is defined for all values of x. So the only places we're going to find critical points is when this thing is equal to 0. So let's set it equal to 0."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to get a critical point when this thing right over here, for some value of x, is either undefined or 0. Well, this thing is defined for all values of x. So the only places we're going to find critical points is when this thing is equal to 0. So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well, this is going to happen when x is equal to 2 and x is equal to negative 2."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well, this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime of negative 2 is also, same exact reason, is also equal to 0. So we can say, and I'll switch colors here, that f has critical points at x equals 2 and x equals negative 2."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime of negative 2 is also, same exact reason, is also equal to 0. So we can say, and I'll switch colors here, that f has critical points at x equals 2 and x equals negative 2. Well, that's fair enough. But we still don't know whether the function takes on minimum values at those points, maximum values at those points, or neither. To figure that out, we have to figure out whether the derivative changes signs around these points."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can say, and I'll switch colors here, that f has critical points at x equals 2 and x equals negative 2. Well, that's fair enough. But we still don't know whether the function takes on minimum values at those points, maximum values at those points, or neither. To figure that out, we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. So I'll draw an axis right over here."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "To figure that out, we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. So I'll draw an axis right over here. I'll do it down here, because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll draw an axis right over here. I'll do it down here, because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2. 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well, when x is equal to 0 for the derivative, we're at negative 12."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And x is equal to negative 2. 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well, when x is equal to 0 for the derivative, we're at negative 12. So this is a point y is equal to negative 12. So we're graphing y is equal to f prime of x. So it'll look something like this."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, when x is equal to 0 for the derivative, we're at negative 12. So this is a point y is equal to negative 12. So we're graphing y is equal to f prime of x. So it'll look something like this. It will look something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it'll look something like this. It will look something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well, over here, our derivative is crossing from being positive. We have a positive derivative to being a negative derivative."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well, over here, our derivative is crossing from being positive. We have a positive derivative to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative. That was our criteria for a critical point to be a maximum point. Over here, we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have a positive derivative to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative. That was our criteria for a critical point to be a maximum point. Over here, we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point. So a minimum. And I just want to make sure we have the correct intuition. If some function is increasing going into some point, and at that point, we see actually we have a derivative of 0."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Over here, we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point. So a minimum. And I just want to make sure we have the correct intuition. If some function is increasing going into some point, and at that point, we see actually we have a derivative of 0. The derivative could also be undefined. We have a derivative of 0. And then the function begins decreasing."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "If some function is increasing going into some point, and at that point, we see actually we have a derivative of 0. The derivative could also be undefined. We have a derivative of 0. And then the function begins decreasing. That's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember, this is the graph of the derivative."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the function begins decreasing. That's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember, this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation where going into the point, the function has a negative slope."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation where going into the point, the function has a negative slope. We see we have a negative slope here. So the function might look something like this. And then right at this point, the function is either undefined or has 0 slope."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we have a situation where going into the point, the function has a negative slope. We see we have a negative slope here. So the function might look something like this. And then right at this point, the function is either undefined or has 0 slope. So in this case, it has 0 slope. And then after that point, let me do it right under it. So going into it, we have a negative slope."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then right at this point, the function is either undefined or has 0 slope. So in this case, it has 0 slope. And then after that point, let me do it right under it. So going into it, we have a negative slope. And then right over here, we have a 0 slope. Actually, I could draw it even better than that. So if we were to imagine going into it, we have a negative slope."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So going into it, we have a negative slope. And then right over here, we have a 0 slope. Actually, I could draw it even better than that. So if we were to imagine going into it, we have a negative slope. Right at that point, we have a 0 slope. And then we have a positive slope. So the function begins increasing."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we were to imagine going into it, we have a negative slope. Right at that point, we have a 0 slope. And then we have a positive slope. So the function begins increasing. That's why we say we have a minimum point right over there. So what I did right over here is to try to conceptualize what the function itself could look like given the derivative, in this case, switching from a positive derivative to a negative derivative across that critical point or going from a negative derivative to a positive derivative. That's why this is a criteria for a maximum point."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the function begins increasing. That's why we say we have a minimum point right over there. So what I did right over here is to try to conceptualize what the function itself could look like given the derivative, in this case, switching from a positive derivative to a negative derivative across that critical point or going from a negative derivative to a positive derivative. That's why this is a criteria for a maximum point. This is a criteria for a minimum point. Well, with that out of the way, can we use this intuition that we just talked about to at least try to sketch the graph of f of x? So let's try to do it."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's why this is a criteria for a maximum point. This is a criteria for a minimum point. Well, with that out of the way, can we use this intuition that we just talked about to at least try to sketch the graph of f of x? So let's try to do it. And it's just going to be a sketch. It's not going to be very exact. But at least it will give us a sense of the shape of what f of x looks like."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's try to do it. And it's just going to be a sketch. It's not going to be very exact. But at least it will give us a sense of the shape of what f of x looks like. So my best attempt. So it might not be drawn completely to scale. So it's my x-axis."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "But at least it will give us a sense of the shape of what f of x looks like. So my best attempt. So it might not be drawn completely to scale. So it's my x-axis. This is my y-axis. We know we have a critical point at x is equal to positive 2. And we have a critical point at x is equal to negative 2."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's my x-axis. This is my y-axis. We know we have a critical point at x is equal to positive 2. And we have a critical point at x is equal to negative 2. We know just from inspection that the y-intercept right here of the graph of y is equal to f of x, when x is 0, f of x is 2. So we're going to hit right over. Let's say we're going to hit right over."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we have a critical point at x is equal to negative 2. We know just from inspection that the y-intercept right here of the graph of y is equal to f of x, when x is 0, f of x is 2. So we're going to hit right over. Let's say we're going to hit right over. I don't want to draw this completely to the same scale as the x-axis. So let's say that this is 2 right over here. So this is where we're going to cross."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say we're going to hit right over. I don't want to draw this completely to the same scale as the x-axis. So let's say that this is 2 right over here. So this is where we're going to cross. This is going to be our y-intercept. And so we said already that we have a maximum point at x is equal to negative 2. So what is f of negative 2?"}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is where we're going to cross. This is going to be our y-intercept. And so we said already that we have a maximum point at x is equal to negative 2. So what is f of negative 2? f of negative 2 is equal to 8. Or negative 8. Let me be careful."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is f of negative 2? f of negative 2 is equal to 8. Or negative 8. Let me be careful. It's negative 8. And then we're going to have 12 times negative 2, which is negative 24. But then we're going to add it."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me be careful. It's negative 8. And then we're going to have 12 times negative 2, which is negative 24. But then we're going to add it. So we're subtracting negative 24. So this is plus 24. And then we finally have plus 2."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then we're going to add it. So we're subtracting negative 24. So this is plus 24. And then we finally have plus 2. So negative 8 plus 24 plus 2. So that's going to be negative 8 plus 24 is 16 plus 2 is 18. So f of negative 2 is equal to 18."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we finally have plus 2. So negative 8 plus 24 plus 2. So that's going to be negative 8 plus 24 is 16 plus 2 is 18. So f of negative 2 is equal to 18. And I'm not drawing it completely to scale. Let's say that this is 18 right over here. So this is the function."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of negative 2 is equal to 18. And I'm not drawing it completely to scale. Let's say that this is 18 right over here. So this is the function. This is the point negative 2 comma 18. And we know that it's a maximum point. The derivative going into that point is negative."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the function. This is the point negative 2 comma 18. And we know that it's a maximum point. The derivative going into that point is negative. The derivative going into that point is positive. So we are increasing. The slope is positive."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative going into that point is negative. The derivative going into that point is positive. So we are increasing. The slope is positive. And then after we cross that point, the slope becomes negative. The derivative crossed the x-axis. The slope becomes negative."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope is positive. And then after we cross that point, the slope becomes negative. The derivative crossed the x-axis. The slope becomes negative. Actually, I want to use that same color. It looks like this. And then, of course, the graph is going to have a y-intercept, something like that."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope becomes negative. Actually, I want to use that same color. It looks like this. And then, of course, the graph is going to have a y-intercept, something like that. And then as we approach 2, we are approaching another critical point. Now what is f of 2? f of 2 is going to be equal to positive 8 minus 24 plus 2."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then, of course, the graph is going to have a y-intercept, something like that. And then as we approach 2, we are approaching another critical point. Now what is f of 2? f of 2 is going to be equal to positive 8 minus 24 plus 2. So this is 10 minus 24, which is equal to negative 14. So let's say that this is a point negative 14 right over here. Actually, I could draw it a little bit."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "f of 2 is going to be equal to positive 8 minus 24 plus 2. So this is 10 minus 24, which is equal to negative 14. So let's say that this is a point negative 14 right over here. Actually, I could draw it a little bit. Let's say this is negative 14. So this is f of 2 right over there. And we saw already that the slope is negative as we approach it."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, I could draw it a little bit. Let's say this is negative 14. So this is f of 2 right over there. And we saw already that the slope is negative as we approach it. So our function is decreasing as we approach it. And then right there, the slope is 0. We figured that out earlier."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we saw already that the slope is negative as we approach it. So our function is decreasing as we approach it. And then right there, the slope is 0. We figured that out earlier. That's how we identified it being a critical point. And then the slope is increasing after that. The derivative is positive."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "We figured that out earlier. That's how we identified it being a critical point. And then the slope is increasing after that. The derivative is positive. The slope is increasing. So this is our sketch of what f of x could look like given that these are the critical points. And we were able to identify 2 as a minimum point."}, {"video_title": "Analyzing a function with its derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative is positive. The slope is increasing. So this is our sketch of what f of x could look like given that these are the critical points. And we were able to identify 2 as a minimum point. So this was a minimum value. The function takes on a minimum value when x is equal to 2. And the function took on a maximum value when f was equal to negative 2."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "What does three plus the definite integral from one to five of R of T DT equals 19 mean? And we have some choices, so like always, pause the video and see if you can work through it. All right, now let's work through this together. So they tell us that she made $3,000 in the first month, and we also see this three here, so that's interesting. Maybe they represent the same thing, we don't know for sure yet. But let's look at this definite integral. The definite integral from one to five of R of T DT."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they tell us that she made $3,000 in the first month, and we also see this three here, so that's interesting. Maybe they represent the same thing, we don't know for sure yet. But let's look at this definite integral. The definite integral from one to five of R of T DT. This is the area under this rate curve. R of T is the rate at which Julia makes revenue on a monthly basis. So if you take the area under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "The definite integral from one to five of R of T DT. This is the area under this rate curve. R of T is the rate at which Julia makes revenue on a monthly basis. So if you take the area under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased. So if you add that to the amount that she made in month one, well that tells you the total she makes from essentially time zero all the way to month five. And they're saying that is equal to 19. So let's see which of these choices are consistent with that."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you take the area under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased. So if you add that to the amount that she made in month one, well that tells you the total she makes from essentially time zero all the way to month five. And they're saying that is equal to 19. So let's see which of these choices are consistent with that. Julia made an additional $19,000 between months one and five. Choice A would be correct if you didn't see this three over here, because just the definite integral is the additional between months one and five, but that's not what this expression says. It says three plus this is equal to 19."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see which of these choices are consistent with that. Julia made an additional $19,000 between months one and five. Choice A would be correct if you didn't see this three over here, because just the definite integral is the additional between months one and five, but that's not what this expression says. It says three plus this is equal to 19. If it said Julia made an additional $16,000, well then that would make sense, because you could subtract three from both sides and you'd get that result, but that's not what they're saying. Julia made an average of $19,000 per month. Well once again, that's also not right, because we already just said from the beginning, from time zero all the way until the fifth month, she made a total of $19,000, not the average per month is 19,000."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "It says three plus this is equal to 19. If it said Julia made an additional $16,000, well then that would make sense, because you could subtract three from both sides and you'd get that result, but that's not what they're saying. Julia made an average of $19,000 per month. Well once again, that's also not right, because we already just said from the beginning, from time zero all the way until the fifth month, she made a total of $19,000, not the average per month is 19,000. Julia made $19,000 in the fifth month. Once again, this is not just saying what happened in the fifth month. This is saying we have the $3,000 from the first month, and then we have the additional from month, between months one and month five."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well once again, that's also not right, because we already just said from the beginning, from time zero all the way until the fifth month, she made a total of $19,000, not the average per month is 19,000. Julia made $19,000 in the fifth month. Once again, this is not just saying what happened in the fifth month. This is saying we have the $3,000 from the first month, and then we have the additional from month, between months one and month five. So that's not that, so this better be our choice. By the end of the fifth month, Julia had made a total of $19,000. Yes, that is correct."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is saying we have the $3,000 from the first month, and then we have the additional from month, between months one and month five. So that's not that, so this better be our choice. By the end of the fifth month, Julia had made a total of $19,000. Yes, that is correct. She made $3,000 in month one, and then as we go between month one to the end of month five, to the end of the fifth month, she has made a total of $19,000. Let's do another one of these. So here we're told the function K of t gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "Yes, that is correct. She made $3,000 in month one, and then as we go between month one to the end of month five, to the end of the fifth month, she has made a total of $19,000. Let's do another one of these. So here we're told the function K of t gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day. So this is really quantity as a function of time. It isn't rate. What does the definite integral from zero to four of K prime of t dt represent?"}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here we're told the function K of t gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day. So this is really quantity as a function of time. It isn't rate. What does the definite integral from zero to four of K prime of t dt represent? Once again, pause the video and see if you can work through it. Well, K of t is the amount of ketchup as a function of time. So K prime of t, that's going to be the rate at which our amount of ketchup is changing as a function of time."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "What does the definite integral from zero to four of K prime of t dt represent? Once again, pause the video and see if you can work through it. Well, K of t is the amount of ketchup as a function of time. So K prime of t, that's going to be the rate at which our amount of ketchup is changing as a function of time. But once again, when you're taking the area under the rate curve, that tells you the net change in the original quantity, in the amount of ketchup. And it's the net change between time zero and time four. So let's see which of these choices match up to that."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So K prime of t, that's going to be the rate at which our amount of ketchup is changing as a function of time. But once again, when you're taking the area under the rate curve, that tells you the net change in the original quantity, in the amount of ketchup. And it's the net change between time zero and time four. So let's see which of these choices match up to that. The average rate of change of the ketchup production over the first four hours, no, that does not tell us the average rate of change. There's other ways to calculate that. The time it takes to produce four kilograms of ketchup."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see which of these choices match up to that. The average rate of change of the ketchup production over the first four hours, no, that does not tell us the average rate of change. There's other ways to calculate that. The time it takes to produce four kilograms of ketchup. So does this represent the time it takes to represent four kilograms of ketchup? No, this four is a time right over here. This is gonna tell you how much ketchup gets produced from time zero to time four."}, {"video_title": "Worked examples interpreting definite integrals in context   AP Calculus AB   Khan Academy.mp3", "Sentence": "The time it takes to produce four kilograms of ketchup. So does this represent the time it takes to represent four kilograms of ketchup? No, this four is a time right over here. This is gonna tell you how much ketchup gets produced from time zero to time four. The instantaneous rate of production at t equals four. No, this would be K prime of four. That's not what this integral represents."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "which of the following functions are continuous at x equals three? Well, as we said in the previous video, in the previous example, in order to be continuous at a point, you at least have to be defined at that point. We saw our definition of continuity. F is continuous at a if and only if the limit of f as x approaches a is equal to f of a. So over here in this case, we could say that a function is continuous at x equals three so f is continuous at x equals three if and only if, if and only if, the limit as x approaches three of f of x is equal to f of three. Now let's look at this first function right over here, natural log of x minus three. Well, try to evaluate, and it's not f now, it's g, try to evaluate g of three."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "F is continuous at a if and only if the limit of f as x approaches a is equal to f of a. So over here in this case, we could say that a function is continuous at x equals three so f is continuous at x equals three if and only if, if and only if, the limit as x approaches three of f of x is equal to f of three. Now let's look at this first function right over here, natural log of x minus three. Well, try to evaluate, and it's not f now, it's g, try to evaluate g of three. G of three, let me write it here, g of three is equal to the natural log of zero, three minus three. This is not defined. You can't raise e to any power to get to zero."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, try to evaluate, and it's not f now, it's g, try to evaluate g of three. G of three, let me write it here, g of three is equal to the natural log of zero, three minus three. This is not defined. You can't raise e to any power to get to zero. You try to go to, you know, you could say negative infinity, but that's not, this is not defined. And so if this isn't even defined at x equals three, there's no way that it's going to be continuous at x equals three. So we could rule this one out."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can't raise e to any power to get to zero. You try to go to, you know, you could say negative infinity, but that's not, this is not defined. And so if this isn't even defined at x equals three, there's no way that it's going to be continuous at x equals three. So we could rule this one out. Now f of x is equal to e to the x minus three. Well, this is just a shifted over version of e to the x. This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could rule this one out. Now f of x is equal to e to the x minus three. Well, this is just a shifted over version of e to the x. This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here. The limit of e to the x minus three as x approaches three, well, that is going to be, that is going to be e to the three minus three, or e to the zero, or one. And so f is the only one that is continuous. And once again, it's good to think about what's going on here visually if you like."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here. The limit of e to the x minus three as x approaches three, well, that is going to be, that is going to be e to the three minus three, or e to the zero, or one. And so f is the only one that is continuous. And once again, it's good to think about what's going on here visually if you like. Both of these are, you could think of them, this is a shifted over version of ln of x, this is a shifted over version of e to the x, and so if we like, we could draw ourselves some axes. So that's our y-axis. This is our x-axis."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, it's good to think about what's going on here visually if you like. Both of these are, you could think of them, this is a shifted over version of ln of x, this is a shifted over version of e to the x, and so if we like, we could draw ourselves some axes. So that's our y-axis. This is our x-axis. And actually, let me draw some points here. So that's one, that is one, that is, let's see, two, three, two, and three. And let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our x-axis. And actually, let me draw some points here. So that's one, that is one, that is, let's see, two, three, two, and three. And let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it. So let me draw it, this is one, two, three, four, five, and six. And on this axis, I want to make them on the same scale, so let's say this is one, two, three. I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it. So let me draw it, this is one, two, three, four, five, and six. And on this axis, I want to make them on the same scale, so let's say this is one, two, three. I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here. So g of x, ln of x minus three is gonna look something like this. If you put three in it, it's not defined. If you put four in it, ln of four, well, that's, sorry, ln of four minus one, so that's gonna be ln of, sorry, ln of four minus three is, actually, let me just draw a table here, I'm confusing you."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here. So g of x, ln of x minus three is gonna look something like this. If you put three in it, it's not defined. If you put four in it, ln of four, well, that's, sorry, ln of four minus one, so that's gonna be ln of, sorry, ln of four minus three is, actually, let me just draw a table here, I'm confusing you. So if I say x, and I say g of x, so at three, you're undefined. At four, this is ln of one, ln of one, which is equal to zero. So it's right over there."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you put four in it, ln of four, well, that's, sorry, ln of four minus one, so that's gonna be ln of, sorry, ln of four minus three is, actually, let me just draw a table here, I'm confusing you. So if I say x, and I say g of x, so at three, you're undefined. At four, this is ln of one, ln of one, which is equal to zero. So it's right over there. So g of x is gonna look something like, something like that. And so you can see, at three, you have this discontinuity there. It's not even defined to the left of three."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's right over there. So g of x is gonna look something like, something like that. And so you can see, at three, you have this discontinuity there. It's not even defined to the left of three. Now f of x is a little bit more straightforward. F of three is gonna be e to the three minus three, or e to the zero, so it's going to be one. So it's gonna look something like this."}, {"video_title": "Functions continuous at specific x-values   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's not even defined to the left of three. Now f of x is a little bit more straightforward. F of three is gonna be e to the three minus three, or e to the zero, so it's going to be one. So it's gonna look something like this. It's gonna look something, something like that. Like that, there's no jumps, there's no gaps. It is going to be continuous at, frankly, all real numbers, so for sure, it's going to be continuous at three."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Thousands of pairs produced. Now let's think about how much money you're going to make per pair. Actually, let me say how much revenue, which is how much do you actually get to sell those shoes for? So let's write a function right here. Revenue as a function of x. Well, you have a wholesaler who's willing to pay you $10 per pair for as many pairs as you're willing to give him. So your revenue as a function of x is going to be 10 times x."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's write a function right here. Revenue as a function of x. Well, you have a wholesaler who's willing to pay you $10 per pair for as many pairs as you're willing to give him. So your revenue as a function of x is going to be 10 times x. And since x is in thousands of pairs produced, if x is one, that means 1,000 pairs produced times 10, which means $10,000. But this will just give you 10. So this right over here is in thousands of dollars."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So your revenue as a function of x is going to be 10 times x. And since x is in thousands of pairs produced, if x is one, that means 1,000 pairs produced times 10, which means $10,000. But this will just give you 10. So this right over here is in thousands of dollars. Thousands of dollars. So if x is one, that means 1,000 pairs produced. 10 times one says r is equal to 10, but that really means $10,000."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is in thousands of dollars. Thousands of dollars. So if x is one, that means 1,000 pairs produced. 10 times one says r is equal to 10, but that really means $10,000. Now, it would be a nice business if all you had was revenue and no costs, but you do have costs. You have materials, you have to build your factory, you have to pay your employees, you have to pay the electricity bill. And so you hire a bunch of consultants to come up with what your cost is as a function of x."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "10 times one says r is equal to 10, but that really means $10,000. Now, it would be a nice business if all you had was revenue and no costs, but you do have costs. You have materials, you have to build your factory, you have to pay your employees, you have to pay the electricity bill. And so you hire a bunch of consultants to come up with what your cost is as a function of x. What your cost is as a function of x. And they come up with a function. They say it is the number of the thousands of pairs you produce cubed minus six times the thousands of pairs you produce squared plus 15 times the thousands of pairs that you produce."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you hire a bunch of consultants to come up with what your cost is as a function of x. What your cost is as a function of x. And they come up with a function. They say it is the number of the thousands of pairs you produce cubed minus six times the thousands of pairs you produce squared plus 15 times the thousands of pairs that you produce. And once again, this is going to be, this is also going to be in thousands of dollars. Now, given these functions of x for revenue and cost, what is profit as a function of x going to be? Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "They say it is the number of the thousands of pairs you produce cubed minus six times the thousands of pairs you produce squared plus 15 times the thousands of pairs that you produce. And once again, this is going to be, this is also going to be in thousands of dollars. Now, given these functions of x for revenue and cost, what is profit as a function of x going to be? Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x. Minus your cost as a function of x. If you produce a certain amount, and let's say you bring in, I don't know, $10,000 of revenue and it costs you $5,000 to produce those shoes, you'll have $5,000 in profit. Those numbers aren't the ones that would actually, you would get from this right here."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x. Minus your cost as a function of x. If you produce a certain amount, and let's say you bring in, I don't know, $10,000 of revenue and it costs you $5,000 to produce those shoes, you'll have $5,000 in profit. Those numbers aren't the ones that would actually, you would get from this right here. I'm just giving you an example. So this is what you want to optimize. You want to optimize p as a function, optimize p as a function of x."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Those numbers aren't the ones that would actually, you would get from this right here. I'm just giving you an example. So this is what you want to optimize. You want to optimize p as a function, optimize p as a function of x. So what is it? I've just said it here in abstract terms, but we know what r of x is and what c of x. This is 10x, this is 10x minus all of this business."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You want to optimize p as a function, optimize p as a function of x. So what is it? I've just said it here in abstract terms, but we know what r of x is and what c of x. This is 10x, this is 10x minus all of this business. So minus x to the third plus 6x squared minus 15x. I just subtracted x squared. You subtract a 6x squared, it becomes positive."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 10x, this is 10x minus all of this business. So minus x to the third plus 6x squared minus 15x. I just subtracted x squared. You subtract a 6x squared, it becomes positive. You subtract a 15x, it becomes negative 15x. And then we can simplify this as, let's see, we have negative x to the third plus 6x squared minus 15x plus 10x, so that is minus 5x. Now if we want to optimize this profit function analytically, the easiest way is to think about what are the critical points of this profit function and are any of those critical points minimum points or maximum points?"}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You subtract a 6x squared, it becomes positive. You subtract a 15x, it becomes negative 15x. And then we can simplify this as, let's see, we have negative x to the third plus 6x squared minus 15x plus 10x, so that is minus 5x. Now if we want to optimize this profit function analytically, the easiest way is to think about what are the critical points of this profit function and are any of those critical points minimum points or maximum points? And if one of them is a maximum point, then we can say, well, let's produce that many. That is going to be, we will have optimized or we'll figure out the quantity we need to produce in order to optimize our profit. So to figure out critical points, we essentially have to find the derivative of our function and figure out when does that derivative equal zero or when is that derivative undefined?"}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now if we want to optimize this profit function analytically, the easiest way is to think about what are the critical points of this profit function and are any of those critical points minimum points or maximum points? And if one of them is a maximum point, then we can say, well, let's produce that many. That is going to be, we will have optimized or we'll figure out the quantity we need to produce in order to optimize our profit. So to figure out critical points, we essentially have to find the derivative of our function and figure out when does that derivative equal zero or when is that derivative undefined? That's the definition of critical points. So p prime of x is going to be equal to negative 3x squared plus 12x minus five. And so this thing is going to be defined for all x."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So to figure out critical points, we essentially have to find the derivative of our function and figure out when does that derivative equal zero or when is that derivative undefined? That's the definition of critical points. So p prime of x is going to be equal to negative 3x squared plus 12x minus five. And so this thing is going to be defined for all x. So the only critical points we're going to have is when the first derivative right over here is equal to zero. So negative 3x squared plus 12x minus five needs to be equal to zero in order for x to be a critical point. So now we just have to solve for x."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this thing is going to be defined for all x. So the only critical points we're going to have is when the first derivative right over here is equal to zero. So negative 3x squared plus 12x minus five needs to be equal to zero in order for x to be a critical point. So now we just have to solve for x. And so we just are essentially solving a quadratic equation just so that I don't have as many negatives. Let's multiply both sides by negative one. I just like to have a clean first coefficient."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we just have to solve for x. And so we just are essentially solving a quadratic equation just so that I don't have as many negatives. Let's multiply both sides by negative one. I just like to have a clean first coefficient. So if we multiply both sides by negative one, we get 3x squared minus 12x plus five is equal to zero. And now we can use the quadratic formula to solve for x. So x is going to be equal to negative b, which is 12 plus or minus the square root."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I just like to have a clean first coefficient. So if we multiply both sides by negative one, we get 3x squared minus 12x plus five is equal to zero. And now we can use the quadratic formula to solve for x. So x is going to be equal to negative b, which is 12 plus or minus the square root. I always need to make my radical signs wide enough. The square root of b squared, which is 144, minus four times a, which is three, times c, which is five, times five. All of that, all of that over 2a."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So x is going to be equal to negative b, which is 12 plus or minus the square root. I always need to make my radical signs wide enough. The square root of b squared, which is 144, minus four times a, which is three, times c, which is five, times five. All of that, all of that over 2a. So two times three is six. So x is equal to 12 plus or minus the square root of, let's see, four times three is 12, times five is 60. 144 minus 60 is 84."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that, all of that over 2a. So two times three is six. So x is equal to 12 plus or minus the square root of, let's see, four times three is 12, times five is 60. 144 minus 60 is 84. All of that over six. So x could be equal to 12 plus the square root of 84 over six or x could be equal to 12 minus the square root of 84 over six. So let's figure out what these two are."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "144 minus 60 is 84. All of that over six. So x could be equal to 12 plus the square root of 84 over six or x could be equal to 12 minus the square root of 84 over six. So let's figure out what these two are. And I'll use the calculator. I'll use the calculator for this one. So I get, let's see, 12 plus the square root of 84 divided by six gives me 3.5, I'll just say 3.53."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's figure out what these two are. And I'll use the calculator. I'll use the calculator for this one. So I get, let's see, 12 plus the square root of 84 divided by six gives me 3.5, I'll just say 3.53. So approximately, actually let me go one more digit because I'm talking about thousands. So let me say 3.528. 3.528, 3.528."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I get, let's see, 12 plus the square root of 84 divided by six gives me 3.5, I'll just say 3.53. So approximately, actually let me go one more digit because I'm talking about thousands. So let me say 3.528. 3.528, 3.528. So this would literally be 3,528 shoes because this is in thousands. And let's do our pairs of shoes. And then let's do the situation where we subtract."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "3.528, 3.528. So this would literally be 3,528 shoes because this is in thousands. And let's do our pairs of shoes. And then let's do the situation where we subtract. And actually we can look at our previous entry and just change this to a subtraction. Change that to not a negative sign, a subtraction. There you go."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let's do the situation where we subtract. And actually we can look at our previous entry and just change this to a subtraction. Change that to not a negative sign, a subtraction. There you go. And we get.4725,.47, let me remember that,.4725. Approximately equal to 0.4725. I have a horrible memory so let me review that I wrote the same thing."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "There you go. And we get.4725,.47, let me remember that,.4725. Approximately equal to 0.4725. I have a horrible memory so let me review that I wrote the same thing. 47, 4725, yep, all right. Now, all we know about these are these are both critical points. These are points at which our derivative is equal to zero."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I have a horrible memory so let me review that I wrote the same thing. 47, 4725, yep, all right. Now, all we know about these are these are both critical points. These are points at which our derivative is equal to zero. But we don't know whether they're minimum points. They're points at which the function takes on a minimum value, a maximum value, or neither. To do that, I'll use the second derivative test to figure out if our function is concave upwards or concave downwards or neither at one of these points."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "These are points at which our derivative is equal to zero. But we don't know whether they're minimum points. They're points at which the function takes on a minimum value, a maximum value, or neither. To do that, I'll use the second derivative test to figure out if our function is concave upwards or concave downwards or neither at one of these points. So let's look at the second derivative. So P prime prime of x is going to be equal to negative six x plus negative six x plus 12. And so if we look at, let me make sure I have enough space."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "To do that, I'll use the second derivative test to figure out if our function is concave upwards or concave downwards or neither at one of these points. So let's look at the second derivative. So P prime prime of x is going to be equal to negative six x plus negative six x plus 12. And so if we look at, let me make sure I have enough space. So if we look at P prime prime, P prime prime of 3.528. So let's see if I can think this through. So this is between three and four."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we look at, let me make sure I have enough space. So if we look at P prime prime, P prime prime of 3.528. So let's see if I can think this through. So this is between three and four. So if we take the lower value, three times negative six is negative 18 plus 12. So that's going to be less than zero. And if this was four, it would be even more negative."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is between three and four. So if we take the lower value, three times negative six is negative 18 plus 12. So that's going to be less than zero. And if this was four, it would be even more negative. So this thing is going to be less than zero. I don't even have to use my calculator to evaluate it. Now what about this thing right over here,.47?"}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if this was four, it would be even more negative. So this thing is going to be less than zero. I don't even have to use my calculator to evaluate it. Now what about this thing right over here,.47? Well,.47, that's gonna get us, that's roughly.5. So negative six times.5 is negative three. This is going to be nowhere close to being negative."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what about this thing right over here,.47? Well,.47, that's gonna get us, that's roughly.5. So negative six times.5 is negative three. This is going to be nowhere close to being negative. This is definitely going to be positive. So P prime prime of 0.4725 is greater than zero. So the fact that the second derivative is less than zero, that means that my derivative is decreasing, my derivative, my first derivative is decreasing when x is equal to this value, which means that our graph, our function is concave downwards here."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be nowhere close to being negative. This is definitely going to be positive. So P prime prime of 0.4725 is greater than zero. So the fact that the second derivative is less than zero, that means that my derivative is decreasing, my derivative, my first derivative is decreasing when x is equal to this value, which means that our graph, our function is concave downwards here. Concave, concave downwards. And concave downwards means it looks something like this. And so, and you can see when it looks something like that, the slope is constantly decreasing."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the fact that the second derivative is less than zero, that means that my derivative is decreasing, my derivative, my first derivative is decreasing when x is equal to this value, which means that our graph, our function is concave downwards here. Concave, concave downwards. And concave downwards means it looks something like this. And so, and you can see when it looks something like that, the slope is constantly decreasing. So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal to 3.528, it must be a maximum. It must be a maximum. So we actually do take on a maximum value when x is 3.528."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so, and you can see when it looks something like that, the slope is constantly decreasing. So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal to 3.528, it must be a maximum. It must be a maximum. So we actually do take on a maximum value when x is 3.528. On the other side, we see that over here we are concave upwards, concave upwards. Concave upwards, the graph will look something like this over here. And if the slope is zero where the graph looks like that, we see that that is a local minimum."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we actually do take on a maximum value when x is 3.528. On the other side, we see that over here we are concave upwards, concave upwards. Concave upwards, the graph will look something like this over here. And if the slope is zero where the graph looks like that, we see that that is a local minimum. That is a local minimum. And so we definitely don't want to do this. We would produce 472 and a half units if we were looking to minimize our profit, maximize our loss."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if the slope is zero where the graph looks like that, we see that that is a local minimum. That is a local minimum. And so we definitely don't want to do this. We would produce 472 and a half units if we were looking to minimize our profit, maximize our loss. So we definitely don't want to do this. But let's actually think about what our profit is going to be if we produce 3.528 thousands of shoes, or 3,528 shoes. Well, to do that, we just have to input it back into our original profit function right over here."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We would produce 472 and a half units if we were looking to minimize our profit, maximize our loss. So we definitely don't want to do this. But let's actually think about what our profit is going to be if we produce 3.528 thousands of shoes, or 3,528 shoes. Well, to do that, we just have to input it back into our original profit function right over here. So let's do that. So get my calculator out. So my original profit function is right over there."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to do that, we just have to input it back into our original profit function right over here. So let's do that. So get my calculator out. So my original profit function is right over there. So I want to be able to see that and that. So I get negative 3.528 to the third power plus 6 times 3.528 squared minus 5 times 3.528 gives me, and we get a drum roll now, gives me a profit of 13.128. So let me write this down."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So my original profit function is right over there. So I want to be able to see that and that. So I get negative 3.528 to the third power plus 6 times 3.528 squared minus 5 times 3.528 gives me, and we get a drum roll now, gives me a profit of 13.128. So let me write this down. The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I produce exactly that many shoes, it's equal to 13.128. Or actually, it's approximately because I'm still rounding. 13.128."}, {"video_title": "Optimization profit   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write this down. The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I produce exactly that many shoes, it's equal to 13.128. Or actually, it's approximately because I'm still rounding. 13.128. So if I produce 3,528 shoes in a given period, I am going to have a profit of $13,128. Remember, this right over here is in thousands. This right over here is 13.128 thousands of dollars in profit, which is $13,128."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we're curious about is what is the derivative of this with respect to x? What is dy dx, which we could also write as y prime? Well, there's a couple of ways to think about it. This isn't a straightforward expression here, but you might notice that I have something being raised to the third power. In fact, if we look at the outside of this expression, we have some business in here as being raised to the third power. And so one way to tackle this is to apply the chain rule. So if we apply the chain rule, it's gonna be the derivative of the outside with respect to the inside, or the something to the third power, the derivative of the something to the third power with respect to that something."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This isn't a straightforward expression here, but you might notice that I have something being raised to the third power. In fact, if we look at the outside of this expression, we have some business in here as being raised to the third power. And so one way to tackle this is to apply the chain rule. So if we apply the chain rule, it's gonna be the derivative of the outside with respect to the inside, or the something to the third power, the derivative of the something to the third power with respect to that something. So it's going to be three times that something squared times the derivative with respect to x of that something. In this case, the something is sine, let me write that in a blue color, it is sine of x squared. It is sine of x squared."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we apply the chain rule, it's gonna be the derivative of the outside with respect to the inside, or the something to the third power, the derivative of the something to the third power with respect to that something. So it's going to be three times that something squared times the derivative with respect to x of that something. In this case, the something is sine, let me write that in a blue color, it is sine of x squared. It is sine of x squared. If I, no matter what was inside of these orange parentheses, I would put it inside of the orange parentheses and these orange brackets right over here. We learned that in the chain rule. So let's see, we know this is just a matter, the first part of the expression is just a matter of algebraic simplification."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is sine of x squared. If I, no matter what was inside of these orange parentheses, I would put it inside of the orange parentheses and these orange brackets right over here. We learned that in the chain rule. So let's see, we know this is just a matter, the first part of the expression is just a matter of algebraic simplification. But the second part, we need to now take the derivative of sine of x squared. Well now, we would want to use the chain rule again. So I'm gonna take the derivative, it's sine of something."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, we know this is just a matter, the first part of the expression is just a matter of algebraic simplification. But the second part, we need to now take the derivative of sine of x squared. Well now, we would want to use the chain rule again. So I'm gonna take the derivative, it's sine of something. So this is going to be, the derivative of this is going to be the sine of something with respect to something. So that is cosine of that something times the derivative with respect to x of the something. In this case, the something is our x squared."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm gonna take the derivative, it's sine of something. So this is going to be, the derivative of this is going to be the sine of something with respect to something. So that is cosine of that something times the derivative with respect to x of the something. In this case, the something is our x squared. And of course, we have all of this out front, which is the three times sine of x squared, and I could write it like this, squared. Alright, so we're getting close. Now we just have to figure out the derivative with respect to x of x squared, and we've seen that many times before."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "In this case, the something is our x squared. And of course, we have all of this out front, which is the three times sine of x squared, and I could write it like this, squared. Alright, so we're getting close. Now we just have to figure out the derivative with respect to x of x squared, and we've seen that many times before. That, we just use the power rule, that's going to be two x. Two x. And so if we wanted to write the dy dx, let me get a little bit of a mini drum roll here, this shouldn't take us too long."}, {"video_title": "Applying the chain rule twice   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we just have to figure out the derivative with respect to x of x squared, and we've seen that many times before. That, we just use the power rule, that's going to be two x. Two x. And so if we wanted to write the dy dx, let me get a little bit of a mini drum roll here, this shouldn't take us too long. Dy dx, I'll multiply the three times the two x, which is going to be six x, so I covered those so far, times sine squared of x squared, times sine squared of x squared, times cosine of x squared. And we are done. So we're applying the chain rule multiple times."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But then we saw that the other way around isn't necessarily true. x equal a being a critical point does not necessarily mean that the function takes on a minimum or maximum value at that point. So what we're going to try to do in this video is try to come up with some criteria, especially involving the derivative of the function around x equals a, to figure out if it is a minimum or a maximum point. So let's look at what we saw in the last video. We saw that this point right over here is where the function takes on a maximum value. So this critical point in particular was x-naught. What made it a critical point was that the derivative is zero."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at what we saw in the last video. We saw that this point right over here is where the function takes on a maximum value. So this critical point in particular was x-naught. What made it a critical point was that the derivative is zero. You have a critical point where either the derivative is zero or the derivative is undefined. So this is a critical point. And let's explore what the derivative is doing as we approach that point."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What made it a critical point was that the derivative is zero. You have a critical point where either the derivative is zero or the derivative is undefined. So this is a critical point. And let's explore what the derivative is doing as we approach that point. So in order for this to be a maximum point, the function is increasing as we approach it. The function is increasing is another way of saying that the slope is positive. The slope is changing, but it stays positive the whole time, which means that the function is increasing."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's explore what the derivative is doing as we approach that point. So in order for this to be a maximum point, the function is increasing as we approach it. The function is increasing is another way of saying that the slope is positive. The slope is changing, but it stays positive the whole time, which means that the function is increasing. And the slope being positive is another way of saying that the derivative is greater than zero as we approach that point. Now what happens as we pass that point? Right at that point, the slope is zero."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope is changing, but it stays positive the whole time, which means that the function is increasing. And the slope being positive is another way of saying that the derivative is greater than zero as we approach that point. Now what happens as we pass that point? Right at that point, the slope is zero. But then as we pass that point, what has to happen in order for that to be a maximum point? Well, the value of the function has to go down. If the value of the function is going down, that means the slope is negative."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right at that point, the slope is zero. But then as we pass that point, what has to happen in order for that to be a maximum point? Well, the value of the function has to go down. If the value of the function is going down, that means the slope is negative. And that's another way of saying that the derivative is negative. So that seems like a pretty good criteria for identifying whether a critical point is a maximum point. So let's say that we have critical point A."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If the value of the function is going down, that means the slope is negative. And that's another way of saying that the derivative is negative. So that seems like a pretty good criteria for identifying whether a critical point is a maximum point. So let's say that we have critical point A. We are at a maximum point if f prime of x switches signs from positive to negative as we cross x equals a. That's exactly what happened right over here. Let's make sure it happened at our other maximum point right over here."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that we have critical point A. We are at a maximum point if f prime of x switches signs from positive to negative as we cross x equals a. That's exactly what happened right over here. Let's make sure it happened at our other maximum point right over here. So right over here, as we approach that point, the function is increasing. The function increasing means that the slope is positive. It's a different positive slope."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's make sure it happened at our other maximum point right over here. So right over here, as we approach that point, the function is increasing. The function increasing means that the slope is positive. It's a different positive slope. The slope is changing. It's actually getting more and more and more positive. But it is definitely positive."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's a different positive slope. The slope is changing. It's actually getting more and more and more positive. But it is definitely positive. So it's positive going into that point, and then it becomes negative after we cross that point. The slope was undefined right at the point, but it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it is definitely positive. So it's positive going into that point, and then it becomes negative after we cross that point. The slope was undefined right at the point, but it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point. So, so far, our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make, and I think we call this, let's see, this was x zero, this was x one, this was x two, this was x one, this was x two, so this is x three. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So these both meet our criteria for being a maximum point. So, so far, our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make, and I think we call this, let's see, this was x zero, this was x one, this was x two, this was x one, this was x two, so this is x three. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative, and then as we cross it, our slope is still negative. We're still decreasing. So we haven't switched signs."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative, and then as we cross it, our slope is still negative. We're still decreasing. So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with a criteria for a minimum point, and I think you could see where this is likely to go. Well, we identified in the last video that this right over here is a minimum point."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with a criteria for a minimum point, and I think you could see where this is likely to go. Well, we identified in the last video that this right over here is a minimum point. We can see that. It's a local minimum just by looking at it. And what's the slope doing as we approach it?"}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we identified in the last video that this right over here is a minimum point. We can see that. It's a local minimum just by looking at it. And what's the slope doing as we approach it? So the function is decreasing. The slope is negative as we approach it. f prime of x is less than zero as we approach that point."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what's the slope doing as we approach it? So the function is decreasing. The slope is negative as we approach it. f prime of x is less than zero as we approach that point. And then right after we cross it, this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "f prime of x is less than zero as we approach that point. And then right after we cross it, this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than zero. So this seems like pretty good criteria for a minimum point."}, {"video_title": "Finding relative extrema (first derivative test)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than zero. So this seems like pretty good criteria for a minimum point. f prime of x switches signs from negative to positive as we cross a. If we have some critical point a, the function takes on a minimum value at a if the derivative of our function switches signs from negative to positive as we cross a, from negative to positive. Now once again, this point right over here, this critical point, x sub 3, does not meet that criteria."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "The rate at which rainwater flows into a drain pipe is modeled by the function r, where r of t is equal to 20 sin of t squared over 35 cubic feet per hour. t is measured in hours, and 0 is less than or equal to t, which is less than or equal to 8, so t is going to go between 0 and 8. The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by d of t. It's equal to negative 0.04 t to the third power, plus 0.4 t squared plus 0.96 t cubic feet per hour. For the same interval, right over here, there are 30 cubic feet of water in the pipe at time t equals 0. Part a. How many cubic feet of rainwater flow into the pipe during the 8 hour time interval 0 is less than or equal to t is less than or equal to 8? Alright, so we know the rate, the rate that things flow into the rainwater pipe."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "For the same interval, right over here, there are 30 cubic feet of water in the pipe at time t equals 0. Part a. How many cubic feet of rainwater flow into the pipe during the 8 hour time interval 0 is less than or equal to t is less than or equal to 8? Alright, so we know the rate, the rate that things flow into the rainwater pipe. In fact, we could, let me draw a little rainwater pipe here just so we can visualize what's going on. So if this is, if that is the pipe right over there, things are flowing in at a rate of r of t, and things are flowing out at a rate of d of t, and they even tell us that there's 30 cubic feet of water right in the beginning. But these are the rates of entering and the rates of exiting."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, so we know the rate, the rate that things flow into the rainwater pipe. In fact, we could, let me draw a little rainwater pipe here just so we can visualize what's going on. So if this is, if that is the pipe right over there, things are flowing in at a rate of r of t, and things are flowing out at a rate of d of t, and they even tell us that there's 30 cubic feet of water right in the beginning. But these are the rates of entering and the rates of exiting. So how much water, they're asking how many cubic feet of water flow into, so enter into the pipe during the 8 hour time interval. So if you have your rate, this is the rate at which things are flowing into it, they give it in cubic feet per hour. If you multiply it times some change in time, even infinitesimally small change in time, so d t, this is the amount that flows in over that very small change in time."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "But these are the rates of entering and the rates of exiting. So how much water, they're asking how many cubic feet of water flow into, so enter into the pipe during the 8 hour time interval. So if you have your rate, this is the rate at which things are flowing into it, they give it in cubic feet per hour. If you multiply it times some change in time, even infinitesimally small change in time, so d t, this is the amount that flows in over that very small change in time. And so what we want to do is we want to sum up these very small, these amounts over very small changes in time to go from time is equal to 0 all the way to time is equal to 8. So this expression right over here, this is going to give us how many cubic feet of water flow into the pipe. Once again, what am I doing?"}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you multiply it times some change in time, even infinitesimally small change in time, so d t, this is the amount that flows in over that very small change in time. And so what we want to do is we want to sum up these very small, these amounts over very small changes in time to go from time is equal to 0 all the way to time is equal to 8. So this expression right over here, this is going to give us how many cubic feet of water flow into the pipe. Once again, what am I doing? r of t times d of t, this is how much flows, what volume flows in over a very small interval d t, and then we're going to sum it up from t equals 0 to t equals 8. That's the power of the definite integral, and so this is going to be equal to the integral from 0 to 8 of 20 sine of t squared over 35 d t. And lucky for us, we can use calculators in this section of the AP exam, so let's bring out a graphing calculator where we can evaluate definite integrals. And so let's see, we want to do definite integrals so that I can click math right over here, move down."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, what am I doing? r of t times d of t, this is how much flows, what volume flows in over a very small interval d t, and then we're going to sum it up from t equals 0 to t equals 8. That's the power of the definite integral, and so this is going to be equal to the integral from 0 to 8 of 20 sine of t squared over 35 d t. And lucky for us, we can use calculators in this section of the AP exam, so let's bring out a graphing calculator where we can evaluate definite integrals. And so let's see, we want to do definite integrals so that I can click math right over here, move down. So this function f n integral, this is a integral of our function or a function integral right over here, so let me press enter. And the way that you do it is you first define the function, then you put a comma, then you say what variable is the variable that you're integrating with respect to, and then you put the bounds of integration. So I'm going to write 20 sine of, and just because it's easier for me to input x than t, I'm going to use x, but if you just use this as sine of x squared over 35 dx, you're going to get the same value."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's see, we want to do definite integrals so that I can click math right over here, move down. So this function f n integral, this is a integral of our function or a function integral right over here, so let me press enter. And the way that you do it is you first define the function, then you put a comma, then you say what variable is the variable that you're integrating with respect to, and then you put the bounds of integration. So I'm going to write 20 sine of, and just because it's easier for me to input x than t, I'm going to use x, but if you just use this as sine of x squared over 35 dx, you're going to get the same value. So you're going to get x squared divided by 35, close that parenthesis, so that is my function there. Actually, I don't know if it's going to understand. Let me put the times, second insert times just to make sure it understands that."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm going to write 20 sine of, and just because it's easier for me to input x than t, I'm going to use x, but if you just use this as sine of x squared over 35 dx, you're going to get the same value. So you're going to get x squared divided by 35, close that parenthesis, so that is my function there. Actually, I don't know if it's going to understand. Let me put the times, second insert times just to make sure it understands that. Okay, so that's my function, and then let me throw a comma here, make it clear that I'm integrating with respect to x. I could have put a t here and integrated with respect to t, we'd get the same value. Comma, my lower bound is zero, and my upper bound is eight, and close the parenthesis and then let the calculator munch on it a little bit. And we get 76.570, so this is approximately 76.570."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me put the times, second insert times just to make sure it understands that. Okay, so that's my function, and then let me throw a comma here, make it clear that I'm integrating with respect to x. I could have put a t here and integrated with respect to t, we'd get the same value. Comma, my lower bound is zero, and my upper bound is eight, and close the parenthesis and then let the calculator munch on it a little bit. And we get 76.570, so this is approximately 76.570. Now let's tackle the next part. Is the amount of water in the pipe increasing or decreasing at time t is equal to three hours? Give a reason for your answer."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we get 76.570, so this is approximately 76.570. Now let's tackle the next part. Is the amount of water in the pipe increasing or decreasing at time t is equal to three hours? Give a reason for your answer. Well, what would make it increasing? Well, if the rate at which things are going in is larger than the rate of things going out, then the amount of water would be increasing. But if it's the other way around, if we're draining faster at t equals three, then things are flowing into the pipe, well then the amount of water would be decreasing."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Give a reason for your answer. Well, what would make it increasing? Well, if the rate at which things are going in is larger than the rate of things going out, then the amount of water would be increasing. But if it's the other way around, if we're draining faster at t equals three, then things are flowing into the pipe, well then the amount of water would be decreasing. So, if r of t greater than, actually let me write it this way, if r of three, when t equals three, because t is given in hour, t is measured in hours, if r of three is greater than d of three, that means water is flowing in at a higher rate than leaving. So that means that water in pipe, right then, then water in pipe increasing. And then if it's the other way around, if d of three is greater than r of three, then water in pipe decreasing."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if it's the other way around, if we're draining faster at t equals three, then things are flowing into the pipe, well then the amount of water would be decreasing. So, if r of t greater than, actually let me write it this way, if r of three, when t equals three, because t is given in hour, t is measured in hours, if r of three is greater than d of three, that means water is flowing in at a higher rate than leaving. So that means that water in pipe, right then, then water in pipe increasing. And then if it's the other way around, if d of three is greater than r of three, then water in pipe decreasing. Then you're draining faster than you're putting into it. Then water in pipe decreasing. So we just have to evaluate these functions at three."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then if it's the other way around, if d of three is greater than r of three, then water in pipe decreasing. Then you're draining faster than you're putting into it. Then water in pipe decreasing. So we just have to evaluate these functions at three. So let's see, r, actually I can do it right over here. So let me make a little line here. So r of three is equal to, well let me get my calculator out, this is going to be, oops, not that calculator, let me get this calculator out."}, {"video_title": "2015 AP Calculus AB BC 1ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to evaluate these functions at three. So let's see, r, actually I can do it right over here. So let me make a little line here. So r of three is equal to, well let me get my calculator out, this is going to be, oops, not that calculator, let me get this calculator out. So, and I'm assuming that things are in radians here, so I already put my calculator in radian mode. So it's going to be 20 times sine of, whoops, 20 times sine of three squared is nine, divided by 35, and it gives us, this is equal to five point, approximately 5.09. This is approximately 5.09, and d of three, d of three is going to be approximately, so calculator back out, so it is, we have negative 0.04 times, times three to the third power, so times 27, plus, plus 0.4 times nine, times nine t squared, plus, plus 0.0, oh no, it's just 0.96, nine six times t, times three."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Part c, find the average value of f on the interval negative one to one. So the average value of a function over an interval is just going to be, so let's just write average. The average value of our function is just going to be the integral over the interval, negative one to one, of f of x, d of x, divided by our change in x. So divided by, sorry this is from negative one to one, negative one to one, f of x, d of x, divided by our change in x. And our change in x is one minus negative one. So this is going to be equal to one half times the integral, and here f of x is piecewise defined. So what we can do is break up this integral into two intervals."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So divided by, sorry this is from negative one to one, negative one to one, f of x, d of x, divided by our change in x. And our change in x is one minus negative one. So this is going to be equal to one half times the integral, and here f of x is piecewise defined. So what we can do is break up this integral into two intervals. We can say the integral of f of x from negative one to zero of f of x, d x, plus, let me just write it this way so we don't have to keep rewriting the one half. Let me write, I'll use brackets. Plus the integral from zero to one of f of x, f of x, d x."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we can do is break up this integral into two intervals. We can say the integral of f of x from negative one to zero of f of x, d x, plus, let me just write it this way so we don't have to keep rewriting the one half. Let me write, I'll use brackets. Plus the integral from zero to one of f of x, f of x, d x. And the reason why I broke it up like this is because this function has a different definition or it takes, it's a different, it's piecewise defined. It's different when we're less than or equal to zero versus when we are greater than zero. So that's why I like to break it up this way."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus the integral from zero to one of f of x, f of x, d x. And the reason why I broke it up like this is because this function has a different definition or it takes, it's a different, it's piecewise defined. It's different when we're less than or equal to zero versus when we are greater than zero. So that's why I like to break it up this way. So then we get this is equal to one half times, and in brackets, in big brackets like this, this first part right over here, we can write as the integral from negative one to zero. What is f of x between negative one and zero? It's one minus two sine of x."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's why I like to break it up this way. So then we get this is equal to one half times, and in brackets, in big brackets like this, this first part right over here, we can write as the integral from negative one to zero. What is f of x between negative one and zero? It's one minus two sine of x. One minus two sine of x, d x. And then plus this thing right over here. Plus the integral from zero to one."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's one minus two sine of x. One minus two sine of x, d x. And then plus this thing right over here. Plus the integral from zero to one. And what is our function when it's between zero and one? It's e to the negative four x. E to the negative four x, d x. And now we can do each of these integrals separately."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus the integral from zero to one. And what is our function when it's between zero and one? It's e to the negative four x. E to the negative four x, d x. And now we can do each of these integrals separately. And so this is going to be equal to, this is going to be equal to one half, and once again I'll do a big open bracket right over here. My pen is getting loose, let me tie on the front a little bit better. There you go."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we can do each of these integrals separately. And so this is going to be equal to, this is going to be equal to one half, and once again I'll do a big open bracket right over here. My pen is getting loose, let me tie on the front a little bit better. There you go. All right, back to work. So one half, open brackets, and now let's take the antiderivative of one minus two sine of x. Antiderivative of one with respect to x is just x. Negative two sine of x."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "There you go. All right, back to work. So one half, open brackets, and now let's take the antiderivative of one minus two sine of x. Antiderivative of one with respect to x is just x. Negative two sine of x. Well the derivative of cosine of x is negative sine of x, so this is just going to be two cosine of x. And you can verify that derivative of cosine of x is negative sine of x, you have the two out front, so it's negative two sine of x. So we're going to evaluate that at zero and at negative one."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative two sine of x. Well the derivative of cosine of x is negative sine of x, so this is just going to be two cosine of x. And you can verify that derivative of cosine of x is negative sine of x, you have the two out front, so it's negative two sine of x. So we're going to evaluate that at zero and at negative one. And to that we're going to add, so let's do this definite integral over here. The antiderivative of e to the negative four x, the antiderivative is going to be equal to negative e to the negative four x over four. And the way you realize that is, if this was just an e to the x, then its antiderivative would just be e to the x."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to evaluate that at zero and at negative one. And to that we're going to add, so let's do this definite integral over here. The antiderivative of e to the negative four x, the antiderivative is going to be equal to negative e to the negative four x over four. And the way you realize that is, if this was just an e to the x, then its antiderivative would just be e to the x. If you have an e to the negative four x, then you know that whatever its antiderivative is, is going to be essentially an e to the negative four x. But when you take its derivative, you're going to have to take the derivative of the negative four x part because of the chain rule. And so you're going to have to have a negative four that comes out."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the way you realize that is, if this was just an e to the x, then its antiderivative would just be e to the x. If you have an e to the negative four x, then you know that whatever its antiderivative is, is going to be essentially an e to the negative four x. But when you take its derivative, you're going to have to take the derivative of the negative four x part because of the chain rule. And so you're going to have to have a negative four that comes out. But we don't see a negative four over here. We don't see a negative four here, so we're obviously going to have to divide by a negative four so it cancels out. Another way to think about it is, we could have rewritten this thing."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you're going to have to have a negative four that comes out. But we don't see a negative four over here. We don't see a negative four here, so we're obviously going to have to divide by a negative four so it cancels out. Another way to think about it is, we could have rewritten this thing. So this is equal to e to the negative four x dx. This is exactly what our problem is, what we need to take the definite integral. And so that it becomes completely clear so that the derivative of this thing is sitting around here."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Another way to think about it is, we could have rewritten this thing. So this is equal to e to the negative four x dx. This is exactly what our problem is, what we need to take the definite integral. And so that it becomes completely clear so that the derivative of this thing is sitting around here. We could put a negative four over here. But you can't just willy nilly throw a negative four there. You would have to put a negative 1 4th outside of it in order for it to be, in order negative 1 4th times negative four, you're just multiplying by one, which doesn't change the value."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that it becomes completely clear so that the derivative of this thing is sitting around here. We could put a negative four over here. But you can't just willy nilly throw a negative four there. You would have to put a negative 1 4th outside of it in order for it to be, in order negative 1 4th times negative four, you're just multiplying by one, which doesn't change the value. And then here you'd clearly see, here you'd clearly see that this is, this right over here is the derivative of e to the negative four x. So you'd have e to the negative four x. The antiderivative of this is e to the negative four x, and then you have this negative 1 4th over here."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "You would have to put a negative 1 4th outside of it in order for it to be, in order negative 1 4th times negative four, you're just multiplying by one, which doesn't change the value. And then here you'd clearly see, here you'd clearly see that this is, this right over here is the derivative of e to the negative four x. So you'd have e to the negative four x. The antiderivative of this is e to the negative four x, and then you have this negative 1 4th over here. So either way, hopefully that makes sense. We go into more detail of that earlier in the calculus playlist. So then this is from, we're gonna evaluate it at one and at zero."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of this is e to the negative four x, and then you have this negative 1 4th over here. So either way, hopefully that makes sense. We go into more detail of that earlier in the calculus playlist. So then this is from, we're gonna evaluate it at one and at zero. Evaluate it at one and at zero. And then we wanna close the brackets. And so what do we have here?"}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So then this is from, we're gonna evaluate it at one and at zero. Evaluate it at one and at zero. And then we wanna close the brackets. And so what do we have here? This is equal to 1 1.5. Once again, open the brackets. If you evaluate all of this at zero, you get zero plus two cosine of zero."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what do we have here? This is equal to 1 1.5. Once again, open the brackets. If you evaluate all of this at zero, you get zero plus two cosine of zero. Cosine of zero is one. So you just get a two when you evaluate it at the zero. So you get a two."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you evaluate all of this at zero, you get zero plus two cosine of zero. Cosine of zero is one. So you just get a two when you evaluate it at the zero. So you get a two. And then from that we want to subtract whatever we get when we evaluate it at negative one. We wanna subtract negative one plus two cosine of negative one. Plus two cosine of negative one."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you get a two. And then from that we want to subtract whatever we get when we evaluate it at negative one. We wanna subtract negative one plus two cosine of negative one. Plus two cosine of negative one. So that's, this whole thing right over here evaluates to this whole thing right over there. And then we have plus, we wanna evaluate this at one. So this gives us negative e to the negative four."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus two cosine of negative one. So that's, this whole thing right over here evaluates to this whole thing right over there. And then we have plus, we wanna evaluate this at one. So this gives us negative e to the negative four. Four times one is just, or negative four x when x is one is negative four over four. And from that we need to subtract this thing evaluated at zero. We need to subtract this thing evaluated at zero."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this gives us negative e to the negative four. Four times one is just, or negative four x when x is one is negative four over four. And from that we need to subtract this thing evaluated at zero. We need to subtract this thing evaluated at zero. So that's going to be negative e to the zero over four. Well that's just, e to the zero is one. So that's just negative one, negative one over four."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "We need to subtract this thing evaluated at zero. So that's going to be negative e to the zero over four. Well that's just, e to the zero is one. So that's just negative one, negative one over four. And once again, all of this is going to be multiplied by 1 1.5. And now we just have to simplify it. So this is equal to, I'll just still throw out the 1 1.5 here."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's just negative one, negative one over four. And once again, all of this is going to be multiplied by 1 1.5. And now we just have to simplify it. So this is equal to, I'll just still throw out the 1 1.5 here. This is equal to 1 1.5 times, so I'll do this in a new color here. Let me do it in, well I don't have too many new colors. Okay, so this is equal to two plus one."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is equal to, I'll just still throw out the 1 1.5 here. This is equal to 1 1.5 times, so I'll do this in a new color here. Let me do it in, well I don't have too many new colors. Okay, so this is equal to two plus one. So this and this together is going to be equal to three. And then you have this negative outside. So that's why it was a plus one."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Okay, so this is equal to two plus one. So this and this together is going to be equal to three. And then you have this negative outside. So that's why it was a plus one. And then you have a negative times a positive two. So minus two cosine of negative one. And then you have plus, or maybe I should say minus, minus e to the negative four over four."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's why it was a plus one. And then you have a negative times a positive two. So minus two cosine of negative one. And then you have plus, or maybe I should say minus, minus e to the negative four over four. And then you have plus 1 4th. Plus 1 4th. And then we want to close our brackets."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you have plus, or maybe I should say minus, minus e to the negative four over four. And then you have plus 1 4th. Plus 1 4th. And then we want to close our brackets. And then we can do one last step of simplification here. Three, we could add the three to the 1 4th. Three is the same thing as 12 over four."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we want to close our brackets. And then we can do one last step of simplification here. Three, we could add the three to the 1 4th. Three is the same thing as 12 over four. 12 over four plus 1 4th is 13 4ths. So you have 13 4ths minus two cosine, two cosine of negative one. Minus e to the negative four over four."}, {"video_title": "2011 Calculus AB free response #6c   AP Calculus AB   Khan Academy.mp3", "Sentence": "Three is the same thing as 12 over four. 12 over four plus 1 4th is 13 4ths. So you have 13 4ths minus two cosine, two cosine of negative one. Minus e to the negative four over four. And then of course you have your 1 1.5 sitting out here. And it's not the most beautiful or simple thing, but this is our answer. This is the average value of f of x over that interval."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And like always, pause this video and see if you could figure it out. Well as you can see here, we have a composite function. So we're taking the secant not just of x, but you could view this as of another expression that I guess you could define, or as of another function. So for example, if we call this right over here u of x. So let's do that. So if we say u of x is equal to three pi over two minus x, we could also figure out u prime of x is going to be equal to derivative of three pi over two, that's just going to be zero, derivative of minus x, so that's just going to be minus one, and you could just view that as a power rule. It's one times negative one times x to the zero power, which is just one."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, if we call this right over here u of x. So let's do that. So if we say u of x is equal to three pi over two minus x, we could also figure out u prime of x is going to be equal to derivative of three pi over two, that's just going to be zero, derivative of minus x, so that's just going to be minus one, and you could just view that as a power rule. It's one times negative one times x to the zero power, which is just one. So there you go. So we could view this as the derivative of secant with respect to u of x, and when we take the derivative, the derivative of secant with respect to u of x times the derivative of u with respect to x. And you might say, well what about the derivative of secant?"}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's one times negative one times x to the zero power, which is just one. So there you go. So we could view this as the derivative of secant with respect to u of x, and when we take the derivative, the derivative of secant with respect to u of x times the derivative of u with respect to x. And you might say, well what about the derivative of secant? Well in other videos we actually prove it out, and you could actually re-derive it. Secant is just one over cosine of x, so it comes straight out of the chain rule. So in other videos we prove that the derivative of the secant of x, is equal to sine of x over cosine of x over cosine of x squared."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you might say, well what about the derivative of secant? Well in other videos we actually prove it out, and you could actually re-derive it. Secant is just one over cosine of x, so it comes straight out of the chain rule. So in other videos we prove that the derivative of the secant of x, is equal to sine of x over cosine of x over cosine of x squared. So if we're trying to find the derivative of y with respect to x, well it's going to be the derivative of secant with respect to u of x times the derivative of u with respect to x. So let's do that. The derivative of secant with respect to u of x, well instead of seeing an x everywhere, you're gonna see a u of x everywhere."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in other videos we prove that the derivative of the secant of x, is equal to sine of x over cosine of x over cosine of x squared. So if we're trying to find the derivative of y with respect to x, well it's going to be the derivative of secant with respect to u of x times the derivative of u with respect to x. So let's do that. The derivative of secant with respect to u of x, well instead of seeing an x everywhere, you're gonna see a u of x everywhere. So this is going to be sine of u of x, sine of u of x, u of x, and I could, I don't have to write u of x, I could write three pi over two minus x, but I'll write u of x right over here just to really visualize what we're doing. So sine of u of x over, over cosine squared of u of x. Cosine squared, let me do those parentheses in the blue color just to make sure that you identify it with the trig function. So cosine squared of u of x, u of x."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of secant with respect to u of x, well instead of seeing an x everywhere, you're gonna see a u of x everywhere. So this is going to be sine of u of x, sine of u of x, u of x, and I could, I don't have to write u of x, I could write three pi over two minus x, but I'll write u of x right over here just to really visualize what we're doing. So sine of u of x over, over cosine squared of u of x. Cosine squared, let me do those parentheses in the blue color just to make sure that you identify it with the trig function. So cosine squared of u of x, u of x. So that's the derivative of secant with respect to u of x, and then the chain rule tells us it's gonna be that times u prime, u prime of x. So what is this going to be equal to? Well, I could just substitute back."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So cosine squared of u of x, u of x. So that's the derivative of secant with respect to u of x, and then the chain rule tells us it's gonna be that times u prime, u prime of x. So what is this going to be equal to? Well, I could just substitute back. This is going to be equal to, I will write it like this, sine of u of x, which is three pi over two minus x, and I'll fill that in in a second, over cosine of u of x squared times u prime of x. U of x is three pi over two minus x. Three pi over two minus x. And then u prime of x, we already figured out is negative one, so I could write times negative one."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I could just substitute back. This is going to be equal to, I will write it like this, sine of u of x, which is three pi over two minus x, and I'll fill that in in a second, over cosine of u of x squared times u prime of x. U of x is three pi over two minus x. Three pi over two minus x. And then u prime of x, we already figured out is negative one, so I could write times negative one. Well, yeah, let me just leave it out there for now. I could have just put a negative out front, but I really want you to be able to see what I'm doing here. So now we want to evaluate at x equals pi over four."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then u prime of x, we already figured out is negative one, so I could write times negative one. Well, yeah, let me just leave it out there for now. I could have just put a negative out front, but I really want you to be able to see what I'm doing here. So now we want to evaluate at x equals pi over four. So when that is equal to pi over four, pi over four. So let's see, this is going to be, this is going to be equal to sine of, what's three pi over two minus pi over four? I'll do that over here."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we want to evaluate at x equals pi over four. So when that is equal to pi over four, pi over four. So let's see, this is going to be, this is going to be equal to sine of, what's three pi over two minus pi over four? I'll do that over here. So if you have a common denominator, that is six pi over four, the same thing as three pi over two, minus pi over four, sorry, minus pi over four is equal to five, five pi over four. So it's sine of five pi over four, five pi over four, over cosine squared of five pi over four, and then times negative one. I could just put that out here."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll do that over here. So if you have a common denominator, that is six pi over four, the same thing as three pi over two, minus pi over four, sorry, minus pi over four is equal to five, five pi over four. So it's sine of five pi over four, five pi over four, over cosine squared of five pi over four, and then times negative one. I could just put that out here. Now what is sine of five pi over four and cosine squared of five pi over four? Well, I don't have that memorized, but let's actually draw a unit circle and we should be able to figure out what that is. So a unit circle, I'm gonna try to hand draw it as best as I can."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could just put that out here. Now what is sine of five pi over four and cosine squared of five pi over four? Well, I don't have that memorized, but let's actually draw a unit circle and we should be able to figure out what that is. So a unit circle, I'm gonna try to hand draw it as best as I can. Please forgive me that this circle does not look really like a circle. All right. Okay, so let me just remember my angles."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So a unit circle, I'm gonna try to hand draw it as best as I can. Please forgive me that this circle does not look really like a circle. All right. Okay, so let me just remember my angles. So I, in my brain, I sometimes convert into degrees. Pi over four is 45 degrees. This is pi over two."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Okay, so let me just remember my angles. So I, in my brain, I sometimes convert into degrees. Pi over four is 45 degrees. This is pi over two. This is three pi over four. This is four pi over four. This is five pi over four."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is pi over two. This is three pi over four. This is four pi over four. This is five pi over four. Lands you right over there. So if you wanted to see where you intersect the unit circle, this is at the point, this is at the point where your x coordinate is negative square root of two over two, negative square root of two over two, and your y coordinate is negative square root of two over two. If you're wondering how I got that, I encourage you to review the unit circle and some of the standard angles around the unit circle."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is five pi over four. Lands you right over there. So if you wanted to see where you intersect the unit circle, this is at the point, this is at the point where your x coordinate is negative square root of two over two, negative square root of two over two, and your y coordinate is negative square root of two over two. If you're wondering how I got that, I encourage you to review the unit circle and some of the standard angles around the unit circle. You'll see that in the trigonometry section of Khan Academy. But this is enough for us because the sine is the y coordinate. It's the y coordinate here."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you're wondering how I got that, I encourage you to review the unit circle and some of the standard angles around the unit circle. You'll see that in the trigonometry section of Khan Academy. But this is enough for us because the sine is the y coordinate. It's the y coordinate here. So negative square root of two over two. This is negative square root of two over two. And then the cosine is the x coordinate, which is also negative square root of two over two, but it's gonna be that squared."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's the y coordinate here. So negative square root of two over two. This is negative square root of two over two. And then the cosine is the x coordinate, which is also negative square root of two over two, but it's gonna be that squared. Negative square root of two over two. We're squaring it. So if we square this, it's gonna become, this is going to become, it's gonna become positive."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the cosine is the x coordinate, which is also negative square root of two over two, but it's gonna be that squared. Negative square root of two over two. We're squaring it. So if we square this, it's gonna become, this is going to become, it's gonna become positive. And then square root of two squared is two. And then two squared is four. So it's 1 1 2."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we square this, it's gonna become, this is going to become, it's gonna become positive. And then square root of two squared is two. And then two squared is four. So it's 1 1 2. So this is, the denominator is equal to 1 1 2. See, the numerator, this negative cancels out with that negative. And so we are left with, and we deserve a little bit of a drum roll, that we are left with square root of two over two, that's the numerator, divided by 1 1 2."}, {"video_title": "Trig functions differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's 1 1 2. So this is, the denominator is equal to 1 1 2. See, the numerator, this negative cancels out with that negative. And so we are left with, and we deserve a little bit of a drum roll, that we are left with square root of two over two, that's the numerator, divided by 1 1 2. Well, that's the same thing as multiplying by two. So we are left with positive square root of two is the slope of the tangent line to the graph of y is equal to this when x is equal to pi over four. Pi over four, pretty exciting."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left-hand side, we essentially are just going to apply the chain rule. First, we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so first, on the left-hand side, we essentially are just going to apply the chain rule. First, we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1. And the derivative of y with respect to x, that's what we're trying to solve."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1. And the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. This right over here is the derivative of x minus y squared with respect to x minus y."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. This right over here is the derivative of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right-hand side of this equation."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here is the derivative of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right-hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x, we're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's go to the right-hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x, we're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do, let's make it clear. This right over here I can rewrite as 2x minus 2y."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do, let's make it clear. This right over here I can rewrite as 2x minus 2y. So let me do that. So I can save some space. This is 2x minus 2y if I just distribute the 2."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right over here I can rewrite as 2x minus 2y. So let me do that. So I can save some space. This is 2x minus 2y if I just distribute the 2. And now I can distribute the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 2x minus 2y if I just distribute the 2. And now I can distribute the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. Is equal to 1 plus."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. Is equal to 1 plus. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is equal to 1 plus. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides. So over here we're going to subtract minus. We're going to subtract 2x minus 2y from that side. And then we could also subtract a dy dx from both sides."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's subtract 2x minus 2y from both sides. So over here we're going to subtract minus. We're going to subtract 2x minus 2y from that side. And then we could also subtract a dy dx from both sides. So that all of our dy dx's are on the left hand side and all of our non-dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we could also subtract a dy dx from both sides. So that all of our dy dx's are on the left hand side and all of our non-dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? What are we left with? Well, on the left hand side, these cancel out."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? What are we left with? Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx. Or just minus a dy dx. Let me make it clear."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx. Or just minus a dy dx. Let me make it clear. We could write this as a minus 1 dy dx. So we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to, on this side, this cancels out."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make it clear. We could write this as a minus 1 dy dx. So we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to, on this side, this cancels out. We are left with 1 minus 2x plus 2y. So let me write it that way. We could write this as negative 2y is just a positive 2y."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to, on this side, this cancels out. We are left with 1 minus 2x plus 2y. So let me write it that way. We could write this as negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1 plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1."}, {"video_title": "Worked example Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could write this as negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1 plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1. And we are left with, we deserve a little bit of a drum roll at this point. As you can see, the hardest part was really the algebra to solve for dy dx. We get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It gives the corresponding f of x. What is a reasonable estimate for the limit of f of x as x approaches one from the left? So pause this video and see if you can figure it out on your own. Alright, now let's work through this together. So the first thing that is really important to realize is when you see this x approaches one and you see this little negative superscript here, this does not mean approaching negative one. So this does not mean negative one. Sometimes your brain just sees a one and that little negative sign there, and you're like, oh, this must be a weird way of writing negative one, or you don't even think about it."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, now let's work through this together. So the first thing that is really important to realize is when you see this x approaches one and you see this little negative superscript here, this does not mean approaching negative one. So this does not mean negative one. Sometimes your brain just sees a one and that little negative sign there, and you're like, oh, this must be a weird way of writing negative one, or you don't even think about it. But it's not saying that. It's saying, this is saying, let me put a little arrow here, this is the limit of f of x as x approaches one from the left. From the left."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sometimes your brain just sees a one and that little negative sign there, and you're like, oh, this must be a weird way of writing negative one, or you don't even think about it. But it's not saying that. It's saying, this is saying, let me put a little arrow here, this is the limit of f of x as x approaches one from the left. From the left. So from the left, how do we know that? Well, that's what that little negative tells us. It tells us we're approaching one from values less than one."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "From the left. So from the left, how do we know that? Well, that's what that little negative tells us. It tells us we're approaching one from values less than one. If we're approaching one from the right, from values greater than one, that would be a positive sign right over there. So let's think about it. We want the limit as x approaches one from the left."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It tells us we're approaching one from values less than one. If we're approaching one from the right, from values greater than one, that would be a positive sign right over there. So let's think about it. We want the limit as x approaches one from the left. And lucky for us on this table, we have some values of x approaching one from the left. 0.9, which is already pretty close to one, then we get even closer to one from the left. Notice, these are all less than one, but they're getting closer and closer to one."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We want the limit as x approaches one from the left. And lucky for us on this table, we have some values of x approaching one from the left. 0.9, which is already pretty close to one, then we get even closer to one from the left. Notice, these are all less than one, but they're getting closer and closer to one. And so what we really wanna look at is the value, what does f of x approach as x is getting closer and closer, let me write it, as x is getting closer and closer to one from the left, from the left. And a key realization here is, if we're thinking about general limits, not just from one direction, then we might wanna look at from the left and from the right. But they're asking us only from the left, so we should only be looking at these values right over here."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Notice, these are all less than one, but they're getting closer and closer to one. And so what we really wanna look at is the value, what does f of x approach as x is getting closer and closer, let me write it, as x is getting closer and closer to one from the left, from the left. And a key realization here is, if we're thinking about general limits, not just from one direction, then we might wanna look at from the left and from the right. But they're asking us only from the left, so we should only be looking at these values right over here. In fact, we shouldn't even let the value of f of x at x equal one confuse us. Sometimes, and oftentimes, the limit is approaching a different value than the value of the function at that point. So let's look at this."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But they're asking us only from the left, so we should only be looking at these values right over here. In fact, we shouldn't even let the value of f of x at x equal one confuse us. Sometimes, and oftentimes, the limit is approaching a different value than the value of the function at that point. So let's look at this. At 0.9, f of x is 2.5. When we get even closer to one from the left, we go to 2.1. When we get even closer to one from the left, we're getting even closer to two."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at this. At 0.9, f of x is 2.5. When we get even closer to one from the left, we go to 2.1. When we get even closer to one from the left, we're getting even closer to two. So a reasonable estimate for the limit as x approaches one from the left of f of x, it looks like x, it looks like f of x, right over here, is approaching two. We don't know for sure. That's why they're saying what is a reasonable estimate."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "When we get even closer to one from the left, we're getting even closer to two. So a reasonable estimate for the limit as x approaches one from the left of f of x, it looks like x, it looks like f of x, right over here, is approaching two. We don't know for sure. That's why they're saying what is a reasonable estimate. It might be approaching 2.01, or it might be approaching 1.999. On Khan Academy, these will often be multiple choice questions, so you have to pick the most reasonable one. It would not be fair if they gave 1.999 as a choice and 2.01."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's why they're saying what is a reasonable estimate. It might be approaching 2.01, or it might be approaching 1.999. On Khan Academy, these will often be multiple choice questions, so you have to pick the most reasonable one. It would not be fair if they gave 1.999 as a choice and 2.01. But if you were saying, hey, maybe this is approaching a whole number, then two could be a reasonable estimate right over here, although it doesn't have to be two. It could be 2.01258. It might be what it is actually approaching."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would not be fair if they gave 1.999 as a choice and 2.01. But if you were saying, hey, maybe this is approaching a whole number, then two could be a reasonable estimate right over here, although it doesn't have to be two. It could be 2.01258. It might be what it is actually approaching. So let's try another example here. Here it does look like there's a reasonable estimate for the limit as we approach this value from the left. So now it says the function f is defined over the real numbers."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It might be what it is actually approaching. So let's try another example here. Here it does look like there's a reasonable estimate for the limit as we approach this value from the left. So now it says the function f is defined over the real numbers. This table gives select values of f, similar to the last question. What is a reasonable estimate for the limit as x approaches negative two from the left? So this is confusing."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now it says the function f is defined over the real numbers. This table gives select values of f, similar to the last question. What is a reasonable estimate for the limit as x approaches negative two from the left? So this is confusing. You see these two negative signs. This first negative sign tells us we're approaching negative two. We wanna say what happens when we're approaching negative two and we're gonna approach, once again, from the left."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is confusing. You see these two negative signs. This first negative sign tells us we're approaching negative two. We wanna say what happens when we're approaching negative two and we're gonna approach, once again, from the left. So lucky for us, they have values of x that are approaching negative two from the left. So this is x approaches negative two from the left. And so that is happening right over here."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We wanna say what happens when we're approaching negative two and we're gonna approach, once again, from the left. So lucky for us, they have values of x that are approaching negative two from the left. So this is x approaches negative two from the left. And so that is happening right over here. So that's these values. So notice, this is negative 2.05. Then we get even closer, negative 2.01."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so that is happening right over here. So that's these values. So notice, this is negative 2.05. Then we get even closer, negative 2.01. Then we get even closer, negative 2.002. And these are from the left because these are values less than negative two, but they're getting closer and closer to negative two. And so let's see."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then we get even closer, negative 2.01. Then we get even closer, negative 2.002. And these are from the left because these are values less than negative two, but they're getting closer and closer to negative two. And so let's see. When we're a little bit further, f of x is negative 20. We get a little bit closer, it's negative 100. We get even a little bit closer, it goes to negative 500."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's see. When we're a little bit further, f of x is negative 20. We get a little bit closer, it's negative 100. We get even a little bit closer, it goes to negative 500. So it looks it would be reasonable, and we don't know for sure. This is just giving us a few sample points for this function. But if we follow this trend, as we get closer and closer to two, as we get closer and closer to negative two without getting there, it looks like this is getting unbounded."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get even a little bit closer, it goes to negative 500. So it looks it would be reasonable, and we don't know for sure. This is just giving us a few sample points for this function. But if we follow this trend, as we get closer and closer to two, as we get closer and closer to negative two without getting there, it looks like this is getting unbounded. It looks like it's becoming infinitely negative. And so technically, it looks like this is, I would write this is unbounded. Unbounded."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if we follow this trend, as we get closer and closer to two, as we get closer and closer to negative two without getting there, it looks like this is getting unbounded. It looks like it's becoming infinitely negative. And so technically, it looks like this is, I would write this is unbounded. Unbounded. And so if this was a multiple choice question, technically you would say this, the limit as x approaches negative two from the left does not exist. Does not exist. If someone asked the other question, if they said what is the limit as x approaches negative two from the right of f of x, well then you would say, all right, well here are values approaching negative two from the right."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Unbounded. And so if this was a multiple choice question, technically you would say this, the limit as x approaches negative two from the left does not exist. Does not exist. If someone asked the other question, if they said what is the limit as x approaches negative two from the right of f of x, well then you would say, all right, well here are values approaching negative two from the right. So this is x approaching negative two from the right, right over here. And remember, when you're looking at a limit, sometimes it might be distracting to look at the actual value of the function at that point. So you wanna think about what is the value of the function approaching as your x is approaching that value, as x is approaching, in this case, negative two from the right."}, {"video_title": "One-sided limits from tables   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If someone asked the other question, if they said what is the limit as x approaches negative two from the right of f of x, well then you would say, all right, well here are values approaching negative two from the right. So this is x approaching negative two from the right, right over here. And remember, when you're looking at a limit, sometimes it might be distracting to look at the actual value of the function at that point. So you wanna think about what is the value of the function approaching as your x is approaching that value, as x is approaching, in this case, negative two from the right. So as we're getting closer and closer to negative two from values larger than negative two, it looks like f of x is getting closer and closer to negative four, which is f of negative two. But that actually seems like a reasonable estimate. Once again, we don't know absolutely for sure just by sampling some points, but this would be a reasonable estimate."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in this video, we're going to think about what the derivative with respect to x of the natural log of x is. And I'm gonna go straight to the punchline. It is equal to one over x. In a future video, I'm actually going to prove this. It's a little bit involved. But in this one, we're just going to appreciate that this seems like it is actually true. So right here is the graph of y is equal to the natural log of x."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "In a future video, I'm actually going to prove this. It's a little bit involved. But in this one, we're just going to appreciate that this seems like it is actually true. So right here is the graph of y is equal to the natural log of x. And just to feel good about the statement, let's take the slope, let's try to approximate what the slope of the tangent line is at different points. So let's say right over here, when x is equal to one, what does the slope of the tangent line look like? Well, it looks like here, the slope looks like it is equal, pretty close to being equal to one, which is consistent with this statement."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So right here is the graph of y is equal to the natural log of x. And just to feel good about the statement, let's take the slope, let's try to approximate what the slope of the tangent line is at different points. So let's say right over here, when x is equal to one, what does the slope of the tangent line look like? Well, it looks like here, the slope looks like it is equal, pretty close to being equal to one, which is consistent with this statement. If x is equal to one, one over one is still one. And that seems like what we see right over there. What about when x is equal to two?"}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it looks like here, the slope looks like it is equal, pretty close to being equal to one, which is consistent with this statement. If x is equal to one, one over one is still one. And that seems like what we see right over there. What about when x is equal to two? Well, this point right over here is the natural log of two. But more interestingly, what's the slope here? Well, it looks like, let's see, if I try to draw a tangent line, the slope of the tangent line looks pretty close to 1 1."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What about when x is equal to two? Well, this point right over here is the natural log of two. But more interestingly, what's the slope here? Well, it looks like, let's see, if I try to draw a tangent line, the slope of the tangent line looks pretty close to 1 1. Well, once again, that is one over x. One over two is 1 1. Let's keep doing this."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it looks like, let's see, if I try to draw a tangent line, the slope of the tangent line looks pretty close to 1 1. Well, once again, that is one over x. One over two is 1 1. Let's keep doing this. If I go right over here, when x is equal to four, this point is four comma natural log of four. But the slope of the tangent line here looks pretty close to 1 4. And if you accept this, it is exactly 1 4."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's keep doing this. If I go right over here, when x is equal to four, this point is four comma natural log of four. But the slope of the tangent line here looks pretty close to 1 4. And if you accept this, it is exactly 1 4. And you could even go to values less than one. Right over here, when x is equal to 1 1, one over 1 1, the slope should be two. And it does indeed, let me do this in a slightly different color, it does indeed look like the slope is two over there."}, {"video_title": "Derivative of ln(x)   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you accept this, it is exactly 1 4. And you could even go to values less than one. Right over here, when x is equal to 1 1, one over 1 1, the slope should be two. And it does indeed, let me do this in a slightly different color, it does indeed look like the slope is two over there. So, once again, you take the derivative with respect to x of the natural log of x, it is one over x. And hopefully you get a sense that that is actually true here. And a future video will actually prove it."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "This seems all very abstract. I want to somehow use this thing. What we will do in this video is use it and to rigorously prove that a limit actually exists. Right over here, I've defined a function f of x. It's equal to 2x everywhere except for x equals 5. It's 2x everywhere for all the other values of x, but when x is equal to 5, it's just equal to x. I could have really just written 5. I could have just written 5 there."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right over here, I've defined a function f of x. It's equal to 2x everywhere except for x equals 5. It's 2x everywhere for all the other values of x, but when x is equal to 5, it's just equal to x. I could have really just written 5. I could have just written 5 there. It's equal to 5 when x is equal to 5. It's equal to itself. We've drawn the graph here."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could have just written 5 there. It's equal to 5 when x is equal to 5. It's equal to itself. We've drawn the graph here. Else it looks just like 2x. At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've drawn the graph here. Else it looks just like 2x. At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there. If I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. Well, let's see. The closer I get to 5, the closer f of x seems to be getting to 10."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead, the function is defined to be that point right over there. If I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. Well, let's see. The closer I get to 5, the closer f of x seems to be getting to 10. The closer I get to 5, the closer f of x seems to be getting to 10. You might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "The closer I get to 5, the closer f of x seems to be getting to 10. The closer I get to 5, the closer f of x seems to be getting to 10. You might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way. What we're going to do is use the epsilon delta definition to actually prove it. The way that most of these proofs typically go is we define delta in the abstract. We essentially try to come up with a way that given any epsilon, we can always come up with a delta."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks that way. What we're going to do is use the epsilon delta definition to actually prove it. The way that most of these proofs typically go is we define delta in the abstract. We essentially try to come up with a way that given any epsilon, we can always come up with a delta. Another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. Maybe I shouldn't use f again. Delta equals function of epsilon that is defined for any positive epsilon."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "We essentially try to come up with a way that given any epsilon, we can always come up with a delta. Another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. Maybe I shouldn't use f again. Delta equals function of epsilon that is defined for any positive epsilon. You give me an epsilon, I just put it into our little formula, our little function box, and I will always get you a delta. If I can do that for any epsilon, that will always give you a delta where this is true, that if x is within that range of delta of c, then the corresponding f of x is going to be within epsilon of l, then the limit definitely exists. Let's try to do that."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "Delta equals function of epsilon that is defined for any positive epsilon. You give me an epsilon, I just put it into our little formula, our little function box, and I will always get you a delta. If I can do that for any epsilon, that will always give you a delta where this is true, that if x is within that range of delta of c, then the corresponding f of x is going to be within epsilon of l, then the limit definitely exists. Let's try to do that. Let's think about being within delta of our c. Let's think about this right over here as 5 plus delta. This is 5 minus delta. That's our range we're going to think about."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's try to do that. Let's think about being within delta of our c. Let's think about this right over here as 5 plus delta. This is 5 minus delta. That's our range we're going to think about. We're going to think about it in the abstract at first, and then we're going to try to come up with a formula for delta in terms of epsilon. How can we describe all of the x's that are in this range but not equal to 5 itself? We really care about the things that are within delta of 5 but not necessarily equal to 5."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's our range we're going to think about. We're going to think about it in the abstract at first, and then we're going to try to come up with a formula for delta in terms of epsilon. How can we describe all of the x's that are in this range but not equal to 5 itself? We really care about the things that are within delta of 5 but not necessarily equal to 5. This is just a strictly less than. They're within a range of c but not equal to c. That's going to be all of the x's that satisfy x minus 5 is less than. Less than is less than delta."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "We really care about the things that are within delta of 5 but not necessarily equal to 5. This is just a strictly less than. They're within a range of c but not equal to c. That's going to be all of the x's that satisfy x minus 5 is less than. Less than is less than delta. That describes all of these x's right over here. What we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality so it starts to look something like this, or it starts to look exactly like that. As we do that, the right-hand side of the inequality is going to be expressed in terms of delta."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "Less than is less than delta. That describes all of these x's right over here. What we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality so it starts to look something like this, or it starts to look exactly like that. As we do that, the right-hand side of the inequality is going to be expressed in terms of delta. Then we can essentially say, well, look, if the right-hand side looks like delta, or is in terms of delta, and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me. I'm about to do it."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "As we do that, the right-hand side of the inequality is going to be expressed in terms of delta. Then we can essentially say, well, look, if the right-hand side looks like delta, or is in terms of delta, and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me. I'm about to do it. If we want x minus 5 to look a lot more like this, when x is not equal to 5, and all of this whole interval, x is not equal to 5, f of x is equal to 2x. Our limit, our proposed limit, is equal to 10. If we could somehow get this to be 2x minus 10, then we're in good shape."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm about to do it. If we want x minus 5 to look a lot more like this, when x is not equal to 5, and all of this whole interval, x is not equal to 5, f of x is equal to 2x. Our limit, our proposed limit, is equal to 10. If we could somehow get this to be 2x minus 10, then we're in good shape. The easiest way to do that is to multiply both sides of this inequality by 2. You multiply both sides of this inequality by 2. Two times the absolute value of something, that's the same thing as the absolute value of 2 times that thing."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we could somehow get this to be 2x minus 10, then we're in good shape. The easiest way to do that is to multiply both sides of this inequality by 2. You multiply both sides of this inequality by 2. Two times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. On the left-hand side, right over here, this is just going to be the absolute value of 2x minus 10. It's going to be less than on the right-hand side."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. On the left-hand side, right over here, this is just going to be the absolute value of 2x minus 10. It's going to be less than on the right-hand side. You just end up with a 2 delta. What do we have here on the left-hand side? This is f of x as long as x does not equal 5."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be less than on the right-hand side. You just end up with a 2 delta. What do we have here on the left-hand side? This is f of x as long as x does not equal 5. This is our limit. We can rewrite this as f of x minus L is less than 2 delta. This is for x does not equal 5."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is f of x as long as x does not equal 5. This is our limit. We can rewrite this as f of x minus L is less than 2 delta. This is for x does not equal 5. This is f of x. This literally is our limit. This is interesting."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is for x does not equal 5. This is f of x. This literally is our limit. This is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. This has in terms of epsilon. This has it in terms of delta."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. This has in terms of epsilon. This has it in terms of delta. How can we define delta so that 2 delta is essentially going to be epsilon? This is our chance. We will just define, we will make, and this is where we're defining delta as a function of epsilon."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "This has it in terms of delta. How can we define delta so that 2 delta is essentially going to be epsilon? This is our chance. We will just define, we will make, and this is where we're defining delta as a function of epsilon. We're going to make 2 delta equal epsilon, or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. If you make delta equal epsilon over 2, if you make, let me switch colors just to ease the monotony, if you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2 is just going to be less than epsilon. This is the key."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "We will just define, we will make, and this is where we're defining delta as a function of epsilon. We're going to make 2 delta equal epsilon, or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. If you make delta equal epsilon over 2, if you make, let me switch colors just to ease the monotony, if you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2 is just going to be less than epsilon. This is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going to be within epsilon of our limit. This tells you, say the epsilon, and remember, it has to be true for any positive epsilon, but you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going to be within epsilon of our limit. This tells you, say the epsilon, and remember, it has to be true for any positive epsilon, but you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. Our limit is up here, so our epsilon is 0.5, so it would literally be the range I want to be between 10 plus epsilon would be 10.5, and then 10 minus epsilon would be 9.5. Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25, so that'll give us a range between 4.75 and 5.25."}, {"video_title": "Formal definition of limits Part 4 using the definition   AP Calculus AB   Khan Academy.mp3", "Sentence": "If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. Our limit is up here, so our epsilon is 0.5, so it would literally be the range I want to be between 10 plus epsilon would be 10.5, and then 10 minus epsilon would be 9.5. Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25, so that'll give us a range between 4.75 and 5.25. As long as we pick an x between 4.75 and 5.25, then the corresponding f of x, but not x equals 5, the corresponding f of x will be between 9.5 and 10.5. You give me any epsilon, I can just apply this formula right over here to come up with the delta. This would apply for any real number, especially any positive number."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We're first exposed to the idea of a slope of a line early on in our algebra careers, but I figure it never hurts to review it a bit. So let me draw some axes. That is my y-axis. Or maybe I should call it my f of x-axis. y is equal to f of x. And let me draw my x-axis just like that. That is my x-axis."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Or maybe I should call it my f of x-axis. y is equal to f of x. And let me draw my x-axis just like that. That is my x-axis. And let me draw a line. So let me draw a line like this. And what we want to do is remind ourselves, how do we find the slope of that line?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That is my x-axis. And let me draw a line. So let me draw a line like this. And what we want to do is remind ourselves, how do we find the slope of that line? And what we do is we take two points on the line. So let's say we take this point right here. Let's say that that is the point x is equal to a."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And what we want to do is remind ourselves, how do we find the slope of that line? And what we do is we take two points on the line. So let's say we take this point right here. Let's say that that is the point x is equal to a. And then what would this be? This would be the point f of a, where the function is going to be some line. We could write f of x is going to be equal to mx plus b."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that that is the point x is equal to a. And then what would this be? This would be the point f of a, where the function is going to be some line. We could write f of x is going to be equal to mx plus b. We don't know what m and b are, but this is all a little bit of a review. So this is a. And then the y value is what happens to the function when you evaluate it at a."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We could write f of x is going to be equal to mx plus b. We don't know what m and b are, but this is all a little bit of a review. So this is a. And then the y value is what happens to the function when you evaluate it at a. So that's that point right there. And then we could take another point on this line. Let's say we take the point b right there."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then the y value is what happens to the function when you evaluate it at a. So that's that point right there. And then we could take another point on this line. Let's say we take the point b right there. And then this coordinate up here is going to be the point b, bf of b. Because this is just the point when you evaluate the function at b. You put b in here, you're going to get that point right there."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say we take the point b right there. And then this coordinate up here is going to be the point b, bf of b. Because this is just the point when you evaluate the function at b. You put b in here, you're going to get that point right there. So let me just draw a little line right there. So that is f of b right there. Actually, let me make it clear that this coordinate right here is the point a, f of a."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You put b in here, you're going to get that point right there. So let me just draw a little line right there. So that is f of b right there. Actually, let me make it clear that this coordinate right here is the point a, f of a. So how do we find the slope between these two points? Or more generally, of this entire line? Because a line has the slope is consistent the whole way through it."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Actually, let me make it clear that this coordinate right here is the point a, f of a. So how do we find the slope between these two points? Or more generally, of this entire line? Because a line has the slope is consistent the whole way through it. And we know that once we find the slope, that's actually going to be the value of this f. That's all a review of your algebra. But how do we do it? Well, a couple of ways to think about it."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Because a line has the slope is consistent the whole way through it. And we know that once we find the slope, that's actually going to be the value of this f. That's all a review of your algebra. But how do we do it? Well, a couple of ways to think about it. Slope is equal to rise over run. You might have seen that when you first learned algebra. Or another way of writing it, it's change in y over change in x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, a couple of ways to think about it. Slope is equal to rise over run. You might have seen that when you first learned algebra. Or another way of writing it, it's change in y over change in x. So let's figure out what the change in y over the change in x is for this particular case. So the change in y is equal to what? Well, you can take this guy as being the first point or that guy as being the first point."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Or another way of writing it, it's change in y over change in x. So let's figure out what the change in y over the change in x is for this particular case. So the change in y is equal to what? Well, you can take this guy as being the first point or that guy as being the first point. But since this guy has a larger x and a larger y, let's start with him. So the change in y between that guy and that guy is this distance right here. So let me draw a little triangle."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, you can take this guy as being the first point or that guy as being the first point. But since this guy has a larger x and a larger y, let's start with him. So the change in y between that guy and that guy is this distance right here. So let me draw a little triangle. That distance right there is the change in y. Or I could just transfer it to the y-axis. This is the change in y."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me draw a little triangle. That distance right there is the change in y. Or I could just transfer it to the y-axis. This is the change in y. That is your change in y, that distance. So what is that distance? It's f of b minus f of a."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the change in y. That is your change in y, that distance. So what is that distance? It's f of b minus f of a. So it equals f of b minus f of a. That is your change in y right here. That's your change in y."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It's f of b minus f of a. So it equals f of b minus f of a. That is your change in y right here. That's your change in y. Now what is your change in x? We have the slope is change in y over change in x. Well, what's our change in x?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's your change in y. Now what is your change in x? We have the slope is change in y over change in x. Well, what's our change in x? Well, it's this distance. Remember, we're taking this to be the first point. So we took its y minus the other point's y."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, what's our change in x? Well, it's this distance. Remember, we're taking this to be the first point. So we took its y minus the other point's y. So to be consistent, we're going to have to take this point's x minus this point's x. So this point's x-coordinate is b. So it's going to be b minus a."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So we took its y minus the other point's y. So to be consistent, we're going to have to take this point's x minus this point's x. So this point's x-coordinate is b. So it's going to be b minus a. And just like that, if you knew the equation of this line or if you had the coordinates of these two points, you would just plug them in right here and you would get your slope. That straightforward. And that comes straight out of your Algebra 1 class."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's going to be b minus a. And just like that, if you knew the equation of this line or if you had the coordinates of these two points, you would just plug them in right here and you would get your slope. That straightforward. And that comes straight out of your Algebra 1 class. And let me just, you know, just to make sure it's concrete for you. If this was the point, let's say this was the point 2, 3. And let's say that this up here was the point 5, 7."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And that comes straight out of your Algebra 1 class. And let me just, you know, just to make sure it's concrete for you. If this was the point, let's say this was the point 2, 3. And let's say that this up here was the point 5, 7. Then if we wanted to find the slope of this line, we would do 7 minus 3. So 7 minus 3. That would be our change in y."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And let's say that this up here was the point 5, 7. Then if we wanted to find the slope of this line, we would do 7 minus 3. So 7 minus 3. That would be our change in y. This would be 7. And this would be 3. And then we would do that over 5 minus 2."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That would be our change in y. This would be 7. And this would be 3. And then we would do that over 5 minus 2. Because this would be a 5 and this would be a 2. And so this would be your change in x. 5 minus 2."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then we would do that over 5 minus 2. Because this would be a 5 and this would be a 2. And so this would be your change in x. 5 minus 2. So 7 minus 3 is 4. And 5 minus 2 is 3. So your slope would be 4 over 3."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "5 minus 2. So 7 minus 3 is 4. And 5 minus 2 is 3. So your slope would be 4 over 3. Now let's see if we can generalize this. And this is the new concept that we're going to be learning as we delve into calculus. Let's see if we can generalize this somehow to a curve."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So your slope would be 4 over 3. Now let's see if we can generalize this. And this is the new concept that we're going to be learning as we delve into calculus. Let's see if we can generalize this somehow to a curve. So let's say I have a curve. We have to have a curve before we can generalize it to a curve. Let me scroll down a little."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's see if we can generalize this somehow to a curve. So let's say I have a curve. We have to have a curve before we can generalize it to a curve. Let me scroll down a little. Actually, I want to leave this up here. Just show you the similarity. So let's say I have, I'll keep it pretty general right now."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me scroll down a little. Actually, I want to leave this up here. Just show you the similarity. So let's say I have, I'll keep it pretty general right now. Let's say I have a curve. I'll make it a familiar looking curve. Let's say it's the curve y is equal to x squared, which looks something like that."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say I have, I'll keep it pretty general right now. Let's say I have a curve. I'll make it a familiar looking curve. Let's say it's the curve y is equal to x squared, which looks something like that. And I want to find the slope. Let's say I want to find the slope at some point. And actually, before even talking about it, let's even think about what it means to find the slope of a curve."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say it's the curve y is equal to x squared, which looks something like that. And I want to find the slope. Let's say I want to find the slope at some point. And actually, before even talking about it, let's even think about what it means to find the slope of a curve. Here the slope was the same the whole time. But on a curve, your slope is changing. And just to get an intuition for what that means is, what's the slope over here?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And actually, before even talking about it, let's even think about what it means to find the slope of a curve. Here the slope was the same the whole time. But on a curve, your slope is changing. And just to get an intuition for what that means is, what's the slope over here? Your slope over here is the slope of the tangent line. The line just barely touches it. That's the slope over there."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And just to get an intuition for what that means is, what's the slope over here? Your slope over here is the slope of the tangent line. The line just barely touches it. That's the slope over there. It's a negative slope. Then over here, your slope is still negative, but it's a little bit less negative. It goes like that."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's the slope over there. It's a negative slope. Then over here, your slope is still negative, but it's a little bit less negative. It goes like that. I don't know if I did that, drew that. Let me do it in a different color. Let me do it in purple."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It goes like that. I don't know if I did that, drew that. Let me do it in a different color. Let me do it in purple. So over here, your slope is slightly less negative. It's a slightly less downward sloping line. And then when you go over here, at the zero point right here, your slope is pretty much flat, because the horizontal line, y equals 0, is tangent to this curve."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do it in purple. So over here, your slope is slightly less negative. It's a slightly less downward sloping line. And then when you go over here, at the zero point right here, your slope is pretty much flat, because the horizontal line, y equals 0, is tangent to this curve. And then as you go to more positive x's, then your slope starts increasing. I'm trying to draw a tangent line. And here it's increasing even more."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then when you go over here, at the zero point right here, your slope is pretty much flat, because the horizontal line, y equals 0, is tangent to this curve. And then as you go to more positive x's, then your slope starts increasing. I'm trying to draw a tangent line. And here it's increasing even more. So your slope is changing the entire time. And this is kind of the big change that happens when you go from a line to a curve. A line, your slope is the same the entire time."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And here it's increasing even more. So your slope is changing the entire time. And this is kind of the big change that happens when you go from a line to a curve. A line, your slope is the same the entire time. You can take any two points in the line, take the change in y over the change in x, and you get the slope for the entire line. But as you can see already, it's going to be a little bit more nuanced when we do it for a curve, because it depends what point we're talking about. We can't just say, what is the slope for this curve?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "A line, your slope is the same the entire time. You can take any two points in the line, take the change in y over the change in x, and you get the slope for the entire line. But as you can see already, it's going to be a little bit more nuanced when we do it for a curve, because it depends what point we're talking about. We can't just say, what is the slope for this curve? The slope is different at every point along the curve. It changes. If we go up here, it's going to be even steeper."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We can't just say, what is the slope for this curve? The slope is different at every point along the curve. It changes. If we go up here, it's going to be even steeper. It's going to look something like that. So let's try a bit of an experiment. And I know how this experiment turns out, so it won't be too much of a risk."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If we go up here, it's going to be even steeper. It's going to look something like that. So let's try a bit of an experiment. And I know how this experiment turns out, so it won't be too much of a risk. Let me draw better than that. So that is my y-axis. And that's my x-axis."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And I know how this experiment turns out, so it won't be too much of a risk. Let me draw better than that. So that is my y-axis. And that's my x-axis. And let's call this, we could call this y, or we could call this the f of x-axis. Either way. And let me draw my curve again."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And that's my x-axis. And let's call this, we could call this y, or we could call this the f of x-axis. Either way. And let me draw my curve again. So I'll just draw it in the positive coordinate like that. That's my curve. And what if I want to find the slope right there?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And let me draw my curve again. So I'll just draw it in the positive coordinate like that. That's my curve. And what if I want to find the slope right there? What can I do? Well, based on our definition of a slope, we need two points to find a slope. Right here, I don't know how to find the slope."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And what if I want to find the slope right there? What can I do? Well, based on our definition of a slope, we need two points to find a slope. Right here, I don't know how to find the slope. Put just one point. So let's just call this one, let's just call this point right here. Let's call that, well that's going to be x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Right here, I don't know how to find the slope. Put just one point. So let's just call this one, let's just call this point right here. Let's call that, well that's going to be x. We're going to be general. This is going to be our point x. But to find our slope according to our traditional algebra 1 definition of a slope, we need two points."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's call that, well that's going to be x. We're going to be general. This is going to be our point x. But to find our slope according to our traditional algebra 1 definition of a slope, we need two points. So let's get another point in here. Let's just take a slightly larger version of this x. So let's say we have that point right there."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But to find our slope according to our traditional algebra 1 definition of a slope, we need two points. So let's get another point in here. Let's just take a slightly larger version of this x. So let's say we have that point right there. Actually, let me do it even further out, just because it's going to get messy otherwise. So let's say we have this point right here. And it's just h bigger than x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say we have that point right there. Actually, let me do it even further out, just because it's going to get messy otherwise. So let's say we have this point right here. And it's just h bigger than x. Or actually, instead of saying h bigger, let's just, well, let me just say h bigger. So this is x plus h. That's what that point is right there. So what are going to be their corresponding y-coordinates on the curve?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And it's just h bigger than x. Or actually, instead of saying h bigger, let's just, well, let me just say h bigger. So this is x plus h. That's what that point is right there. So what are going to be their corresponding y-coordinates on the curve? Well, this is the curve of y is equal to f of x. So this point right here is going to be f of our particular x right here. And maybe to show you that I'm taking a particular x, maybe I'll do a little 0 here."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So what are going to be their corresponding y-coordinates on the curve? Well, this is the curve of y is equal to f of x. So this point right here is going to be f of our particular x right here. And maybe to show you that I'm taking a particular x, maybe I'll do a little 0 here. This is x naught. This is x naught plus h. This is f of x naught. And then what is this going to be up here?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And maybe to show you that I'm taking a particular x, maybe I'll do a little 0 here. This is x naught. This is x naught plus h. This is f of x naught. And then what is this going to be up here? This point up here. That point up here. Its y-coordinate is going to be f of this x-coordinate, which I shifted over a little bit."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And then what is this going to be up here? This point up here. That point up here. Its y-coordinate is going to be f of this x-coordinate, which I shifted over a little bit. It's right there. f of this x-coordinate, which is f of x naught plus h. That's its y-coordinate. So what is the slope going to be between these two points that are relatively close to each other?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Its y-coordinate is going to be f of this x-coordinate, which I shifted over a little bit. It's right there. f of this x-coordinate, which is f of x naught plus h. That's its y-coordinate. So what is the slope going to be between these two points that are relatively close to each other? Remember, this isn't going to be the slope just at this point. This is the slope of the line between these two points. And if I were to actually draw it out, it would actually be a secant line to the curve."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So what is the slope going to be between these two points that are relatively close to each other? Remember, this isn't going to be the slope just at this point. This is the slope of the line between these two points. And if I were to actually draw it out, it would actually be a secant line to the curve. So it would intersect the curve twice. Once at this point and once at this point. You can't see it."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And if I were to actually draw it out, it would actually be a secant line to the curve. So it would intersect the curve twice. Once at this point and once at this point. You can't see it. If I blew it up a little bit, it would look something like this. It would look something like this. It would look like that, where this is our coordinate x naught."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "You can't see it. If I blew it up a little bit, it would look something like this. It would look something like this. It would look like that, where this is our coordinate x naught. f of x naught. And up here is our coordinate for this point, which would be the x-coordinate would be x naught plus h. And the y-coordinate would be f of x naught plus h. Just whatever this function is, we're evaluating it at this x-coordinate. That's all it is."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It would look like that, where this is our coordinate x naught. f of x naught. And up here is our coordinate for this point, which would be the x-coordinate would be x naught plus h. And the y-coordinate would be f of x naught plus h. Just whatever this function is, we're evaluating it at this x-coordinate. That's all it is. So these are the two points. So maybe a good start is just say, hey, what is the slope of this secant line? And just like we did in the previous example, you find the change in x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That's all it is. So these are the two points. So maybe a good start is just say, hey, what is the slope of this secant line? And just like we did in the previous example, you find the change in x. This is, sorry, the change in y. That'll be your change in y. And you divide that by your change in x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And just like we did in the previous example, you find the change in x. This is, sorry, the change in y. That'll be your change in y. And you divide that by your change in x. Let me draw it here. Your change in y would be that right here. Change in y."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And you divide that by your change in x. Let me draw it here. Your change in y would be that right here. Change in y. And then your change in x would be that right there. So what is the slope going to be of the secant line? The slope is going to be equal to, let's start with this point up here, just because it seems to be larger."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Change in y. And then your change in x would be that right there. So what is the slope going to be of the secant line? The slope is going to be equal to, let's start with this point up here, just because it seems to be larger. So we want a change in y. So this value right here, this y value, is f of x naught plus h. I just evaluated this guy up here. It looks like a fancy term."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope is going to be equal to, let's start with this point up here, just because it seems to be larger. So we want a change in y. So this value right here, this y value, is f of x naught plus h. I just evaluated this guy up here. It looks like a fancy term. But all it means is, look, the slightly larger x, I'll evaluate its y-coordinate. Where the curve is at that value of x. So that is going to be, so the change in y is going to be f of x naught plus h. That's just the y-coordinate up here."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It looks like a fancy term. But all it means is, look, the slightly larger x, I'll evaluate its y-coordinate. Where the curve is at that value of x. So that is going to be, so the change in y is going to be f of x naught plus h. That's just the y-coordinate up here. Minus this y-coordinate over here. So minus f of x naught. And that's our change in, so that equals our change in y."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So that is going to be, so the change in y is going to be f of x naught plus h. That's just the y-coordinate up here. Minus this y-coordinate over here. So minus f of x naught. And that's our change in, so that equals our change in y. And you want to divide that by your change in x. So what is this? This is the larger x value."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And that's our change in, so that equals our change in y. And you want to divide that by your change in x. So what is this? This is the larger x value. We started with this coordinate, so we start with its x-coordinate. So it's x naught plus h. Minus this x-coordinate. Well, we just picked a general number."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is the larger x value. We started with this coordinate, so we start with its x-coordinate. So it's x naught plus h. Minus this x-coordinate. Well, we just picked a general number. It's x naught. So that is over your change in x. Just like that."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, we just picked a general number. It's x naught. So that is over your change in x. Just like that. So this is a slope of the secant line. We still haven't answered what the slope is right at that point. But maybe this will help us get there."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Just like that. So this is a slope of the secant line. We still haven't answered what the slope is right at that point. But maybe this will help us get there. So if we simplify this, so let me write it down like this. The slope of the secant, let me write that properly. The slope of the secant line is equal to the value of the function at this point, f of x naught plus h, minus the value of the function here, minus f of x naught."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But maybe this will help us get there. So if we simplify this, so let me write it down like this. The slope of the secant, let me write that properly. The slope of the secant line is equal to the value of the function at this point, f of x naught plus h, minus the value of the function here, minus f of x naught. So that just tells us the change in y. It's the exact same definition of slope we've always used. Over the change in x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope of the secant line is equal to the value of the function at this point, f of x naught plus h, minus the value of the function here, minus f of x naught. So that just tells us the change in y. It's the exact same definition of slope we've always used. Over the change in x. And we can simplify this. We have x naught plus h, minus x naught. So x naught minus x naught cancel out."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Over the change in x. And we can simplify this. We have x naught plus h, minus x naught. So x naught minus x naught cancel out. So you have that over h. So this is equal to our change in y over change in x. Fair enough. But I started off saying, I want to find the slope of the line at that point."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So x naught minus x naught cancel out. So you have that over h. So this is equal to our change in y over change in x. Fair enough. But I started off saying, I want to find the slope of the line at that point. At this point right here. This is the zoomed out version of it. So what can I do?"}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But I started off saying, I want to find the slope of the line at that point. At this point right here. This is the zoomed out version of it. So what can I do? Well, I defined this second point here as just the first point, plus some h. And we have something in our toolkit called a limit. This h is just a general number. This h is just, it could be 10, it could be 2, it could be 0.02, it could be 1 times 10 to the negative 100."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So what can I do? Well, I defined this second point here as just the first point, plus some h. And we have something in our toolkit called a limit. This h is just a general number. This h is just, it could be 10, it could be 2, it could be 0.02, it could be 1 times 10 to the negative 100. It could be an arbitrarily small number. So what happens, what would happen, at least just theoretically, if I were to take the limit as h approaches 0? So if first, maybe h is this fairly large number over here."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This h is just, it could be 10, it could be 2, it could be 0.02, it could be 1 times 10 to the negative 100. It could be an arbitrarily small number. So what happens, what would happen, at least just theoretically, if I were to take the limit as h approaches 0? So if first, maybe h is this fairly large number over here. And then if I take h a little bit smaller, then I'd be finding the slope of this secant line. If I took h to be even a little bit smaller, I'd be finding the slope of that secant line. If h is a little bit smaller, I'd be finding the slope of that line."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So if first, maybe h is this fairly large number over here. And then if I take h a little bit smaller, then I'd be finding the slope of this secant line. If I took h to be even a little bit smaller, I'd be finding the slope of that secant line. If h is a little bit smaller, I'd be finding the slope of that line. So as h approaches 0, I'll be getting closer and closer to finding the slope of the line right at my point in question. Obviously, if h is a large number, my secant line is going to be way off from the slope at exactly that point right there. But if h is 0.0000001, if it's an infinitesimally small number, then I'm going to get pretty close."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If h is a little bit smaller, I'd be finding the slope of that line. So as h approaches 0, I'll be getting closer and closer to finding the slope of the line right at my point in question. Obviously, if h is a large number, my secant line is going to be way off from the slope at exactly that point right there. But if h is 0.0000001, if it's an infinitesimally small number, then I'm going to get pretty close. So what happens if I take the limit as h approaches 0 of this? So the limit as h approaches 0 of my secant slope of, let me switch to green, f of x naught plus h minus f of x naught, that was my change in y, over my change in h, which is my change in x. And now just to clarify something, and sometimes you'll see it in different calculus books, sometimes instead of an h, they'll write a delta x here, where this second point would have been defined as x naught plus delta x."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But if h is 0.0000001, if it's an infinitesimally small number, then I'm going to get pretty close. So what happens if I take the limit as h approaches 0 of this? So the limit as h approaches 0 of my secant slope of, let me switch to green, f of x naught plus h minus f of x naught, that was my change in y, over my change in h, which is my change in x. And now just to clarify something, and sometimes you'll see it in different calculus books, sometimes instead of an h, they'll write a delta x here, where this second point would have been defined as x naught plus delta x. And then this would have simplified to just delta x over there, and we'd be taking the limit as delta x approaches 0. The exact same thing. h delta x doesn't matter."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And now just to clarify something, and sometimes you'll see it in different calculus books, sometimes instead of an h, they'll write a delta x here, where this second point would have been defined as x naught plus delta x. And then this would have simplified to just delta x over there, and we'd be taking the limit as delta x approaches 0. The exact same thing. h delta x doesn't matter. We're taking h as the difference between one x point and then the higher x point, and then we're just going to take the limit as that approaches 0. We could have called that delta x just as easily. But I'm going to call this thing, which we're saying it equals the slope of the tangent line, and it does equal the slope of the tangent line, I'm going to call this the derivative of f. Let me write that down."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "h delta x doesn't matter. We're taking h as the difference between one x point and then the higher x point, and then we're just going to take the limit as that approaches 0. We could have called that delta x just as easily. But I'm going to call this thing, which we're saying it equals the slope of the tangent line, and it does equal the slope of the tangent line, I'm going to call this the derivative of f. Let me write that down. The derivative of f. And I'm going to say that this is equal to f prime of x. And this is going to be another function. Because remember, the slope changes at every x value."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "But I'm going to call this thing, which we're saying it equals the slope of the tangent line, and it does equal the slope of the tangent line, I'm going to call this the derivative of f. Let me write that down. The derivative of f. And I'm going to say that this is equal to f prime of x. And this is going to be another function. Because remember, the slope changes at every x value. No matter what x value you pick, the slope is going to be different. It doesn't have to be, but the way I drew this curve, it is different. It can be different."}, {"video_title": "Derivative as slope of a tangent line   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Because remember, the slope changes at every x value. No matter what x value you pick, the slope is going to be different. It doesn't have to be, but the way I drew this curve, it is different. It can be different. So now you give me an x value in here. I'll apply this formula over here, and then I can tell you the slope at that point. That all seems very confusing and maybe abstract at this point."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is his solution. And then later we are asked, is Robert's work correct? If not, what's his mistake? So pause this video and try to figure it out on your own. All right, now let's work through this together. So our original g of x is equal to the cube root of x, which is the same thing as x to the 1 3rd. So in step one, it looks like Robert's trying to find the first and second derivative."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video and try to figure it out on your own. All right, now let's work through this together. So our original g of x is equal to the cube root of x, which is the same thing as x to the 1 3rd. So in step one, it looks like Robert's trying to find the first and second derivative. So the first derivative, we just do the power rule. So it'll be 1 3rd x to the decrement the exponent. So this is looking good."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in step one, it looks like Robert's trying to find the first and second derivative. So the first derivative, we just do the power rule. So it'll be 1 3rd x to the decrement the exponent. So this is looking good. Second derivative, we take this, multiply this times 1 3rd, which would be negative 2 9ths, and then decrement negative 2 3rds, which would indeed be negative 5 3rds. So that looks right. And then it looks like Robert's trying to rewrite it."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is looking good. Second derivative, we take this, multiply this times 1 3rd, which would be negative 2 9ths, and then decrement negative 2 3rds, which would indeed be negative 5 3rds. So that looks right. And then it looks like Robert's trying to rewrite it. So we have the negative 2 9ths still. But then he recognized that this is the same thing as x to the 5 3rds in the denominator. And x to the 5 3rds is the same thing as x, as the cube root of x to the 5th."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it looks like Robert's trying to rewrite it. So we have the negative 2 9ths still. But then he recognized that this is the same thing as x to the 5 3rds in the denominator. And x to the 5 3rds is the same thing as x, as the cube root of x to the 5th. So this is all looking good. Step one looks good. And then step two, it looks like he's trying to find the solution, or he's trying to find x values where the second derivative is equal to zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And x to the 5 3rds is the same thing as x, as the cube root of x to the 5th. So this is all looking good. Step one looks good. And then step two, it looks like he's trying to find the solution, or he's trying to find x values where the second derivative is equal to zero. And it is indeed true that this has no solution, that you can never make this second derivative equal to zero. In order to be zero, the numerator would have to be zero, and well, two is never going to be equal to zero. So this is correct."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then step two, it looks like he's trying to find the solution, or he's trying to find x values where the second derivative is equal to zero. And it is indeed true that this has no solution, that you can never make this second derivative equal to zero. In order to be zero, the numerator would have to be zero, and well, two is never going to be equal to zero. So this is correct. And then step three, he says g doesn't have any inflection points. Now this is a little bit suspect. It is, in many cases, our inflection point is a situation where our second derivative is equal to zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is correct. And then step three, he says g doesn't have any inflection points. Now this is a little bit suspect. It is, in many cases, our inflection point is a situation where our second derivative is equal to zero. And even then, we don't know it's an inflection point. It would be a candidate inflection point. We would have to confirm that our second derivative crosses signs, or switches signs, as we cross that x value."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is, in many cases, our inflection point is a situation where our second derivative is equal to zero. And even then, we don't know it's an inflection point. It would be a candidate inflection point. We would have to confirm that our second derivative crosses signs, or switches signs, as we cross that x value. But here we can't find a situation where our second derivative is equal to zero. But we have to remind ourselves that other candidate inflection points are where our second derivative is undefined. And so he can't make this statement without seeing where our second derivative could be undefined."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "We would have to confirm that our second derivative crosses signs, or switches signs, as we cross that x value. But here we can't find a situation where our second derivative is equal to zero. But we have to remind ourselves that other candidate inflection points are where our second derivative is undefined. And so he can't make this statement without seeing where our second derivative could be undefined. So for example, he could say that g prime prime is undefined, undefined when what? Well, this is going to be undefined when x is equal to zero. X zero to the fifth, cube root of that, that's gonna be zero but then you're dividing by zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so he can't make this statement without seeing where our second derivative could be undefined. So for example, he could say that g prime prime is undefined, undefined when what? Well, this is going to be undefined when x is equal to zero. X zero to the fifth, cube root of that, that's gonna be zero but then you're dividing by zero. So g prime prime undefined when x is equal to zero. So therefore, x equals, so we could say, candidate, candidate, candidate inflection point when x equals zero. And so then we would want to test it."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "X zero to the fifth, cube root of that, that's gonna be zero but then you're dividing by zero. So g prime prime undefined when x is equal to zero. So therefore, x equals, so we could say, candidate, candidate, candidate inflection point when x equals zero. And so then we would want to test it. And we could set up a traditional table that you might have seen before where we have our interval or intervals. We could have test values in our intervals. We have to be careful with those, make sure that they are indicative."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so then we would want to test it. And we could set up a traditional table that you might have seen before where we have our interval or intervals. We could have test values in our intervals. We have to be careful with those, make sure that they are indicative. And then we would say the sine of our second derivative of g prime prime. And then we would have our concavity, concavity of g. And in order for x equals zero to be an inflection point, we would have to switch sines as, or our second derivative would have to switch sines as we cross x equals zero, and which would mean our concavity of g switches sines as we go, as we cross x equals zero. So let's do values less than zero, negative infinity to zero, and then values greater than zero, infinity, zero to infinity."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have to be careful with those, make sure that they are indicative. And then we would say the sine of our second derivative of g prime prime. And then we would have our concavity, concavity of g. And in order for x equals zero to be an inflection point, we would have to switch sines as, or our second derivative would have to switch sines as we cross x equals zero, and which would mean our concavity of g switches sines as we go, as we cross x equals zero. So let's do values less than zero, negative infinity to zero, and then values greater than zero, infinity, zero to infinity. I could do test values, let's say, I'll use negative one and one. And you have to be careful when you use these, you have to make sure that we are close enough that nothing unusual happens between these test values up until we get to that candidate inflection point. And now what's the sine of our second derivative when x is equal to negative one?"}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do values less than zero, negative infinity to zero, and then values greater than zero, infinity, zero to infinity. I could do test values, let's say, I'll use negative one and one. And you have to be careful when you use these, you have to make sure that we are close enough that nothing unusual happens between these test values up until we get to that candidate inflection point. And now what's the sine of our second derivative when x is equal to negative one? When x equals negative one, so let's see, negative one to the fifth power is negative one cube root of negative one is negative one. And so we're gonna have negative 2 9ths divided by negative one is gonna be positive 2 9ths. So our sine right over here is gonna be positive."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now what's the sine of our second derivative when x is equal to negative one? When x equals negative one, so let's see, negative one to the fifth power is negative one cube root of negative one is negative one. And so we're gonna have negative 2 9ths divided by negative one is gonna be positive 2 9ths. So our sine right over here is gonna be positive. And when, and this is gonna be in general when we're dealing with any negative value. Because if you take any negative value to the fifth power, it's gonna be negative. And then you take that, the cube root of that, you're gonna have negative."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our sine right over here is gonna be positive. And when, and this is gonna be in general when we're dealing with any negative value. Because if you take any negative value to the fifth power, it's gonna be negative. And then you take that, the cube root of that, you're gonna have negative. And then you have a negative value divided by that, you're gonna get a positive value. So you can feel good that this test value is indicative of actually this entire interval. And if you're dealing with a positive value, well, that to the fifth power is gonna be positive cube root of that's still going to be positive."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you take that, the cube root of that, you're gonna have negative. And then you have a negative value divided by that, you're gonna get a positive value. So you can feel good that this test value is indicative of actually this entire interval. And if you're dealing with a positive value, well, that to the fifth power is gonna be positive cube root of that's still going to be positive. But then you're gonna have negative 2 9ths divided by that positive value. So this is going to be negative. So it is indeed the case that our concavity of g switches as we cross x equals zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you're dealing with a positive value, well, that to the fifth power is gonna be positive cube root of that's still going to be positive. But then you're gonna have negative 2 9ths divided by that positive value. So this is going to be negative. So it is indeed the case that our concavity of g switches as we cross x equals zero. We're concave upwards when x is less than zero, our second derivative is positive, and we are concave downwards when x is greater than zero. Let me write that a little bit. Downwards, downwards when x is greater than zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it is indeed the case that our concavity of g switches as we cross x equals zero. We're concave upwards when x is less than zero, our second derivative is positive, and we are concave downwards when x is greater than zero. Let me write that a little bit. Downwards, downwards when x is greater than zero. So we are switching concavity as we cross x equals zero, and so this tells us that x, so let's see, we are switching signs, switching, let me say g prime prime, switching signs as we cross x equals zero, and our function is defined at x equals zero, and function defined at x equals zero. So we have an inflection point at x equals zero. So inflection point at x is equal to zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "Downwards, downwards when x is greater than zero. So we are switching concavity as we cross x equals zero, and so this tells us that x, so let's see, we are switching signs, switching, let me say g prime prime, switching signs as we cross x equals zero, and our function is defined at x equals zero, and function defined at x equals zero. So we have an inflection point at x equals zero. So inflection point at x is equal to zero. And if you're familiar with the graph of the cube root, you would indeed see an inflection point at that point. So there we go. He was wrong in step three."}, {"video_title": "Mistakes when finding inflection points second derivative undefined   AP Calculus AB   Khan Academy.mp3", "Sentence": "So inflection point at x is equal to zero. And if you're familiar with the graph of the cube root, you would indeed see an inflection point at that point. So there we go. He was wrong in step three. There actually is an inflection point. It's not when the second derivative is equal to zero. It's actually where the second derivative is undefined."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this region right over here is the base of a three-dimensional figure. And what we know about this three-dimensional figure is if we were to take cross sections that are perpendicular to the x-axis, so cross sections, I guess we could say, that are parallel to the y-axis. So let's say we were to take a cross section like that. We know that this will be a semicircle. So if we take a slightly different view of the same three-dimensional figure, we would see something like this. I've kind of laid the coordinate plane down flat and I'm looking at it from above. So this cross section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross section right over here."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that this will be a semicircle. So if we take a slightly different view of the same three-dimensional figure, we would see something like this. I've kind of laid the coordinate plane down flat and I'm looking at it from above. So this cross section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross section right over here. It would be that cross section right over here, which is a semicircle. If we were to take this cross section right over here along the y-axis, that would be this cross section, this larger semicircle. So given that, given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've, I could shade it in a little bit."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this cross section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross section right over here. It would be that cross section right over here, which is a semicircle. If we were to take this cross section right over here along the y-axis, that would be this cross section, this larger semicircle. So given that, given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've, I could shade it in a little bit. The volume of this thing that I've just started, this three-dimensional figure that we are attempting to visualize. So I'm assuming you've had a go at it. And one way to think about it is, well, let's just think about each of these."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So given that, given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've, I could shade it in a little bit. The volume of this thing that I've just started, this three-dimensional figure that we are attempting to visualize. So I'm assuming you've had a go at it. And one way to think about it is, well, let's just think about each of these. Let's split up the figure into a bunch of disks. Into a bunch of disks. If you figure out the volume of each of those disks, if you figure out the volume of each of those disks and sum them up, then that would be a pretty good approximation for the volume of the whole thing."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "And one way to think about it is, well, let's just think about each of these. Let's split up the figure into a bunch of disks. Into a bunch of disks. If you figure out the volume of each of those disks, if you figure out the volume of each of those disks and sum them up, then that would be a pretty good approximation for the volume of the whole thing. And then if you took the limit as you get an infinite number of disks that are infinitely thin, then you're going to get the exact volume. So let's just say, let's just take the approximating case first. So let's say that right over here at x, what is going to be, what is going to be the diameter, what is going to be the diameter of the disk at x?"}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you figure out the volume of each of those disks, if you figure out the volume of each of those disks and sum them up, then that would be a pretty good approximation for the volume of the whole thing. And then if you took the limit as you get an infinite number of disks that are infinitely thin, then you're going to get the exact volume. So let's just say, let's just take the approximating case first. So let's say that right over here at x, what is going to be, what is going to be the diameter, what is going to be the diameter of the disk at x? Well, to think about that, we just have to re-express x plus y is equal to one. That's the same thing as saying that the function that f of x, or y is equal to f of x, is equal to one minus x. And so the diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that right over here at x, what is going to be, what is going to be the diameter, what is going to be the diameter of the disk at x? Well, to think about that, we just have to re-express x plus y is equal to one. That's the same thing as saying that the function that f of x, or y is equal to f of x, is equal to one minus x. And so the diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height. It's going to be the difference between one minus x and the x-axis, or between one minus x and x equals zero. And so this is just going to be, as a function of x, the diameter is going to be one minus x. Now, if you want to find the surface area for a circle, or if we want to find the area, I guess you could say, of a circle, we know that area is pi r squared."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height. It's going to be the difference between one minus x and the x-axis, or between one minus x and x equals zero. And so this is just going to be, as a function of x, the diameter is going to be one minus x. Now, if you want to find the surface area for a circle, or if we want to find the area, I guess you could say, of a circle, we know that area is pi r squared. Now, for our semicircle, you're going to divide by two. So what's the radius going to be? So let me zoom in a little bit."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, if you want to find the surface area for a circle, or if we want to find the area, I guess you could say, of a circle, we know that area is pi r squared. Now, for our semicircle, you're going to divide by two. So what's the radius going to be? So let me zoom in a little bit. So what is the radius of one of these semicircles going to be? So the radius, I could draw it like this. My best attempt to draw one of these things."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me zoom in a little bit. So what is the radius of one of these semicircles going to be? So the radius, I could draw it like this. My best attempt to draw one of these things. So it's going to look something like this. I'm trying to draw it at an angle. So the radius, what's going to be half the diameter?"}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "My best attempt to draw one of these things. So it's going to look something like this. I'm trying to draw it at an angle. So the radius, what's going to be half the diameter? The diameter is one minus x. The radius is going to be one minus x over two. That distance is one minus x over two."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the radius, what's going to be half the diameter? The diameter is one minus x. The radius is going to be one minus x over two. That distance is one minus x over two. That distance is one minus x over two. That distance is one minus x over two. So that is one minus x over two is equal to the radius."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "That distance is one minus x over two. That distance is one minus x over two. That distance is one minus x over two. So that is one minus x over two is equal to the radius. And so what would be the area of, what would be the area of this side right over here? Well, if it was a full circle, it would be pi r squared, but it's a semicircle, so it's pi r squared over two. So let me write this."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is one minus x over two is equal to the radius. And so what would be the area of, what would be the area of this side right over here? Well, if it was a full circle, it would be pi r squared, but it's a semicircle, so it's pi r squared over two. So let me write this. So the area is equal to pi r squared, r squared over two, because it's a semicircle. So in terms of x, I guess I could say our area as a function of x right over there. Let me write it that way."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write this. So the area is equal to pi r squared, r squared over two, because it's a semicircle. So in terms of x, I guess I could say our area as a function of x right over there. Let me write it that way. Our area as a function of x is going to be pi over two. Let me just write that. It's going to be pi over two times one minus x over two squared."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write it that way. Our area as a function of x is going to be pi over two. Let me just write that. It's going to be pi over two times one minus x over two squared. So it's one minus x over two, over two squared. And so if I wanted a volume of just this disc right over here, I'd then multiply that area times the depth. I'd just multiply the area times this depth here."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be pi over two times one minus x over two squared. So it's one minus x over two, over two squared. And so if I wanted a volume of just this disc right over here, I'd then multiply that area times the depth. I'd just multiply the area times this depth here. And we could call that dx, or we could call that delta x. So we could call the depth there delta x. Now, that's the, actually let me be clear."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'd just multiply the area times this depth here. And we could call that dx, or we could call that delta x. So we could call the depth there delta x. Now, that's the, actually let me be clear. That's the area, the volume, the volume of one of those shells is going to be equal to, and I'll just write this in one color. It's going to be pi over two times one minus x over two squared times the depth, the area times the depth. So this is volume of one of these half discs."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, that's the, actually let me be clear. That's the area, the volume, the volume of one of those shells is going to be equal to, and I'll just write this in one color. It's going to be pi over two times one minus x over two squared times the depth, the area times the depth. So this is volume of one of these half discs. Half discs. So the volume of the whole thing we can approximate as the sum of these, or we could take the limit as our delta x's get much, much, much, much smaller, and we have an infinite number of these things, which is essentially, if we're taking that limit, we're gonna take the definite integral. So we could take the definite integral, so the volume that we care about, the volume of this figure is going to be equal to the definite integral from x equals zero to x equals one."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is volume of one of these half discs. Half discs. So the volume of the whole thing we can approximate as the sum of these, or we could take the limit as our delta x's get much, much, much, much smaller, and we have an infinite number of these things, which is essentially, if we're taking that limit, we're gonna take the definite integral. So we could take the definite integral, so the volume that we care about, the volume of this figure is going to be equal to the definite integral from x equals zero to x equals one. That's where we intersect the x-axis. X equals zero to x equals one of, we're integrating an infinite number of these things that are infinitely, I guess, thin. So it's going to be, let me write it."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could take the definite integral, so the volume that we care about, the volume of this figure is going to be equal to the definite integral from x equals zero to x equals one. That's where we intersect the x-axis. X equals zero to x equals one of, we're integrating an infinite number of these things that are infinitely, I guess, thin. So it's going to be, let me write it. It's gonna be pi over two times, I'll just write it, what's one minus x squared? Let me just expand it out for fun right over here. So that's going to be, that's the same thing as x minus one squared, so it's gonna be x squared minus two x plus one, and then two squared is four, over four, and instead of delta x, I'm gonna write dx now."}, {"video_title": "Volume with cross sections semicircle   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be, let me write it. It's gonna be pi over two times, I'll just write it, what's one minus x squared? Let me just expand it out for fun right over here. So that's going to be, that's the same thing as x minus one squared, so it's gonna be x squared minus two x plus one, and then two squared is four, over four, and instead of delta x, I'm gonna write dx now. I'm gonna write dx, because I'm taking the limit as these become infinitely small, and I have an infinite, and I'm summing an infinite number of them. So the volume is just going to be, we just have to evaluate this definite integral, and if you feel so inspired, if you feel inspired at the moment, feel free to pause and try to evaluate this definite integral. So let's just take some of these constants out."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are told the differentiable function h and its derivative h prime are graphed. And you can see it here, h is in blue, and then its derivative h prime is in this orange color. Four students were asked to give an appropriate calculus-based justification for the fact that h is increasing when x is greater than zero. Can you match the teacher's comments to the justifications? So, before I even look at what the students wrote, you might say, hey, look, I can just look at this and see that h is increasing when x is greater than zero. But just by looking at the graph of h, that by itself is not a calculus-based justification. We're not using calculus."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can you match the teacher's comments to the justifications? So, before I even look at what the students wrote, you might say, hey, look, I can just look at this and see that h is increasing when x is greater than zero. But just by looking at the graph of h, that by itself is not a calculus-based justification. We're not using calculus. We're just using our knowledge of what it means for a graph to be increasing. In order for it to be a calculus-based justification, we should use calculus in some way, and so maybe use the derivative in some way. Now, you might recognize that you know that a function is increasing if its derivative is positive."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're not using calculus. We're just using our knowledge of what it means for a graph to be increasing. In order for it to be a calculus-based justification, we should use calculus in some way, and so maybe use the derivative in some way. Now, you might recognize that you know that a function is increasing if its derivative is positive. So, before I even look at what the students said, what I would say, my calculus-based justification, and I wouldn't even have to see the graph of h, I would just have to see the graph of h prime, is to say, look, h prime is greater than, h prime is positive when x is greater than zero. If the derivative is positive, then that means that the slope of the tangent line is positive, and that means that the graph of the original function is going to be increasing. Now, let's see whether one of the students said that or what some of the other students wrote."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, you might recognize that you know that a function is increasing if its derivative is positive. So, before I even look at what the students said, what I would say, my calculus-based justification, and I wouldn't even have to see the graph of h, I would just have to see the graph of h prime, is to say, look, h prime is greater than, h prime is positive when x is greater than zero. If the derivative is positive, then that means that the slope of the tangent line is positive, and that means that the graph of the original function is going to be increasing. Now, let's see whether one of the students said that or what some of the other students wrote. So, can you match the teacher's comments to the justifications? So, one student wrote, the derivative of h is increasing when x is greater than zero. So, it is indeed the case that the derivative is increasing when x is greater than zero, but that's not the justification for why h is increasing."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, let's see whether one of the students said that or what some of the other students wrote. So, can you match the teacher's comments to the justifications? So, one student wrote, the derivative of h is increasing when x is greater than zero. So, it is indeed the case that the derivative is increasing when x is greater than zero, but that's not the justification for why h is increasing. For example, the derivative could be increasing while still being negative, in which case h would be decreasing. The appropriate justification is that h prime is positive, not that it's necessarily increasing, because you could be increasing and still not be positive. So, let's see."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, it is indeed the case that the derivative is increasing when x is greater than zero, but that's not the justification for why h is increasing. For example, the derivative could be increasing while still being negative, in which case h would be decreasing. The appropriate justification is that h prime is positive, not that it's necessarily increasing, because you could be increasing and still not be positive. So, let's see. I would say that this does justify why h is increasing. When x is greater than zero, as the x values increase, the function values also increase. Well, that is a justification for why h is increasing, but that's not calculus-based, and no way are you using a derivative."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, let's see. I would say that this does justify why h is increasing. When x is greater than zero, as the x values increase, the function values also increase. Well, that is a justification for why h is increasing, but that's not calculus-based, and no way are you using a derivative. So, this isn't a calculus-based justification. It's above the x-axis. So, this one, what is it?"}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that is a justification for why h is increasing, but that's not calculus-based, and no way are you using a derivative. So, this isn't a calculus-based justification. It's above the x-axis. So, this one, what is it? Are they talking about h? Are they talking about h prime? If they're saying that h prime is above the x-axis when x is greater than zero, then that would be a good answer, but this is just, you know, what is above the x-axis and over what interval?"}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, this one, what is it? Are they talking about h? Are they talking about h prime? If they're saying that h prime is above the x-axis when x is greater than zero, then that would be a good answer, but this is just, you know, what is above the x-axis and over what interval? So, I would actually, let's scroll down a little bit. This looks like a good thing for the teacher to write. Please use more precise language."}, {"video_title": "Calculus based justification for function increasing   AP Calculus AB   Khan Academy.mp3", "Sentence": "If they're saying that h prime is above the x-axis when x is greater than zero, then that would be a good answer, but this is just, you know, what is above the x-axis and over what interval? So, I would actually, let's scroll down a little bit. This looks like a good thing for the teacher to write. Please use more precise language. This cannot be accepted as a correct justification. And then finally, this last student wrote the derivative of h is positive when x is greater than zero, and it is indeed the case. If your derivative is positive, that means that your original function is going to be increasing over that interval."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is get some practice finding general solutions to separable differential equations so let's say that I had the differential equation dy dx, the derivative of y with respect to x is equal to e to the x over y. See if you can find the general solution to this differential equation. I'm giving you a huge hint. It is a separable differential equation. All right, so when we're dealing with a separable differential equation, what we want to do is get the y's and the dy's on one side and then the x's and the dx's on the other side and we really treat these differentials kind of like variables which is a little hand wavy with the mathematics but that's what we will do. So let's see, if we multiply both sides times y, so we're gonna multiply both sides times y, what are we going to get? We're gonna get y times the derivative of y with respect to x is equal to e to the x and now we can multiply both sides by the differential dx, multiply both of them by dx, those cancel out and we are left with y times dy is equal to e to the x dx and now we can take the integral of both sides so let us do that."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is a separable differential equation. All right, so when we're dealing with a separable differential equation, what we want to do is get the y's and the dy's on one side and then the x's and the dx's on the other side and we really treat these differentials kind of like variables which is a little hand wavy with the mathematics but that's what we will do. So let's see, if we multiply both sides times y, so we're gonna multiply both sides times y, what are we going to get? We're gonna get y times the derivative of y with respect to x is equal to e to the x and now we can multiply both sides by the differential dx, multiply both of them by dx, those cancel out and we are left with y times dy is equal to e to the x dx and now we can take the integral of both sides so let us do that. So what is the integral of y dy? Well, here we would just use the reverse power rule. We would increment the exponent so it's y to the first but so now when we take the antiderivative, it would be y squared and then we divide by that incremented exponent is equal to, well, the exciting thing about e to the x is its antiderivative is, and its derivative is e to the x is equal to e to the x plus, is equal to e to the x plus c and so we can leave it like this if we like."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna get y times the derivative of y with respect to x is equal to e to the x and now we can multiply both sides by the differential dx, multiply both of them by dx, those cancel out and we are left with y times dy is equal to e to the x dx and now we can take the integral of both sides so let us do that. So what is the integral of y dy? Well, here we would just use the reverse power rule. We would increment the exponent so it's y to the first but so now when we take the antiderivative, it would be y squared and then we divide by that incremented exponent is equal to, well, the exciting thing about e to the x is its antiderivative is, and its derivative is e to the x is equal to e to the x plus, is equal to e to the x plus c and so we can leave it like this if we like. In fact, this right over here, this isn't an explicit function. Y here isn't an explicit function of x. You could actually say y is equal to the plus or minus square root of two times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "We would increment the exponent so it's y to the first but so now when we take the antiderivative, it would be y squared and then we divide by that incremented exponent is equal to, well, the exciting thing about e to the x is its antiderivative is, and its derivative is e to the x is equal to e to the x plus, is equal to e to the x plus c and so we can leave it like this if we like. In fact, this right over here, this isn't an explicit function. Y here isn't an explicit function of x. You could actually say y is equal to the plus or minus square root of two times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation. Let's do another example. So let's say that we have the derivative of y with respect to x is equal to, let's say it's equal to y squared times sine of x. Pause the video and see if you can find the general solution here."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could actually say y is equal to the plus or minus square root of two times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation. Let's do another example. So let's say that we have the derivative of y with respect to x is equal to, let's say it's equal to y squared times sine of x. Pause the video and see if you can find the general solution here. So once again, we want to separate our y's and our x's. So let's see, we can multiply both sides times y to the negative two power. Y to the negative two, y to the negative two."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and see if you can find the general solution here. So once again, we want to separate our y's and our x's. So let's see, we can multiply both sides times y to the negative two power. Y to the negative two, y to the negative two. These become one and then we can also multiply both sides times dx. So if we multiply dx here, those cancel out and then we multiply dx here. And so we're left with y to the negative two power times dy is equal to sine of x dx and now we just can integrate both sides."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Y to the negative two, y to the negative two. These become one and then we can also multiply both sides times dx. So if we multiply dx here, those cancel out and then we multiply dx here. And so we're left with y to the negative two power times dy is equal to sine of x dx and now we just can integrate both sides. Now what is the antiderivative of y to the negative two? Well once again, we use the reverse power rule. We increment the exponent, so it's gonna be y to the negative one and then we divide by that newly incremented exponent."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we're left with y to the negative two power times dy is equal to sine of x dx and now we just can integrate both sides. Now what is the antiderivative of y to the negative two? Well once again, we use the reverse power rule. We increment the exponent, so it's gonna be y to the negative one and then we divide by that newly incremented exponent. So we divide by negative one. Well that would just make this thing negative. That is going to be equal to, so what's the antiderivative of sine of x?"}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "We increment the exponent, so it's gonna be y to the negative one and then we divide by that newly incremented exponent. So we divide by negative one. Well that would just make this thing negative. That is going to be equal to, so what's the antiderivative of sine of x? Well it is, you might recognize it if I put a negative there and a negative there. The antiderivative of negative sine of x, well that's cosine of x. So this whole thing is gonna be negative cosine of x or another way to write this, I could multiply both sides times the negative one and so these would both become positive and so I could write one over y is equal to cosine of x and actually let me write it this way, plus c. Don't wanna forget my plus c's."}, {"video_title": "Worked example separable differential equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is going to be equal to, so what's the antiderivative of sine of x? Well it is, you might recognize it if I put a negative there and a negative there. The antiderivative of negative sine of x, well that's cosine of x. So this whole thing is gonna be negative cosine of x or another way to write this, I could multiply both sides times the negative one and so these would both become positive and so I could write one over y is equal to cosine of x and actually let me write it this way, plus c. Don't wanna forget my plus c's. Plus c or I can take the reciprocal of both sides. If I wanna solve explicitly for y, I could get y is equal to one over cosine of x plus c as our general solution and we're done. That was strangely fun."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "But instead of rotating around the x-axis this time, I want to rotate around the y-axis. And instead of going between zero and some point, I'm going to go between y is equal to one and y is equal to four. So what I'm going to do is I'm going to take this graph right over here, I'm going to take this curve, instead of rotating it around the x-axis like we did in the last few videos, I'm going to rotate it around the y-axis. So I'm going to rotate it around just like that. So what's the shape that we would get? Let me see if we can visualize this. So the base is going to look something like that."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm going to rotate it around just like that. So what's the shape that we would get? Let me see if we can visualize this. So the base is going to look something like that. We could see through it. It's going to look something like that. And then this up here, the top of it, would look something like that."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the base is going to look something like that. We could see through it. It's going to look something like that. And then this up here, the top of it, would look something like that. And we care about the stuff in between. So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this up here, the top of it, would look something like that. And we care about the stuff in between. So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit. So it would look something like that. So let me draw it separately just so we can visualize it. So I'll draw it at different angles."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me shade it in a little bit. So it would look something like that. So let me draw it separately just so we can visualize it. So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It looks something like it's a little bit smaller like that. And then it gets cut off right over here like that."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It looks something like it's a little bit smaller like that. And then it gets cut off right over here like that. So it looks, I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do that same yellow color."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it gets cut off right over here like that. So it looks, I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do that same yellow color. The visualization here is probably the hardest part. But as we can see, it's not too bad. So it looks something like this."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me do that same yellow color. The visualization here is probably the hardest part. But as we can see, it's not too bad. So it looks something like this. It looks like maybe a truffle, an upside down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks something like this. It looks like maybe a truffle, an upside down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that. Then it goes down over here. And then the x-axis is going like this. So I just tilted this over."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the y-axis is popping out in this example like that. Then it goes down over here. And then the x-axis is going like this. So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing. Well, what we can do instead of creating disks where the depth is in little dx's, what if we created disks where the depth is in dy's? So let's think about that a little bit."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing. Well, what we can do instead of creating disks where the depth is in little dx's, what if we created disks where the depth is in dy's? So let's think about that a little bit. So let's think about constructing a disk at a certain y value. So let's think about a certain y value. We're going to construct a disk right over there that has the same radius of the shape at that point."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. So let's think about constructing a disk at a certain y value. So let's think about a certain y value. We're going to construct a disk right over there that has the same radius of the shape at that point. So that's our disk right over here. And then it has a depth. Instead of saying it has a depth of dx, let's say it has a depth of dy."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to construct a disk right over there that has the same radius of the shape at that point. So that's our disk right over here. And then it has a depth. Instead of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disk in terms of y? And as you can imagine, we're going to do this definite integral with respect to y."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Instead of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disk in terms of y? And as you can imagine, we're going to do this definite integral with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these disks, or I guess you could say the face of this coin. Well, to find that area, it's pi r squared."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as you can imagine, we're going to do this definite integral with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these disks, or I guess you could say the face of this coin. Well, to find that area, it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to find that area, it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, but that's OK, because we are in the positive x-axis right over here. So we can also call this function right over here x is equal to the square root of y."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, but that's OK, because we are in the positive x-axis right over here. So we can also call this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can also call this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y. It's going to be the square root of y. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x and I'm saying this is f of y. It's going to be a function of y."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, our radius right over here is going to be f of y. It's going to be the square root of y. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x and I'm saying this is f of y. It's going to be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius squared."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius squared. Our radius is square root of y. So this thing is going to be equal to pi squared of y squared is just pi times y. Now, if we want the volume, we just have to multiply the area of the surface times the depth, times dy."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius squared. Our radius is square root of y. So this thing is going to be equal to pi squared of y squared is just pi times y. Now, if we want the volume, we just have to multiply the area of the surface times the depth, times dy. So the volume of each of those disks is going to be pi y times dy. This gives you the volume of a disk. Now, if we want the volume of this entire thing, we just have to sum all of these disks for all the y values between y is equal to 1 and y is equal to 4."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, if we want the volume, we just have to multiply the area of the surface times the depth, times dy. So the volume of each of those disks is going to be pi y times dy. This gives you the volume of a disk. Now, if we want the volume of this entire thing, we just have to sum all of these disks for all the y values between y is equal to 1 and y is equal to 4. So let's do that. So we just take the definite integral from y is equal to 1 to y equals 4. And just as a reminder, definite integral is a very special type of sum."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, if we want the volume of this entire thing, we just have to sum all of these disks for all the y values between y is equal to 1 and y is equal to 4. So let's do that. So we just take the definite integral from y is equal to 1 to y equals 4. And just as a reminder, definite integral is a very special type of sum. We're summing up all of these things, but we're taking the limit of that sum as these dy's get shorter, or gets, I guess, squatter and squatter, smaller and smaller. And we have a larger and larger number of these disks. Really, as these become infinitely small and we have an infinite number of disks, so that our sum doesn't just approximate the volume."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just as a reminder, definite integral is a very special type of sum. We're summing up all of these things, but we're taking the limit of that sum as these dy's get shorter, or gets, I guess, squatter and squatter, smaller and smaller. And we have a larger and larger number of these disks. Really, as these become infinitely small and we have an infinite number of disks, so that our sum doesn't just approximate the volume. It actually is the volume at the limit. So to figure out the volume of this entire thing, we just have to evaluate this definite integral in terms of y. And so how do we do that?"}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Really, as these become infinitely small and we have an infinite number of disks, so that our sum doesn't just approximate the volume. It actually is the volume at the limit. So to figure out the volume of this entire thing, we just have to evaluate this definite integral in terms of y. And so how do we do that? What's going to be equal to? Well, we could take the pi outside. It's going to be pi times the antiderivative of y, which is just y squared over 2, evaluated from 1 to 4, which is equal to pi times, well, you evaluate it at 4, you get 16 over 2."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so how do we do that? What's going to be equal to? Well, we could take the pi outside. It's going to be pi times the antiderivative of y, which is just y squared over 2, evaluated from 1 to 4, which is equal to pi times, well, you evaluate it at 4, you get 16 over 2. Let me just write it out like this. 4 squared over 2 minus 1 squared over 2, which is equal to pi times 16 over 2 is 8 minus 1 half. And so we could do this as 16 halves minus 1 half, which is equal to 15 halves."}, {"video_title": "Disc method around y-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be pi times the antiderivative of y, which is just y squared over 2, evaluated from 1 to 4, which is equal to pi times, well, you evaluate it at 4, you get 16 over 2. Let me just write it out like this. 4 squared over 2 minus 1 squared over 2, which is equal to pi times 16 over 2 is 8 minus 1 half. And so we could do this as 16 halves minus 1 half, which is equal to 15 halves. So this is equal to 15 halves times pi. Or another way of thinking of it is 7 and 1 half times pi. But this is a little bit clearer."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The differentiable functions x and y are related by the following equation. The sine of x plus cosine of y is equal to square root of two. They also tell us that the derivative of x with respect to t is equal to five. They also ask us, find the derivative of y with respect to t when y is equal to pi over four and zero is less than x is less than pi over two. So given that they are telling us the derivative of x with respect to t and we wanna find the derivative of y with respect to t, it's a safe assumption that both x and y are functions of t. So you could even rewrite this equation right over here. You could rewrite it as sine of x, which is a function of t, plus cosine of y, which is a function of t, is equal to square root of two. Now it might confuse you a little bit."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "They also ask us, find the derivative of y with respect to t when y is equal to pi over four and zero is less than x is less than pi over two. So given that they are telling us the derivative of x with respect to t and we wanna find the derivative of y with respect to t, it's a safe assumption that both x and y are functions of t. So you could even rewrite this equation right over here. You could rewrite it as sine of x, which is a function of t, plus cosine of y, which is a function of t, is equal to square root of two. Now it might confuse you a little bit. You're not used to seeing x as a function of a third variable or y as a function of something other than x. But remember, x and y are just variables. This could be f of t and this could be g of t instead of x of t and y of t, and that might feel a little bit more natural to you."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now it might confuse you a little bit. You're not used to seeing x as a function of a third variable or y as a function of something other than x. But remember, x and y are just variables. This could be f of t and this could be g of t instead of x of t and y of t, and that might feel a little bit more natural to you. But needless to say, if we wanna find dy dt, what we wanna do is take the derivative with respect to t of both sides of this equation. So let's do that. So we're gonna do it on the left-hand side."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "This could be f of t and this could be g of t instead of x of t and y of t, and that might feel a little bit more natural to you. But needless to say, if we wanna find dy dt, what we wanna do is take the derivative with respect to t of both sides of this equation. So let's do that. So we're gonna do it on the left-hand side. So it's gonna be, we're gonna take that with respect to t, derivative of that with respect to t. We're gonna take the derivative of that with respect to t. And then we're gonna take the derivative of the right-hand side, this constant, with respect to t. So let's think about each of these things. So what is this, let me do this in a new color, the stuff that I'm doing in this aqua color right over here. How could I write that?"}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna do it on the left-hand side. So it's gonna be, we're gonna take that with respect to t, derivative of that with respect to t. We're gonna take the derivative of that with respect to t. And then we're gonna take the derivative of the right-hand side, this constant, with respect to t. So let's think about each of these things. So what is this, let me do this in a new color, the stuff that I'm doing in this aqua color right over here. How could I write that? So I'm taking the derivative with respect to t. I have sine of something, which is itself a function of t. So I would just apply the chain rule here. I'm first going to take the derivative with respect to x of sine of x. I could write sine of x of t, but I'll just revert back to the sine of x here for simplicity. And then I would then multiply that times the derivative of the inside, you could say, with respect to t, times the derivative of x with respect to t. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with x's and y's, but all that's happening, I'm taking the derivative of the outside of the sine of something with respect to the something, in this case it is x, and then I'm taking the derivative of the something, in this case x, with respect to t. Well, we can do the same thing here for this second term here."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "How could I write that? So I'm taking the derivative with respect to t. I have sine of something, which is itself a function of t. So I would just apply the chain rule here. I'm first going to take the derivative with respect to x of sine of x. I could write sine of x of t, but I'll just revert back to the sine of x here for simplicity. And then I would then multiply that times the derivative of the inside, you could say, with respect to t, times the derivative of x with respect to t. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with x's and y's, but all that's happening, I'm taking the derivative of the outside of the sine of something with respect to the something, in this case it is x, and then I'm taking the derivative of the something, in this case x, with respect to t. Well, we can do the same thing here for this second term here. So I'd wanna take the derivative with respect to y of, I guess you could say the outside, of cosine of y, and then I would multiply that times the derivative of y with respect to t. And then all of that is going to be equal to what? Well, the derivative with respect to t of a constant, square root of two is a constant, it's not gonna change as t changes, so its derivative, its rate of change is zero. All right, so now we just have to figure out all of these things."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I would then multiply that times the derivative of the inside, you could say, with respect to t, times the derivative of x with respect to t. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with x's and y's, but all that's happening, I'm taking the derivative of the outside of the sine of something with respect to the something, in this case it is x, and then I'm taking the derivative of the something, in this case x, with respect to t. Well, we can do the same thing here for this second term here. So I'd wanna take the derivative with respect to y of, I guess you could say the outside, of cosine of y, and then I would multiply that times the derivative of y with respect to t. And then all of that is going to be equal to what? Well, the derivative with respect to t of a constant, square root of two is a constant, it's not gonna change as t changes, so its derivative, its rate of change is zero. All right, so now we just have to figure out all of these things. So first of all, the derivative with respect to x of sine of x is cosine of x times derivative of x with respect to t, I'll just write that out here, derivative of x with respect to t. And then we're going to have, it's gonna be a plus here, the derivative of y with respect to t, so plus the derivative of y with respect to t, I'm just swapping the order here so that this goes out front. Now what's the derivative of cosine of y with respect to y? Well, that is negative sine of y."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so now we just have to figure out all of these things. So first of all, the derivative with respect to x of sine of x is cosine of x times derivative of x with respect to t, I'll just write that out here, derivative of x with respect to t. And then we're going to have, it's gonna be a plus here, the derivative of y with respect to t, so plus the derivative of y with respect to t, I'm just swapping the order here so that this goes out front. Now what's the derivative of cosine of y with respect to y? Well, that is negative sine of y. And so, actually let me just put a sine of y here, and then I wanna have a negative. So let me erase this and put a negative there. And that is all going to be equal to zero."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that is negative sine of y. And so, actually let me just put a sine of y here, and then I wanna have a negative. So let me erase this and put a negative there. And that is all going to be equal to zero. And so, what can we figure out now? They've told us that the derivative of x with respect to t is equal to five. They tell us that right over here."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that is all going to be equal to zero. And so, what can we figure out now? They've told us that the derivative of x with respect to t is equal to five. They tell us that right over here. So this is equal to five. We wanna find the derivative of y with respect to t. They tell us what y is. Y is pi over four."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "They tell us that right over here. So this is equal to five. We wanna find the derivative of y with respect to t. They tell us what y is. Y is pi over four. This, y is pi over four, so we know this is pi over four. And let's see, we have to figure out what, we still have two unknowns here. We don't know what x is, and we don't know what the derivative of y with respect to t is."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Y is pi over four. This, y is pi over four, so we know this is pi over four. And let's see, we have to figure out what, we still have two unknowns here. We don't know what x is, and we don't know what the derivative of y with respect to t is. And this is what we need to figure out. So what would x be? What would x be when y is pi over four?"}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know what x is, and we don't know what the derivative of y with respect to t is. And this is what we need to figure out. So what would x be? What would x be when y is pi over four? Well, to figure that out, we can go back to this original equation right over here. So when y is pi over four, you get, let me write it down, sine of x plus cosine of pi over four is equal to square root of two. Cosine of pi over four, we revert to our unit, or we think about our unit circle."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "What would x be when y is pi over four? Well, to figure that out, we can go back to this original equation right over here. So when y is pi over four, you get, let me write it down, sine of x plus cosine of pi over four is equal to square root of two. Cosine of pi over four, we revert to our unit, or we think about our unit circle. We're in the first quadrant. If you think in degrees, it's a 45 degree angle. That's going to be square root of two over two."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of pi over four, we revert to our unit, or we think about our unit circle. We're in the first quadrant. If you think in degrees, it's a 45 degree angle. That's going to be square root of two over two. And so we can subtract square root of two over two from both sides, which is going to give us sine of x is equal to, well, if you take square root of two over two from square root of two, you're taking half of it away, so you're gonna have half of it less. So square root of two over two. And so what x value, when I take the sine of it, and remember, where the angle, if we're thinking of the unit circle, it's going to be in that first quadrant."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's going to be square root of two over two. And so we can subtract square root of two over two from both sides, which is going to give us sine of x is equal to, well, if you take square root of two over two from square root of two, you're taking half of it away, so you're gonna have half of it less. So square root of two over two. And so what x value, when I take the sine of it, and remember, where the angle, if we're thinking of the unit circle, it's going to be in that first quadrant. X is an angle in this case right over here. Well, that's going to be, once again, pi over four. So this tells us that x is equal to pi over four when y is equal to pi over four."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what x value, when I take the sine of it, and remember, where the angle, if we're thinking of the unit circle, it's going to be in that first quadrant. X is an angle in this case right over here. Well, that's going to be, once again, pi over four. So this tells us that x is equal to pi over four when y is equal to pi over four. And so we know that this is pi over four as well. So let me just rewrite this because it's getting a little bit messy. So we know that five times cosine of pi over four minus dy dt, the derivative of y with respect to t, which is what we want to figure out, times sine of pi over four is equal to zero."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this tells us that x is equal to pi over four when y is equal to pi over four. And so we know that this is pi over four as well. So let me just rewrite this because it's getting a little bit messy. So we know that five times cosine of pi over four minus dy dt, the derivative of y with respect to t, which is what we want to figure out, times sine of pi over four is equal to zero. Is equal to zero, and let me put some parentheses here just to clarify things a little bit. All right, so let's see. Now it's just a little bit of algebra."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that five times cosine of pi over four minus dy dt, the derivative of y with respect to t, which is what we want to figure out, times sine of pi over four is equal to zero. Is equal to zero, and let me put some parentheses here just to clarify things a little bit. All right, so let's see. Now it's just a little bit of algebra. Cosine of pi over four, we already know, is square root of two over two. Sine of pi over four is also square root of two over two. And let's see, what if we divide both sides of this equation by square root of two over two?"}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now it's just a little bit of algebra. Cosine of pi over four, we already know, is square root of two over two. Sine of pi over four is also square root of two over two. And let's see, what if we divide both sides of this equation by square root of two over two? Well, what's that going to give us? Well, then, square root of two over two divided by square root of two over, square root of two over two divided by square root of two over two is gonna be one. Square root of two over two divided by square root of two over two is gonna be one."}, {"video_title": "Worked example Differentiating related functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, what if we divide both sides of this equation by square root of two over two? Well, what's that going to give us? Well, then, square root of two over two divided by square root of two over, square root of two over two divided by square root of two over two is gonna be one. Square root of two over two divided by square root of two over two is gonna be one. And then zero divided by square root of two over two is just still going to be zero. And so this whole thing simplifies to five times one, which is just five, minus the derivative of y with respect to t is equal to zero. And so there you have it."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "And in this video in particular, we will explore modeling population. Modeling population. We're actually gonna go into some depth on this eventually, but here we're gonna start with simpler models. And we'll see, we will stumble on, using the logic of differential equations, things that you might have seen in your algebra or your pre-calculus class. So on some level, what we're going to do here is going to be review, but we're gonna get there using the power of modeling with differential equations. So let's just define some variables. Let's say that P is equal to our population."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "And we'll see, we will stumble on, using the logic of differential equations, things that you might have seen in your algebra or your pre-calculus class. So on some level, what we're going to do here is going to be review, but we're gonna get there using the power of modeling with differential equations. So let's just define some variables. Let's say that P is equal to our population. And let's say that T is, let's say that T is equal to the time that has passed in days. In days. It could have been years or months."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "Let's say that P is equal to our population. And let's say that T is, let's say that T is equal to the time that has passed in days. In days. It could have been years or months. But let's say we're doing the population of insects that reproduce quite quickly. So days seem like a nice time span to care about. Now, what would be a reasonable model?"}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "It could have been years or months. But let's say we're doing the population of insects that reproduce quite quickly. So days seem like a nice time span to care about. Now, what would be a reasonable model? Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population. The actual population. Why is that reasonable?"}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "Now, what would be a reasonable model? Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population. The actual population. Why is that reasonable? Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing is going to be more, or a thousand insects, is going to be more insects per second or per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "Why is that reasonable? Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing is going to be more, or a thousand insects, is going to be more insects per second or per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population. And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population. And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution. So how do we do that? And I encourage you to pause the video at any time and see if you can solve this differential equation. So assuming you at least maybe have had an attempt at it."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution. So how do we do that? And I encourage you to pause the video at any time and see if you can solve this differential equation. So assuming you at least maybe have had an attempt at it. And you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and all the differentials involving that variable on one side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the derivative of P with respect to t, this isn't quite a fraction."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "So assuming you at least maybe have had an attempt at it. And you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and all the differentials involving that variable on one side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change in P over change in time. This is our instantaneous change. But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "And once again, dP, the derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change in P over change in time. This is our instantaneous change. But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So we want to put all the P's and dP's on one side, and all the things that involve t or that I guess don't involve P on the other side. So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So we want to put all the P's and dP's on one side, and all the things that involve t or that I guess don't involve P on the other side. So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel. And then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity, which it really isn't a quantity."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel. And then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity, which it really isn't a quantity. You really have to view this as a limit of that change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But once again, for the sake of this, we can do this."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "Once again, treating the differential like a quantity, which it really isn't a quantity. You really have to view this as a limit of that change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But once again, for the sake of this, we can do this. And when we do that, we would be left with one over P dP is equal to K dt. Now we can integrate both sides. Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "But once again, for the sake of this, we can do this. And when we do that, we would be left with one over P dP is equal to K dt. Now we can integrate both sides. Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's. And then if we integrate this side, we would get the natural log of the absolute value of our population. And we could say plus some constant if we want, but we're gonna get a constant on this side as well. So we could just say that's going to be equal to K times T plus some constant."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's. And then if we integrate this side, we would get the natural log of the absolute value of our population. And we could say plus some constant if we want, but we're gonna get a constant on this side as well. So we could just say that's going to be equal to K times T plus some constant. I'll just call that C1. And once again, I could have put a plus C2 here, but I could have then subtracted the constant from both sides, and I would just get the constant on the right-hand side. Now, how can I solve for P?"}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "So we could just say that's going to be equal to K times T plus some constant. I'll just call that C1. And once again, I could have put a plus C2 here, but I could have then subtracted the constant from both sides, and I would just get the constant on the right-hand side. Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means, that's the same thing, that means that the absolute value of P is equal to E to all of this business. E to the, let me do the same colors, KT plus C1."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means, that's the same thing, that means that the absolute value of P is equal to E to all of this business. E to the, let me do the same colors, KT plus C1. Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1. Now, this is just E to some constant, so we could just call this, let's just call that the constant C. So, this is all simplified to CE, CE to the KT, to the KT. And if we assume our population at any given time is positive, then we could get rid of this absolute value sign."}, {"video_title": "Modeling population with simple differential equation   Khan Academy.mp3", "Sentence": "E to the, let me do the same colors, KT plus C1. Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1. Now, this is just E to some constant, so we could just call this, let's just call that the constant C. So, this is all simplified to CE, CE to the KT, to the KT. And if we assume our population at any given time is positive, then we could get rid of this absolute value sign. Now, we have a general solution to this, frankly, fairly general differential equation, we just said proportional, we haven't given what the proportionality constant is, but we could say if we assume positive population, that the population is going to be equal to some constant C times E to the KT power. And the reason why I said that you've seen this before is this is just an exponential function, and it's very likely that in algebra or in pre-calculus class, you have modeled things with exponential functions, and my guess is that you've modeled things with populated, modeled things like population. The reason why this is interesting is you now see where this is coming from, the underlying logic that's just driven by the actual differential equation, the rate of change with respect to time of the population, well, maybe it's just proportional to population."}, {"video_title": "Justification with the mean value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can we use the mean value theorem to say that the equation g prime of x is equal to 1 1 2 has a solution where negative one is less than x is less than two? If so, write a justification. All right, pause this video and see if you can figure that out. So the key to using the mean value theorem to even before you even think about using it, you have to make sure that you are continuous over the closed interval and differentiable over the open interval. So this is the open interval here and then the closed interval would include the endpoints. But you might immediately realize that both of these intervals contain x equals zero and at x equals zero, the function is undefined. And if it's undefined there, well, it's not going to be continuous or differentiable at that point."}, {"video_title": "Justification with the mean value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the key to using the mean value theorem to even before you even think about using it, you have to make sure that you are continuous over the closed interval and differentiable over the open interval. So this is the open interval here and then the closed interval would include the endpoints. But you might immediately realize that both of these intervals contain x equals zero and at x equals zero, the function is undefined. And if it's undefined there, well, it's not going to be continuous or differentiable at that point. And so no, not continuous or differentiable or differentiable over the interval. Over the interval. All right, let's do the second part."}, {"video_title": "Justification with the mean value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if it's undefined there, well, it's not going to be continuous or differentiable at that point. And so no, not continuous or differentiable or differentiable over the interval. Over the interval. All right, let's do the second part. Can we use the mean value theorem to say that there is a value c such that g prime of c is equal to negative 1 1 2 and one is less than c is less than two? If so, write a justification. So pause the video again."}, {"video_title": "Justification with the mean value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, let's do the second part. Can we use the mean value theorem to say that there is a value c such that g prime of c is equal to negative 1 1 2 and one is less than c is less than two? If so, write a justification. So pause the video again. All right, so in this situation, between one and two on both the open and the closed intervals, well, this is a rational function and a rational function is going to be continuous and differentiable at every point in its domain and its domain completely contains this open and closed interval. Or another way to think about it, every point on this open interval and on the closed interval is in the domain. So we can write g of x is a rational function which lets us know that it is continuous and differentiable at every point in its domain."}, {"video_title": "Justification with the mean value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause the video again. All right, so in this situation, between one and two on both the open and the closed intervals, well, this is a rational function and a rational function is going to be continuous and differentiable at every point in its domain and its domain completely contains this open and closed interval. Or another way to think about it, every point on this open interval and on the closed interval is in the domain. So we can write g of x is a rational function which lets us know that it is continuous and differentiable at every point in its domain. At every point in its domain. The closed interval from one to two is in domain. And so now let's see what the average rate of change is from one to two."}, {"video_title": "Justification with the mean value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can write g of x is a rational function which lets us know that it is continuous and differentiable at every point in its domain. At every point in its domain. The closed interval from one to two is in domain. And so now let's see what the average rate of change is from one to two. And so we get g of two minus g of one over two minus one is equal to 1 1\u20442 minus one over one, which is equal to negative 1 1\u20442. Therefore, therefore, by the mean value theorem, there must be a c where one is less than c is less than two and g prime of c is equal to the average rate of change between the endpoints, negative 1 1\u20442. And we're done."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If these things look unfamiliar to you, I encourage you maybe to not watch this video, because in this video we're gonna think about when do we apply these rules, what strategies, and can we algebraically convert expressions so that we can use a simpler rule? But just as a quick review, this is, of course, the power rule right over here. Very handy for taking derivatives of x raised to some power. It's also, we can use that with the derivative properties of sums of derivatives or differences of derivatives to take derivatives of polynomials. This right over here is the product rule. If I have an expression that I wanna take the derivative of and I can think of it as the product of two functions, well then the derivative is gonna be the derivative of the first function times the second function plus the first function times the derivative of the second. Once again, if this looks completely unfamiliar to you or you're a little shaky, go watch the videos, do the practice on the power rule or the product rule, or in this case, the quotient rule."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's also, we can use that with the derivative properties of sums of derivatives or differences of derivatives to take derivatives of polynomials. This right over here is the product rule. If I have an expression that I wanna take the derivative of and I can think of it as the product of two functions, well then the derivative is gonna be the derivative of the first function times the second function plus the first function times the derivative of the second. Once again, if this looks completely unfamiliar to you or you're a little shaky, go watch the videos, do the practice on the power rule or the product rule, or in this case, the quotient rule. And the quotient rule is a little bit more involved and we have practice and videos on that. And I always have mixed feelings about it, because if you don't remember the quotient rule, you can usually, or you can always convert a quotient into a product by expressing this thing at the bottom as f of x, or by expressing this as f of x times g of x to the negative one, so you could take the derivative with a combination of the products and this fourth rule over here, the chain rule. And if any of this is looking unfamiliar again, don't watch this video."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, if this looks completely unfamiliar to you or you're a little shaky, go watch the videos, do the practice on the power rule or the product rule, or in this case, the quotient rule. And the quotient rule is a little bit more involved and we have practice and videos on that. And I always have mixed feelings about it, because if you don't remember the quotient rule, you can usually, or you can always convert a quotient into a product by expressing this thing at the bottom as f of x, or by expressing this as f of x times g of x to the negative one, so you could take the derivative with a combination of the products and this fourth rule over here, the chain rule. And if any of this is looking unfamiliar again, don't watch this video. This video is for folks who are familiar with each of these derivative rules or derivative techniques and now wanna think about, well, what are strategies for deciding when to apply which? So let's do that. Let's say that I have the expression, let's say I'm interested in taking the derivative of x squared plus x minus two over x minus one."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if any of this is looking unfamiliar again, don't watch this video. This video is for folks who are familiar with each of these derivative rules or derivative techniques and now wanna think about, well, what are strategies for deciding when to apply which? So let's do that. Let's say that I have the expression, let's say I'm interested in taking the derivative of x squared plus x minus two over x minus one. Which of these rules or techniques would you use? Well, you might immediately say, hey, look, this looks like a rational expression. I could say, look, I could say this is my f of x right over here."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that I have the expression, let's say I'm interested in taking the derivative of x squared plus x minus two over x minus one. Which of these rules or techniques would you use? Well, you might immediately say, hey, look, this looks like a rational expression. I could say, look, I could say this is my f of x right over here. I could say this is my g of x right over here, and I could apply the quotient rule. This looks like a quotient of two expressions. And you could do that, and if you do all the mathematics correctly, you will get the correct answer."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could say, look, I could say this is my f of x right over here. I could say this is my g of x right over here, and I could apply the quotient rule. This looks like a quotient of two expressions. And you could do that, and if you do all the mathematics correctly, you will get the correct answer. But in this case, it's good to just take a little time to realize, well, can I simplify this algebraically so maybe I can do a little bit less work? And if you look at it that way, you might realize, wait, what if I factored this numerator? I can factor it as x plus two times x minus one, and then I could cancel these two characters out, and I could say, hey, you know what?"}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you could do that, and if you do all the mathematics correctly, you will get the correct answer. But in this case, it's good to just take a little time to realize, well, can I simplify this algebraically so maybe I can do a little bit less work? And if you look at it that way, you might realize, wait, what if I factored this numerator? I can factor it as x plus two times x minus one, and then I could cancel these two characters out, and I could say, hey, you know what? This is gonna be the same thing as the derivative with respect to x of x plus two. Derivative with respect to x of x plus two, which is much, much, much, much, much more straightforward than trying to apply the quotient rule. Here, you would just take the derivative with respect to x of x, which is just gonna be one, and the derivative with respect to x of two is just gonna be zero, and so all of this is just going to simplify to one."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I can factor it as x plus two times x minus one, and then I could cancel these two characters out, and I could say, hey, you know what? This is gonna be the same thing as the derivative with respect to x of x plus two. Derivative with respect to x of x plus two, which is much, much, much, much, much more straightforward than trying to apply the quotient rule. Here, you would just take the derivative with respect to x of x, which is just gonna be one, and the derivative with respect to x of two is just gonna be zero, and so all of this is just going to simplify to one. For taking the derivative of that, you're essentially just using the power rule. And so once again, just a simple algebraic recognition, things become much, much, much, much more simple. Let's do another example."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Here, you would just take the derivative with respect to x of x, which is just gonna be one, and the derivative with respect to x of two is just gonna be zero, and so all of this is just going to simplify to one. For taking the derivative of that, you're essentially just using the power rule. And so once again, just a simple algebraic recognition, things become much, much, much, much more simple. Let's do another example. So let's say that you were to see, or someone were to ask you to take the derivative with respect to x of, let me see. So let's say you had x squared plus two x minus five over, over x. So once again, you might be tempted to use the quotient rule."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do another example. So let's say that you were to see, or someone were to ask you to take the derivative with respect to x of, let me see. So let's say you had x squared plus two x minus five over, over x. So once again, you might be tempted to use the quotient rule. This looks like the quotient of two expressions, but then you might realize, well, there's some algebraic manipulations I can do to make this simpler. You could express this as a product. You could say that this is the same thing as, and I'm just gonna focus on what's inside the parentheses or inside the brackets."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So once again, you might be tempted to use the quotient rule. This looks like the quotient of two expressions, but then you might realize, well, there's some algebraic manipulations I can do to make this simpler. You could express this as a product. You could say that this is the same thing as, and I'm just gonna focus on what's inside the parentheses or inside the brackets. This is the same thing as x to the negative one times x squared plus two x minus five. And then you might want to apply the product rule, but there's even a better simplification here. You could just divide each of these terms by x, or one way to think about it, distribute this one over x across all the terms."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say that this is the same thing as, and I'm just gonna focus on what's inside the parentheses or inside the brackets. This is the same thing as x to the negative one times x squared plus two x minus five. And then you might want to apply the product rule, but there's even a better simplification here. You could just divide each of these terms by x, or one way to think about it, distribute this one over x across all the terms. x to the negative one is the same thing as one over x. And if you do that, x squared divided by x is going to be x. Two x divided by x is going to be two."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could just divide each of these terms by x, or one way to think about it, distribute this one over x across all the terms. x to the negative one is the same thing as one over x. And if you do that, x squared divided by x is going to be x. Two x divided by x is going to be two. And then negative five divided by x, well, you could write that as negative five over x or negative five x to the negative one. And now, taking the derivative of this with respect to x is much easier than using either the quotient or the power rule. This is going to be, let's see, derivative of that is just gonna be one, derivative of two is just gonna be zero."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two x divided by x is going to be two. And then negative five divided by x, well, you could write that as negative five over x or negative five x to the negative one. And now, taking the derivative of this with respect to x is much easier than using either the quotient or the power rule. This is going to be, let's see, derivative of that is just gonna be one, derivative of two is just gonna be zero. And here, even though you have a negative exponent, it might look a little intimidating. This is just taking it using the power rule. So negative one times negative five is positive five."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be, let's see, derivative of that is just gonna be one, derivative of two is just gonna be zero. And here, even though you have a negative exponent, it might look a little intimidating. This is just taking it using the power rule. So negative one times negative five is positive five. X to the, if we take one less than negative one, we're gonna go to the negative two power. So once again, making this algebraic recognition simplified things a good bit. Let's do a few more examples of just starting to recognize when we might be able to simplify things to do things a little bit easier."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So negative one times negative five is positive five. X to the, if we take one less than negative one, we're gonna go to the negative two power. So once again, making this algebraic recognition simplified things a good bit. Let's do a few more examples of just starting to recognize when we might be able to simplify things to do things a little bit easier. So let's say that someone said, hey, you, take the derivative with respect to x. And I'm using x as our variable that we're taking the derivative with respect to, but obviously this works for any variables that we are using. So let's say we're saying square root of x over x squared."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's do a few more examples of just starting to recognize when we might be able to simplify things to do things a little bit easier. So let's say that someone said, hey, you, take the derivative with respect to x. And I'm using x as our variable that we're taking the derivative with respect to, but obviously this works for any variables that we are using. So let's say we're saying square root of x over x squared. Pause this video and think about how would you approach this if you wanna take the derivative with respect to x of the square root of x over x squared. Well, once again, you might say this is a quotient of two expressions, might try to apply the quotient rule, or you might recognize, well, look, this is the same thing. Let me just focus on what's inside the brackets."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say we're saying square root of x over x squared. Pause this video and think about how would you approach this if you wanna take the derivative with respect to x of the square root of x over x squared. Well, once again, you might say this is a quotient of two expressions, might try to apply the quotient rule, or you might recognize, well, look, this is the same thing. Let me just focus on what's inside the brackets. You could view this as x to the negative two times the square root of x, times the square root of x. And then you might wanna use a product rule, but you could simplify this even better. You could say this is the same thing as x to the negative two times x to the 1 1\u20442 power."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me just focus on what's inside the brackets. You could view this as x to the negative two times the square root of x, times the square root of x. And then you might wanna use a product rule, but you could simplify this even better. You could say this is the same thing as x to the negative two times x to the 1 1\u20442 power. And now you're just using our exponent properties, negative two plus 1 1\u20442 is negative 3 1\u20442. So this is the derivative of x to the negative 3 1\u20442 power. And so here, once again, we took something that we thought we might have to use a quotient rule or use the product rule, and now this just becomes a straightforward using the power rule."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say this is the same thing as x to the negative two times x to the 1 1\u20442 power. And now you're just using our exponent properties, negative two plus 1 1\u20442 is negative 3 1\u20442. So this is the derivative of x to the negative 3 1\u20442 power. And so here, once again, we took something that we thought we might have to use a quotient rule or use the product rule, and now this just becomes a straightforward using the power rule. So this is just going to be equal to, so bring the negative 3 1\u20442 out front, negative 3 1\u20442, x to the negative 3 1\u20442 minus one is negative 5 1\u20442 power. So once again, just before you just, especially if you're about to apply the quotient rule and sometimes even the product rule, just see, is there an algebraic simplification, sometimes a trigonometric simplification, that you can make that eases your job, that makes things less hairy? And as a general tip, I can't say this is gonna be always true, but if you're taking some type of an exam and you're going down some really hairy route, which the quotient rule will often take you, it's a good sign that, hey, take a pause before trying to run through all of that algebra to apply the quotient rule and see if you can simplify things."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so here, once again, we took something that we thought we might have to use a quotient rule or use the product rule, and now this just becomes a straightforward using the power rule. So this is just going to be equal to, so bring the negative 3 1\u20442 out front, negative 3 1\u20442, x to the negative 3 1\u20442 minus one is negative 5 1\u20442 power. So once again, just before you just, especially if you're about to apply the quotient rule and sometimes even the product rule, just see, is there an algebraic simplification, sometimes a trigonometric simplification, that you can make that eases your job, that makes things less hairy? And as a general tip, I can't say this is gonna be always true, but if you're taking some type of an exam and you're going down some really hairy route, which the quotient rule will often take you, it's a good sign that, hey, take a pause before trying to run through all of that algebra to apply the quotient rule and see if you can simplify things. So let's give another example. And this one, there's not a must, there's not an obvious way, and it really depends on what folks' preferences are. But let's say you wanna take the derivative with respect to x of one over two x to the negative five."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as a general tip, I can't say this is gonna be always true, but if you're taking some type of an exam and you're going down some really hairy route, which the quotient rule will often take you, it's a good sign that, hey, take a pause before trying to run through all of that algebra to apply the quotient rule and see if you can simplify things. So let's give another example. And this one, there's not a must, there's not an obvious way, and it really depends on what folks' preferences are. But let's say you wanna take the derivative with respect to x of one over two x to the negative five. Sorry, one over two x minus five, I should say. Well, here, you could immediately, you could apply the quotient rule here, or the numerator you view that as f of x. You could view this as the same thing as the derivative with respect to x instead of two x minus five."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's say you wanna take the derivative with respect to x of one over two x to the negative five. Sorry, one over two x minus five, I should say. Well, here, you could immediately, you could apply the quotient rule here, or the numerator you view that as f of x. You could view this as the same thing as the derivative with respect to x instead of two x minus five. Let me do that in the blue color. Two x minus five to the negative one power. And now this, in this situation, you would use a combination of the power rule and the chain rule."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could view this as the same thing as the derivative with respect to x instead of two x minus five. Let me do that in the blue color. Two x minus five to the negative one power. And now this, in this situation, you would use a combination of the power rule and the chain rule. You'd say, okay, my g of x is two x minus five, and f of g of x is going to be this whole expression. And so if you did, if you applied the chain rule, this is going to be the derivative of the outside function, our f of x with respect to the inside function, so, or the derivative of f of g of x with respect to g of x. So it's going to be negative, we'll bring that negative out front, so we're essentially just gonna use the power rule here."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now this, in this situation, you would use a combination of the power rule and the chain rule. You'd say, okay, my g of x is two x minus five, and f of g of x is going to be this whole expression. And so if you did, if you applied the chain rule, this is going to be the derivative of the outside function, our f of x with respect to the inside function, so, or the derivative of f of g of x with respect to g of x. So it's going to be negative, we'll bring that negative out front, so we're essentially just gonna use the power rule here. Negative two x minus five to the negative two. And then we multiply that times the derivative of, times the derivative of the inside function. So the inside function's derivative, the derivative of two x is two, derivative of negative five is zero, so it's gonna be times two."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be negative, we'll bring that negative out front, so we're essentially just gonna use the power rule here. Negative two x minus five to the negative two. And then we multiply that times the derivative of, times the derivative of the inside function. So the inside function's derivative, the derivative of two x is two, derivative of negative five is zero, so it's gonna be times two. And of course you can simplify it, so it's negative two times all of this business. Let me do one more example here, just to hit the point home. And once again, there isn't a must way, there isn't a way that you have to do this, but just let you appreciate that there's multiple ways to approach these types of derivatives."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the inside function's derivative, the derivative of two x is two, derivative of negative five is zero, so it's gonna be times two. And of course you can simplify it, so it's negative two times all of this business. Let me do one more example here, just to hit the point home. And once again, there isn't a must way, there isn't a way that you have to do this, but just let you appreciate that there's multiple ways to approach these types of derivatives. So let's say someone said, take the derivative of two x plus one squared. Pause the video and think about how you would do that. Well one way to do it is just to apply the chain rule just like we just did."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, there isn't a must way, there isn't a way that you have to do this, but just let you appreciate that there's multiple ways to approach these types of derivatives. So let's say someone said, take the derivative of two x plus one squared. Pause the video and think about how you would do that. Well one way to do it is just to apply the chain rule just like we just did. So you could say, all right, it's gonna be the derivative of the outside with respect to the inside. So it's gonna be two times two x plus one to the first power, taking one less than that, times the derivative of the inside, which is just going to be two. And so this is going to be equal to four times two x plus one which is equal to, if we wanna distribute the four, we could say it's eight x plus four."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well one way to do it is just to apply the chain rule just like we just did. So you could say, all right, it's gonna be the derivative of the outside with respect to the inside. So it's gonna be two times two x plus one to the first power, taking one less than that, times the derivative of the inside, which is just going to be two. And so this is going to be equal to four times two x plus one which is equal to, if we wanna distribute the four, we could say it's eight x plus four. That's a completely legitimate way of doing it. Now there are other ways of doing it. You could expand out two x plus one squared."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to four times two x plus one which is equal to, if we wanna distribute the four, we could say it's eight x plus four. That's a completely legitimate way of doing it. Now there are other ways of doing it. You could expand out two x plus one squared. You could say, hey, this is the same thing as the derivative with respect to x of two x squared is going to be four x squared, and then two times the product of these terms is gonna be plus four x plus one. And now you would just apply the power rule. So a little bit of extra algebra up front, but you can just go straight forward with the power rule, and you're gonna get this exact same thing."}, {"video_title": "Manipulating functions before differentiation   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could expand out two x plus one squared. You could say, hey, this is the same thing as the derivative with respect to x of two x squared is going to be four x squared, and then two times the product of these terms is gonna be plus four x plus one. And now you would just apply the power rule. So a little bit of extra algebra up front, but you can just go straight forward with the power rule, and you're gonna get this exact same thing. So the whole takeaway here is, pause, look at your expression. See if there's a way to simplify it. And it's especially a good thing if you can get out of using the quotient rule because that sometimes is just hard to know or remember."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And at first it could look intimidating. We have a sine of x here, we have a cosine of x, we have this crazy expression here, we have a pi over a cube root of x, we're squaring the whole thing, and at first it might seem intimidating. But as we'll see in this video, we can actually do this with the tools already in our toolkit, using our existing derivative properties, using what we know about the power rule, which tells us that the derivative with respect to x of x to the n is equal to n times x to the n minus one. We've seen that multiple times. We also need to use the fact that the derivative of cosine of x is equal to negative sine of x. And the other way around, the derivative with respect to x of sine of x is equal to positive cosine of x. So using just that, we can actually evaluate this, or evaluate g prime of x."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've seen that multiple times. We also need to use the fact that the derivative of cosine of x is equal to negative sine of x. And the other way around, the derivative with respect to x of sine of x is equal to positive cosine of x. So using just that, we can actually evaluate this, or evaluate g prime of x. So pause the video and see if you can do it. So probably the most intimidating part of this, because we know the derivative of sine of x and cosine of x is this expression here, and we can just rewrite this, or simplify it a little bit, so it takes a form that you might be a little bit more familiar with. So, actually let me just do this on the side here."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So using just that, we can actually evaluate this, or evaluate g prime of x. So pause the video and see if you can do it. So probably the most intimidating part of this, because we know the derivative of sine of x and cosine of x is this expression here, and we can just rewrite this, or simplify it a little bit, so it takes a form that you might be a little bit more familiar with. So, actually let me just do this on the side here. So pi, pi over the cube root of x squared. Well that's the same thing. This is equal to pi squared over the cube root of x squared."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, actually let me just do this on the side here. So pi, pi over the cube root of x squared. Well that's the same thing. This is equal to pi squared over the cube root of x squared. This is just exponent properties that we're dealing with. And so this is the same thing. We're gonna take x to the 1 3rd power, and then raise that to the 2nd power."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is equal to pi squared over the cube root of x squared. This is just exponent properties that we're dealing with. And so this is the same thing. We're gonna take x to the 1 3rd power, and then raise that to the 2nd power. So this is equal to pi squared over, let me write it this way, I'm not gonna skip any steps, because this is a good review of exponent properties. X to the 1 3rd squared, which is the same thing as pi squared over x to the 2 3rd power, which is the same thing as pi squared times x to the negative 2 3rd power. So when you write it like this, it starts to get into a form, you're like, oh, I can see how the power rule could apply there."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna take x to the 1 3rd power, and then raise that to the 2nd power. So this is equal to pi squared over, let me write it this way, I'm not gonna skip any steps, because this is a good review of exponent properties. X to the 1 3rd squared, which is the same thing as pi squared over x to the 2 3rd power, which is the same thing as pi squared times x to the negative 2 3rd power. So when you write it like this, it starts to get into a form, you're like, oh, I can see how the power rule could apply there. So this thing is just pi squared times x to the negative 2 3rd power. So actually let me delete this. So this thing can be rewritten, this thing can be rewritten as pi squared times x to the negative 2 3rd power."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when you write it like this, it starts to get into a form, you're like, oh, I can see how the power rule could apply there. So this thing is just pi squared times x to the negative 2 3rd power. So actually let me delete this. So this thing can be rewritten, this thing can be rewritten as pi squared times x to the negative 2 3rd power. So now let's take the derivative of each of these, each of these pieces of this expression. So we're gonna take, we wanna evaluate what g prime of x is. So g prime of x is going to be equal to, you could view it as a derivative with respect to x of seven sine of x."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this thing can be rewritten, this thing can be rewritten as pi squared times x to the negative 2 3rd power. So now let's take the derivative of each of these, each of these pieces of this expression. So we're gonna take, we wanna evaluate what g prime of x is. So g prime of x is going to be equal to, you could view it as a derivative with respect to x of seven sine of x. So we could take, let's do the derivative operator on both sides here, just to make it clear what we're doing. So we're gonna apply it there, we're gonna apply it there, and we're going to apply it there. So this derivative, this is the same thing as this is going to be seven times the derivative of sine of x, so this is just going to be seven times cosine of x."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g prime of x is going to be equal to, you could view it as a derivative with respect to x of seven sine of x. So we could take, let's do the derivative operator on both sides here, just to make it clear what we're doing. So we're gonna apply it there, we're gonna apply it there, and we're going to apply it there. So this derivative, this is the same thing as this is going to be seven times the derivative of sine of x, so this is just going to be seven times cosine of x. This one over here, this is going to be three, or we're subtracting, so it's gonna be this minus, we can bring the constant out that we're multiplying the expression by, and the derivative of cosine of x, so it's minus three times, the derivative of cosine of x is negative sine of x, negative sine of x. And then finally, here in the yellow, we just apply the power rule. So we have the negative 2 3rds, actually let's not forget this minus sign, I'm gonna write it out here."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this derivative, this is the same thing as this is going to be seven times the derivative of sine of x, so this is just going to be seven times cosine of x. This one over here, this is going to be three, or we're subtracting, so it's gonna be this minus, we can bring the constant out that we're multiplying the expression by, and the derivative of cosine of x, so it's minus three times, the derivative of cosine of x is negative sine of x, negative sine of x. And then finally, here in the yellow, we just apply the power rule. So we have the negative 2 3rds, actually let's not forget this minus sign, I'm gonna write it out here. So you have the negative 2 3rds, you multiply the exponent times the coefficient, it might look confusing, pi squared, but that's just a number. So it's gonna be negative, and then you have negative 2 3rds times pi squared, times pi squared, times x to the negative 2 3rds minus one power. Negative 2 3rds minus one power."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have the negative 2 3rds, actually let's not forget this minus sign, I'm gonna write it out here. So you have the negative 2 3rds, you multiply the exponent times the coefficient, it might look confusing, pi squared, but that's just a number. So it's gonna be negative, and then you have negative 2 3rds times pi squared, times pi squared, times x to the negative 2 3rds minus one power. Negative 2 3rds minus one power. So what is this going to be? So we get g prime of x is equal to, is equal to seven cosine of x, and let's see, we have a negative three times a negative sine of x, so that's a positive three sine of x. And then, we're subtracting, but then this is going to be a negative, so that's going to be a positive."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative 2 3rds minus one power. So what is this going to be? So we get g prime of x is equal to, is equal to seven cosine of x, and let's see, we have a negative three times a negative sine of x, so that's a positive three sine of x. And then, we're subtracting, but then this is going to be a negative, so that's going to be a positive. So we could say plus two pi squared over three, two pi squared over three, that's that part there, times x to the, so negative 2 3rds minus one, we could say negative one and 2 3rds, or we could say negative 5 3rds power. Negative 5 3rds power. And there you have it, we were able to tackle this thing that looked a little bit hairy, but all we had to use was the power rule and what we knew to be the derivatives of sine and cosine."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is equal to three, the value of the function is six, f of three is six, you could view it that way, h of three is zero, f prime of three is six, and h prime of three is four. And now they want us to evaluate the derivative with respect to x of the product of f of x and h of x when x is equal to three. So one way you could view this is, if we viewed some function, if we viewed some function g of x, g of x as being equal to the product of f of x and h of x, this expression is the derivative of g of x. So we could write g prime of x is equal to the derivative with respect to x of f of x times h of x, which is what we see right here which is what we want to evaluate at x equals three. So we essentially want to evaluate g prime of three. This is what they're asking us to do. Well, to do that, let's go first up here."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could write g prime of x is equal to the derivative with respect to x of f of x times h of x, which is what we see right here which is what we want to evaluate at x equals three. So we essentially want to evaluate g prime of three. This is what they're asking us to do. Well, to do that, let's go first up here. Let's just think about what they're doing. They're asking us to take the derivative with respect to x of the product of two functions that we have some information about. Well, if we're taking the derivative of the product of two functions, you could imagine that the product rule could prove useful here."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, to do that, let's go first up here. Let's just think about what they're doing. They're asking us to take the derivative with respect to x of the product of two functions that we have some information about. Well, if we're taking the derivative of the product of two functions, you could imagine that the product rule could prove useful here. So I'm just gonna restate the product rule. This is going to be equal to the derivative of the first function, f prime of x, times the second function, not taking its derivative, plus the first function, not taking its derivative, f of x, times the derivative of the second function, h prime of x. So if you're trying to find g prime of three, well, that's just going to be f prime of three times h of three plus f of three times h prime of three."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if we're taking the derivative of the product of two functions, you could imagine that the product rule could prove useful here. So I'm just gonna restate the product rule. This is going to be equal to the derivative of the first function, f prime of x, times the second function, not taking its derivative, plus the first function, not taking its derivative, f of x, times the derivative of the second function, h prime of x. So if you're trying to find g prime of three, well, that's just going to be f prime of three times h of three plus f of three times h prime of three. And lucky for us, they give us what all of these things evaluate to. f prime of three, right over here, they tell us. f prime, when x is equal to three, is equal to six."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you're trying to find g prime of three, well, that's just going to be f prime of three times h of three plus f of three times h prime of three. And lucky for us, they give us what all of these things evaluate to. f prime of three, right over here, they tell us. f prime, when x is equal to three, is equal to six. So this right over here is six. H of three, they give us that too. H of three, when x is three, the value of our function is zero."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "f prime, when x is equal to three, is equal to six. So this right over here is six. H of three, they give us that too. H of three, when x is three, the value of our function is zero. So this is zero. So this first term, you just get six times zero, which is gonna be zero, but we'll get to that. Now, f of three."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "H of three, when x is three, the value of our function is zero. So this is zero. So this first term, you just get six times zero, which is gonna be zero, but we'll get to that. Now, f of three. F of three, well, the function, when x is equal to three, y, or the f of three is equal to six. So that is six. And then finally, the h prime evaluated at three, h prime of x, when x is equal to three, h prime of x is equal to four."}, {"video_title": "Worked example Product rule with table   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, f of three. F of three, well, the function, when x is equal to three, y, or the f of three is equal to six. So that is six. And then finally, the h prime evaluated at three, h prime of x, when x is equal to three, h prime of x is equal to four. Or you could say this is h prime of three. So this is four. And so there you have it."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say we've got some function f that is continuous over the interval between c and d. And the reason why I'm using c and d instead of a and b is so I can use a and b for later. And let's say we set up some function capital F of x, which is defined as the area under the curve between c and some value x, where x is in this interval where f is continuous, under the curve, so it's the area under the curve between c and x. So if this is x right over here, under the curve, f of t, dt. So this right over here, f of x, is that area. That right over there is what f of x is. Now the fundamental theorem of calculus tells us that if f is continuous over this interval, then f of x is differentiable at every x in the interval and the derivative, the derivative of capital F of x. And let me be clear."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here, f of x, is that area. That right over there is what f of x is. Now the fundamental theorem of calculus tells us that if f is continuous over this interval, then f of x is differentiable at every x in the interval and the derivative, the derivative of capital F of x. And let me be clear. Capital F of x is differentiable at every possible x between c and d, and the derivative of capital F of x is going to be equal to lowercase f of x. Fair enough. Now what I wanna do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me be clear. Capital F of x is differentiable at every possible x between c and d, and the derivative of capital F of x is going to be equal to lowercase f of x. Fair enough. Now what I wanna do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. So let's think about, let's think about what f of, f of b, what f of b minus f of a is, what this is, where both b and a are also in this interval. So f of b, and we're gonna assume that b is larger than a. So let's say that b is this right over here."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what I wanna do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. So let's think about, let's think about what f of, f of b, what f of b minus f of a is, what this is, where both b and a are also in this interval. So f of b, and we're gonna assume that b is larger than a. So let's say that b is this right over here. Let me do that in the same color. So let's say that b is right over here. F of b is going to be equal to, we just literally replaced the b where you see the x, it's going to be equal to the definite integral between c and b of f of t, dt, which is just another way of saying, it's just another way of saying is the area under the curve between c and b."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that b is this right over here. Let me do that in the same color. So let's say that b is right over here. F of b is going to be equal to, we just literally replaced the b where you see the x, it's going to be equal to the definite integral between c and b of f of t, dt, which is just another way of saying, it's just another way of saying is the area under the curve between c and b. So this f of b, capital F of b, is all of this business, all of this business right over here. And from that, we are going to want to subtract, we are going to subtract capital F of a, which is just the integral between c and lowercase a of lowercase f of t, dt. So let's say that this is, let's say that this is a right over here."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "F of b is going to be equal to, we just literally replaced the b where you see the x, it's going to be equal to the definite integral between c and b of f of t, dt, which is just another way of saying, it's just another way of saying is the area under the curve between c and b. So this f of b, capital F of b, is all of this business, all of this business right over here. And from that, we are going to want to subtract, we are going to subtract capital F of a, which is just the integral between c and lowercase a of lowercase f of t, dt. So let's say that this is, let's say that this is a right over here. Capital F of a is just literally the area between c and a under the curve, lowercase f of t. So it's this right over here. It's all of this business, all of this right over here. So what is, if you have this blue area, which is all of this, and you subtract out this red, this magenta area, what are you left with?"}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that this is, let's say that this is a right over here. Capital F of a is just literally the area between c and a under the curve, lowercase f of t. So it's this right over here. It's all of this business, all of this right over here. So what is, if you have this blue area, which is all of this, and you subtract out this red, this magenta area, what are you left with? Well, you're left with this green area. You're left with this green area right over here. And how would we represent that?"}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what is, if you have this blue area, which is all of this, and you subtract out this red, this magenta area, what are you left with? Well, you're left with this green area. You're left with this green area right over here. And how would we represent that? How would we denote that? Well, we could denote that as the definite integral between a and b of f of t, of f of t, dt. And there you have it."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And how would we represent that? How would we denote that? Well, we could denote that as the definite integral between a and b of f of t, of f of t, dt. And there you have it. This right over here is the second fundamental theorem of calculus. It tells us that this, if f is continuous on the interval, that this is going to be equal to the antiderivative or an antiderivative of f. And we see right over here that capital F is the antiderivative of f. So we could view this as capital F antiderivative. This is how we define capital F. The antiderivative, or we didn't define it that way, but the fundamental theorem of calculus tells us that capital F is an antiderivative of lowercase f. So right over here, this tells you, if you have a definite integral like this, it completely equivalent to an antiderivative of it evaluated at f and from that, or evaluated at b, and from that, you subtract it, evaluated at a."}, {"video_title": "Fundamental theorem of calculus (Part 2)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there you have it. This right over here is the second fundamental theorem of calculus. It tells us that this, if f is continuous on the interval, that this is going to be equal to the antiderivative or an antiderivative of f. And we see right over here that capital F is the antiderivative of f. So we could view this as capital F antiderivative. This is how we define capital F. The antiderivative, or we didn't define it that way, but the fundamental theorem of calculus tells us that capital F is an antiderivative of lowercase f. So right over here, this tells you, if you have a definite integral like this, it completely equivalent to an antiderivative of it evaluated at f and from that, or evaluated at b, and from that, you subtract it, evaluated at a. So normally it looks like this. I just switched the order. The definite integral from a to b of f of t dt is equal to an antiderivative of f, so capital F evaluated at b, and from that, the antiderivative, from that, subtract the antiderivative evaluated at a."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of the material for the cheapest container. So let's draw this open storage container, this open rectangular storage container. So it's going to have an open top. So let me draw its open top as good as I can."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Find the cost of the material for the cheapest container. So let's draw this open storage container, this open rectangular storage container. So it's going to have an open top. So let me draw its open top as good as I can. So it's going to have an open top. That's the top of my container. And then let me draw the sides just like that."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw its open top as good as I can. So it's going to have an open top. That's the top of my container. And then let me draw the sides just like that. So it might look something like that. And then I could draw, and since it's open top, I can see through. I can see the inside of the container as well."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let me draw the sides just like that. So it might look something like that. And then I could draw, and since it's open top, I can see through. I can see the inside of the container as well. So the container would look something like that. And so what do they tell us? They tell us that the volume needs to be 10 cubic meters."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I can see the inside of the container as well. So the container would look something like that. And so what do they tell us? They tell us that the volume needs to be 10 cubic meters. So let me write that down. The volume needs to be equal to 10 meters cubed. The length of its base is twice the width."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "They tell us that the volume needs to be 10 cubic meters. So let me write that down. The volume needs to be equal to 10 meters cubed. The length of its base is twice the width. So the length, let's call this the width x. So the length is going to be twice that. It's going to be 2x."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The length of its base is twice the width. So the length, let's call this the width x. So the length is going to be twice that. It's going to be 2x. That's what they tell us right over here. They tell us material for the base costs $10 per square meter. So the base, so this area right over here, if I was transparent, I could continue to draw it down here."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be 2x. That's what they tell us right over here. They tell us material for the base costs $10 per square meter. So the base, so this area right over here, if I was transparent, I could continue to draw it down here. But this right over here, that material costs $10 per square meter. Let me label that. $10 per square meter."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the base, so this area right over here, if I was transparent, I could continue to draw it down here. But this right over here, that material costs $10 per square meter. Let me label that. $10 per square meter. And then they say material for the sides costs $6 per square meter. So the material over here costs $6 per meter squared. So let's see if we can come up with a value or how much this box would cost to make as a function of x."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "$10 per square meter. And then they say material for the sides costs $6 per square meter. So the material over here costs $6 per meter squared. So let's see if we can come up with a value or how much this box would cost to make as a function of x. But x only gives us the dimensions of the base. We also need a dimension for height. So it'll be a function of x and height for now."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can come up with a value or how much this box would cost to make as a function of x. But x only gives us the dimensions of the base. We also need a dimension for height. So it'll be a function of x and height for now. So let's write h as the height right over here. So what is the cost of this container going to be? So the cost is going to be equal to the cost of the base."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it'll be a function of x and height for now. So let's write h as the height right over here. So what is the cost of this container going to be? So the cost is going to be equal to the cost of the base. Well, the cost of the base is going to be $10 times, I'll just write 10. It's just going to be 10 times the area of the base. Well, what's the area of the base?"}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the cost is going to be equal to the cost of the base. Well, the cost of the base is going to be $10 times, I'll just write 10. It's just going to be 10 times the area of the base. Well, what's the area of the base? Well, it's going to be the width times the length. So 10 times x times 2x. That is the cost of base."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's the area of the base? Well, it's going to be the width times the length. So 10 times x times 2x. That is the cost of base. And now what's going to be the cost of the sides? Well, the different sides are going to have different dimensions. You have this side over here and this side over here, which have the same dimension."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is the cost of base. And now what's going to be the cost of the sides? Well, the different sides are going to have different dimensions. You have this side over here and this side over here, which have the same dimension. They both have an area of x times h. You have x times h. And then our material is $6 per square meter. So 6 times x times h would be the cost of one of these side panels. But so for two of them, we have to multiply by 2."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have this side over here and this side over here, which have the same dimension. They both have an area of x times h. You have x times h. And then our material is $6 per square meter. So 6 times x times h would be the cost of one of these side panels. But so for two of them, we have to multiply by 2. So plus 2 times 6 times h. And then we have these two side panels. We have this side panel right over here. And we have this side panel right over here."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But so for two of them, we have to multiply by 2. So plus 2 times 6 times h. And then we have these two side panels. We have this side panel right over here. And we have this side panel right over here. The area of each of them is going to be 2x times h. So it's going to be 2x times h. The cost of the material is going to be 6. So the cost of one of the panels is going to be $6 per square meters times 2xh meters squared. But we have two of these panels, one panel and two panels."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we have this side panel right over here. The area of each of them is going to be 2x times h. So it's going to be 2x times h. The cost of the material is going to be 6. So the cost of one of the panels is going to be $6 per square meters times 2xh meters squared. But we have two of these panels, one panel and two panels. So we have to multiply by 2. And so we will get, so this is right over here. This is the cost of the sides."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we have two of these panels, one panel and two panels. So we have to multiply by 2. And so we will get, so this is right over here. This is the cost of the sides. And so let's see if we can simplify this. And I'll write it all in a neutral color. So this is going to be equal to 10."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the cost of the sides. And so let's see if we can simplify this. And I'll write it all in a neutral color. So this is going to be equal to 10. Let's see, 10 times 2 is 20. x times x is x squared. And you have 2 times 6 times xh. So this is going to be plus 12xh."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to 10. Let's see, 10 times 2 is 20. x times x is x squared. And you have 2 times 6 times xh. So this is going to be plus 12xh. And then this is going to be 2 times 6, which is 12 times 2 is 24xh. Plus 24xh. So this is going to be equal to 20x squared plus 36xh."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be plus 12xh. And then this is going to be 2 times 6, which is 12 times 2 is 24xh. Plus 24xh. So this is going to be equal to 20x squared plus 36xh. So this is going to be my cost. But I'm not ready to optimize it yet. We don't know how to optimize with respect to two variables."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to 20x squared plus 36xh. So this is going to be my cost. But I'm not ready to optimize it yet. We don't know how to optimize with respect to two variables. We only know how to optimize with respect to one variable. And maybe I'll say, let's optimize with respect to x. But if we want to optimize with respect to x, we have to express h as a function of x."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know how to optimize with respect to two variables. We only know how to optimize with respect to one variable. And maybe I'll say, let's optimize with respect to x. But if we want to optimize with respect to x, we have to express h as a function of x. So how can we do that? How can we express h as a function of x? Well, we know that the volume has to be 10 cubic meters."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if we want to optimize with respect to x, we have to express h as a function of x. So how can we do that? How can we express h as a function of x? Well, we know that the volume has to be 10 cubic meters. So we know that x, the width times the length times 2x times the height times h needs to be equal to 10. Or another way of saying that, this tells us that 2x squared times h needs to be equal to 10. And so if we want h as a function of x, we just divide both sides by 2x squared."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we know that the volume has to be 10 cubic meters. So we know that x, the width times the length times 2x times the height times h needs to be equal to 10. Or another way of saying that, this tells us that 2x squared times h needs to be equal to 10. And so if we want h as a function of x, we just divide both sides by 2x squared. And we get h is equal to 10 over 2x squared. Or we could say that h is equal to 5 over x squared. And then we can substitute back right over here."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we want h as a function of x, we just divide both sides by 2x squared. And we get h is equal to 10 over 2x squared. Or we could say that h is equal to 5 over x squared. And then we can substitute back right over here. h is equal to 5 over x squared. So all of this business is going to be equal to 20 times x squared plus 36 times x times 5 over x squared. So our cost as a function of x is going to be 20x squared."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can substitute back right over here. h is equal to 5 over x squared. So all of this business is going to be equal to 20 times x squared plus 36 times x times 5 over x squared. So our cost as a function of x is going to be 20x squared. 36 times 5. Let's see, 30 times 5 is 150. Plus another 30 is going to be 180."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So our cost as a function of x is going to be 20x squared. 36 times 5. Let's see, 30 times 5 is 150. Plus another 30 is going to be 180. So it's going to be plus 180 times x times x to the negative 2, 180x to the negative 1 power. So we finally have cost as a function of x. Now we're ready to optimize."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus another 30 is going to be 180. So it's going to be plus 180 times x times x to the negative 2, 180x to the negative 1 power. So we finally have cost as a function of x. Now we're ready to optimize. To optimize, we just have to figure out what are the critical points here and whether those critical points are minimum or maximum value. So let's see what we can do. So to find a critical point, we take the derivative, figure out where the derivative is undefined equal to 0."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we're ready to optimize. To optimize, we just have to figure out what are the critical points here and whether those critical points are minimum or maximum value. So let's see what we can do. So to find a critical point, we take the derivative, figure out where the derivative is undefined equal to 0. And those are our candidate critical points. And then from the critical points we find, they might be minimum or maximum values. So the derivative of c of our cost with respect to x is going to be equal to 40 times x minus 180 times x to the negative 2 power."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So to find a critical point, we take the derivative, figure out where the derivative is undefined equal to 0. And those are our candidate critical points. And then from the critical points we find, they might be minimum or maximum values. So the derivative of c of our cost with respect to x is going to be equal to 40 times x minus 180 times x to the negative 2 power. Now this seems, well, it's defined for all x except for x equaling 0. But x equaling 0 is not interesting to us as a critical point because then we're going to have a degenerate. This is going to have no base at all."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of c of our cost with respect to x is going to be equal to 40 times x minus 180 times x to the negative 2 power. Now this seems, well, it's defined for all x except for x equaling 0. But x equaling 0 is not interesting to us as a critical point because then we're going to have a degenerate. This is going to have no base at all. So we don't want to worry about that critical point. We would have no volume at all. So it would not work out."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to have no base at all. So we don't want to worry about that critical point. We would have no volume at all. So it would not work out. And actually, if x equals 0, then our height is undefined as well. So let's see. So this was defined for everything else, for anything other than x equals 0."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would not work out. And actually, if x equals 0, then our height is undefined as well. So let's see. So this was defined for everything else, for anything other than x equals 0. So let's see when this derivative is equal to 0 in our search for potential critical points. So when does? I'll do it over here."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this was defined for everything else, for anything other than x equals 0. So let's see when this derivative is equal to 0 in our search for potential critical points. So when does? I'll do it over here. When does 40x minus 180x to the negative 2 equal 0? Well, we could add the 180x to the negative 2 to both sides. We get 40x is equal to 180."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll do it over here. When does 40x minus 180x to the negative 2 equal 0? Well, we could add the 180x to the negative 2 to both sides. We get 40x is equal to 180. And I could write it as 180 over x squared. Now let's see. We could multiply both sides of this equation by x squared."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We get 40x is equal to 180. And I could write it as 180 over x squared. Now let's see. We could multiply both sides of this equation by x squared. And we would get 40x to the third is equal to 180. Divide both sides by 40. You get x to the third is equal to 180 over 40, which is the same thing as 18 over 4, which is the same thing as 9 over 2."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could multiply both sides of this equation by x squared. And we would get 40x to the third is equal to 180. Divide both sides by 40. You get x to the third is equal to 180 over 40, which is the same thing as 18 over 4, which is the same thing as 9 over 2. And so if we want to solve for x, we get that x is equal to a critical point. We get a critical point at x is equal to 9 halves to the 1 third power, the cube root of 9 halves. So let's see."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You get x to the third is equal to 180 over 40, which is the same thing as 18 over 4, which is the same thing as 9 over 2. And so if we want to solve for x, we get that x is equal to a critical point. We get a critical point at x is equal to 9 halves to the 1 third power, the cube root of 9 halves. So let's see. Let's get an approximate value for what that is. So if we take 9 halves, 9 divided by 2, I guess we could call that 4.5. And we want to raise it to the 1 third power."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see. Let's get an approximate value for what that is. So if we take 9 halves, 9 divided by 2, I guess we could call that 4.5. And we want to raise it to the 1 third power. We get 1.65. So it's approximately equal to 1.65 as our critical point. Now the way the problem is asked, we're only getting one legitimate critical point here."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we want to raise it to the 1 third power. We get 1.65. So it's approximately equal to 1.65 as our critical point. Now the way the problem is asked, we're only getting one legitimate critical point here. So that's probably going to be the x at which we achieve a minimum value. But let's use our second derivative test just in case to make sure that we're definitely concave upwards over here, in which case that this will definitely be the x value at which we achieve a minimum value. So the second derivative of our cost function is just the derivative of this, which is going to be equal to 40 minus 180 times negative 2, which is negative 360."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now the way the problem is asked, we're only getting one legitimate critical point here. So that's probably going to be the x at which we achieve a minimum value. But let's use our second derivative test just in case to make sure that we're definitely concave upwards over here, in which case that this will definitely be the x value at which we achieve a minimum value. So the second derivative of our cost function is just the derivative of this, which is going to be equal to 40 minus 180 times negative 2, which is negative 360. So it's going to be plus 360 over x to the third. The derivative of this is negative 2 times negative 180, which is positive 360, x to the negative 3 power, which is exactly this right over here. So when x is equal to 1.65, this is going to be positive."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the second derivative of our cost function is just the derivative of this, which is going to be equal to 40 minus 180 times negative 2, which is negative 360. So it's going to be plus 360 over x to the third. The derivative of this is negative 2 times negative 180, which is positive 360, x to the negative 3 power, which is exactly this right over here. So when x is equal to 1.65, this is going to be positive. This is going to be positive. So let me write this down. c prime prime of 1.65 is definitely greater than 0."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x is equal to 1.65, this is going to be positive. This is going to be positive. So let me write this down. c prime prime of 1.65 is definitely greater than 0. So we're definitely concave upwards when x is 1.65, concave upwards, which means that our graph is going to look something like this. And so where the derivative equal to 0, which is right over there, we are at a minimum point. We are minimizing our cost."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "c prime prime of 1.65 is definitely greater than 0. So we're definitely concave upwards when x is 1.65, concave upwards, which means that our graph is going to look something like this. And so where the derivative equal to 0, which is right over there, we are at a minimum point. We are minimizing our cost. And so if we go back to the question, the only thing that we have to do now, we know the x value that minimizes our cost. We now have to find the cost of the material for the cheapest container. So we just have to figure out what our cost is."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are minimizing our cost. And so if we go back to the question, the only thing that we have to do now, we know the x value that minimizes our cost. We now have to find the cost of the material for the cheapest container. So we just have to figure out what our cost is. And we already know what our cost is as a function of x. So we just have to put 1.65 into this equation, evaluate the function at 1.65. So let's do that."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to figure out what our cost is. And we already know what our cost is as a function of x. So we just have to put 1.65 into this equation, evaluate the function at 1.65. So let's do that. We get our cost is going to be equal to 20 times 1.65. I should say approximately equal to, because I'm using an approximation of this original value. 1.65 squared plus 180."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. We get our cost is going to be equal to 20 times 1.65. I should say approximately equal to, because I'm using an approximation of this original value. 1.65 squared plus 180. I could say divided by 1.65. That's the same thing as multiplying by 1.65 to the negative 1. So divided by 1.65, which is equal to $163.5."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "1.65 squared plus 180. I could say divided by 1.65. That's the same thing as multiplying by 1.65 to the negative 1. So divided by 1.65, which is equal to $163.5. So it's approximately. So the cost when, let me do this in a new color. We deserve a drum roll now."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So divided by 1.65, which is equal to $163.5. So it's approximately. So the cost when, let me do this in a new color. We deserve a drum roll now. The cost when x is 1.65 is approximately equal to $163.54, which is quite an expensive box. So this is fairly expensive material here, although it's a fairly large box. 1.65 meters in width."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We deserve a drum roll now. The cost when x is 1.65 is approximately equal to $163.54, which is quite an expensive box. So this is fairly expensive material here, although it's a fairly large box. 1.65 meters in width. It's going to be twice that in length. And then you could figure out what its height is going to be, although it's not going to be too tall. 5 divided by 1.65 squared."}, {"video_title": "Optimization cost of materials   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "1.65 meters in width. It's going to be twice that in length. And then you could figure out what its height is going to be, although it's not going to be too tall. 5 divided by 1.65 squared. I don't know. It'll be roughly a little under 2 meters tall. So it actually is quite a large box made out of quite expensive material."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to create a solid of revolution, but we're not going to do it by rotating this around the x or the y-axis. Instead, we're going to rotate it around another somewhat arbitrary line. And in this case, I will rotate it around the line y is equal to 1. So let's say that this right over here is the line y equals 1. So this is what we're going to rotate it around, y equals 1. So the first thing we want to do is visualize what we're doing. And we actually care about the interval."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that this right over here is the line y equals 1. So this is what we're going to rotate it around, y equals 1. So the first thing we want to do is visualize what we're doing. And we actually care about the interval. So let's say that the interval is between this point right over here, where the two points intersect. And let's say between that and x is equal to 4. So let's say that this right over here, so this is x equals 4."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we actually care about the interval. So let's say that the interval is between this point right over here, where the two points intersect. And let's say between that and x is equal to 4. So let's say that this right over here, so this is x equals 4. So it's going to be just right over here. So this is the interval that we're rotating. We're going to rotate it around y equals 1, not around the x-axis."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that this right over here, so this is x equals 4. So it's going to be just right over here. So this is the interval that we're rotating. We're going to rotate it around y equals 1, not around the x-axis. So what would our figure look like? Well, we're going to rotate it around like this. We're going to rotate it around something like this."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to rotate it around y equals 1, not around the x-axis. So what would our figure look like? Well, we're going to rotate it around like this. We're going to rotate it around something like this. And so your figure is going to look, I guess you could call it, it'll almost look like a cone head if you view it on the side, or something like a bullet, but not quite looking like a bullet. But it would be a shape that looks something like that. So hopefully we can visualize this."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to rotate it around something like this. And so your figure is going to look, I guess you could call it, it'll almost look like a cone head if you view it on the side, or something like a bullet, but not quite looking like a bullet. But it would be a shape that looks something like that. So hopefully we can visualize this. We've done this several times already, attempting to visualize these shapes. But let's think about how we could actually figure out the volume of this solid of revolution. Well, let's just think about it disk by disk."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So hopefully we can visualize this. We've done this several times already, attempting to visualize these shapes. But let's think about how we could actually figure out the volume of this solid of revolution. Well, let's just think about it disk by disk. So let's construct a disk right over here. And we've done this many, many, many times already, where we essentially want the volume of each of these disks, and then we're going to take this for each of these x's in our interval, and then we're going to take the sum of the volumes of all of the disks and we're going to find the volume. We're going to stack them all up, or I guess put them next to each other in this case, and find the volume of our entire figure."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's just think about it disk by disk. So let's construct a disk right over here. And we've done this many, many, many times already, where we essentially want the volume of each of these disks, and then we're going to take this for each of these x's in our interval, and then we're going to take the sum of the volumes of all of the disks and we're going to find the volume. We're going to stack them all up, or I guess put them next to each other in this case, and find the volume of our entire figure. So to figure out the volume of each disk, we just have to figure out the area of its face, and I'll do that in purple. We just have to figure out the area of the face and multiply it by the depth. So what's the area of this face?"}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to stack them all up, or I guess put them next to each other in this case, and find the volume of our entire figure. So to figure out the volume of each disk, we just have to figure out the area of its face, and I'll do that in purple. We just have to figure out the area of the face and multiply it by the depth. So what's the area of this face? Well, it's going to be pi times the radius of the face squared. Now what's the radius of the face here? Well, it's not just square root of x."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's the area of this face? Well, it's going to be pi times the radius of the face squared. Now what's the radius of the face here? Well, it's not just square root of x. Square root of x would tell us the distance between the x-axis and our function. It's now square root of x minus 1. This length right over here is square root of x minus 1 for any given x in our interval."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's not just square root of x. Square root of x would tell us the distance between the x-axis and our function. It's now square root of x minus 1. This length right over here is square root of x minus 1 for any given x in our interval. So it's going to be equal to square root of... Let me do that same orange color just to make it clear where I'm getting it from. It's going to be equal to square root of x minus 1. It's essentially our function minus what we are rotating around."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This length right over here is square root of x minus 1 for any given x in our interval. So it's going to be equal to square root of... Let me do that same orange color just to make it clear where I'm getting it from. It's going to be equal to square root of x minus 1. It's essentially our function minus what we are rotating around. That gives us the radius, and so this will give us the area of each of our faces, and then we just multiply that times the depth. The depth, of course, is dx. We've done that multiple times, so times dx."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's essentially our function minus what we are rotating around. That gives us the radius, and so this will give us the area of each of our faces, and then we just multiply that times the depth. The depth, of course, is dx. We've done that multiple times, so times dx. This is the thing that we want to sum up. This is the thing that we want to sum up over our interval. Our interval, let's see, this point right over here where the square root of x is equal to 1, that's just going to be x equals 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've done that multiple times, so times dx. This is the thing that we want to sum up. This is the thing that we want to sum up over our interval. Our interval, let's see, this point right over here where the square root of x is equal to 1, that's just going to be x equals 1. This is just x equals 1 over here. We said that we would do this all the way until x equals 4. This was kind of our end of our interval."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our interval, let's see, this point right over here where the square root of x is equal to 1, that's just going to be x equals 1. This is just x equals 1 over here. We said that we would do this all the way until x equals 4. This was kind of our end of our interval. This is until x is equal to... Let me be careful. This is the x-axis right over here. This is the x-axis right over here, all the way until x equals 4."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This was kind of our end of our interval. This is until x is equal to... Let me be careful. This is the x-axis right over here. This is the x-axis right over here, all the way until x equals 4. You get confused. This is the x-axis. We're going from x equals 1 to x equals 4, and we're essentially taking this area, one way to think about it, between our square root of x and 1, and we're rotating it around 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the x-axis right over here, all the way until x equals 4. You get confused. This is the x-axis. We're going from x equals 1 to x equals 4, and we're essentially taking this area, one way to think about it, between our square root of x and 1, and we're rotating it around 1. We're using these little disks right over here, so the interval between 1 and 4. This is going to give us the volume of our solid of revolution. Now we just have to evaluate this definite integral."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going from x equals 1 to x equals 4, and we're essentially taking this area, one way to think about it, between our square root of x and 1, and we're rotating it around 1. We're using these little disks right over here, so the interval between 1 and 4. This is going to give us the volume of our solid of revolution. Now we just have to evaluate this definite integral. Let's give a shot at it. This is going to be equal to the integral from 1 to 4, and we can factor out, or we can put the pi outside the integral sign, and then we can expand this out. Square root of x squared."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we just have to evaluate this definite integral. Let's give a shot at it. This is going to be equal to the integral from 1 to 4, and we can factor out, or we can put the pi outside the integral sign, and then we can expand this out. Square root of x squared. All we're going to do is expand this binomial. Square root of x minus 1 times square root of x minus 1. Square root of x times square root of x is x."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Square root of x squared. All we're going to do is expand this binomial. Square root of x minus 1 times square root of x minus 1. Square root of x times square root of x is x. Square root of x times negative 1 is negative square root of x. Negative 1 times square root of x is another negative square root of x. The negative 1 times negative 1 is equal to positive 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Square root of x times square root of x is x. Square root of x times negative 1 is negative square root of x. Negative 1 times square root of x is another negative square root of x. The negative 1 times negative 1 is equal to positive 1. This part right over here will simplify to x minus 2 square roots of x. This is minus 2 square roots of x. Minus 2 square roots of x, and then you have plus 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "The negative 1 times negative 1 is equal to positive 1. This part right over here will simplify to x minus 2 square roots of x. This is minus 2 square roots of x. Minus 2 square roots of x, and then you have plus 1. Then all of that times dx. This is going to be equal to, let's put our pi out there, the antiderivative of this business. The antiderivative of x is x squared over 2."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Minus 2 square roots of x, and then you have plus 1. Then all of that times dx. This is going to be equal to, let's put our pi out there, the antiderivative of this business. The antiderivative of x is x squared over 2. The antiderivative, so minus 2 times the antiderivative of square root of x. Square root of x is just x to the 1 half power, so we increment the power by 1. It's going to be x to the 3 half power times 2 thirds."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of x is x squared over 2. The antiderivative, so minus 2 times the antiderivative of square root of x. Square root of x is just x to the 1 half power, so we increment the power by 1. It's going to be x to the 3 half power times 2 thirds. We could say it's going to be times 2 thirds x to the 3 halves. Let me make it clear. This negative 2 right here is this negative 2."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be x to the 3 half power times 2 thirds. We could say it's going to be times 2 thirds x to the 3 halves. Let me make it clear. This negative 2 right here is this negative 2. This expression right over here is the antiderivative of square root of x. You can verify this. The antiderivative of this 3 halves times 2 thirds is 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This negative 2 right here is this negative 2. This expression right over here is the antiderivative of square root of x. You can verify this. The antiderivative of this 3 halves times 2 thirds is 1. Decrement the 3 halves, you get x to the 1 half. Now let's take the antiderivative of 1. That's just going to be equal to x."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of this 3 halves times 2 thirds is 1. Decrement the 3 halves, you get x to the 1 half. Now let's take the antiderivative of 1. That's just going to be equal to x. We're going to evaluate this from 1 to 4. We're in the home stretch. This is going to be equal to pi times, first let's do it at 4."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's just going to be equal to x. We're going to evaluate this from 1 to 4. We're in the home stretch. This is going to be equal to pi times, first let's do it at 4. See if 4 squared, I'll write it all out, is 1 squared over 2 minus 4 to the 3, let me do this part first, minus 4 thirds, 4 thirds, and that was 4 to the 3 halves. 4 to the 1 half is 2, and then you raise that to the 3rd power, you get 8. Times 8 plus 4, and then you subtract out all of this stuff evaluated at 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be equal to pi times, first let's do it at 4. See if 4 squared, I'll write it all out, is 1 squared over 2 minus 4 to the 3, let me do this part first, minus 4 thirds, 4 thirds, and that was 4 to the 3 halves. 4 to the 1 half is 2, and then you raise that to the 3rd power, you get 8. Times 8 plus 4, and then you subtract out all of this stuff evaluated at 1. 1 squared over 2, that's just going to be minus 1 half. Then we're subtracting this now. Actually, I don't want to skip too many steps."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Times 8 plus 4, and then you subtract out all of this stuff evaluated at 1. 1 squared over 2, that's just going to be minus 1 half. Then we're subtracting this now. Actually, I don't want to skip too many steps. Let's do, it's going to be this. This is when we evaluated at 4, and then we're going to subtract when we evaluate this whole thing at 1. When we evaluate it at 1, we do this in this green color."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, I don't want to skip too many steps. Let's do, it's going to be this. This is when we evaluated at 4, and then we're going to subtract when we evaluate this whole thing at 1. When we evaluate it at 1, we do this in this green color. That's not the green color I thought I was going to do. When you evaluate it at 1, you get 1 squared over 2, which is 1 half, and then you get 1 to the 3 halves, which is just 1. This becomes minus 4 thirds, and then you have plus 1."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "When we evaluate it at 1, we do this in this green color. That's not the green color I thought I was going to do. When you evaluate it at 1, you get 1 squared over 2, which is 1 half, and then you get 1 to the 3 halves, which is just 1. This becomes minus 4 thirds, and then you have plus 1. Let's simplify this, and I'll do it all in the same color now. This is going to be equal to pi times 4 squared over 2, that is 16 over 2, which is equal to 8. Then you have 4 times 8, which is 32 over 3, so minus 32 over 3 plus 4."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This becomes minus 4 thirds, and then you have plus 1. Let's simplify this, and I'll do it all in the same color now. This is going to be equal to pi times 4 squared over 2, that is 16 over 2, which is equal to 8. Then you have 4 times 8, which is 32 over 3, so minus 32 over 3 plus 4. Then you have minus 1 half, we're just distributing the negative, plus 4 thirds, plus 4 thirds, and then you have minus 1. Now we just have to add up a bunch of fractions to simplify this thing. What do we get?"}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then you have 4 times 8, which is 32 over 3, so minus 32 over 3 plus 4. Then you have minus 1 half, we're just distributing the negative, plus 4 thirds, plus 4 thirds, and then you have minus 1. Now we just have to add up a bunch of fractions to simplify this thing. What do we get? This is going to be equal to pi times, see our least common multiple looks like 6. We're going to put everything over a denominator of 6. 8 is the same thing as 48 over 6."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "What do we get? This is going to be equal to pi times, see our least common multiple looks like 6. We're going to put everything over a denominator of 6. 8 is the same thing as 48 over 6. 32 over 3 is the same thing as 64 over 6, so minus 64 over 6. 4 is the same thing as 24 over 6. 1 half is the same thing as 3 over 6, so this is negative 3 over 6, the same thing as negative 1 half."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "8 is the same thing as 48 over 6. 32 over 3 is the same thing as 64 over 6, so minus 64 over 6. 4 is the same thing as 24 over 6. 1 half is the same thing as 3 over 6, so this is negative 3 over 6, the same thing as negative 1 half. 4 thirds is the same thing as 8 over 6, and then negative 1 is the same thing as negative 6 over 6. A little bit of arithmetic here, and let's see what we get. If we take 48 and we subtract 64 from 48, we get negative 16, is that right?"}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "1 half is the same thing as 3 over 6, so this is negative 3 over 6, the same thing as negative 1 half. 4 thirds is the same thing as 8 over 6, and then negative 1 is the same thing as negative 6 over 6. A little bit of arithmetic here, and let's see what we get. If we take 48 and we subtract 64 from 48, we get negative 16, is that right? Yes, we get negative 16. Then if we add 24 to negative 16, you get positive 8. Positive 8 minus 3 is 5."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we take 48 and we subtract 64 from 48, we get negative 16, is that right? Yes, we get negative 16. Then if we add 24 to negative 16, you get positive 8. Positive 8 minus 3 is 5. 5 plus 8 is 13. 13 minus 6 is 7. This whole numerator simplifies to 7, so we get 7 pi over 6 as our volume."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Positive 8 minus 3 is 5. 5 plus 8 is 13. 13 minus 6 is 7. This whole numerator simplifies to 7, so we get 7 pi over 6 as our volume. Let me just verify I did that right. This is going to be negative 16. We get to positive 8."}, {"video_title": "Disc method rotation around horizontal line   AP Calculus AB   Khan Academy.mp3", "Sentence": "This whole numerator simplifies to 7, so we get 7 pi over 6 as our volume. Let me just verify I did that right. This is going to be negative 16. We get to positive 8. Positive 8 plus positive 8 is 16. 16 minus 9 is, yes, it is indeed 7. We get 7 pi over 6, and we're done."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the graph of y is equal to f of x. This is the graph of y is equal to g of x. And we already know something, or we know ways to represent the area under the curve y equals f of x between these two points, x is equal to a and x equal to b. So this area right over here, this area right over here, between the curve and the x axis, between x equals a and x equals b, we know we can write that as the definite integral from a to b of f of x dx. And we can do the same thing over here. We could call this area, let me pick a color that I have not used. Well, this is a slightly different green."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this area right over here, this area right over here, between the curve and the x axis, between x equals a and x equals b, we know we can write that as the definite integral from a to b of f of x dx. And we can do the same thing over here. We could call this area, let me pick a color that I have not used. Well, this is a slightly different green. I could call this area right over here, the area under the curve y is equal to g of x, and above the positive x axis between x equals a and x equals b, we could call that the definite integral from a to b of g of x dx. Now, given these two things, let's actually think about the area under the curve of the function created by the sum of these two functions. So what do I mean by that?"}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is a slightly different green. I could call this area right over here, the area under the curve y is equal to g of x, and above the positive x axis between x equals a and x equals b, we could call that the definite integral from a to b of g of x dx. Now, given these two things, let's actually think about the area under the curve of the function created by the sum of these two functions. So what do I mean by that? So let me, this is actually a fun thing to do. Let me start again. That's exactly what we have over here."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what do I mean by that? So let me, this is actually a fun thing to do. Let me start again. That's exactly what we have over here. This is the graph of y is equal to f of x. But what I want to do is I want to approximate the graph of y is equal to, so my goal is to graph y is equal to f of x, f of x plus g of x, plus g of x, plus g of x. So for any given x, it's going to be f of x, so that's the f of x, and I'm going to add the g of x to it."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's exactly what we have over here. This is the graph of y is equal to f of x. But what I want to do is I want to approximate the graph of y is equal to, so my goal is to graph y is equal to f of x, f of x plus g of x, plus g of x, plus g of x. So for any given x, it's going to be f of x, so that's the f of x, and I'm going to add the g of x to it. So what would that look like? So that's going to look like, let's see, at when x is zero, g of x looks like it's about that length right there. I'm obviously approximating it, so I'm going to have to add that length right over here, so it'd probably be right around there."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for any given x, it's going to be f of x, so that's the f of x, and I'm going to add the g of x to it. So what would that look like? So that's going to look like, let's see, at when x is zero, g of x looks like it's about that length right there. I'm obviously approximating it, so I'm going to have to add that length right over here, so it'd probably be right around there. At x equals a, it's a little bit more, but now my f of x curve has gone more, has increased, but if I take that same distance above it, if I add the g of x there, it gets me right about there. Once again, I'm just eyeballing it, trying to get an approximation. Give you an intuition, actually, for what f of x plus g of x is."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm obviously approximating it, so I'm going to have to add that length right over here, so it'd probably be right around there. At x equals a, it's a little bit more, but now my f of x curve has gone more, has increased, but if I take that same distance above it, if I add the g of x there, it gets me right about there. Once again, I'm just eyeballing it, trying to get an approximation. Give you an intuition, actually, for what f of x plus g of x is. I'm just trying to add g of x for a given x. Now let's see, if I'm a little bit, let's say that I'm between a and b, g of x is about that distance right over there. So if I wanted to put that same distance right over here, it gets me right about, it gets me right about there."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Give you an intuition, actually, for what f of x plus g of x is. I'm just trying to add g of x for a given x. Now let's see, if I'm a little bit, let's say that I'm between a and b, g of x is about that distance right over there. So if I wanted to put that same distance right over here, it gets me right about, it gets me right about there. And then when x is equal to b, g of x is about that big, so I have to add that length, which is about, which looks something like that. That actually looks like a little bit too much. Maybe something like that."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I wanted to put that same distance right over here, it gets me right about, it gets me right about there. And then when x is equal to b, g of x is about that big, so I have to add that length, which is about, which looks something like that. That actually looks like a little bit too much. Maybe something like that. So if I were to add the two, if I were to add the two, I get a curve that looks something like this. That looks something like this, and maybe it just keeps on going higher and higher. So this is the curve, or it's a pretty good approximation of the curve of f of x plus g of x."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe something like that. So if I were to add the two, if I were to add the two, I get a curve that looks something like this. That looks something like this, and maybe it just keeps on going higher and higher. So this is the curve, or it's a pretty good approximation of the curve of f of x plus g of x. Now an interesting question is, is what would be, well, we know how we can represent this area. So the area under the curve, f of x plus g of x, above the positive x-axis, between x equals a and x equals b, we know we can represent that as, let me see, I have not used pink yet. So this area right over here, we know that that could be represented as the definite integral from a to b of f of x plus g of x dx."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the curve, or it's a pretty good approximation of the curve of f of x plus g of x. Now an interesting question is, is what would be, well, we know how we can represent this area. So the area under the curve, f of x plus g of x, above the positive x-axis, between x equals a and x equals b, we know we can represent that as, let me see, I have not used pink yet. So this area right over here, we know that that could be represented as the definite integral from a to b of f of x plus g of x dx. Now the question is, how does this thing relate to d, or how does this area relate to these areas right over here? Well, the important thing to realize is this area that we have in yellow, that's going to be this area right over here. That one's pretty clear."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this area right over here, we know that that could be represented as the definite integral from a to b of f of x plus g of x dx. Now the question is, how does this thing relate to d, or how does this area relate to these areas right over here? Well, the important thing to realize is this area that we have in yellow, that's going to be this area right over here. That one's pretty clear. But how does this area in green relate to this area there? And to think about that, we just have to think about what does an integral mean? Remember, what does it represent?"}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "That one's pretty clear. But how does this area in green relate to this area there? And to think about that, we just have to think about what does an integral mean? Remember, what does it represent? We've already thought about these like these really small rectangles. We're taking the sum of an infinitely, or the limit as we get an infinite number of these infinitely thin rectangles. But when we're thinking about Riemann sums, we're thinking about, okay, we have some change in x, and then you multiply it times essentially the height, which is going to be the value of the function at that point."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, what does it represent? We've already thought about these like these really small rectangles. We're taking the sum of an infinitely, or the limit as we get an infinite number of these infinitely thin rectangles. But when we're thinking about Riemann sums, we're thinking about, okay, we have some change in x, and then you multiply it times essentially the height, which is going to be the value of the function at that point. Well, over here, you could have the same change in x, you could have the exact same change in x, and what is the height right over here? Well, that's going to be this exact height right over here. We saw that when we constructed it."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But when we're thinking about Riemann sums, we're thinking about, okay, we have some change in x, and then you multiply it times essentially the height, which is going to be the value of the function at that point. Well, over here, you could have the same change in x, you could have the exact same change in x, and what is the height right over here? Well, that's going to be this exact height right over here. We saw that when we constructed it. This is going to be the g of x at that value. So even though the rectangles look like they're kind of shifted around a little bit, and they're actually all shifted up by the f of x, the heights of these rectangles that I'm drawing right over here are exactly the same thing as the heights of the rectangles that I'm drawing over here. They, once again, they are all just shifted up and down by this f of x function, but these are the exact same rectangles, or they have the exact same heights."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "We saw that when we constructed it. This is going to be the g of x at that value. So even though the rectangles look like they're kind of shifted around a little bit, and they're actually all shifted up by the f of x, the heights of these rectangles that I'm drawing right over here are exactly the same thing as the heights of the rectangles that I'm drawing over here. They, once again, they are all just shifted up and down by this f of x function, but these are the exact same rectangles, or they have the exact same heights. And the limit as you get more and more of these by making them thinner and thinner is going to be the same as the limit as you get more and more of these as you get thinner and thinner. And so this area right over here, and I'm obviously not doing a rigorous proof, I'm giving you the intuition for it, is the exact same thing as this area right over here. So the area under this curve, the definite integral from a to b of f of x plus g of x dx, is just going to be the sum of these two, is just going to be the sum of these two definite integrals."}, {"video_title": "Integrating sums of functions   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "They, once again, they are all just shifted up and down by this f of x function, but these are the exact same rectangles, or they have the exact same heights. And the limit as you get more and more of these by making them thinner and thinner is going to be the same as the limit as you get more and more of these as you get thinner and thinner. And so this area right over here, and I'm obviously not doing a rigorous proof, I'm giving you the intuition for it, is the exact same thing as this area right over here. So the area under this curve, the definite integral from a to b of f of x plus g of x dx, is just going to be the sum of these two, is just going to be the sum of these two definite integrals. And you might say, oh, this is obvious, or maybe it's not so obvious, but when is this actually useful? Well, as you later actually learn to evaluate these integrals, you'll see that one of the most powerful ideas is being able to decompose them in this way, to say, okay, if I'm taking the definite integral from zero to one of x squared plus sine of x, which you may or may not have learned to do so far, you can at least start to break this down. You can say, okay, well, this is going to be the same thing as the integral from zero to one of x squared dx, plus the integral from zero to one of sine of x dx."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and see which of these are actually separable. Now, the way that I approach this is I try to solve for the derivative, and if when I solve for the derivative, if I get dy dx is equal to some function of y times some other function of x, then I say, okay, this is separable, because I could rewrite this as, I could divide both sides by g of y, and I get one over g of y, which is itself a function of y, times dy is equal to h of x dx. You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx, and then it's clear you have a separable equation, you can integrate both sides. But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually, I'm gonna write it this way, I'm gonna write it three minus y times one over x. And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually, I'm gonna write it this way, I'm gonna write it three minus y times one over x. And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly, this one right over here is separable. Now let's do the second one."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly, this one right over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides. I'm gonna add one to both sides."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides. I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y."}, {"video_title": "Worked example identifying separable equations   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y. And here, my algebraic toolkit of how do I separate x and y so I can write this as a function of x times a function of y, not obvious to me here. So this one is not separable. There's only the first one and the third one."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And of course you could always graph a function just by trying out a bunch of points. But we want to really focus on the points that are interesting to us and then just to get the general shape of the function, especially we want to focus on the things that we can take out from this function using our calculus toolkit or our derivative toolkit. So the first thing we probably want to do is figure out the critical points. We want to figure out, I'll write it here, critical points. And just as a refresher of what critical points means, it's the points where the derivative of f of x is 0. So critical points are f prime of x is either equal to 0 or it's undefined. This function looks differentiable everywhere, so the critical points that we're worried about are probably, well I can tell you, they're definitely just the points where f prime of x is going to be equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We want to figure out, I'll write it here, critical points. And just as a refresher of what critical points means, it's the points where the derivative of f of x is 0. So critical points are f prime of x is either equal to 0 or it's undefined. This function looks differentiable everywhere, so the critical points that we're worried about are probably, well I can tell you, they're definitely just the points where f prime of x is going to be equal to 0. This derivative, f prime of x, is going to actually be defined over the entire domain. So let's actually write down the derivative right now. So the derivative of this, f prime of x, this is pretty straightforward, the derivative of 3x to the fourth, 4 times 3 is 12."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This function looks differentiable everywhere, so the critical points that we're worried about are probably, well I can tell you, they're definitely just the points where f prime of x is going to be equal to 0. This derivative, f prime of x, is going to actually be defined over the entire domain. So let's actually write down the derivative right now. So the derivative of this, f prime of x, this is pretty straightforward, the derivative of 3x to the fourth, 4 times 3 is 12. 12x to the, we'll just decrement the 4 by 1, 3. You just multiply times the exponent and then decrease the new exponent by 1. Minus 3 times 4 is 12, times x to the, 1 less than 3 is 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So the derivative of this, f prime of x, this is pretty straightforward, the derivative of 3x to the fourth, 4 times 3 is 12. 12x to the, we'll just decrement the 4 by 1, 3. You just multiply times the exponent and then decrease the new exponent by 1. Minus 3 times 4 is 12, times x to the, 1 less than 3 is 2. And then the derivative of a constant, the slope of a constant you can almost imagine, is 0. It's not changing, a constant by definition isn't changing. So that's f prime of x."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Minus 3 times 4 is 12, times x to the, 1 less than 3 is 2. And then the derivative of a constant, the slope of a constant you can almost imagine, is 0. It's not changing, a constant by definition isn't changing. So that's f prime of x. So let's figure out the critical points. The critical points are where this thing is either going to be equal to 0 or it's undefined. Now, I could look over the entire domain of real numbers and this thing is defined pretty much anywhere."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's f prime of x. So let's figure out the critical points. The critical points are where this thing is either going to be equal to 0 or it's undefined. Now, I could look over the entire domain of real numbers and this thing is defined pretty much anywhere. I could put any number here and it's not going to blow up. It's going to give me an answer to what the function is. So it's defined everywhere, so let's just figure out where it's equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, I could look over the entire domain of real numbers and this thing is defined pretty much anywhere. I could put any number here and it's not going to blow up. It's going to give me an answer to what the function is. So it's defined everywhere, so let's just figure out where it's equal to 0. So f prime of x is equal to 0. So let's solve which x is, so let's solve, well I don't have to rewrite that, I just wrote that. Let's solve for whether this is equal to 0, and I'll do it in the same color."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So it's defined everywhere, so let's just figure out where it's equal to 0. So f prime of x is equal to 0. So let's solve which x is, so let's solve, well I don't have to rewrite that, I just wrote that. Let's solve for whether this is equal to 0, and I'll do it in the same color. So 12x to the 3rd minus 12x squared is equal to 0. And so let's see what we can do, what can we do to solve this. Well we could factor out a 12x."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's solve for whether this is equal to 0, and I'll do it in the same color. So 12x to the 3rd minus 12x squared is equal to 0. And so let's see what we can do, what can we do to solve this. Well we could factor out a 12x. So if we factor out a 12x, then this term becomes just x, and then actually let's factor out a 12x squared. We factor out a 12x squared, if we divide both of these by 12x squared, this term just becomes an x, and then minus 12x squared divided by 12x squared is just 1, is equal to 0. I just rewrote this top thing like this."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well we could factor out a 12x. So if we factor out a 12x, then this term becomes just x, and then actually let's factor out a 12x squared. We factor out a 12x squared, if we divide both of these by 12x squared, this term just becomes an x, and then minus 12x squared divided by 12x squared is just 1, is equal to 0. I just rewrote this top thing like this. You could go the other way, if I distributed this 12x squared times this entire quantity, you would get my derivative right there. So the reason why I did that is because to solve for 0, or if I want all of the x's that make this equation equal to 0, I now have written it in a form where I'm multiplying one thing by another thing. And in order for this to be 0, one or both of these things must be equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I just rewrote this top thing like this. You could go the other way, if I distributed this 12x squared times this entire quantity, you would get my derivative right there. So the reason why I did that is because to solve for 0, or if I want all of the x's that make this equation equal to 0, I now have written it in a form where I'm multiplying one thing by another thing. And in order for this to be 0, one or both of these things must be equal to 0. So 12x squared or equal to 0, which means that x is equal to 0 will make this quantity equal to 0. And then the other thing that would make this quantity 0 is if x minus 1 is equal to 0. So x minus 1 is equal to 0 when x is equal to 1."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And in order for this to be 0, one or both of these things must be equal to 0. So 12x squared or equal to 0, which means that x is equal to 0 will make this quantity equal to 0. And then the other thing that would make this quantity 0 is if x minus 1 is equal to 0. So x minus 1 is equal to 0 when x is equal to 1. So these are our two critical points. Our two critical points are x is equal to 0 and x is equal to 1. And remember, those are just the points where our first derivative is equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So x minus 1 is equal to 0 when x is equal to 1. So these are our two critical points. Our two critical points are x is equal to 0 and x is equal to 1. And remember, those are just the points where our first derivative is equal to 0. Where the slope is 0. They might be maximum points, they might be minimum points, they might be inflection points, we don't know. They might be, you know, if this was a constant function, they could just be anything."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And remember, those are just the points where our first derivative is equal to 0. Where the slope is 0. They might be maximum points, they might be minimum points, they might be inflection points, we don't know. They might be, you know, if this was a constant function, they could just be anything. So we really can't say a lot about them just yet, but they are points of interest. I guess that's all we can say, that they are definitely points of interest. But let's keep going and let's try to understand the concativity and maybe we can get a better sense of this graph."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "They might be, you know, if this was a constant function, they could just be anything. So we really can't say a lot about them just yet, but they are points of interest. I guess that's all we can say, that they are definitely points of interest. But let's keep going and let's try to understand the concativity and maybe we can get a better sense of this graph. So let's figure out the second derivative. So the second, I'll do that in, let me do it in this orange color. So the second derivative of my function f, let's see, 3 times 12 is 36x squared minus 24x."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But let's keep going and let's try to understand the concativity and maybe we can get a better sense of this graph. So let's figure out the second derivative. So the second, I'll do that in, let me do it in this orange color. So the second derivative of my function f, let's see, 3 times 12 is 36x squared minus 24x. So let's see, well there's a couple of things we can do. Now that we know the second derivative, we can answer the question, is my graph concave upwards or downwards at either of these points? So let's figure out what, at either of these critical points."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So the second derivative of my function f, let's see, 3 times 12 is 36x squared minus 24x. So let's see, well there's a couple of things we can do. Now that we know the second derivative, we can answer the question, is my graph concave upwards or downwards at either of these points? So let's figure out what, at either of these critical points. And it'll all fit together. Remember, if it's concave upwards, then we're kind of in a U shape. If it's concave downwards, then we're in a kind of upside down U shape."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's figure out what, at either of these critical points. And it'll all fit together. Remember, if it's concave upwards, then we're kind of in a U shape. If it's concave downwards, then we're in a kind of upside down U shape. So f prime prime, our second derivative at x is equal to 0 is equal to what? It's equal to 36 zero squared minus 24 times 0. So that's just 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "If it's concave downwards, then we're in a kind of upside down U shape. So f prime prime, our second derivative at x is equal to 0 is equal to what? It's equal to 36 zero squared minus 24 times 0. So that's just 0. So f prime prime, so it's just equal to 0. So we're in neither concave upwards nor concave downwards here. It might be a transition point, it may not."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's just 0. So f prime prime, so it's just equal to 0. So we're in neither concave upwards nor concave downwards here. It might be a transition point, it may not. If it is a transition point, then we're dealing with an inflection point. We're not sure yet. Now let's see what f prime prime, our second derivative, evaluated at 1 is."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It might be a transition point, it may not. If it is a transition point, then we're dealing with an inflection point. We're not sure yet. Now let's see what f prime prime, our second derivative, evaluated at 1 is. So that's 36 times 1. Let me write it down. That's equal to 36 times 1 squared, which is just 36, minus 24 times 1."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Now let's see what f prime prime, our second derivative, evaluated at 1 is. So that's 36 times 1. Let me write it down. That's equal to 36 times 1 squared, which is just 36, minus 24 times 1. So it's 36 minus 24, so it's equal to 12. So this is positive. Our second derivative is positive here."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That's equal to 36 times 1 squared, which is just 36, minus 24 times 1. So it's 36 minus 24, so it's equal to 12. So this is positive. Our second derivative is positive here. It's equal to 12, which means our first derivative, our slope, is increasing. The rate of change of our slope is positive here. So at this point, right here, we are concave upwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Our second derivative is positive here. It's equal to 12, which means our first derivative, our slope, is increasing. The rate of change of our slope is positive here. So at this point, right here, we are concave upwards. Which tells me that this is probably a minimum point. The slope is 0 here, but we are concave upwards at that point. So that's interesting."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So at this point, right here, we are concave upwards. Which tells me that this is probably a minimum point. The slope is 0 here, but we are concave upwards at that point. So that's interesting. So let's see if there are any other potential inflection points here. We already know that this is a potential inflection point. Let me circle it in red."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's interesting. So let's see if there are any other potential inflection points here. We already know that this is a potential inflection point. Let me circle it in red. It's a potential inflection point. We don't know whether our function actually transitions at that point. We'll have to experiment a little bit to see if that's really the case."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me circle it in red. It's a potential inflection point. We don't know whether our function actually transitions at that point. We'll have to experiment a little bit to see if that's really the case. But let's see if there are any other inflection points, or potential inflection points. So let's see if this equals 0 anywhere else. So 36x squared minus 24x is equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We'll have to experiment a little bit to see if that's really the case. But let's see if there are any other inflection points, or potential inflection points. So let's see if this equals 0 anywhere else. So 36x squared minus 24x is equal to 0. Let's solve for x. Let us factor out 12x times 3x. 3x times 12x is 36x squared."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So 36x squared minus 24x is equal to 0. Let's solve for x. Let us factor out 12x times 3x. 3x times 12x is 36x squared. Minus 2 is equal to 0. These two are equivalent expressions. If you multiply this out, you'll get this thing up here."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "3x times 12x is 36x squared. Minus 2 is equal to 0. These two are equivalent expressions. If you multiply this out, you'll get this thing up here. So this thing is going to be equal to 0 either if 12x is equal to 0. So 12x is equal to 0. That gives us x is equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "If you multiply this out, you'll get this thing up here. So this thing is going to be equal to 0 either if 12x is equal to 0. So 12x is equal to 0. That gives us x is equal to 0. So at x equals 0, this thing equals 0. So the second derivative is 0 there. And we already knew that because we tested that number out."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That gives us x is equal to 0. So at x equals 0, this thing equals 0. So the second derivative is 0 there. And we already knew that because we tested that number out. Or this thing. If this expression was 0, then the entire second derivative would also be 0. So let's write that."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And we already knew that because we tested that number out. Or this thing. If this expression was 0, then the entire second derivative would also be 0. So let's write that. So 3x minus 2 is equal to 0. 3x is equal to 2. Just adding 2 to both sides."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's write that. So 3x minus 2 is equal to 0. 3x is equal to 2. Just adding 2 to both sides. 3x is equal to 2 thirds. So this is another interesting point that we haven't really hit upon before. That might be an inflection point."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Just adding 2 to both sides. 3x is equal to 2 thirds. So this is another interesting point that we haven't really hit upon before. That might be an inflection point. And the reason why I say it might be is because the second derivative is definitely 0 here. You put 2 thirds here, you're going to get 0. What we have to do is see whether the second derivative is positive or negative on either sides of 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That might be an inflection point. And the reason why I say it might be is because the second derivative is definitely 0 here. You put 2 thirds here, you're going to get 0. What we have to do is see whether the second derivative is positive or negative on either sides of 2 thirds. And we already have a sense of that. We could try out a couple of numbers. We know that if we take x is greater than 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "What we have to do is see whether the second derivative is positive or negative on either sides of 2 thirds. And we already have a sense of that. We could try out a couple of numbers. We know that if we take x is greater than 2 thirds. Let me scroll down a little bit just so we have some space. So let's see what happens. When x is greater than 2 thirds, what is f prime prime?"}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We know that if we take x is greater than 2 thirds. Let me scroll down a little bit just so we have some space. So let's see what happens. When x is greater than 2 thirds, what is f prime prime? What is the second derivative? So let's try out a value that's pretty close just to get a sense of things. So let me rewrite it."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "When x is greater than 2 thirds, what is f prime prime? What is the second derivative? So let's try out a value that's pretty close just to get a sense of things. So let me rewrite it. f prime prime of x is equal to 12x times 3x minus 2. So if x is greater than 2 thirds, this term right here is going to be positive. That's definitely, you know, any positive number times 12 is going to be positive."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me rewrite it. f prime prime of x is equal to 12x times 3x minus 2. So if x is greater than 2 thirds, this term right here is going to be positive. That's definitely, you know, any positive number times 12 is going to be positive. But what about this term right here? 3 times 2 thirds minus 2 is exactly 0. That's 2 minus 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That's definitely, you know, any positive number times 12 is going to be positive. But what about this term right here? 3 times 2 thirds minus 2 is exactly 0. That's 2 minus 2. But anything larger than that, 3 times, you know, if I had 2.1 thirds here, this is going to be a positive quantity. Any value of x greater than 2 thirds will make this thing right here positive. This thing is also going to be positive."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That's 2 minus 2. But anything larger than that, 3 times, you know, if I had 2.1 thirds here, this is going to be a positive quantity. Any value of x greater than 2 thirds will make this thing right here positive. This thing is also going to be positive. So that means that when x is greater than 2 thirds, that tells us that the second derivative is positive. It is greater than 0. So in our domain, as long as x is larger than 2 thirds, we are concave upwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This thing is also going to be positive. So that means that when x is greater than 2 thirds, that tells us that the second derivative is positive. It is greater than 0. So in our domain, as long as x is larger than 2 thirds, we are concave upwards. And we saw that here. At x is equal to 1, we were concave upwards. But what about x being less than 2 thirds?"}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So in our domain, as long as x is larger than 2 thirds, we are concave upwards. And we saw that here. At x is equal to 1, we were concave upwards. But what about x being less than 2 thirds? So when x is less than 2 thirds, let me write it, let me scroll down a little bit. When x is less than 2 thirds, what's going on? I'll rewrite it."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But what about x being less than 2 thirds? So when x is less than 2 thirds, let me write it, let me scroll down a little bit. When x is less than 2 thirds, what's going on? I'll rewrite it. f prime prime of x, second derivative, 12x times 3x minus 2. Well, if we go, you know, really far less, we're going to get a negative number here. And this might be negative."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I'll rewrite it. f prime prime of x, second derivative, 12x times 3x minus 2. Well, if we go, you know, really far less, we're going to get a negative number here. And this might be negative. But let's see if we just go really, like, you know, right below 2 thirds. We're still in the positive domain. So, you know, if this was like 1.9 thirds, which is, you know, a mixture of a decimal and a fraction, or even 1 third, this thing is still going to be positive."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And this might be negative. But let's see if we just go really, like, you know, right below 2 thirds. We're still in the positive domain. So, you know, if this was like 1.9 thirds, which is, you know, a mixture of a decimal and a fraction, or even 1 third, this thing is still going to be positive. Right below 2 thirds, this thing is still going to be positive. We're going to be multiplying 12 by a positive number. But what's going to be going on, what's going on right here?"}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So, you know, if this was like 1.9 thirds, which is, you know, a mixture of a decimal and a fraction, or even 1 third, this thing is still going to be positive. Right below 2 thirds, this thing is still going to be positive. We're going to be multiplying 12 by a positive number. But what's going to be going on, what's going on right here? At 2 thirds, we're exactly 0. But as you go, anything less than 2 thirds, 3 times 1 third is only 1. 1 minus 2, you're going to get negative numbers."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But what's going to be going on, what's going on right here? At 2 thirds, we're exactly 0. But as you go, anything less than 2 thirds, 3 times 1 third is only 1. 1 minus 2, you're going to get negative numbers. So when x is less than 2 thirds, this thing right here is going to be negative. So the second derivative, if x is less than 2 thirds, the second derivative right to the left, or right when you go less than 2 thirds, the second derivative of x is less than 0. Now the fact that we have this transition from when we're less than 2 thirds, we have a negative second derivative, and when we're greater than 2 thirds, we have a positive second derivative, that tells us that this indeed is an inflection point."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "1 minus 2, you're going to get negative numbers. So when x is less than 2 thirds, this thing right here is going to be negative. So the second derivative, if x is less than 2 thirds, the second derivative right to the left, or right when you go less than 2 thirds, the second derivative of x is less than 0. Now the fact that we have this transition from when we're less than 2 thirds, we have a negative second derivative, and when we're greater than 2 thirds, we have a positive second derivative, that tells us that this indeed is an inflection point. That x is equal to 2 thirds is definitely an inflection point for our original function up here. Now we have one more candidate inflection point, and then we're ready to graph. Then once you know all the inflection points and the maximum and minimum, you are ready to graph the function."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Now the fact that we have this transition from when we're less than 2 thirds, we have a negative second derivative, and when we're greater than 2 thirds, we have a positive second derivative, that tells us that this indeed is an inflection point. That x is equal to 2 thirds is definitely an inflection point for our original function up here. Now we have one more candidate inflection point, and then we're ready to graph. Then once you know all the inflection points and the maximum and minimum, you are ready to graph the function. So let's see if x is equal to 0 is an inflection point. We know that the second derivative is 0 at 0, but what happens above and below the second derivative? So let me do our little test here."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Then once you know all the inflection points and the maximum and minimum, you are ready to graph the function. So let's see if x is equal to 0 is an inflection point. We know that the second derivative is 0 at 0, but what happens above and below the second derivative? So let me do our little test here. Let me draw a line so we don't get confused with all of the stuff that I wrote here. So when x is greater than 0, what's happening to the second derivative? Remember, the second derivative was equal to 12x times 3x minus 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me do our little test here. Let me draw a line so we don't get confused with all of the stuff that I wrote here. So when x is greater than 0, what's happening to the second derivative? Remember, the second derivative was equal to 12x times 3x minus 2. I like writing it this way because you've kind of decomposed it into two linear expressions, and you can see whether each of them are positive or negative. So if x is greater than 0, this thing right here is definitely going to be positive. And then this thing right here, right when you go right above x is greater than 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Remember, the second derivative was equal to 12x times 3x minus 2. I like writing it this way because you've kind of decomposed it into two linear expressions, and you can see whether each of them are positive or negative. So if x is greater than 0, this thing right here is definitely going to be positive. And then this thing right here, right when you go right above x is greater than 0. So you have to make sure to be very close to this number. So if this number is like point 1, so you're right above 0. So this isn't going to be true for all of x greater than 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then this thing right here, right when you go right above x is greater than 0. So you have to make sure to be very close to this number. So if this number is like point 1, so you're right above 0. So this isn't going to be true for all of x greater than 0. We just want to test exactly what happens right when we go right above 0. So if this was point 1, you would have point 3 minus 2. That would be a negative number."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this isn't going to be true for all of x greater than 0. We just want to test exactly what happens right when we go right above 0. So if this was point 1, you would have point 3 minus 2. That would be a negative number. So right as x goes right above 0, this thing right here is negative. So as x is greater than 0, you will have your second derivative is going to be less than 0, your concave downwards, which makes sense because at some point we're going to be hitting a transition. Remember, we were concave downwards before we got to 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That would be a negative number. So right as x goes right above 0, this thing right here is negative. So as x is greater than 0, you will have your second derivative is going to be less than 0, your concave downwards, which makes sense because at some point we're going to be hitting a transition. Remember, we were concave downwards before we got to 2 thirds. So this is consistent. From 0 to 2 thirds, we are concave downwards. And then at 2 thirds, we become concave upwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Remember, we were concave downwards before we got to 2 thirds. So this is consistent. From 0 to 2 thirds, we are concave downwards. And then at 2 thirds, we become concave upwards. Now let's see what happens when x is right less than, when x is just barely less than 0. So once again, f prime, the second derivative of x is equal to 12x times 3x minus 2. Well, right if x was minus point 1 or minus point 001, no matter what, this thing is going to be negative."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then at 2 thirds, we become concave upwards. Now let's see what happens when x is right less than, when x is just barely less than 0. So once again, f prime, the second derivative of x is equal to 12x times 3x minus 2. Well, right if x was minus point 1 or minus point 001, no matter what, this thing is going to be negative. This expression right here is going to be negative. The 12x, right, you have some negative value here times 12, it's going to be negative. And then what's this going to be?"}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, right if x was minus point 1 or minus point 001, no matter what, this thing is going to be negative. This expression right here is going to be negative. The 12x, right, you have some negative value here times 12, it's going to be negative. And then what's this going to be? Well, 3 times minus point 1 is going to be minus point 3. Minus 2 is minus 2 point 3. You're definitely going to have a negative."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then what's this going to be? Well, 3 times minus point 1 is going to be minus point 3. Minus 2 is minus 2 point 3. You're definitely going to have a negative. This value right here is going to be negative. And then when you subtract from a negative, it's definitely going to be negative. So that is also going to be negative."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "You're definitely going to have a negative. This value right here is going to be negative. And then when you subtract from a negative, it's definitely going to be negative. So that is also going to be negative. But if you multiply a negative times a negative, you're going to get a positive. So actually, right below x is less than 0, the second derivative is positive. Now, this all might have been a little bit confusing, but we should now have the payoff."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that is also going to be negative. But if you multiply a negative times a negative, you're going to get a positive. So actually, right below x is less than 0, the second derivative is positive. Now, this all might have been a little bit confusing, but we should now have the payoff. We now have the payoff. We have all of the interesting things going on. We know that at x is equal to 1, let me write it over here."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, this all might have been a little bit confusing, but we should now have the payoff. We now have the payoff. We have all of the interesting things going on. We know that at x is equal to 1, let me write it over here. We figured out at x is equal to 1, the slope is 0. So f prime prime is, sorry, let me write it this way. I should have said, we know that the slope is 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We know that at x is equal to 1, let me write it over here. We figured out at x is equal to 1, the slope is 0. So f prime prime is, sorry, let me write it this way. I should have said, we know that the slope is 0. Slope is equal to 0, and we figured that out because the first derivative was 0. This was a critical point. And we know that we're dealing with, the function is concave upwards at this point."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I should have said, we know that the slope is 0. Slope is equal to 0, and we figured that out because the first derivative was 0. This was a critical point. And we know that we're dealing with, the function is concave upwards at this point. And that tells us that this is going to be a minimum point. So what is f of, and we should actually get the coordinates so we can actually graph it. That was the whole point of this video."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And we know that we're dealing with, the function is concave upwards at this point. And that tells us that this is going to be a minimum point. So what is f of, and we should actually get the coordinates so we can actually graph it. That was the whole point of this video. And f of 1 is equal to what? f of 1, let's go back to our original function, is 3 times 1. 1 to the 4th is just 1."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That was the whole point of this video. And f of 1 is equal to what? f of 1, let's go back to our original function, is 3 times 1. 1 to the 4th is just 1. 3 times 1 minus 4 plus 2. So it's 3 times 1 minus 4 times 1, which is minus 1 plus 2. Well, that's just a positive 1."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "1 to the 4th is just 1. 3 times 1 minus 4 plus 2. So it's 3 times 1 minus 4 times 1, which is minus 1 plus 2. Well, that's just a positive 1. So f of 1 is 1. And then we know at x is equal to 0, we also figured out that the slope is equal to 0. But we figured out that this was an inflection point."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, that's just a positive 1. So f of 1 is 1. And then we know at x is equal to 0, we also figured out that the slope is equal to 0. But we figured out that this was an inflection point. The concativity switches before and after. So this is an inflection point. And we are concave below 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But we figured out that this was an inflection point. The concativity switches before and after. So this is an inflection point. And we are concave below 0. So when x is less than 0, we are upwards. Our second derivative is positive. And when x is greater than 0, we are downwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And we are concave below 0. So when x is less than 0, we are upwards. Our second derivative is positive. And when x is greater than 0, we are downwards. We are concave downwards. Right above. Not for all of the domain x and 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And when x is greater than 0, we are downwards. We are concave downwards. Right above. Not for all of the domain x and 0. Just right above 0. Downwards. And then what is f of 0?"}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Not for all of the domain x and 0. Just right above 0. Downwards. And then what is f of 0? Just so we know. We want to graph that point. f of 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then what is f of 0? Just so we know. We want to graph that point. f of 0. Let's see. f of 0, this is easy. 3 times 0 minus 4 times 0 plus 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "f of 0. Let's see. f of 0, this is easy. 3 times 0 minus 4 times 0 plus 2. Well, that's just 2. f of 0 is 2. And then finally, we got the point x is equal to 2 thirds. x is equal to, let me do that in another color."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "3 times 0 minus 4 times 0 plus 2. Well, that's just 2. f of 0 is 2. And then finally, we got the point x is equal to 2 thirds. x is equal to, let me do that in another color. We had the point x is equal to 2 thirds. We figured out that this was an inflection point. The slope definitely isn't 0 there."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "x is equal to, let me do that in another color. We had the point x is equal to 2 thirds. We figured out that this was an inflection point. The slope definitely isn't 0 there. Because it wasn't one of the critical points. And we know that we are downwards. We know that when x is less than 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope definitely isn't 0 there. Because it wasn't one of the critical points. And we know that we are downwards. We know that when x is less than 2 thirds. Or right less than 2 thirds. We are concave downwards. And when x is greater than 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We know that when x is less than 2 thirds. Or right less than 2 thirds. We are concave downwards. And when x is greater than 2 thirds. We saw it up here. When x was greater than 2 thirds. Right up here, we were concave upwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And when x is greater than 2 thirds. We saw it up here. When x was greater than 2 thirds. Right up here, we were concave upwards. The second derivative was positive. We were upwards. And we can actually figure out what's f of 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Right up here, we were concave upwards. The second derivative was positive. We were upwards. And we can actually figure out what's f of 2 thirds. That's actually a little bit complicated. We don't even have to figure that out to graph. I think we can do a pretty good job of graphing it."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And we can actually figure out what's f of 2 thirds. That's actually a little bit complicated. We don't even have to figure that out to graph. I think we can do a pretty good job of graphing it. Just with what we know right now. So that's our take away. Let me do a rough graph."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I think we can do a pretty good job of graphing it. Just with what we know right now. So that's our take away. Let me do a rough graph. Let's see. So let me do my axes. Like that."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do a rough graph. Let's see. So let me do my axes. Like that. So let's see. We're going to want to graph the point 0, 2. So let's say that the point 0, 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Like that. So let's see. We're going to want to graph the point 0, 2. So let's say that the point 0, 2. So this is x is equal to 0. And we go up 1, 2. So this is the point 0, 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say that the point 0, 2. So this is x is equal to 0. And we go up 1, 2. So this is the point 0, 2. Maybe I'll do it in that color. The color I was using. So that's this color."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is the point 0, 2. Maybe I'll do it in that color. The color I was using. So that's this color. So that's that point right there. Then we have the point x. We have f of 1."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So that's this color. So that's that point right there. Then we have the point x. We have f of 1. Which is the point 1, 1. Right. So this point right here."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We have f of 1. Which is the point 1, 1. Right. So this point right here. Is we'll go up 1. So that's the point 1, 1. This was the point 0, 2."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this point right here. Is we'll go up 1. So that's the point 1, 1. This was the point 0, 2. And then we have the x is equal to 2 thirds. Which is our inflection point. So when x is 2 thirds."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This was the point 0, 2. And then we have the x is equal to 2 thirds. Which is our inflection point. So when x is 2 thirds. We don't know exactly what number f of 2 thirds is. Might be here someplace. Let's say f of 2 thirds is right there."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So when x is 2 thirds. We don't know exactly what number f of 2 thirds is. Might be here someplace. Let's say f of 2 thirds is right there. So that's the point 2 thirds. And then whatever f of 2 thirds is. It looks like it's going to be 1 point something."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say f of 2 thirds is right there. So that's the point 2 thirds. And then whatever f of 2 thirds is. It looks like it's going to be 1 point something. f of 2 thirds. You could calculate it if you like. I'm going to substitute back in the function."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It looks like it's going to be 1 point something. f of 2 thirds. You could calculate it if you like. I'm going to substitute back in the function. But we're ready to graph this. We're ready to graph this thing. So we know that at x is equal to 1."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "I'm going to substitute back in the function. But we're ready to graph this. We're ready to graph this thing. So we know that at x is equal to 1. The slope is 0. We know that the slope is 0. It's flat here."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So we know that at x is equal to 1. The slope is 0. We know that the slope is 0. It's flat here. And we know it's concave upwards. So we're dealing. It looks like this."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It's flat here. And we know it's concave upwards. So we're dealing. It looks like this. It looks like that over that interval. We're concave upwards. And we know we're concave upwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It looks like this. It looks like that over that interval. We're concave upwards. And we know we're concave upwards. From x is equal to 2 thirds and on. Right. Let me do it in that color."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And we know we're concave upwards. From x is equal to 2 thirds and on. Right. Let me do it in that color. We know x is equal to 2 thirds and on. And we're concave upwards. And so that's why I was able to draw this u shape."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me do it in that color. We know x is equal to 2 thirds and on. And we're concave upwards. And so that's why I was able to draw this u shape. Now we know that when x is less than 2 thirds. And greater than 0. We're concave downwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And so that's why I was able to draw this u shape. Now we know that when x is less than 2 thirds. And greater than 0. We're concave downwards. So the graph would look something like this. Over this interval. We'll be concave downwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We're concave downwards. So the graph would look something like this. Over this interval. We'll be concave downwards. Let me draw it nicely. Over this interval. The slope is decreasing."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We'll be concave downwards. Let me draw it nicely. Over this interval. The slope is decreasing. And you could see it. If you keep drawing tangent lines. It's flattish there."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope is decreasing. And you could see it. If you keep drawing tangent lines. It's flattish there. It gets negative. More negative. More negative."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It's flattish there. It gets negative. More negative. More negative. More negative. Until the inflection point. And then it starts increasing again."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "More negative. More negative. Until the inflection point. And then it starts increasing again. Because we go back to concave upwards. And finally. The last interval is below 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then it starts increasing again. Because we go back to concave upwards. And finally. The last interval is below 0. And we know below 0. When x is less than 0. We're concave upwards."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The last interval is below 0. And we know below 0. When x is less than 0. We're concave upwards. So the graph looks like this. The graph looks like that. And we also know that x is equal to 0."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We're concave upwards. So the graph looks like this. The graph looks like that. And we also know that x is equal to 0. Was a critical point. The slope was 0. So this graph is actually flat right there too."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And we also know that x is equal to 0. Was a critical point. The slope was 0. So this graph is actually flat right there too. So this is an inflection point. Where the slope was also 0. So this is our final graph."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this graph is actually flat right there too. So this is an inflection point. Where the slope was also 0. So this is our final graph. We're done. After all that work. We were able to use our calculus skills."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is our final graph. We're done. After all that work. We were able to use our calculus skills. And our knowledge of inflection points. And concativity. And transitions in concativity."}, {"video_title": "Graphing using derivatives   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "We were able to use our calculus skills. And our knowledge of inflection points. And concativity. And transitions in concativity. To actually graph this fairly hairy looking graph. This should be kind of what it looks like. If you graph it on your calculator."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just remind ourselves what it means for c to be the number that satisfies the mean value theorem for f. This means that over this interval, c is a point, x equals c where the slope of the tangent line at x equals c so I could write f prime of c, so that is the slope of the tangent line when x is equal to c, this is equal to the slope of the secant line that connects these two points so this is going to be equal to, see the slope of the secant line that connects the points three f of three and one f of one so it's going to be f of three minus f of one over three minus one and if you wanted to think about what this means visually it would look something like this. So this is our x axis and this is one, two actually let me spread it out a little bit more, one, two and three and so you have one comma f of one right over there, so that is at the point one comma f of one and we could evaluate that actually what, that's one comma one, right? So that's going to be the point one comma one and then you have the point three comma, let's see, you're going to have four times three is twelve minus three is nine, so it's going to be three comma three three comma three, so maybe it's right over there three comma three and it might look, the curve might look something like this, so it might look something like that. So if you think about the slope of the line that connects these two points, so this line that connects those two points all the mean value theorem, I'm going to do that in a different color, all the mean value theorem tells us is that there's a point between one and three where the slope of the tangent line has the exact same slope, so if I were to eyeball it, it looks like it's right around there although we are actually going to solve for it, so some point where the slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values, so that is c, that would be c right over there. So really we just have to solve this, so let's first just find out what f prime of x is and then we could substitute a c in there and then we can evaluate this on the right hand side. So I'm going to rewrite f of x f of x is equal to and I'm going to write it as four x to the minus three to the one half power, it makes it a little bit more obvious that we can apply the power rule and the chain rule here, so f prime of x is going to be the derivative of four x minus three to the one half with respect to four x minus three, so that is going to be one half times four x minus three to the negative one half and then we're going to multiply that times the derivative of four x minus three with respect to x, well, derivative of four x with respect to x is just four and the derivative of negative three with respect word x well that's just going to be zero, so the derivative of four x minus three is four, so times four. So f prime of x, f prime of x is equal to four times 1 1\u20442, which is two over the square root of four x minus three."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you think about the slope of the line that connects these two points, so this line that connects those two points all the mean value theorem, I'm going to do that in a different color, all the mean value theorem tells us is that there's a point between one and three where the slope of the tangent line has the exact same slope, so if I were to eyeball it, it looks like it's right around there although we are actually going to solve for it, so some point where the slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values, so that is c, that would be c right over there. So really we just have to solve this, so let's first just find out what f prime of x is and then we could substitute a c in there and then we can evaluate this on the right hand side. So I'm going to rewrite f of x f of x is equal to and I'm going to write it as four x to the minus three to the one half power, it makes it a little bit more obvious that we can apply the power rule and the chain rule here, so f prime of x is going to be the derivative of four x minus three to the one half with respect to four x minus three, so that is going to be one half times four x minus three to the negative one half and then we're going to multiply that times the derivative of four x minus three with respect to x, well, derivative of four x with respect to x is just four and the derivative of negative three with respect word x well that's just going to be zero, so the derivative of four x minus three is four, so times four. So f prime of x, f prime of x is equal to four times 1 1\u20442, which is two over the square root of four x minus three. Four x minus three to the 1 1\u20442 would just be the square root of four x minus three, but it's the negative 1 1\u20442, so we're gonna put it in the denominator right over here. And so f prime of c, we could rewrite this as two over the square root of four c minus three. And what is that going to be equal to?"}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f prime of x, f prime of x is equal to four times 1 1\u20442, which is two over the square root of four x minus three. Four x minus three to the 1 1\u20442 would just be the square root of four x minus three, but it's the negative 1 1\u20442, so we're gonna put it in the denominator right over here. And so f prime of c, we could rewrite this as two over the square root of four c minus three. And what is that going to be equal to? That is going to be equal to, let's see, f of three we already figured out is three. F of one we already figured out is one. And so we get three minus one over three minus one."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what is that going to be equal to? That is going to be equal to, let's see, f of three we already figured out is three. F of one we already figured out is one. And so we get three minus one over three minus one. Well, that's gonna be two over two, which is equal to one. So there's some point between one and three where the derivative at that point, the slope of the tangent line, is equal to one. So let's see if we can solve this thing right over here."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we get three minus one over three minus one. Well, that's gonna be two over two, which is equal to one. So there's some point between one and three where the derivative at that point, the slope of the tangent line, is equal to one. So let's see if we can solve this thing right over here. Well, we can multiply both sides of this by the square root of four c minus three. And so then we are going to get, we're going to get two is equal to the square root of four c minus three. All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can solve this thing right over here. Well, we can multiply both sides of this by the square root of four c minus three. And so then we are going to get, we're going to get two is equal to the square root of four c minus three. All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator. And so, let's see, now to get rid of the radical, we can square both sides. And so, actually let me just show that. So now we can square both sides."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator. And so, let's see, now to get rid of the radical, we can square both sides. And so, actually let me just show that. So now we can square both sides. So we get four is equal to four c minus three. Add three to both sides. Seven is equal to four c. And then divide both sides by four."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now we can square both sides. So we get four is equal to four c minus three. Add three to both sides. Seven is equal to four c. And then divide both sides by four. I'll go right here to do it. You're going to get c is equal to 7 4ths. C is equal to, is equal to seven over four, which is equal to 1 3 4ths."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Seven is equal to four c. And then divide both sides by four. I'll go right here to do it. You're going to get c is equal to 7 4ths. C is equal to, is equal to seven over four, which is equal to 1 3 4ths. Or we could view this as 1.75. So actually, the c value is a little bit closer. I hand drew this."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "C is equal to, is equal to seven over four, which is equal to 1 3 4ths. Or we could view this as 1.75. So actually, the c value is a little bit closer. I hand drew this. It's closer to about right over there on our diagram. And actually that looks pretty, pretty close. That actually looks pretty good."}, {"video_title": "Mean value theorem example square root function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I hand drew this. It's closer to about right over there on our diagram. And actually that looks pretty, pretty close. That actually looks pretty good. I just hand drew this curve, so it's definitely not exact. But anyway, hopefully that gives you a sense of what's going on here. We're just saying, hey, the mean value theorem gives us some c where the slope of the tangent line is the same as the slope of the line that connects three, one f of one and three f of three."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then they give us her steps, and at the end they say, is Erin's work correct? If not, what's her mistake? So pause this video and see if you can figure it out yourself, is Erin correct? Or did she make a mistake, and where was that mistake? Alright, now let's just do it together. So she says that this is the derivative. I'm just going to reevaluate it here to the right of her work."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or did she make a mistake, and where was that mistake? Alright, now let's just do it together. So she says that this is the derivative. I'm just going to reevaluate it here to the right of her work. So let's see, f prime of x is just going to be the chain rule. I'm going to take the derivative of the outside with respect to the inside. So this is going to be two-thirds times x squared minus one to the two-thirds minus one, so to the negative one-third power, times the derivative of the inside with respect to x."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm just going to reevaluate it here to the right of her work. So let's see, f prime of x is just going to be the chain rule. I'm going to take the derivative of the outside with respect to the inside. So this is going to be two-thirds times x squared minus one to the two-thirds minus one, so to the negative one-third power, times the derivative of the inside with respect to x. So the derivative of x squared minus one with respect to x is two x. There's a fire hydrant. A fire, fire, fire, not a hydrant."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be two-thirds times x squared minus one to the two-thirds minus one, so to the negative one-third power, times the derivative of the inside with respect to x. So the derivative of x squared minus one with respect to x is two x. There's a fire hydrant. A fire, fire, fire, not a hydrant. That would be a noisy hydrant. There's a fire truck outside. But okay, I think it's passed."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "A fire, fire, fire, not a hydrant. That would be a noisy hydrant. There's a fire truck outside. But okay, I think it's passed. But this looks like what she got for the derivative, because if you multiply two times two x, you do indeed get four x. You have this three right over here in the denominator, and x squared minus one to the negative one-third, that's the same thing as x squared minus one to the one-third in the denominator, which is the same thing as the cube root of x squared minus one. So all of this is looking good."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "But okay, I think it's passed. But this looks like what she got for the derivative, because if you multiply two times two x, you do indeed get four x. You have this three right over here in the denominator, and x squared minus one to the negative one-third, that's the same thing as x squared minus one to the one-third in the denominator, which is the same thing as the cube root of x squared minus one. So all of this is looking good. That is indeed the derivative. Step two, the critical point is x equals zero. So let's see."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So all of this is looking good. That is indeed the derivative. Step two, the critical point is x equals zero. So let's see. A critical point is where our first derivative is either equal to zero or it is undefined. And so it does indeed seem that f prime of zero is going to be four times zero. It's gonna be zero over three times the cube root of zero minus one of negative one."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see. A critical point is where our first derivative is either equal to zero or it is undefined. And so it does indeed seem that f prime of zero is going to be four times zero. It's gonna be zero over three times the cube root of zero minus one of negative one. And so this is three times negative one, or zero over negative three. So this is indeed equal to zero. So this is true."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's gonna be zero over three times the cube root of zero minus one of negative one. And so this is three times negative one, or zero over negative three. So this is indeed equal to zero. So this is true. A critical point is at x equals zero. But a question is, is this the only critical point? Well, as we've mentioned, a critical point is where a function's derivative is either equal to zero or it's undefined."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is true. A critical point is at x equals zero. But a question is, is this the only critical point? Well, as we've mentioned, a critical point is where a function's derivative is either equal to zero or it's undefined. This is the only one where the derivative is equal to zero. But can you find some x values where the derivative is undefined? Well, what if we make the derivative, what would make the denominator of the derivative equal to zero?"}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, as we've mentioned, a critical point is where a function's derivative is either equal to zero or it's undefined. This is the only one where the derivative is equal to zero. But can you find some x values where the derivative is undefined? Well, what if we make the derivative, what would make the denominator of the derivative equal to zero? Well, if x squared minus one is equal to zero, you take the cube root of zero, you're gonna get zero in the denominator. So what would make x squared minus one equal to zero? Well, x is equal to plus or minus one."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what if we make the derivative, what would make the denominator of the derivative equal to zero? Well, if x squared minus one is equal to zero, you take the cube root of zero, you're gonna get zero in the denominator. So what would make x squared minus one equal to zero? Well, x is equal to plus or minus one. These are also critical points because they make f prime of x undefined. So I'm not feeling good about step two. It is true that a critical point is x equals zero, but it is not the only critical point."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, x is equal to plus or minus one. These are also critical points because they make f prime of x undefined. So I'm not feeling good about step two. It is true that a critical point is x equals zero, but it is not the only critical point. So I would put that there. And the reason why it's important, you know, you might say, well, what's the harm in not noticing these other critical points? She identified one, maybe this is the relative maximum point."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is true that a critical point is x equals zero, but it is not the only critical point. So I would put that there. And the reason why it's important, you know, you might say, well, what's the harm in not noticing these other critical points? She identified one, maybe this is the relative maximum point. But as we talk about in other videos, in order to use the first derivative test, so to speak, and find this place where the first derivative is zero, in order to test whether it is a maximum or a minimum point is you have to sample values on either side of it to make sure that you have a change in sign of the derivative. But you have to make sure that when you test on either side that you're not going beyond another critical point because critical points are places where you can change direction. And so let's see what she does in step three right over here."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "She identified one, maybe this is the relative maximum point. But as we talk about in other videos, in order to use the first derivative test, so to speak, and find this place where the first derivative is zero, in order to test whether it is a maximum or a minimum point is you have to sample values on either side of it to make sure that you have a change in sign of the derivative. But you have to make sure that when you test on either side that you're not going beyond another critical point because critical points are places where you can change direction. And so let's see what she does in step three right over here. Well, it is indeed in step three that she's testing, she's trying to test values on either side of the critical point that she, that the one critical point that she identified. But the problem here, the reason why this is a little shady is this is beyond another critical point that is less than zero, and this is beyond, this is greater than another critical point that is greater than zero. This is larger than the critical point one, and this is less than the critical point negative one."}, {"video_title": "Analyzing mistakes when finding extrema example 2   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's see what she does in step three right over here. Well, it is indeed in step three that she's testing, she's trying to test values on either side of the critical point that she, that the one critical point that she identified. But the problem here, the reason why this is a little shady is this is beyond another critical point that is less than zero, and this is beyond, this is greater than another critical point that is greater than zero. This is larger than the critical point one, and this is less than the critical point negative one. What she should have tried is x equals 0.5 and x equals negative 0.5. So this is what she should have done, is try maybe negative two, negative one, negative 1.5, zero, 1.5, and then one we know is undefined, and then positive two. Because this is a candidate extremum, this is a candidate extremum, and this is a candidate extremum right over here."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Consider the differential equation, the derivative of y with respect to x is equal to two x minus y. On the axes provided, sketch a slope field for the given differential equation at the six points indicated. And we see one, two, three, four, five, six points. And so what I can do is, let me set up a table actually. Let me set up a table with three columns. One for x, one for y, and then one for the derivative of y with respect to x, which this differential equation tells us in terms of x and y. And so we can look at the different points."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what I can do is, let me set up a table actually. Let me set up a table with three columns. One for x, one for y, and then one for the derivative of y with respect to x, which this differential equation tells us in terms of x and y. And so we can look at the different points. This point right over here, when x is one and y is two, x is one, y is two. Well, the derivative, they tell us, is two times x minus y. So this is two x minus y."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can look at the different points. This point right over here, when x is one and y is two, x is one, y is two. Well, the derivative, they tell us, is two times x minus y. So this is two x minus y. So it's gonna be two times one minus two, which is equal to zero. And so if the derivative there is equal to zero, then if I were to draw a line that indicates the slope at that point, well, I would draw a line with a zero slope. So it would look something like, let me draw it a little bit."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is two x minus y. So it's gonna be two times one minus two, which is equal to zero. And so if the derivative there is equal to zero, then if I were to draw a line that indicates the slope at that point, well, I would draw a line with a zero slope. So it would look something like, let me draw it a little bit. It would look something like that. All right, let's keep going. Let's look at this point, where x is equal to zero, y is equal to two."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would look something like, let me draw it a little bit. It would look something like that. All right, let's keep going. Let's look at this point, where x is equal to zero, y is equal to two. x is equal to zero, y is equal to two. The derivative is going to be two times x minus y. So it's going to be zero minus two, which is equal to negative two."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's look at this point, where x is equal to zero, y is equal to two. x is equal to zero, y is equal to two. The derivative is going to be two times x minus y. So it's going to be zero minus two, which is equal to negative two. So how do we draw a slope of negative two? Well, it's going to go from the top left to the bottom right. It's going to be pretty steep."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be zero minus two, which is equal to negative two. So how do we draw a slope of negative two? Well, it's going to go from the top left to the bottom right. It's going to be pretty steep. So it might look something like that. That looks like a slope of negative, let me try to draw it a little bit better. So I could draw it, well, that's pretty good."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be pretty steep. So it might look something like that. That looks like a slope of negative, let me try to draw it a little bit better. So I could draw it, well, that's pretty good. Maybe I shouldn't use my line tool. Let me just try to draw a reasonable, so if I have a slope of negative two, as I move one to the right, I move two down. So it should look something like that."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I could draw it, well, that's pretty good. Maybe I shouldn't use my line tool. Let me just try to draw a reasonable, so if I have a slope of negative two, as I move one to the right, I move two down. So it should look something like that. All right, let's keep going. x one, y one. x is one, y is one."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it should look something like that. All right, let's keep going. x one, y one. x is one, y is one. It's going to be two times one minus one. So this is two minus one. It's going to have a slope of one."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "x is one, y is one. It's going to be two times one minus one. So this is two minus one. It's going to have a slope of one. And so that would look something like a slope of one. It looks something like that, or I could, let me just draw it by hand. That's actually easier."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to have a slope of one. And so that would look something like a slope of one. It looks something like that, or I could, let me just draw it by hand. That's actually easier. And then we have x is zero, y is one. x is zero, y is one. Slope is going to be two times zero minus one, which is equal to negative one."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's actually easier. And then we have x is zero, y is one. x is zero, y is one. Slope is going to be two times zero minus one, which is equal to negative one. And so now the slope is negative one, and just like that. Notice it's less negative than up here. It's less steep."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Slope is going to be two times zero minus one, which is equal to negative one. And so now the slope is negative one, and just like that. Notice it's less negative than up here. It's less steep. And then let's go to x is one, y is negative one. x is one, y is negative one. So it's going to be two times one minus negative one."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's less steep. And then let's go to x is one, y is negative one. x is one, y is negative one. So it's going to be two times one minus negative one. So this is two, two plus one. This is equal to positive three. So here the slope, it's going to be even steeper, but now in the positive sense."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be two times one minus negative one. So this is two, two plus one. This is equal to positive three. So here the slope, it's going to be even steeper, but now in the positive sense. So slope of three would look something like that. And then finally, we have x is zero, y is negative one. It's going to be two times zero minus negative one, which is going to be equal to a slope of one, which is going to look like this."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here the slope, it's going to be even steeper, but now in the positive sense. So slope of three would look something like that. And then finally, we have x is zero, y is negative one. It's going to be two times zero minus negative one, which is going to be equal to a slope of one, which is going to look like this. It should be the same, it should be parallel to what we have right over there. And we're done. Alright, let's do part b now."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be two times zero minus negative one, which is going to be equal to a slope of one, which is going to look like this. It should be the same, it should be parallel to what we have right over there. And we're done. Alright, let's do part b now. Find the second derivative of y with respect to x in terms of x and y. Determine the concavity of all solutions for the given differential equation in quadrant two. Give a reason for your answer."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, let's do part b now. Find the second derivative of y with respect to x in terms of x and y. Determine the concavity of all solutions for the given differential equation in quadrant two. Give a reason for your answer. Alright, so first let's just find the second derivative. So we already know that dy, let me give myself a little bit more space actually. We already know that the derivative of y with respect to x is equal to two x minus y."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Give a reason for your answer. Alright, so first let's just find the second derivative. So we already know that dy, let me give myself a little bit more space actually. We already know that the derivative of y with respect to x is equal to two x minus y. Now to find the second derivative, I just want to take the derivative of both sides of this with respect to x. So let's do that. So I could take the derivative of the left-hand side with respect to x, and the derivative of the right-hand side with respect to x."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already know that the derivative of y with respect to x is equal to two x minus y. Now to find the second derivative, I just want to take the derivative of both sides of this with respect to x. So let's do that. So I could take the derivative of the left-hand side with respect to x, and the derivative of the right-hand side with respect to x. And so what is this going to be? On the left-hand side, the notation would just be the second derivative of y with respect to x. And then over here, let's take the derivative of each of these with respect to x."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I could take the derivative of the left-hand side with respect to x, and the derivative of the right-hand side with respect to x. And so what is this going to be? On the left-hand side, the notation would just be the second derivative of y with respect to x. And then over here, let's take the derivative of each of these with respect to x. The derivative of two x with respect to x is going to be equal to two, and then minus the derivative of y with respect to x was just going to be minus the derivative of y with respect to x. And so we could say, okay, we found the second derivative, but remember, they're saying in terms of x and y, right now I found the second derivative in terms of, well, in terms of a constant and in terms of the first derivative. And so we can substitute our expression for the first derivative back here to have this expression in terms of x and y."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then over here, let's take the derivative of each of these with respect to x. The derivative of two x with respect to x is going to be equal to two, and then minus the derivative of y with respect to x was just going to be minus the derivative of y with respect to x. And so we could say, okay, we found the second derivative, but remember, they're saying in terms of x and y, right now I found the second derivative in terms of, well, in terms of a constant and in terms of the first derivative. And so we can substitute our expression for the first derivative back here to have this expression in terms of x and y. And so this is all going to be equal to, let me write it over here, the second derivative of y with respect to x is going to be equal to two minus, well, the derivative of y with respect to x, we already know, is two x minus y. Two x minus y. And so this is going to be equal to two minus two x plus y."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we can substitute our expression for the first derivative back here to have this expression in terms of x and y. And so this is all going to be equal to, let me write it over here, the second derivative of y with respect to x is going to be equal to two minus, well, the derivative of y with respect to x, we already know, is two x minus y. Two x minus y. And so this is going to be equal to two minus two x plus y. Plus y. And then they say, determine the concavity of all solutions for the given differential equation in quadrant two. Remember, if you're thinking about our coordinate axes, if you think about our coordinate axes, so if that's our y-axis, that is our x-axis, this is quadrant one, this is quadrant two, this is quadrant three, and this is quadrant four."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to two minus two x plus y. Plus y. And then they say, determine the concavity of all solutions for the given differential equation in quadrant two. Remember, if you're thinking about our coordinate axes, if you think about our coordinate axes, so if that's our y-axis, that is our x-axis, this is quadrant one, this is quadrant two, this is quadrant three, and this is quadrant four. So they're talking about quadrant two. So what do we know about x and y in quadrant two? We know x is less than zero, and we know y is greater than zero."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Remember, if you're thinking about our coordinate axes, if you think about our coordinate axes, so if that's our y-axis, that is our x-axis, this is quadrant one, this is quadrant two, this is quadrant three, and this is quadrant four. So they're talking about quadrant two. So what do we know about x and y in quadrant two? We know x is less than zero, and we know y is greater than zero. So quadrant two, let's write it like this. Quadrant two, we know x is less than zero, and y is greater than zero. And so if that is the case, so if we get, so we have two minus two x plus y."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know x is less than zero, and we know y is greater than zero. So quadrant two, let's write it like this. Quadrant two, we know x is less than zero, and y is greater than zero. And so if that is the case, so if we get, so we have two minus two x plus y. So this is going to be, this right over here is going to be greater than zero. And negative two times a negative value, well this is also going to be greater than zero. So quadrant two, because of this, that means if this whole expression is going to be positive, negative two times a negative number is going to be positive, plus a positive, plus a positive, it's going to be positive."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if that is the case, so if we get, so we have two minus two x plus y. So this is going to be, this right over here is going to be greater than zero. And negative two times a negative value, well this is also going to be greater than zero. So quadrant two, because of this, that means if this whole expression is going to be positive, negative two times a negative number is going to be positive, plus a positive, plus a positive, it's going to be positive. So the second derivative, second derivative with respect to x is positive, positive, which means that our slope is increasing over that interval, which means that we have positive, we could say that we, our concavity, I always have trouble saying that. Concavity, concavity is upwards, upwards. And if you ever forget whether, okay, a second derivative being positive, is that concavity upwards or downwards, I always just like to draw kind of our, you know, the canonical concave upwards or concave downwards."}, {"video_title": "2015 AP Calculus AB BC 4ab   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So quadrant two, because of this, that means if this whole expression is going to be positive, negative two times a negative number is going to be positive, plus a positive, plus a positive, it's going to be positive. So the second derivative, second derivative with respect to x is positive, positive, which means that our slope is increasing over that interval, which means that we have positive, we could say that we, our concavity, I always have trouble saying that. Concavity, concavity is upwards, upwards. And if you ever forget whether, okay, a second derivative being positive, is that concavity upwards or downwards, I always just like to draw kind of our, you know, the canonical concave upwards or concave downwards. And you can see here that your slope, your slope here is turning less negative, or you could say it's becoming more positive, it's increasing, and it continues to increase when you are concave upwards. So second derivative positive, you are concave upwards. Second derivative negative, you are going to be concave downwards."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is her solution. So step one, it looks like she tried to take the derivative. Step two, she tries to find the solution, find where the derivative is equal to zero, and she found that it happens at x equals two, so she says that's a critical point. And step three, she says she makes a conclusion that therefore h has a relative extremum there. Is Pamela's work correct? If not, what's her mistake? So pause this video and try to work through it yourself and see if Pamela's work is correct."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And step three, she says she makes a conclusion that therefore h has a relative extremum there. Is Pamela's work correct? If not, what's her mistake? So pause this video and try to work through it yourself and see if Pamela's work is correct. All right, well I'm just gonna try to do it again in parallel, so first let me just take the derivative here. So h prime of x, just use the power rule multiple times, is gonna be three x squared for the x to the third. Two times negative six is negative 12 or minus 12x, and the derivative of 12x is plus 12."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video and try to work through it yourself and see if Pamela's work is correct. All right, well I'm just gonna try to do it again in parallel, so first let me just take the derivative here. So h prime of x, just use the power rule multiple times, is gonna be three x squared for the x to the third. Two times negative six is negative 12 or minus 12x, and the derivative of 12x is plus 12. And let's see, you can factor out a three here, so it's three times x squared minus four x plus four, and this part is indeed equal to x minus two squared, so this is equal to three times x minus two squared. So her step one looks right on target. Okay, step two, the solution of h prime of x is equal to zero is equal to x equals two."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two times negative six is negative 12 or minus 12x, and the derivative of 12x is plus 12. And let's see, you can factor out a three here, so it's three times x squared minus four x plus four, and this part is indeed equal to x minus two squared, so this is equal to three times x minus two squared. So her step one looks right on target. Okay, step two, the solution of h prime of x is equal to zero is equal to x equals two. Yeah, that works out. If you were to say three times x minus two squared, which is h prime of x, the first derivative, and set that equal to zero, this is going to be true when x is equal to two. And so any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Okay, step two, the solution of h prime of x is equal to zero is equal to x equals two. Yeah, that works out. If you were to say three times x minus two squared, which is h prime of x, the first derivative, and set that equal to zero, this is going to be true when x is equal to two. And so any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point. So this step looks good so far. And then step three, h has a relative extremum at x equals two. All right, so she made a big conclusion here."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point. So this step looks good so far. And then step three, h has a relative extremum at x equals two. All right, so she made a big conclusion here. She assumed that because the derivative was zero, that we have a relative extremum. So let's just see if you can even just make that conclusion. In order to have a relative extremum, your curve is gonna look something like this, and then you would have a relative extremum right over here."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so she made a big conclusion here. She assumed that because the derivative was zero, that we have a relative extremum. So let's just see if you can even just make that conclusion. In order to have a relative extremum, your curve is gonna look something like this, and then you would have a relative extremum right over here. And over here, your slope goes from being positive, and then it hits zero, and then it goes to being negative. Or you could have a relative extremum like this. This would be a maximum point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "In order to have a relative extremum, your curve is gonna look something like this, and then you would have a relative extremum right over here. And over here, your slope goes from being positive, and then it hits zero, and then it goes to being negative. Or you could have a relative extremum like this. This would be a maximum point. This would be a minimum point right over here. And then in a minimum point, your slope is zero right over there, but right before it, your slope was negative, and it goes to being positive. But you actually have cases where your derivative, your first derivative is zero, but you don't have an extremum."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This would be a maximum point. This would be a minimum point right over here. And then in a minimum point, your slope is zero right over there, but right before it, your slope was negative, and it goes to being positive. But you actually have cases where your derivative, your first derivative is zero, but you don't have an extremum. So for example, you could have a point like this, where right over here, your slope or your derivative could be equal to zero, and so your first derivative would be equal to zero, but notice, your slope is positive, it hits zero, and then it goes back to being positive again. And so you can't make the conclusion just because your derivative is zero that it's definitely an extremum. You could say it's a critical point, and so step two is correct."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "But you actually have cases where your derivative, your first derivative is zero, but you don't have an extremum. So for example, you could have a point like this, where right over here, your slope or your derivative could be equal to zero, and so your first derivative would be equal to zero, but notice, your slope is positive, it hits zero, and then it goes back to being positive again. And so you can't make the conclusion just because your derivative is zero that it's definitely an extremum. You could say it's a critical point, and so step two is correct. In order to make this conclusion, you would have to test what the derivative is doing before that point and after that point, and verify that it is switching sides. And we could try to do that. So let's make a little table here."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say it's a critical point, and so step two is correct. In order to make this conclusion, you would have to test what the derivative is doing before that point and after that point, and verify that it is switching sides. And we could try to do that. So let's make a little table here. Make a little table, a little bit neater. So x, x, h prime of x right over here. We know at x equals two, h prime of two is zero, that's our critical point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's make a little table here. Make a little table, a little bit neater. So x, x, h prime of x right over here. We know at x equals two, h prime of two is zero, that's our critical point. But let's try, I don't know, let's see what happens when x is equal to one, and then let's see what happens when x equals three. We're just sampling points on either side of two. And let's see, we are going to have, when x is equal to one, h prime of one is three times one minus two squared."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know at x equals two, h prime of two is zero, that's our critical point. But let's try, I don't know, let's see what happens when x is equal to one, and then let's see what happens when x equals three. We're just sampling points on either side of two. And let's see, we are going to have, when x is equal to one, h prime of one is three times one minus two squared. One minus two, negative one squared is positive one times three is still positive. And then three, well, three minus two squared times three, that's also gonna be three. So this is actually a situation where, like I just drawn it, where our slope is positive before we hit the critical point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we are going to have, when x is equal to one, h prime of one is three times one minus two squared. One minus two, negative one squared is positive one times three is still positive. And then three, well, three minus two squared times three, that's also gonna be three. So this is actually a situation where, like I just drawn it, where our slope is positive before we hit the critical point. It gets to zero, but then it starts becoming positive again. And so that's why you actually have to do this test in order to identify whether it's an extremum. It turns out that this is not an extremum."}, {"video_title": "Analyzing mistakes when finding extrema (example 1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is actually a situation where, like I just drawn it, where our slope is positive before we hit the critical point. It gets to zero, but then it starts becoming positive again. And so that's why you actually have to do this test in order to identify whether it's an extremum. It turns out that this is not an extremum. This is not a maximum or a minimum point here. So Pamela's work is not correct, and her mistake is step three. In order to make this conclusion, you would have to test on either side of that critical point, test the first derivative."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's think about the area under this curve above the x-axis from x equals negative one to x equals two. So that would be this area right over here. And there's many ways that I could tackle this, but what I'm going to do is I'm going to break up this interval into three equal sections that are really the bases of rectangles. And then we're gonna think about the different ways to define the heights of those rectangles. So once again, I'm going to approximate using three rectangles of equal width. And then we'll think about the different ways that we can define the heights of the rectangle. So let's first define the height of each rectangle by the value of the function at the midpoint."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're gonna think about the different ways to define the heights of those rectangles. So once again, I'm going to approximate using three rectangles of equal width. And then we'll think about the different ways that we can define the heights of the rectangle. So let's first define the height of each rectangle by the value of the function at the midpoint. So we see that right over here. So let's just make sure that it actually makes sense to us. So if we look at our first rectangle right over here, actually let's just first appreciate we have split up the interval from x equals negative one to x equals two into three equal sections."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's first define the height of each rectangle by the value of the function at the midpoint. So we see that right over here. So let's just make sure that it actually makes sense to us. So if we look at our first rectangle right over here, actually let's just first appreciate we have split up the interval from x equals negative one to x equals two into three equal sections. And then each of them have a width of one. So if we wanted a better approximation, we could do more sections or more rectangles. But let's just see how we would compute this."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we look at our first rectangle right over here, actually let's just first appreciate we have split up the interval from x equals negative one to x equals two into three equal sections. And then each of them have a width of one. So if we wanted a better approximation, we could do more sections or more rectangles. But let's just see how we would compute this. Well, the width of each of these is one. The height is based on the value of the function at the midpoint. The midpoint here is negative 1 1 2."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's just see how we would compute this. Well, the width of each of these is one. The height is based on the value of the function at the midpoint. The midpoint here is negative 1 1 2. The midpoint here is 1 1 2. The midpoint here is 3 1 2. And so this height is going to be negative 1 1 2 squared plus one."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The midpoint here is negative 1 1 2. The midpoint here is 1 1 2. The midpoint here is 3 1 2. And so this height is going to be negative 1 1 2 squared plus one. So negative 1 1 2 squared is 1 1 4 plus one. So that's 5 4ths. So the height here is 5 4ths."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this height is going to be negative 1 1 2 squared plus one. So negative 1 1 2 squared is 1 1 4 plus one. So that's 5 4ths. So the height here is 5 4ths. So you take 5 4ths times one. This area is 5 4ths. Let me write that down."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the height here is 5 4ths. So you take 5 4ths times one. This area is 5 4ths. Let me write that down. So if we're doing the midpoint to define the height of each rectangle, this first one has an area of 5 4ths. Let me do that in a color you can see. Five over four."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write that down. So if we're doing the midpoint to define the height of each rectangle, this first one has an area of 5 4ths. Let me do that in a color you can see. Five over four. The second one, same idea. 1 1 2 squared plus one is 5 4ths times the width of one. So 5 4ths there."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Five over four. The second one, same idea. 1 1 2 squared plus one is 5 4ths times the width of one. So 5 4ths there. So let me add that. Plus 5 4ths. And then this third rectangle, what's its height?"}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So 5 4ths there. So let me add that. Plus 5 4ths. And then this third rectangle, what's its height? Well, we're gonna take the height at the midpoint. So 3 1 2 squared is 9 4ths plus one, which is the same thing as 13 4ths. So it has a height of 13 4ths, and then a width of one, so times one, would just give us 13 4ths."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this third rectangle, what's its height? Well, we're gonna take the height at the midpoint. So 3 1 2 squared is 9 4ths plus one, which is the same thing as 13 4ths. So it has a height of 13 4ths, and then a width of one, so times one, would just give us 13 4ths. So plus 13 4ths, which would give us 23 over four, which is the same thing as 5 3 4ths. And so this is often known as a midpoint approximation, where we're using the midpoint of each interval to define the height of our rectangle. But this isn't the only way to do it."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it has a height of 13 4ths, and then a width of one, so times one, would just give us 13 4ths. So plus 13 4ths, which would give us 23 over four, which is the same thing as 5 3 4ths. And so this is often known as a midpoint approximation, where we're using the midpoint of each interval to define the height of our rectangle. But this isn't the only way to do it. We could look at the left endpoint or the right endpoint, and we do that in other videos. And then if we wanna do it just for kicks here, let's just do that really fast. So if we wanna look at the left endpoints of our interval, well, here our left endpoint is negative one."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this isn't the only way to do it. We could look at the left endpoint or the right endpoint, and we do that in other videos. And then if we wanna do it just for kicks here, let's just do that really fast. So if we wanna look at the left endpoints of our interval, well, here our left endpoint is negative one. Negative one squared plus one is two. Two times one gives us two. And then here, the left part of this interval is x equals zero."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we wanna look at the left endpoints of our interval, well, here our left endpoint is negative one. Negative one squared plus one is two. Two times one gives us two. And then here, the left part of this interval is x equals zero. Zero squared plus one is one. One times one is one. And now here, our left endpoint is one."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then here, the left part of this interval is x equals zero. Zero squared plus one is one. One times one is one. And now here, our left endpoint is one. One squared plus one is equal to two times one. Our base is equal to two. So here we have a situation where we take our left endpoints, where it is equal to two plus one plus two or five."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now here, our left endpoint is one. One squared plus one is equal to two times one. Our base is equal to two. So here we have a situation where we take our left endpoints, where it is equal to two plus one plus two or five. But we could also look at the right endpoints of our intervals. So this first rectangle here, clearly under-approximating the area over this first interval. Its right endpoint is zero."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here we have a situation where we take our left endpoints, where it is equal to two plus one plus two or five. But we could also look at the right endpoints of our intervals. So this first rectangle here, clearly under-approximating the area over this first interval. Its right endpoint is zero. Zero squared plus one is one. So height of one, width of one has an area of one. Second rectangle here, it has a height of, we look at our right endpoint, one squared plus one is two times our width of one."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Its right endpoint is zero. Zero squared plus one is one. So height of one, width of one has an area of one. Second rectangle here, it has a height of, we look at our right endpoint, one squared plus one is two times our width of one. Well, that's just gonna give us two. And then here, our right endpoint is two squared plus one is five times our width of one gives us five. So in this case, we get, when we look at our right endpoints of our intervals, we get one plus two plus five is equal to eight."}, {"video_title": "Midpoint sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Second rectangle here, it has a height of, we look at our right endpoint, one squared plus one is two times our width of one. Well, that's just gonna give us two. And then here, our right endpoint is two squared plus one is five times our width of one gives us five. So in this case, we get, when we look at our right endpoints of our intervals, we get one plus two plus five is equal to eight. And eyeballing this, it looks like we're definitely over-counting more than under-counting. And so this looks like an over-approximation. So the whole idea here is just to appreciate how we can compute these approximations using rectangles."}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So here we wanna figure out what is the limit as x approaches negative one of g of h of x. And the function g, we see it defined graphically here on the left, and the function h, we see it defined graphically here on the right. Pause this video and have a go at this. All right, now your first temptation might be to say, all right, well, what is the limit as x approaches negative one of h of x? And if that limit exists, then input that into g. So if you take the limit as x approaches negative one of h of x, you see that you have a different limit as you approach from the right than when you approach from the left. So your temptation might be to give up at this point, but what we'll do in this video is to realize that this composite limit actually exists even though the limit as x approaches negative one of h of x does not exist. So how do we figure this out?"}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "All right, now your first temptation might be to say, all right, well, what is the limit as x approaches negative one of h of x? And if that limit exists, then input that into g. So if you take the limit as x approaches negative one of h of x, you see that you have a different limit as you approach from the right than when you approach from the left. So your temptation might be to give up at this point, but what we'll do in this video is to realize that this composite limit actually exists even though the limit as x approaches negative one of h of x does not exist. So how do we figure this out? Well, what we could do is take right-handed and left-handed limits. Let's first figure out what is the limit as x approaches negative one from the right-hand side of g of h of x, h of x. Well, to think about that, what is the limit of h as x approaches negative one from the right-hand side?"}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So how do we figure this out? Well, what we could do is take right-handed and left-handed limits. Let's first figure out what is the limit as x approaches negative one from the right-hand side of g of h of x, h of x. Well, to think about that, what is the limit of h as x approaches negative one from the right-hand side? So as we approach negative one from the right-hand side, it looks like h is approaching negative two. So another way to think about it is this is going to be equal to the limit as h of x approaches negative two. And what direction is it approaching negative two from?"}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "Well, to think about that, what is the limit of h as x approaches negative one from the right-hand side? So as we approach negative one from the right-hand side, it looks like h is approaching negative two. So another way to think about it is this is going to be equal to the limit as h of x approaches negative two. And what direction is it approaching negative two from? Well, it's approaching negative two from values larger than negative two. H of x is decreasing down to negative two as x approaches negative one from the right. So it's approaching from values larger than negative two of g of h of x, g of h of x. I'm color-coding it to be able to keep track of things."}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "And what direction is it approaching negative two from? Well, it's approaching negative two from values larger than negative two. H of x is decreasing down to negative two as x approaches negative one from the right. So it's approaching from values larger than negative two of g of h of x, g of h of x. I'm color-coding it to be able to keep track of things. And so this is analogous to saying what is the limit as if you think about it as x approaches negative two from the positive direction of g. Here, h is just the input into g. So the input into g is approaching negative two from above, from the right, I should say, from values larger than negative two. And we can see that g is approaching three. So this right over here is going to be equal to three."}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So it's approaching from values larger than negative two of g of h of x, g of h of x. I'm color-coding it to be able to keep track of things. And so this is analogous to saying what is the limit as if you think about it as x approaches negative two from the positive direction of g. Here, h is just the input into g. So the input into g is approaching negative two from above, from the right, I should say, from values larger than negative two. And we can see that g is approaching three. So this right over here is going to be equal to three. Now let's take the limit as x approaches. So now let's take the limit as x approaches negative one from the left of g of h of x. So what we could do is first think about, well, what is h approaching as x approaches negative one from the left?"}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So this right over here is going to be equal to three. Now let's take the limit as x approaches. So now let's take the limit as x approaches negative one from the left of g of h of x. So what we could do is first think about, well, what is h approaching as x approaches negative one from the left? So as x approaches negative one from the left, it looks like h is approaching negative three. So we could say this is the limit as h of x is approaching negative three, and it is approaching negative three from values greater than negative three. It's going, h of x is approaching negative three from above, or we could say from values greater than negative three."}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "So what we could do is first think about, well, what is h approaching as x approaches negative one from the left? So as x approaches negative one from the left, it looks like h is approaching negative three. So we could say this is the limit as h of x is approaching negative three, and it is approaching negative three from values greater than negative three. It's going, h of x is approaching negative three from above, or we could say from values greater than negative three. And then of g of h of x. So another way to think about it, what is the limit as the input to g approaches negative three from the right? So as we approach negative three from the right, g is right here at three."}, {"video_title": "Limits of composite functions internal limit doesn't exist   AP Calculus   Khan Academy.mp3", "Sentence": "It's going, h of x is approaching negative three from above, or we could say from values greater than negative three. And then of g of h of x. So another way to think about it, what is the limit as the input to g approaches negative three from the right? So as we approach negative three from the right, g is right here at three. And so this is going to be equal to three again. And so notice the right-hand limit and the left-hand limit in this case are both equal to three. And so when the right-hand and the left-hand limit is equal to the same thing, we know that the limit is equal to that thing."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're going to approximate it by constructing four rectangles under the curve of equal width. So let's first think about what those rectangles look like. So four rectangles of equal width. So it would look like that, like that, and like that. And I haven't really defined the top of the rectangles just yet. But let's think about what those widths have to be if they're going to be equal width. And we can call that width delta x."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would look like that, like that, and like that. And I haven't really defined the top of the rectangles just yet. But let's think about what those widths have to be if they're going to be equal width. And we can call that width delta x. So this distance right over here, we're going to call that delta x. So delta x is going to have to be the total distance that we're traveling in x. So we finish at 3."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can call that width delta x. So this distance right over here, we're going to call that delta x. So delta x is going to have to be the total distance that we're traveling in x. So we finish at 3. We started at 1. And we want four equal width rectangles. So it's going to be equal to 1 half."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we finish at 3. We started at 1. And we want four equal width rectangles. So it's going to be equal to 1 half. So for example, this first interval between the boundary between the first rectangle and the second is going to be 1.5. Then we go 1 half to 2. Then we go to 2.5."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be equal to 1 half. So for example, this first interval between the boundary between the first rectangle and the second is going to be 1.5. Then we go 1 half to 2. Then we go to 2.5. And then we go 1 half to 3. Now let's think about how we'll define the height of the rectangles. For the sake of this video, we'll see in future videos that there's other ways of doing this, I'm going to use the left boundary of the rectangle to define the height, or the function, I should say."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then we go to 2.5. And then we go 1 half to 3. Now let's think about how we'll define the height of the rectangles. For the sake of this video, we'll see in future videos that there's other ways of doing this, I'm going to use the left boundary of the rectangle to define the height, or the function, I should say. I'm going to use the function evaluated at the left boundary to define the height. So for example, for the first rectangle, this point right over here is f of 1. And so I will say that that is the height of our first rectangle."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "For the sake of this video, we'll see in future videos that there's other ways of doing this, I'm going to use the left boundary of the rectangle to define the height, or the function, I should say. I'm going to use the function evaluated at the left boundary to define the height. So for example, for the first rectangle, this point right over here is f of 1. And so I will say that that is the height of our first rectangle. Then we go over here, the left boundary of the second rectangle. We're now looking at the function evaluated at 1.5, so that is f of 1.5. That's the height."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so I will say that that is the height of our first rectangle. Then we go over here, the left boundary of the second rectangle. We're now looking at the function evaluated at 1.5, so that is f of 1.5. That's the height. And so we get our second rectangle. Then we get, I could keep going like this, we get for this third rectangle, we have the function evaluated at 2. So we have the function evaluated at 2."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the height. And so we get our second rectangle. Then we get, I could keep going like this, we get for this third rectangle, we have the function evaluated at 2. So we have the function evaluated at 2. So that's right over here. That's f of 2. And so then we get our third rectangle."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have the function evaluated at 2. So that's right over here. That's f of 2. And so then we get our third rectangle. And then finally, we have our fourth rectangle, the function evaluated at 2.5. So the function evaluated at 2.5 is the height. So this is f of 2.5."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so then we get our third rectangle. And then finally, we have our fourth rectangle, the function evaluated at 2.5. So the function evaluated at 2.5 is the height. So this is f of 2.5. Remember, in each of these, I am just looking at the left boundary of the rectangle and evaluating the function there to get the height of the rectangle. Now that I set it up in this way, what is the total approximate area using the sum of these rectangles? And clearly, this isn't going to be a perfect approximation."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is f of 2.5. Remember, in each of these, I am just looking at the left boundary of the rectangle and evaluating the function there to get the height of the rectangle. Now that I set it up in this way, what is the total approximate area using the sum of these rectangles? And clearly, this isn't going to be a perfect approximation. I'm giving up on a bunch of area here. Let me see if I can color that in with a color that I have not used. So I'm giving up this area."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And clearly, this isn't going to be a perfect approximation. I'm giving up on a bunch of area here. Let me see if I can color that in with a color that I have not used. So I'm giving up this area. I'm giving up this area. I'm giving up that area. I'm giving up that area there."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm giving up this area. I'm giving up this area. I'm giving up that area. I'm giving up that area there. But this is just an approximation. And maybe if I had many more rectangles, it would be a better approximation. So let's figure out what the areas of each of the rectangles are."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm giving up that area there. But this is just an approximation. And maybe if I had many more rectangles, it would be a better approximation. So let's figure out what the areas of each of the rectangles are. So the area of this first rectangle is going to be the height, which is f of 1 times the base, which is delta x. The area of the second rectangle is going to be the height, which we already said is f of 1.5 times the base times delta x. The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2, so plus f of 2 times the base times delta x."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's figure out what the areas of each of the rectangles are. So the area of this first rectangle is going to be the height, which is f of 1 times the base, which is delta x. The area of the second rectangle is going to be the height, which we already said is f of 1.5 times the base times delta x. The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2, so plus f of 2 times the base times delta x. And then finally, the area of the third rectangle is the function. The height is a function evaluated at 2.5. So plus, that's a different color than what I wanted to use."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2, so plus f of 2 times the base times delta x. And then finally, the area of the third rectangle is the function. The height is a function evaluated at 2.5. So plus, that's a different color than what I wanted to use. Let me use that orange color. So plus the function evaluated at 2.5 times the base. This is going to be equal to our approximate area."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So plus, that's a different color than what I wanted to use. Let me use that orange color. So plus the function evaluated at 2.5 times the base. This is going to be equal to our approximate area. Let's make it clear. Approximate area under the curve, just the sum of these rectangles. So let's evaluate this."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be equal to our approximate area. Let's make it clear. Approximate area under the curve, just the sum of these rectangles. So let's evaluate this. So this is going to be equal to f of, it's going to be equal to the function evaluated at 1. 1 squared plus 1 is just 2. So it's going to be 2 times 1 half plus the function evaluated at 1.25."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's evaluate this. So this is going to be equal to f of, it's going to be equal to the function evaluated at 1. 1 squared plus 1 is just 2. So it's going to be 2 times 1 half plus the function evaluated at 1.25. 1.25 squared is 2.25. And then you add 1 to it. It becomes 3.25."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be 2 times 1 half plus the function evaluated at 1.25. 1.25 squared is 2.25. And then you add 1 to it. It becomes 3.25. So plus 3.25 times 1 half. And then we have the function evaluated at 2. Well, 2 squared plus 1 is 5."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "It becomes 3.25. So plus 3.25 times 1 half. And then we have the function evaluated at 2. Well, 2 squared plus 1 is 5. So it's 5 times 1 half. And then finally, you have the function evaluated at 2.5. 2.5 squared is 6.25 times 1 half."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, 2 squared plus 1 is 5. So it's 5 times 1 half. And then finally, you have the function evaluated at 2.5. 2.5 squared is 6.25 times 1 half. 6.25 plus 1. So that's 7.25 times 1 half. And just to make the math simpler, we can factor out the 1 half."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "2.5 squared is 6.25 times 1 half. 6.25 plus 1. So that's 7.25 times 1 half. And just to make the math simpler, we can factor out the 1 half. So this is going to be equal to 1 half times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1 half times. Let's see if I can do this in my head. 2 plus 5 is easy."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just to make the math simpler, we can factor out the 1 half. So this is going to be equal to 1 half times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1 half times. Let's see if I can do this in my head. 2 plus 5 is easy. That's 7. 3 plus 7 is 10. And then we have 0.25 plus 0.25."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "2 plus 5 is easy. That's 7. 3 plus 7 is 10. And then we have 0.25 plus 0.25. So it's going to be 10.5 plus 7 is 17.5. So 1 half times 17.5, which is equal to 8.75, which once again gives us an approximation. Clearly, the way I've drawn it right over here for the function we're using, it's going to be an underestimate because we've given up all of that pink area that I had colored in before."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we have 0.25 plus 0.25. So it's going to be 10.5 plus 7 is 17.5. So 1 half times 17.5, which is equal to 8.75, which once again gives us an approximation. Clearly, the way I've drawn it right over here for the function we're using, it's going to be an underestimate because we've given up all of that pink area that I had colored in before. It's an underestimate, but it's an approximation of the area under the curve. In the next few videos, we're going to try to generalize this to situations where we have an arbitrary function and we have an arbitrary number of rectangles. And we'll also start, in videos after that, we'll look at rectangles where we define the height not by the left boundary, but by the right boundary, or by the midpoint."}, {"video_title": "Riemann approximation introduction   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Clearly, the way I've drawn it right over here for the function we're using, it's going to be an underestimate because we've given up all of that pink area that I had colored in before. It's an underestimate, but it's an approximation of the area under the curve. In the next few videos, we're going to try to generalize this to situations where we have an arbitrary function and we have an arbitrary number of rectangles. And we'll also start, in videos after that, we'll look at rectangles where we define the height not by the left boundary, but by the right boundary, or by the midpoint. Or maybe we don't use rectangles at all. Maybe we might use things like trapezoids. Anyway, have fun."}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say we want to take the indefinite integral of two x plus one times the square root of x squared plus x dx. Does u-substitution apply here? And if it does, how would you define the u? Pause the video and try to think about that. Well, we just have to remind ourselves that u-substitution is really trying to undo the chain rule. If we remind ourselves what the chain rule tells us, it says, look, if we have a composite function, let's say f of g of x, f of g of x, and we take the derivative of that with respect to x, that that is going to be equal to the derivative of the outside function with respect to the inside function, so f prime of g of x times the derivative of the inside function, times the derivative of the inside function. And so u-substitution is all about, well, do we see a pattern like that inside the integral?"}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "Pause the video and try to think about that. Well, we just have to remind ourselves that u-substitution is really trying to undo the chain rule. If we remind ourselves what the chain rule tells us, it says, look, if we have a composite function, let's say f of g of x, f of g of x, and we take the derivative of that with respect to x, that that is going to be equal to the derivative of the outside function with respect to the inside function, so f prime of g of x times the derivative of the inside function, times the derivative of the inside function. And so u-substitution is all about, well, do we see a pattern like that inside the integral? Do we see a potential inside function, a g of x, where I see its derivative being multiplied? Well, we see that over here. If I look at x squared plus x, if I make that the u, what's the derivative of that?"}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so u-substitution is all about, well, do we see a pattern like that inside the integral? Do we see a potential inside function, a g of x, where I see its derivative being multiplied? Well, we see that over here. If I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one. So we should make that substitution. If we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x, is equal to two x plus one."}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one. So we should make that substitution. If we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x, is equal to two x plus one. If we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand-wavy mathematics, but it's appropriate here. So we could say two x plus one times dx. And now what's really interesting is here is we have our u right over there, and notice we have our two x plus one times dx."}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x, is equal to two x plus one. If we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand-wavy mathematics, but it's appropriate here. So we could say two x plus one times dx. And now what's really interesting is here is we have our u right over there, and notice we have our two x plus one times dx. In fact, it's not conventional to see an integral rewritten the way I'm about to write it, but I will. I could rewrite this integral. You should really view this as the product of three things."}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now what's really interesting is here is we have our u right over there, and notice we have our two x plus one times dx. In fact, it's not conventional to see an integral rewritten the way I'm about to write it, but I will. I could rewrite this integral. You should really view this as the product of three things. Oftentimes, people just view the dx as somehow part of the integral operator, but you could rearrange it. This would actually be legitimate. You could say the integral of the square root of x squared plus x times two x plus one dx."}, {"video_title": "_-substitution defining _   AP Calculus AB   Khan Academy.mp3", "Sentence": "You should really view this as the product of three things. Oftentimes, people just view the dx as somehow part of the integral operator, but you could rearrange it. This would actually be legitimate. You could say the integral of the square root of x squared plus x times two x plus one dx. And if you wanted to be really clear, you could even put all of those things in parentheses or something like that. And so here, this is our u, and this right over here is our du. And so we could rewrite this as being equal to the integral of the square root of u, because x squared plus x is u, times du, which is much easier to evaluate."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that we have a cardboard, a sheet of cardboard that is 20 inches by 30 inches. Let me draw the cardboard as neatly as I can. So it might look something like that. So that is my sheet of cardboard. And just to make sure we know the dimensions, it is 20 inches by 30 inches. What we're going to do is cut out the corners of this cardboard. And all the cut out corners are going to be squares."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is my sheet of cardboard. And just to make sure we know the dimensions, it is 20 inches by 30 inches. What we're going to do is cut out the corners of this cardboard. And all the cut out corners are going to be squares. And we're going to cut out an x by x corner from each of the corners of this piece of cardboard. x by x. Over here, x by x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And all the cut out corners are going to be squares. And we're going to cut out an x by x corner from each of the corners of this piece of cardboard. x by x. Over here, x by x. And then over here, x by x. And what we'll do is after we cut out those corners, we can essentially fold down the flaps. Let me draw the flaps so you can imagine."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Over here, x by x. And then over here, x by x. And what we'll do is after we cut out those corners, we can essentially fold down the flaps. Let me draw the flaps so you can imagine. We can fold right there. We could fold right there. We could fold right there."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me draw the flaps so you can imagine. We can fold right there. We could fold right there. We could fold right there. And we would form a box. I guess you could imagine a box without a bottom to it. Or you could view a box without a top to it."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could fold right there. And we would form a box. I guess you could imagine a box without a bottom to it. Or you could view a box without a top to it. So if we were to fold everything up, we would get a container that looks something like this. Let me make my best attempt to draw it. So it would look like this."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could view a box without a top to it. So if we were to fold everything up, we would get a container that looks something like this. Let me make my best attempt to draw it. So it would look like this. This is one flap folded up. You can imagine this flap right over here. If I were to fold it up like that, it now would look like this."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would look like this. This is one flap folded up. You can imagine this flap right over here. If I were to fold it up like that, it now would look like this. The height of the flap is x. So this distance right over here is x. And then if I were to fold this flap, let me do that a little bit neater."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I were to fold it up like that, it now would look like this. The height of the flap is x. So this distance right over here is x. And then if I were to fold this flap, let me do that a little bit neater. If I were to fold this flap right over here, if I were to fold that up, then it would look like this. My best attempt to draw it. It would look like that."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then if I were to fold this flap, let me do that a little bit neater. If I were to fold this flap right over here, if I were to fold that up, then it would look like this. My best attempt to draw it. It would look like that. And then I would fold that back flap up. So the back flap would look something like that. That would be the back flap."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would look like that. And then I would fold that back flap up. So the back flap would look something like that. That would be the back flap. It would look something like that. And then this flap over here, if I fold it up, would look something like that. And then, of course, my base of my whole thing."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "That would be the back flap. It would look something like that. And then this flap over here, if I fold it up, would look something like that. And then, of course, my base of my whole thing. So this whole region right over here of my piece of cardboard would be the floor of this box that I'm constructing. And what I want to do is I want to maximize the volume of this box. I want to maximize how much it can hold."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then, of course, my base of my whole thing. So this whole region right over here of my piece of cardboard would be the floor of this box that I'm constructing. And what I want to do is I want to maximize the volume of this box. I want to maximize how much it can hold. And I want to maximize it by picking my x appropriately. So let's think about what the volume of this box is as a function of x. In order to do that, we have to figure out all the dimensions of this box as a function of x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I want to maximize how much it can hold. And I want to maximize it by picking my x appropriately. So let's think about what the volume of this box is as a function of x. In order to do that, we have to figure out all the dimensions of this box as a function of x. We already know that this corner right over here, which is made up of when this side and this side connect, when you fold these two flaps up, that's going to be the same height over there. That's going to be the same height over there. The height of this box is going to be x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "In order to do that, we have to figure out all the dimensions of this box as a function of x. We already know that this corner right over here, which is made up of when this side and this side connect, when you fold these two flaps up, that's going to be the same height over there. That's going to be the same height over there. The height of this box is going to be x. But what's the width? What is the width of this box going to be? Well, the width of the box is going to be this distance right over here."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The height of this box is going to be x. But what's the width? What is the width of this box going to be? Well, the width of the box is going to be this distance right over here. And this distance is going to be 20 inches minus not one x, but minus two x's. So this is going to be 20 minus two x. You see it right over here."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the width of the box is going to be this distance right over here. And this distance is going to be 20 inches minus not one x, but minus two x's. So this is going to be 20 minus two x. You see it right over here. This whole distance is 20. You subtract this x, you subtract this x, and you get this distance right over here. So it's 20 minus two x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You see it right over here. This whole distance is 20. You subtract this x, you subtract this x, and you get this distance right over here. So it's 20 minus two x. Now the same logic, what is the depth of the box? What is that distance right over there? Well, that distance is this distance right over here."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's 20 minus two x. Now the same logic, what is the depth of the box? What is that distance right over there? Well, that distance is this distance right over here. We know that this entire distance is 30 inches. If we subtract out this x and we subtract out this x, we get the distance that we care about. So this is going to be 30 minus two x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that distance is this distance right over here. We know that this entire distance is 30 inches. If we subtract out this x and we subtract out this x, we get the distance that we care about. So this is going to be 30 minus two x. So now we have all of the dimensions. So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus two x, times the depth, which is 30 minus two x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be 30 minus two x. So now we have all of the dimensions. So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus two x, times the depth, which is 30 minus two x. Now what are possible values of x that give us a valid volume? Well, x can't be less than zero. You can't make a negative cut here."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus two x, times the depth, which is 30 minus two x. Now what are possible values of x that give us a valid volume? Well, x can't be less than zero. You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to zero. So let me write this down."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to zero. So let me write this down. x is going to be greater than or equal to zero. And what does it have to be less than? Well, I can cut at most, we can see here the length right over here, this pink color, this mauve color, is 20 minus two x."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write this down. x is going to be greater than or equal to zero. And what does it have to be less than? Well, I can cut at most, we can see here the length right over here, this pink color, this mauve color, is 20 minus two x. So this has got to be greater than zero. This is always going to be shorter than the 30 minus two x. But the 20 minus two x has to be greater than or equal to zero."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I can cut at most, we can see here the length right over here, this pink color, this mauve color, is 20 minus two x. So this has got to be greater than zero. This is always going to be shorter than the 30 minus two x. But the 20 minus two x has to be greater than or equal to zero. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to two x. Or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the 20 minus two x has to be greater than or equal to zero. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to two x. Or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. It's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between zero and 10."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. It's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between zero and 10. Otherwise, we've cut too much or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our domain, of what x can be for our volume. Well, our volume when x is equal to zero is equal to what?"}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So x has got to be between zero and 10. Otherwise, we've cut too much or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our domain, of what x can be for our volume. Well, our volume when x is equal to zero is equal to what? We're going to have zero times all of this stuff, and it's pretty obvious. You're not going to have any height here, so you're not going to have any volume, so our volume would be zero. What is our volume when x is equal to 10?"}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, our volume when x is equal to zero is equal to what? We're going to have zero times all of this stuff, and it's pretty obvious. You're not going to have any height here, so you're not going to have any volume, so our volume would be zero. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be zero. So once again, we would have no volume. And this term right over here, if we just look at it algebraically, would also be equal to zero, so this whole thing would be equal to zero."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be zero. So once again, we would have no volume. And this term right over here, if we just look at it algebraically, would also be equal to zero, so this whole thing would be equal to zero. So someplace in between x equals zero and x equals 10, we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this term right over here, if we just look at it algebraically, would also be equal to zero, so this whole thing would be equal to zero. So someplace in between x equals zero and x equals 10, we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. First let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'll get my handy TI-85 out. First let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. My minimum x value, let me make that zero. We know that x cannot be less than zero. My maximum x value, well, 10 seems pretty good."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me set my range. My minimum x value, let me make that zero. We know that x cannot be less than zero. My maximum x value, well, 10 seems pretty good. My minimum y value, this is essentially going to be my volume. I'm not going to have negative volume, so let me set that equal zero. And my maximum y value, let's see what would be reasonable here."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "My maximum x value, well, 10 seems pretty good. My minimum y value, this is essentially going to be my volume. I'm not going to have negative volume, so let me set that equal zero. And my maximum y value, let's see what would be reasonable here. I'm just going to pick a random x and see what type of a volume I get. If x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be, did I do that right?"}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And my maximum y value, let's see what would be reasonable here. I'm just going to pick a random x and see what type of a volume I get. If x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be, did I do that right? Yeah, 20 minus 2 times 5, so that would be 10. And then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would be, did I do that right? Yeah, 20 minus 2 times 5, so that would be 10. And then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches, and I just randomly picked the number 5. Let me give my maximum y value a little higher than that, just in case that that isn't the maximum value. I just randomly picked."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches, and I just randomly picked the number 5. Let me give my maximum y value a little higher than that, just in case that that isn't the maximum value. I just randomly picked. So let's say y max is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our y max even larger. So I think this is going to be a decent range. Now let's actually input the function itself."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I just randomly picked. So let's say y max is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our y max even larger. So I think this is going to be a decent range. Now let's actually input the function itself. So our volume is equal to x times 20 minus 2x times 30 minus 2x. And that looks about right, so now I think we can graph it. So second, and I want to select that up there, so I'll do graph."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's actually input the function itself. So our volume is equal to x times 20 minus 2x times 30 minus 2x. And that looks about right, so now I think we can graph it. So second, and I want to select that up there, so I'll do graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10. And it does look like we hit a maximum point right around there."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So second, and I want to select that up there, so I'll do graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10. And it does look like we hit a maximum point right around there. So what I'm going to do is I'm going to use the trace function to figure out roughly what that maximum point is. So let me trace this function so I can still go higher, higher. So over there my volume is 1055.5."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it does look like we hit a maximum point right around there. So what I'm going to do is I'm going to use the trace function to figure out roughly what that maximum point is. So let me trace this function so I can still go higher, higher. So over there my volume is 1055.5. Then I can get to 1056. So see, this was 1056.20. This is 1056.24."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over there my volume is 1055.5. Then I can get to 1056. So see, this was 1056.20. This is 1056.24. Then I go back to 1055. So at least based on the level of zoom that I have in my calculator right now, this is a pretty good approximation for the maximum value that my function actually takes on. So it looks like my maximum value is around 1056."}, {"video_title": "Optimization box volume (Part 1)   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is 1056.24. Then I go back to 1055. So at least based on the level of zoom that I have in my calculator right now, this is a pretty good approximation for the maximum value that my function actually takes on. So it looks like my maximum value is around 1056. And it happens at around x equals 3.89. So it looks like my volume at 3.89 is approximately equal to 1056 cubic inches. Or you could say that we hit a maximum when x is approximately equal to 3.89."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are told the differentiable functions x and y are related by the following equation. Y is equal to square root of x. It's interesting, they're telling us that they're both differentiable functions. Even x is a function, must be a function of something else. Well, they tell us that the derivative of x with respect to t is 12. And they want us to find the derivative of y with respect to t when x is equal to nine. So let's just make sure we can understand this."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Even x is a function, must be a function of something else. Well, they tell us that the derivative of x with respect to t is 12. And they want us to find the derivative of y with respect to t when x is equal to nine. So let's just make sure we can understand this. So they're telling us that both x and y are functions. Arguably, they're both functions of t. Y is a function of x, but then x is a function of t, so y could also be a function of t. One way to think about it is if x is equal to f of t, then y is equal to the square root of x, which would just be f of t. Another way to think about it, if you took t as your input into your function f, you're going to produce x. And then if you took that as your input into the square root function, you are going to produce y."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just make sure we can understand this. So they're telling us that both x and y are functions. Arguably, they're both functions of t. Y is a function of x, but then x is a function of t, so y could also be a function of t. One way to think about it is if x is equal to f of t, then y is equal to the square root of x, which would just be f of t. Another way to think about it, if you took t as your input into your function f, you're going to produce x. And then if you took that as your input into the square root function, you are going to produce y. So you could just view this as just one big box here, that y is a function of t. But now let's actually answer their question. To tackle it, we just have to apply the chain rule. The chain rule tells us that the derivative of y with respect to t is going to be equal to the derivative of y with respect to x times the derivative of x with respect to t. So let's apply it to this particular situation."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then if you took that as your input into the square root function, you are going to produce y. So you could just view this as just one big box here, that y is a function of t. But now let's actually answer their question. To tackle it, we just have to apply the chain rule. The chain rule tells us that the derivative of y with respect to t is going to be equal to the derivative of y with respect to x times the derivative of x with respect to t. So let's apply it to this particular situation. We're gonna have the derivative of y with respect to t is equal to the derivative of y with respect to x. Well, what's that? Well, y is equal to the principal root of x."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The chain rule tells us that the derivative of y with respect to t is going to be equal to the derivative of y with respect to x times the derivative of x with respect to t. So let's apply it to this particular situation. We're gonna have the derivative of y with respect to t is equal to the derivative of y with respect to x. Well, what's that? Well, y is equal to the principal root of x. You could also write this as y is equal to x to the 1 1 2 power. We could just use the power rule. The derivative of y with respect to x is 1 1 2 x to the negative 1 1 2."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, y is equal to the principal root of x. You could also write this as y is equal to x to the 1 1 2 power. We could just use the power rule. The derivative of y with respect to x is 1 1 2 x to the negative 1 1 2. So let me write that down. 1 1 2 x to the negative 1 1 2. And then times the derivative of x with respect to t. Times the derivative of x with respect to t. So let's see, we wanna find what we have here in orange."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of y with respect to x is 1 1 2 x to the negative 1 1 2. So let me write that down. 1 1 2 x to the negative 1 1 2. And then times the derivative of x with respect to t. Times the derivative of x with respect to t. So let's see, we wanna find what we have here in orange. That's what the question asks us. They tell us when x is equal to nine and the derivative of x with respect to t is equal to 12. So we have all of the information necessary to solve for this."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then times the derivative of x with respect to t. Times the derivative of x with respect to t. So let's see, we wanna find what we have here in orange. That's what the question asks us. They tell us when x is equal to nine and the derivative of x with respect to t is equal to 12. So we have all of the information necessary to solve for this. So this is going to be equal to 1 1 2 times nine to the negative 1 1 2. Nine to the negative 1 1 2. Times dx dt."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have all of the information necessary to solve for this. So this is going to be equal to 1 1 2 times nine to the negative 1 1 2. Nine to the negative 1 1 2. Times dx dt. The derivative of x with respect to t is equal to 12. Times 12. So let's see, nine to the 1 1 2 would be three."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Times dx dt. The derivative of x with respect to t is equal to 12. Times 12. So let's see, nine to the 1 1 2 would be three. Nine to the negative 1 1 2 would be 1 1 3. So this is 1 1 3. So this will all simplify to 1 1 2 times 1 1 3 is 1 1 6."}, {"video_title": "Differentiating related functions intro   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see, nine to the 1 1 2 would be three. Nine to the negative 1 1 2 would be 1 1 3. So this is 1 1 3. So this will all simplify to 1 1 2 times 1 1 3 is 1 1 6. So we can have a six in the denominator. And then we are going to have a 12 in the numerator. So 12 6."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "I encourage you, like always, pause this video and try to do it yourself before we do it together. So when you do this first one, you might just try to find the limit as x approaches negative two of f of x, and then the limit as x approaches negative two of g of x, and then add those two limits together. But you will quickly find a problem, because when you find the limit as x approaches negative two of f of x, it looks as we are approaching negative two from the left, it looks like we're approaching one. As we approach x equals negative two from the right, it looks like we're approaching three. So it looks like the limit as x approaches negative two of f of x doesn't exist, and the same thing's true of g of x. If we approach from the left, it looks like we're approaching three. If we approach from the right, it looks like we're approaching one."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "As we approach x equals negative two from the right, it looks like we're approaching three. So it looks like the limit as x approaches negative two of f of x doesn't exist, and the same thing's true of g of x. If we approach from the left, it looks like we're approaching three. If we approach from the right, it looks like we're approaching one. But it turns out that this limit can still exist, as long as the limit as x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two from the right of the sum, f of x plus g of x. So what are these things? Well, as we approach negative two from the left, f of x is approaching, looks like one, and g of x is approaching three."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we approach from the right, it looks like we're approaching one. But it turns out that this limit can still exist, as long as the limit as x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two from the right of the sum, f of x plus g of x. So what are these things? Well, as we approach negative two from the left, f of x is approaching, looks like one, and g of x is approaching three. So it looks like we're approaching one and three, so it looks like this is approaching, the sum is going to approach four. And if we're coming from the right, f of x looks like it's approaching three, and g of x looks like it is approaching one. And so once again, this is equal to four."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, as we approach negative two from the left, f of x is approaching, looks like one, and g of x is approaching three. So it looks like we're approaching one and three, so it looks like this is approaching, the sum is going to approach four. And if we're coming from the right, f of x looks like it's approaching three, and g of x looks like it is approaching one. And so once again, this is equal to four. And since the left and right-handed limits are approaching the same thing, we would say that this limit exists, and it is equal to four. Now let's do this next example, as x approaches one. Well, we'll do the exact same exercise."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so once again, this is equal to four. And since the left and right-handed limits are approaching the same thing, we would say that this limit exists, and it is equal to four. Now let's do this next example, as x approaches one. Well, we'll do the exact same exercise. And once again, if you look at the individual limits for f of x from the left and the right, as we approach one, this limit doesn't exist, but the limit as x approaches one of the sum might exist. So let's try that out. So the limit as x approaches one from the left-hand side of f of x plus g of x, what is that going to be equal to?"}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we'll do the exact same exercise. And once again, if you look at the individual limits for f of x from the left and the right, as we approach one, this limit doesn't exist, but the limit as x approaches one of the sum might exist. So let's try that out. So the limit as x approaches one from the left-hand side of f of x plus g of x, what is that going to be equal to? As we approach, so f of x, as we approach one from the left, looks like this is approaching two. I'm just doing this for shorthand. And g of x, as we approach one from the left, looks like it is approaching zero, so this will be approaching two plus zero, which is two."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the limit as x approaches one from the left-hand side of f of x plus g of x, what is that going to be equal to? As we approach, so f of x, as we approach one from the left, looks like this is approaching two. I'm just doing this for shorthand. And g of x, as we approach one from the left, looks like it is approaching zero, so this will be approaching two plus zero, which is two. And then the limit as x approaches one from the right-hand side of f of x plus g of x is going to be equal to? Well, for f of x, as we're approaching one from the right-hand side, looks like it's approaching negative one. And for g of x, as we're approaching one from the right-hand side, looks like we're approaching zero again."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And g of x, as we approach one from the left, looks like it is approaching zero, so this will be approaching two plus zero, which is two. And then the limit as x approaches one from the right-hand side of f of x plus g of x is going to be equal to? Well, for f of x, as we're approaching one from the right-hand side, looks like it's approaching negative one. And for g of x, as we're approaching one from the right-hand side, looks like we're approaching zero again. And so here it looks like we're approaching negative one. So the left and right-hand limits aren't approaching the same value, so this one does not exist. And then last but not least, x approaches one of f of x times g of x."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And for g of x, as we're approaching one from the right-hand side, looks like we're approaching zero again. And so here it looks like we're approaching negative one. So the left and right-hand limits aren't approaching the same value, so this one does not exist. And then last but not least, x approaches one of f of x times g of x. So we'll do the same drill. Limit as x approaches one from the left-hand side of f of x times g of x. Well, here, and we could even use the values here, we see we was approaching one from the left."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then last but not least, x approaches one of f of x times g of x. So we'll do the same drill. Limit as x approaches one from the left-hand side of f of x times g of x. Well, here, and we could even use the values here, we see we was approaching one from the left. We are approaching two, so this is two. And when we're approaching one from the left here, we're approaching zero. And so this is gonna be two times, we're gonna be approaching two times zero, which is zero."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, here, and we could even use the values here, we see we was approaching one from the left. We are approaching two, so this is two. And when we're approaching one from the left here, we're approaching zero. And so this is gonna be two times, we're gonna be approaching two times zero, which is zero. And then we approach from the right, x approaches one from the right of f of x times g of x. Well, we already saw, when we're approaching one from the right of f of x, we are approaching negative one, but g of x approaching one from the right is still approaching zero, so this is going to be zero again. So this limit exists."}, {"video_title": "Limits of combined functions piecewise functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is gonna be two times, we're gonna be approaching two times zero, which is zero. And then we approach from the right, x approaches one from the right of f of x times g of x. Well, we already saw, when we're approaching one from the right of f of x, we are approaching negative one, but g of x approaching one from the right is still approaching zero, so this is going to be zero again. So this limit exists. We get the same limit when we approach from the left and the right. It is equal to zero. So these are pretty interesting examples, because sometimes when you think that the component limits don't exist, that that means that the sum or the product might not exist, but this shows at least two examples where that is not the case."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we were to graph all of the points x and y that satisfy this relationship, we get a unit circle like this. And what I'm curious about in this video is how we can figure out the slope of the tangent line at any point of this unit circle. And what immediately might be jumping out in your brain is well, a circle defined this way, this isn't a function. It's not y explicitly defined as a function of x. For any x value, you actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of one minus x squared."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's not y explicitly defined as a function of x. For any x value, you actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of one minus x squared. And you could say y is equal to the negative square root of one minus x squared. Take the derivatives of each of these separately, and you would be able to find the derivative for any x, or the derivative of the slope of the tangent line at any point. But what I wanna do in this video is literally leverage the chain rule to take the derivative implicitly so that I don't have to explicitly define y as a function of x either way."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could say y is equal to the positive square root of one minus x squared. And you could say y is equal to the negative square root of one minus x squared. Take the derivatives of each of these separately, and you would be able to find the derivative for any x, or the derivative of the slope of the tangent line at any point. But what I wanna do in this video is literally leverage the chain rule to take the derivative implicitly so that I don't have to explicitly define y as a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation and then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x and explicitly getting y is equal to f prime of x, they call this, which is really just an application of the chain rule, we call it implicit differentiation. Implicit differentiation."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what I wanna do in this video is literally leverage the chain rule to take the derivative implicitly so that I don't have to explicitly define y as a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation and then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x and explicitly getting y is equal to f prime of x, they call this, which is really just an application of the chain rule, we call it implicit differentiation. Implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, x squared plus y squared on the left-hand side of our equation, and then that's going to be equal to the derivative with respect to x on the right-hand side."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, x squared plus y squared on the left-hand side of our equation, and then that's going to be equal to the derivative with respect to x on the right-hand side. I'm just doing the same exact thing to both sides of this equation. Now, if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared plus the derivative with respect to x of y squared."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's the derivative with respect to x of x squared plus y squared, x squared plus y squared on the left-hand side of our equation, and then that's going to be equal to the derivative with respect to x on the right-hand side. I'm just doing the same exact thing to both sides of this equation. Now, if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. So let's see, this is going to be x squared, it's gonna be y squared, and then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be the same thing as the derivative with respect to x of x squared plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. So let's see, this is going to be x squared, it's gonna be y squared, and then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get zero. Now, this first term right over here, we have done many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This isn't changing with respect to x. So we just get zero. Now, this first term right over here, we have done many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be two times x to the first power, or we could just say two x. Now, what's interesting is what we're doing right over here. The derivative with respect to x of y squared."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative with respect to x of x squared is just the power rule here. It's going to be two times x to the first power, or we could just say two x. Now, what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative, let me make it clear, we're just gonna take the derivative of our something, the derivative of y squared, that's what we're taking, you can kind of view that as a function, with respect to y, with respect to y, and then multiply that times the derivative of y, the derivative of y with respect to x. We're assuming that y does change with respect to x. Y is not some type of a constant that we're writing just in abstract terms."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative, let me make it clear, we're just gonna take the derivative of our something, the derivative of y squared, that's what we're taking, you can kind of view that as a function, with respect to y, with respect to y, and then multiply that times the derivative of y, the derivative of y with respect to x. We're assuming that y does change with respect to x. Y is not some type of a constant that we're writing just in abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y with respect to x."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're assuming that y does change with respect to x. Y is not some type of a constant that we're writing just in abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y with respect to x. It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y as a function of x. Y as a function of x. This might be, or y as a function of x squared, which is essentially another way of writing what we had here. This might be a little bit clearer in terms of the chain rule."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we're taking the derivative of y with respect to x. It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y as a function of x. Y as a function of x. This might be, or y as a function of x squared, which is essentially another way of writing what we had here. This might be a little bit clearer in terms of the chain rule. The derivative of y as a function of x squared with respect to y of x, so the derivative of something squared, the derivative of something squared with respect to that something, times the derivative of that something with respect to x. This is just the chain rule. I wanna say it over and over again."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This might be a little bit clearer in terms of the chain rule. The derivative of y as a function of x squared with respect to y of x, so the derivative of something squared, the derivative of something squared with respect to that something, times the derivative of that something with respect to x. This is just the chain rule. I wanna say it over and over again. This is just the chain rule. So let's do that. What do we get on the right-hand side over here?"}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "I wanna say it over and over again. This is just the chain rule. So let's do that. What do we get on the right-hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y is just going to be two times y, two times y, just an application of the chain rule. And the derivative of y with respect to x, well, we don't know what that is, so we're just gonna leave that as times the derivative of y times the derivative of y with respect to x."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "What do we get on the right-hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y is just going to be two times y, two times y, just an application of the chain rule. And the derivative of y with respect to x, well, we don't know what that is, so we're just gonna leave that as times the derivative of y times the derivative of y with respect to x. So let's just write this down over here. So what we have is two x, two x plus, two x plus derivative of something squared with respect to that something is two times the something. In this case, the something is y, so two times y, and then times the derivative of y, the derivative of y with respect to x."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the derivative of y with respect to x, well, we don't know what that is, so we're just gonna leave that as times the derivative of y times the derivative of y with respect to x. So let's just write this down over here. So what we have is two x, two x plus, two x plus derivative of something squared with respect to that something is two times the something. In this case, the something is y, so two times y, and then times the derivative of y, the derivative of y with respect to x. And this is all going to be equal to, all going to be equal to zero. Now that was interesting. Now we have an equation that has the derivative of y with respect to x in it."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "In this case, the something is y, so two times y, and then times the derivative of y, the derivative of y with respect to x. And this is all going to be equal to, all going to be equal to zero. Now that was interesting. Now we have an equation that has the derivative of y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x, solve this equation."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we have an equation that has the derivative of y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x, solve this equation. So let's do that. And actually, just so we all, so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off, and let's continue there."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So all we have to do at this point is solve for the derivative of y with respect to x, solve this equation. So let's do that. And actually, just so we all, so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off, and let's continue there. So let's say, let's subtract two x from both sides. So we're left with two y times the derivative of y with respect to x is equal to, we're subtracting two x from both sides, so it's equal to negative two x. And then if we really want to solve for the derivative of y with respect to x, we can just divide both sides by two y."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is where we left off, and let's continue there. So let's say, let's subtract two x from both sides. So we're left with two y times the derivative of y with respect to x is equal to, we're subtracting two x from both sides, so it's equal to negative two x. And then if we really want to solve for the derivative of y with respect to x, we can just divide both sides by two y. We just divide both sides by two y. And we are left with the derivative of y with respect to x. Let's scroll down a little bit."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then if we really want to solve for the derivative of y with respect to x, we can just divide both sides by two y. We just divide both sides by two y. And we are left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, is equal to, well, the twos cancel out. We are left with negative x, negative x over, over y. So this is interesting."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's scroll down a little bit. The derivative of y with respect to x is equal to, is equal to, well, the twos cancel out. We are left with negative x, negative x over, over y. So this is interesting. We didn't have to explicitly define y as a function of x here, but we got our derivative in terms of an x and a y, not just only in terms of an x. But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point, this point right over here, which if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of two over two, comma, the square root of two over two."}, {"video_title": "Implicit differentiation   Advanced derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is interesting. We didn't have to explicitly define y as a function of x here, but we got our derivative in terms of an x and a y, not just only in terms of an x. But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point, this point right over here, which if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of two over two, comma, the square root of two over two. What is the slope of the tangent line there? Well, we figured it out. It's going to be negative x over y."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "A 20 meter ladder is leaning against a wall. The distance x of t between the bottom of the ladder and the wall is increasing at a rate of three meters per minute. At a certain instant, t sub zero, the top of the ladder is a distance y of t sub zero of 15 meters from the ground. What is the rate of change of the angle, theta of t between the ground and the ladder at that instant? What I'm gonna do is draw this out, and really the first step is to think about, well, what equation will be helpful for us to solve this problem? Then we might just go ahead and actually solve the problem. A 20 meter ladder is leaning against a wall."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is the rate of change of the angle, theta of t between the ground and the ladder at that instant? What I'm gonna do is draw this out, and really the first step is to think about, well, what equation will be helpful for us to solve this problem? Then we might just go ahead and actually solve the problem. A 20 meter ladder is leaning against a wall. Let me draw ourselves a wall here. That is my wall. Now let me draw our 20 meter ladder."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "A 20 meter ladder is leaning against a wall. Let me draw ourselves a wall here. That is my wall. Now let me draw our 20 meter ladder. Maybe it looks something like that. That is 20 meters. They say the distance x of t between the bottom of the ladder and the wall, so it's this distance right over here, this distance right over here is x of t. They say it's increasing at a rate of three meters per minute."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let me draw our 20 meter ladder. Maybe it looks something like that. That is 20 meters. They say the distance x of t between the bottom of the ladder and the wall, so it's this distance right over here, this distance right over here is x of t. They say it's increasing at a rate of three meters per minute. We know that we could either say x prime of t, which is the same thing as dx dt, is equal to three meters. I'll write it out, because it's hard if I just said m per m. It might not be that clear. Meters per minute, so they give us that piece of information."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "They say the distance x of t between the bottom of the ladder and the wall, so it's this distance right over here, this distance right over here is x of t. They say it's increasing at a rate of three meters per minute. We know that we could either say x prime of t, which is the same thing as dx dt, is equal to three meters. I'll write it out, because it's hard if I just said m per m. It might not be that clear. Meters per minute, so they give us that piece of information. The rate of change of x with respect to time, they give us that. At a certain instant, t sub zero, the top of the ladder is a distance of 15 meters. The top of the ladder, so let's make this very clear."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Meters per minute, so they give us that piece of information. The rate of change of x with respect to time, they give us that. At a certain instant, t sub zero, the top of the ladder is a distance of 15 meters. The top of the ladder, so let's make this very clear. This distance right over here is y of t, y of t. They say at time t sub zero, y of t is 15 meters. Let me just write it here. Y of t sub zero is equal to 15 meters."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "The top of the ladder, so let's make this very clear. This distance right over here is y of t, y of t. They say at time t sub zero, y of t is 15 meters. Let me just write it here. Y of t sub zero is equal to 15 meters. Let me write this right over here. This is y of t sub zero. Let's just assume that we're drawing it at that moment, t sub zero, because I think that's going to be important."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Y of t sub zero is equal to 15 meters. Let me write this right over here. This is y of t sub zero. Let's just assume that we're drawing it at that moment, t sub zero, because I think that's going to be important. Y at t sub zero is equal to 15 meters. They want to know what is the rate of change of the angle theta between the ground and the ladder. This is the same."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's just assume that we're drawing it at that moment, t sub zero, because I think that's going to be important. Y at t sub zero is equal to 15 meters. They want to know what is the rate of change of the angle theta between the ground and the ladder. This is the same. Theta is also going to change with respect to time. It's going to be a function of time between the ground and the ladder at that instant. Theta, let me get a new color here."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the same. Theta is also going to change with respect to time. It's going to be a function of time between the ground and the ladder at that instant. Theta, let me get a new color here. Theta is this angle right over here. This is theta, and it's also going to be a function of time. What we'll always want to do in these related rates problems is we want to set up an equation, and really an algebraic equation, maybe a little bit of trigonometry involved, that relates the things that we care about, and then we're likely to have to take the derivative of both sides of that in order to relate the related rates."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Theta, let me get a new color here. Theta is this angle right over here. This is theta, and it's also going to be a function of time. What we'll always want to do in these related rates problems is we want to set up an equation, and really an algebraic equation, maybe a little bit of trigonometry involved, that relates the things that we care about, and then we're likely to have to take the derivative of both sides of that in order to relate the related rates. Let's see. We want to know the rate of change of the angle between the ground and the ladder at that instant. What we need to figure out, we want to figure out theta prime at t sub zero."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we'll always want to do in these related rates problems is we want to set up an equation, and really an algebraic equation, maybe a little bit of trigonometry involved, that relates the things that we care about, and then we're likely to have to take the derivative of both sides of that in order to relate the related rates. Let's see. We want to know the rate of change of the angle between the ground and the ladder at that instant. What we need to figure out, we want to figure out theta prime at t sub zero. This is what we want to figure out. They've given us some interesting things. They've given us, I guess, our rate of change of x with respect to time is constant at three meters per minute, and we know what y is at that moment."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we need to figure out, we want to figure out theta prime at t sub zero. This is what we want to figure out. They've given us some interesting things. They've given us, I guess, our rate of change of x with respect to time is constant at three meters per minute, and we know what y is at that moment. Let's see. Can we create a relationship? Because they gave us dx dt, it'll actually be more useful to find a relationship between x and theta, and then take the derivative of both sides, and then use this information, possibly, to figure out what the appropriate value of x or theta is at that moment."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "They've given us, I guess, our rate of change of x with respect to time is constant at three meters per minute, and we know what y is at that moment. Let's see. Can we create a relationship? Because they gave us dx dt, it'll actually be more useful to find a relationship between x and theta, and then take the derivative of both sides, and then use this information, possibly, to figure out what the appropriate value of x or theta is at that moment. Let's do that. How does x relate to theta? Well, we use a little bit of trigonometry right over here."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because they gave us dx dt, it'll actually be more useful to find a relationship between x and theta, and then take the derivative of both sides, and then use this information, possibly, to figure out what the appropriate value of x or theta is at that moment. Let's do that. How does x relate to theta? Well, we use a little bit of trigonometry right over here. If you took the hypotenuse times the cosine of theta, you would get x. Let me write this right over here. x of t, x of t, is equal to the hypotenuse, 20 meters, that's the length of the ladder, times the cosine, cosine of theta, and I could say the cosine of theta of t just to make it clear that this is a function of time."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we use a little bit of trigonometry right over here. If you took the hypotenuse times the cosine of theta, you would get x. Let me write this right over here. x of t, x of t, is equal to the hypotenuse, 20 meters, that's the length of the ladder, times the cosine, cosine of theta, and I could say the cosine of theta of t just to make it clear that this is a function of time. This comes straight out of trigonometry, actually our basic trigonometric function definitions. Now, why is this useful? Why do I think this is useful?"}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "x of t, x of t, is equal to the hypotenuse, 20 meters, that's the length of the ladder, times the cosine, cosine of theta, and I could say the cosine of theta of t just to make it clear that this is a function of time. This comes straight out of trigonometry, actually our basic trigonometric function definitions. Now, why is this useful? Why do I think this is useful? Well, let's think about what happens when I take the derivative of both sides using the chain rule. On the left-hand side, I am going to have an x prime of t, and then that's going to be equal to, what do I end up on the right-hand side? Well, using the chain rule, first I'll take the derivative with respect to theta, and so that's just going to be negative 20 sine of theta of t, and then I need to multiply that times theta prime of t. So, what I could do is say, hey, look, at t sub zero, I know what x prime of t is, I could try to figure out what sine of theta of t is, and then I'll just solve for this right over there."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Why do I think this is useful? Well, let's think about what happens when I take the derivative of both sides using the chain rule. On the left-hand side, I am going to have an x prime of t, and then that's going to be equal to, what do I end up on the right-hand side? Well, using the chain rule, first I'll take the derivative with respect to theta, and so that's just going to be negative 20 sine of theta of t, and then I need to multiply that times theta prime of t. So, what I could do is say, hey, look, at t sub zero, I know what x prime of t is, I could try to figure out what sine of theta of t is, and then I'll just solve for this right over there. So, let's do that. So, at t sub zero, so at t is equal to t sub zero, t sub zero, what we're gonna have, x prime of t, well, that's at every time, it's three meters per minute, we'll assume that our rates are in meters per minute, and just our values are in meters when we're talking about distance, and our angles are in radians. So, this is going to be equal to three is equal to negative 20 times sine of theta of t times the derivative of theta with respect to time."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, using the chain rule, first I'll take the derivative with respect to theta, and so that's just going to be negative 20 sine of theta of t, and then I need to multiply that times theta prime of t. So, what I could do is say, hey, look, at t sub zero, I know what x prime of t is, I could try to figure out what sine of theta of t is, and then I'll just solve for this right over there. So, let's do that. So, at t sub zero, so at t is equal to t sub zero, t sub zero, what we're gonna have, x prime of t, well, that's at every time, it's three meters per minute, we'll assume that our rates are in meters per minute, and just our values are in meters when we're talking about distance, and our angles are in radians. So, this is going to be equal to three is equal to negative 20 times sine of theta of t times the derivative of theta with respect to time. So, how do we figure out what sine of theta of t is going to be? Well, let's just use that other information they gave us, and I'm gonna scroll down a little bit, get a little bit more real estate. So, sine of theta, let me write it over here, sine of theta at time t sub naught, that's what we care about, t is equal to t sub naught, what's that going to be?"}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, this is going to be equal to three is equal to negative 20 times sine of theta of t times the derivative of theta with respect to time. So, how do we figure out what sine of theta of t is going to be? Well, let's just use that other information they gave us, and I'm gonna scroll down a little bit, get a little bit more real estate. So, sine of theta, let me write it over here, sine of theta at time t sub naught, that's what we care about, t is equal to t sub naught, what's that going to be? Well, sine is opposite over hypotenuse, so that's going to be y at t sub naught over our hypotenuse of 20 meters. Well, that's going to be equal to, that's going to be equal to, they tell us y of t sub naught is 15 meters over 20 meters, which is the same thing as 3 4ths. So, by this yellow information, they actually told us that this right over here is going to be equal to 3 4ths."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, sine of theta, let me write it over here, sine of theta at time t sub naught, that's what we care about, t is equal to t sub naught, what's that going to be? Well, sine is opposite over hypotenuse, so that's going to be y at t sub naught over our hypotenuse of 20 meters. Well, that's going to be equal to, that's going to be equal to, they tell us y of t sub naught is 15 meters over 20 meters, which is the same thing as 3 4ths. So, by this yellow information, they actually told us that this right over here is going to be equal to 3 4ths. So, this times 3 4ths times the rate of change of theta with respect to t. And so, now we just solve for this, and we're done. So, this is going to be, what's negative 20 times 3 4ths? That is negative 15, that is negative 15."}, {"video_title": "Analyzing related rates problems equations (trig)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, by this yellow information, they actually told us that this right over here is going to be equal to 3 4ths. So, this times 3 4ths times the rate of change of theta with respect to t. And so, now we just solve for this, and we're done. So, this is going to be, what's negative 20 times 3 4ths? That is negative 15, that is negative 15. If we divide both sides by negative 15, we get theta prime of t is equal to three over negative 15, three over negative 15, three over negative 15, which is the same thing as being equal to negative 1 5th, and the units here would be in radians per minute, because our rates are all in per minute. So, if I wanted to, I could write radians per minute. Ideally, I would write it right over here, but there you go."}, {"video_title": "Justification with the intermediate value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can we use the intermediate value theorem to say that there is a value c such that g of c is equal to zero and negative one is less than or equal to c is less than or equal to one? If so, write a justification. So in order to even use the intermediate value theorem, you have to be continuous over the interval that you care about, and this interval that we care about is from x equals negative one to one. And one over x is not continuous over that interval. It is not defined when x is equal to zero. And so we could say no, because, because g of x not defined, not defined, or I could say, let me just say not continuous. It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one."}, {"video_title": "Justification with the intermediate value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And one over x is not continuous over that interval. It is not defined when x is equal to zero. And so we could say no, because, because g of x not defined, not defined, or I could say, let me just say not continuous. It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one. And we could even put parentheses not defined, not defined at x is equal to zero. All right, now let's start asking the second question. Can we use the intermediate value theorem to say that the equation g of x is equal to 3 4ths has a solution where one is less than or equal to x is less than or equal to two?"}, {"video_title": "Justification with the intermediate value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one. And we could even put parentheses not defined, not defined at x is equal to zero. All right, now let's start asking the second question. Can we use the intermediate value theorem to say that the equation g of x is equal to 3 4ths has a solution where one is less than or equal to x is less than or equal to two? If so, write a justification. All right, so first let's look at the interval. If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval."}, {"video_title": "Justification with the intermediate value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can we use the intermediate value theorem to say that the equation g of x is equal to 3 4ths has a solution where one is less than or equal to x is less than or equal to two? If so, write a justification. All right, so first let's look at the interval. If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval. So we could say g of x is continuous, is continuous on the closed interval from one to two. And if you wanted to put more justification there, you could say g defined, defined for all real numbers, real numbers, such that x does not equal zero, x does not equal zero. I could write g of x defined for all real numbers such that x does not equal to zero."}, {"video_title": "Justification with the intermediate value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval. So we could say g of x is continuous, is continuous on the closed interval from one to two. And if you wanted to put more justification there, you could say g defined, defined for all real numbers, real numbers, such that x does not equal zero, x does not equal zero. I could write g of x defined for all real numbers such that x does not equal to zero. And you could say rational functions like one over x are continuous, are continuous at all points in their domains, at all points in their domain. That's going, really establishing that g of x is continuous on that interval. And then we wanna see what values does g take over, or what values does g take on at the endpoints, or actually these are the endpoints that we're looking at right over here."}, {"video_title": "Justification with the intermediate value theorem equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could write g of x defined for all real numbers such that x does not equal to zero. And you could say rational functions like one over x are continuous, are continuous at all points in their domains, at all points in their domain. That's going, really establishing that g of x is continuous on that interval. And then we wanna see what values does g take over, or what values does g take on at the endpoints, or actually these are the endpoints that we're looking at right over here. G of one is going to be equal to one over one is one. And g of two is going to be one over two is equal to one over two. So 3 4ths is between, is between g of one and g of two."}, {"video_title": "Introduction to limits at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "We now have a lot of experience taking limits of functions, if I'm taking the limit of f of x, we're gonna think about what does f of x approach as x approaches some value a, and this would be equal to some limit. Now everything we've done up till now is where a is a finite value, but when you look at the graph of the function f right over here, you see something interesting happens. As x gets larger and larger, it looks like our function f is getting closer and closer to two. It looks like we have a horizontal asymptote at y equals two. Similarly, as x gets more and more negative, it also seems like we have a horizontal asymptote at y equals two. So is there some type of notation we can use to think about what is the graph approaching as x gets much larger or as x gets smaller and smaller? And the answer there is limits at infinity."}, {"video_title": "Introduction to limits at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like we have a horizontal asymptote at y equals two. Similarly, as x gets more and more negative, it also seems like we have a horizontal asymptote at y equals two. So is there some type of notation we can use to think about what is the graph approaching as x gets much larger or as x gets smaller and smaller? And the answer there is limits at infinity. So if we wanna think about what is this graph, what is this function approaching as x gets larger and larger, we can think about the limit of f of x as x approaches positive infinity. So that's the notation, and I'm not going to give you the formal definition of this right now. There, in future videos, we might do that, but it's this idea as x gets larger and larger and larger, does it look like that our function is approaching some finite value, that we have a horizontal asymptote there?"}, {"video_title": "Introduction to limits at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the answer there is limits at infinity. So if we wanna think about what is this graph, what is this function approaching as x gets larger and larger, we can think about the limit of f of x as x approaches positive infinity. So that's the notation, and I'm not going to give you the formal definition of this right now. There, in future videos, we might do that, but it's this idea as x gets larger and larger and larger, does it look like that our function is approaching some finite value, that we have a horizontal asymptote there? And in this situation, it looks like it is. It looks like it's approaching the value two. And for this particular function, the limit of f of x as x approaches negative infinity also looks like it is approaching two."}, {"video_title": "Introduction to limits at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "There, in future videos, we might do that, but it's this idea as x gets larger and larger and larger, does it look like that our function is approaching some finite value, that we have a horizontal asymptote there? And in this situation, it looks like it is. It looks like it's approaching the value two. And for this particular function, the limit of f of x as x approaches negative infinity also looks like it is approaching two. This is not always going to be the same. You could have a situation, maybe we had, you could have another function, so let me draw a little horizontal asymptote right over here. You could imagine a function that looks like this."}, {"video_title": "Introduction to limits at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And for this particular function, the limit of f of x as x approaches negative infinity also looks like it is approaching two. This is not always going to be the same. You could have a situation, maybe we had, you could have another function, so let me draw a little horizontal asymptote right over here. You could imagine a function that looks like this. So let me do it like that, and maybe it does something wacky like this, and then it comes down, and it does something like this. Here, our limit as x approaches infinity is still two, but our limit as x approaches negative infinity right over here would be negative two. And of course, there's many situations where as you approach infinity or negative infinity, you aren't actually approaching some finite value."}, {"video_title": "Introduction to limits at infinity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could imagine a function that looks like this. So let me do it like that, and maybe it does something wacky like this, and then it comes down, and it does something like this. Here, our limit as x approaches infinity is still two, but our limit as x approaches negative infinity right over here would be negative two. And of course, there's many situations where as you approach infinity or negative infinity, you aren't actually approaching some finite value. You don't have a horizontal asymptote. But the whole point of this video is just to make you familiar with this notation. And limits at infinity, or you could say limits at negative infinity, they have a different formal definition than some of the limits that we've looked at in the past where we are approaching a finite value, but intuitively, they make sense that these are indeed limits."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "In particular, let's see if we can figure out what the derivative with respect to x of the inverse tangent of x is. I encourage you to pause this video and use a technique similar to the one, or very close to the one, that we've used in the last few videos to figure out what this is. Let's set y equal to the inverse tangent of x. y is equal to inverse tangent of x. That is the same thing as saying that the tangent of y is equal to x. All I've done is, you can kind of think of it as, I've just taken the tangent of both sides right over here. Now we can take the derivative of both sides with respect to x. On the left-hand side, we can just apply the chain rule."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "That is the same thing as saying that the tangent of y is equal to x. All I've done is, you can kind of think of it as, I've just taken the tangent of both sides right over here. Now we can take the derivative of both sides with respect to x. On the left-hand side, we can just apply the chain rule. Derivative of tangent of y with respect to y is going to be secant squared of y, which is the same thing as 1 over cosine of y squared. I like to write it this way. It keeps it a little bit simpler, at least in my brain."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "On the left-hand side, we can just apply the chain rule. Derivative of tangent of y with respect to y is going to be secant squared of y, which is the same thing as 1 over cosine of y squared. I like to write it this way. It keeps it a little bit simpler, at least in my brain. When we're applying the chain rule, it's going to be the derivative of tangent of y with respect to y times the derivative of y with respect to x. On the right-hand side, the derivative of x with respect to x is going to be equal to 1. If we want to solve for the derivative of y with respect to x, we just multiply both sides times the cosine of y squared."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "It keeps it a little bit simpler, at least in my brain. When we're applying the chain rule, it's going to be the derivative of tangent of y with respect to y times the derivative of y with respect to x. On the right-hand side, the derivative of x with respect to x is going to be equal to 1. If we want to solve for the derivative of y with respect to x, we just multiply both sides times the cosine of y squared. We get the derivative of y with respect to x is equal to cosine of y squared. Like we've seen in previous videos, this isn't that satisfying. I've written the derivative of y with respect to x as a function of y."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "If we want to solve for the derivative of y with respect to x, we just multiply both sides times the cosine of y squared. We get the derivative of y with respect to x is equal to cosine of y squared. Like we've seen in previous videos, this isn't that satisfying. I've written the derivative of y with respect to x as a function of y. What we're really interested in is writing it as a function of x. To do that, we need to express this somehow in terms of the tangent of y. The reason why the tangent of y is interesting is because we already know that tangent of y is equal to x."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "I've written the derivative of y with respect to x as a function of y. What we're really interested in is writing it as a function of x. To do that, we need to express this somehow in terms of the tangent of y. The reason why the tangent of y is interesting is because we already know that tangent of y is equal to x. If we can rewrite this using a little bit of trigonometric identities, then with tangent of y, we can substitute all the tangent of y's with an x. Let's see if we can do that. This seems a little bit tricky."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The reason why the tangent of y is interesting is because we already know that tangent of y is equal to x. If we can rewrite this using a little bit of trigonometric identities, then with tangent of y, we can substitute all the tangent of y's with an x. Let's see if we can do that. This seems a little bit tricky. To introduce a tangent of y, we'd want to have a sine divided by a cosine. That's what tangent is. This is just a straight-up cosine squared y."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This seems a little bit tricky. To introduce a tangent of y, we'd want to have a sine divided by a cosine. That's what tangent is. This is just a straight-up cosine squared y. This is really going to take a little bit more experimentation than the last few examples we've done. One thing we could do is say, let's just divide by 1. Dividing by 1 never hurt anyone."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is just a straight-up cosine squared y. This is really going to take a little bit more experimentation than the last few examples we've done. One thing we could do is say, let's just divide by 1. Dividing by 1 never hurt anyone. We could say that this is the same thing as cosine squared y. I'm really doing this to see if I can start to express it as some type of a rational expression, which might involve, at some point, a sine divided by a cosine, and I could have a tangent. Let's divide by 1. We know from the Pythagorean identity that 1 is equal to sine squared of y plus cosine squared of y."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Dividing by 1 never hurt anyone. We could say that this is the same thing as cosine squared y. I'm really doing this to see if I can start to express it as some type of a rational expression, which might involve, at some point, a sine divided by a cosine, and I could have a tangent. Let's divide by 1. We know from the Pythagorean identity that 1 is equal to sine squared of y plus cosine squared of y. Let's write that. We could write cosine squared of y plus sine squared of y. Once again, why was I able to divide by this expression?"}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We know from the Pythagorean identity that 1 is equal to sine squared of y plus cosine squared of y. Let's write that. We could write cosine squared of y plus sine squared of y. Once again, why was I able to divide by this expression? This expression by the Pythagorean identity, which really comes out of the unit circle definition of trig functions, this is equal to 1, so I have not changed the value of this expression. What makes this interesting is if I want to introduce a sine divided by a cosine, I could just divide the numerator and the denominator by cosine squared. Let's do that."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Once again, why was I able to divide by this expression? This expression by the Pythagorean identity, which really comes out of the unit circle definition of trig functions, this is equal to 1, so I have not changed the value of this expression. What makes this interesting is if I want to introduce a sine divided by a cosine, I could just divide the numerator and the denominator by cosine squared. Let's do that. Let's divide the numerator by cosine squared of y and divide the denominator by cosine squared of y, or multiply each of them by 1 over cosine squared of y. What's that going to give us? The numerator, these characters are going to cancel."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's do that. Let's divide the numerator by cosine squared of y and divide the denominator by cosine squared of y, or multiply each of them by 1 over cosine squared of y. What's that going to give us? The numerator, these characters are going to cancel. You're just going to have a 1. The denominator, this times this, that's just going to be equal to 1. Then you're going to have sine squared y over cosine squared y."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "The numerator, these characters are going to cancel. You're just going to have a 1. The denominator, this times this, that's just going to be equal to 1. Then you're going to have sine squared y over cosine squared y. This is the goal that I was trying to achieve. I have a sine divided by cosine squared. This right over here, this is the same thing."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Then you're going to have sine squared y over cosine squared y. This is the goal that I was trying to achieve. I have a sine divided by cosine squared. This right over here, this is the same thing. Let me write it this way. This is the same thing as sine of y over cosine of y whole thing squared, which is, of course, the same thing as 1 over 1 plus tangent of y squared. This is equal to this."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This right over here, this is the same thing. Let me write it this way. This is the same thing as sine of y over cosine of y whole thing squared, which is, of course, the same thing as 1 over 1 plus tangent of y squared. This is equal to this. Why is that useful? We know that x is equal to tangent of y. This is going to be equal to 1 over 1 plus tangent of y is equal to x squared, which is pretty exciting."}, {"video_title": "Derivative of inverse tangent   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is equal to this. Why is that useful? We know that x is equal to tangent of y. This is going to be equal to 1 over 1 plus tangent of y is equal to x squared, which is pretty exciting. We just figured out the derivative of y with respect to x. The derivative of this thing with respect to x is 1 over 1 plus x squared. We could write that right up here."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have already covered the notion of area between a curve and the x-axis using a definite integral. We are now going to then extend this to think about the area between curves. So let's say we care about the region from x equals a to x equals b between y equals f of x and y is equal to g of x. So that would be this area right over here. So based on what you already know about definite integrals, how would you actually try to calculate this? Well, one natural thing that you might say is, well, look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis, and then if I were to subtract from that, if I were to subtract from that this area right over here, which is equal to, that's the definite integral from a to b of g of x dx, well, then I would net out with the original area that I cared about. I would net out with this area right over here, and that indeed would be the case, and we know from our integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx, and now I will make a claim to you, and we'll build a little bit more intuition for this as we go through this video, but over an interval from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x interval that we care about from a to b of f of x minus g of x, so I know what you're thinking."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would be this area right over here. So based on what you already know about definite integrals, how would you actually try to calculate this? Well, one natural thing that you might say is, well, look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis, and then if I were to subtract from that, if I were to subtract from that this area right over here, which is equal to, that's the definite integral from a to b of g of x dx, well, then I would net out with the original area that I cared about. I would net out with this area right over here, and that indeed would be the case, and we know from our integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx, and now I will make a claim to you, and we'll build a little bit more intuition for this as we go through this video, but over an interval from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x interval that we care about from a to b of f of x minus g of x, so I know what you're thinking. You're like, okay, well, that worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So, for example, let's say that we were to think about this interval right over here. Let's say this is the point c, and that's x equals c. This is x equals d right over here, so what if we wanted to calculate this area that I am shading in right over here?"}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "I would net out with this area right over here, and that indeed would be the case, and we know from our integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx, and now I will make a claim to you, and we'll build a little bit more intuition for this as we go through this video, but over an interval from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x interval that we care about from a to b of f of x minus g of x, so I know what you're thinking. You're like, okay, well, that worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So, for example, let's say that we were to think about this interval right over here. Let's say this is the point c, and that's x equals c. This is x equals d right over here, so what if we wanted to calculate this area that I am shading in right over here? You might say, well, does this actually work? Well, let's think about now what the integral, let's think about what the integral from c to d of f of x dx represents. Well, that would represent this area right over here, and what would the integral from c to d of g of x dx represent?"}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say this is the point c, and that's x equals c. This is x equals d right over here, so what if we wanted to calculate this area that I am shading in right over here? You might say, well, does this actually work? Well, let's think about now what the integral, let's think about what the integral from c to d of f of x dx represents. Well, that would represent this area right over here, and what would the integral from c to d of g of x dx represent? Well, you might say it is this area right over here, but remember, over this interval, g of x is below the x-axis, so this would give you a negative value, but if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get the entire positive area. Well, this just amounted to, this is equivalent to the integral from c to d of f of x, of f of x minus g of x again, minus g of x. Let me make it clear, we've got parentheses there, and then we have our dx."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that would represent this area right over here, and what would the integral from c to d of g of x dx represent? Well, you might say it is this area right over here, but remember, over this interval, g of x is below the x-axis, so this would give you a negative value, but if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get the entire positive area. Well, this just amounted to, this is equivalent to the integral from c to d of f of x, of f of x minus g of x again, minus g of x. Let me make it clear, we've got parentheses there, and then we have our dx. So once again, even over this interval, when f of x was above the x-axis and g of x was below the x-axis, it still boiled down to the same thing. Well, let's take another scenario. Let's take the scenario when they are both below the x-axis."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me make it clear, we've got parentheses there, and then we have our dx. So once again, even over this interval, when f of x was above the x-axis and g of x was below the x-axis, it still boiled down to the same thing. Well, let's take another scenario. Let's take the scenario when they are both below the x-axis. Let's say that we wanted to go from x equals, well, I won't use e since that is a loaded letter in mathematics, and so is f and g. Well, let's just say, well, I'm kind of running out of letters now. Let's say that I am gonna go from, I don't know, let's just call this m, and let's call this n right over here. Well, n is getting, let's put n right over here."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's take the scenario when they are both below the x-axis. Let's say that we wanted to go from x equals, well, I won't use e since that is a loaded letter in mathematics, and so is f and g. Well, let's just say, well, I'm kind of running out of letters now. Let's say that I am gonna go from, I don't know, let's just call this m, and let's call this n right over here. Well, n is getting, let's put n right over here. So what I care about is this area, the area, once again, below f. We're assuming that we're looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n? Well, let's think about it a little bit. If we were to evaluate that integral from m to n of, let's put my dx here, of f of x minus, minus g of x, we already know from our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, n is getting, let's put n right over here. So what I care about is this area, the area, once again, below f. We're assuming that we're looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n? Well, let's think about it a little bit. If we were to evaluate that integral from m to n of, let's put my dx here, of f of x minus, minus g of x, we already know from our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx. Now let's think about what each of these represent. So this yellow integral right over here, that would give this, the negative of this area. So that would give a negative value here, but the magnitude of it, the absolute value of it, would be this area right over there."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we were to evaluate that integral from m to n of, let's put my dx here, of f of x minus, minus g of x, we already know from our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx. Now let's think about what each of these represent. So this yellow integral right over here, that would give this, the negative of this area. So that would give a negative value here, but the magnitude of it, the absolute value of it, would be this area right over there. Now what would just the integral, I'm not even thinking about the negative sign here, what would the integral of this g of x of this blue integral give? Well, that would give this the negative of this entire area. But now we're gonna take the negative of that, and so this part right over here, this entire part, including this negative sign, would give us, would give us this entire area, the entire area."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would give a negative value here, but the magnitude of it, the absolute value of it, would be this area right over there. Now what would just the integral, I'm not even thinking about the negative sign here, what would the integral of this g of x of this blue integral give? Well, that would give this the negative of this entire area. But now we're gonna take the negative of that, and so this part right over here, this entire part, including this negative sign, would give us, would give us this entire area, the entire area. This would actually give a positive value because we're taking the negative of a negative. But if with the area that we care about, right over here, the area that we cared about originally, we would want to subtract out this yellow area. Well, this right over here, this yellow integral from the definite integral from m to n of f of x dx, that's exactly that."}, {"video_title": "Area between curves   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "But now we're gonna take the negative of that, and so this part right over here, this entire part, including this negative sign, would give us, would give us this entire area, the entire area. This would actually give a positive value because we're taking the negative of a negative. But if with the area that we care about, right over here, the area that we cared about originally, we would want to subtract out this yellow area. Well, this right over here, this yellow integral from the definite integral from m to n of f of x dx, that's exactly that. That is the negative of that yellow area. So if you add the blue area, and so the negative of a negative, this is gonna be positive, and then this is going to be the negative of the yellow area, you would net out, once again, to the area that we think about. So in every case we saw, if we're talking about an interval where f of x is greater than g of x, the area between the curves is just the definite integral over that interval of f of x minus g of x dx."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that f of x is equal to x. The power rule tells us that f prime of x is going to be equal to what? Well, x is the same thing as x to the first power, so n is implicitly 1 right over here. So we bring the 1 out front, it'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power. x to the 0 is just 1, so it's just going to be equal to 1. Now does that make conceptual sense if we actually try to visualize these functions?"}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we bring the 1 out front, it'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power. x to the 0 is just 1, so it's just going to be equal to 1. Now does that make conceptual sense if we actually try to visualize these functions? So let me draw, let me actually try to graph these functions. So that's my y-axis, this is my x-axis. And let me graph y equals x."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now does that make conceptual sense if we actually try to visualize these functions? So let me draw, let me actually try to graph these functions. So that's my y-axis, this is my x-axis. And let me graph y equals x. So y is equal to f of x here, so y is equal to x. So it looks something like that. So y is equal to x, or this is f of x equals to x, or y is equal to this f of x right over there."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me graph y equals x. So y is equal to f of x here, so y is equal to x. So it looks something like that. So y is equal to x, or this is f of x equals to x, or y is equal to this f of x right over there. Now, the derivative, actually let me just call that f of x, just to make it, not to confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So y is equal to x, or this is f of x equals to x, or y is equal to this f of x right over there. Now, the derivative, actually let me just call that f of x, just to make it, not to confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x. And now let me graph the derivative, let me graph f prime of x. That's saying it's 1, that's saying it's 1 for all x, regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest?"}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "y is equal to f of x, which is equal to x. And now let me graph the derivative, let me graph f prime of x. That's saying it's 1, that's saying it's 1 for all x, regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function. What is the slope of the line, or the tangent line, right at this point? Well, the slope of this right over here, this has slope 1 continuously, or it has a constant slope of 1."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function. What is the slope of the line, or the tangent line, right at this point? Well, the slope of this right over here, this has slope 1 continuously, or it has a constant slope of 1. Slope is equal to 1 no matter what x is, it's a line. And for a line, the slope is constant. So over here, the slope is indeed 1."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the slope of this right over here, this has slope 1 continuously, or it has a constant slope of 1. Slope is equal to 1 no matter what x is, it's a line. And for a line, the slope is constant. So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we got a pretty valid response there."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we got a pretty valid response there. Now let's try something where the slope might change. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what?"}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we got a pretty valid response there. Now let's try something where the slope might change. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2, so it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2x to the first power, it's going to be equal to 2x. So let's see if this makes reasonable sense."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2, so it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2x to the first power, it's going to be equal to 2x. So let's see if this makes reasonable sense. I'm going to try to graph this one a little bit more precisely. So let's see how well I can graph it, how precisely I can graph it. So this is x-axis, y-axis, let me mark some stuff off here."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if this makes reasonable sense. I'm going to try to graph this one a little bit more precisely. So let's see how well I can graph it, how precisely I can graph it. So this is x-axis, y-axis, let me mark some stuff off here. So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is x-axis, y-axis, let me mark some stuff off here. So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4. So g of x, when x is 0, it's 0. When x is 1, it is 1. When x is negative 1, it's 1."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "1, 2, 3, 4. So g of x, when x is 0, it's 0. When x is 1, it is 1. When x is negative 1, it's 1. When x is 2, it is 4. So that puts us right over there, 1, 2, 3, 4. Puts us right over there."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is negative 1, it's 1. When x is 2, it is 4. So that puts us right over there, 1, 2, 3, 4. Puts us right over there. When x is negative 2, you get to 4. It's a parabola. You've seen this for many years."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Puts us right over there. When x is negative 2, you get to 4. It's a parabola. You've seen this for many years. So it looks something like this. Actually, the last two points I graphed were a little bit weird. So this might be right over here."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You've seen this for many years. So it looks something like this. Actually, the last two points I graphed were a little bit weird. So this might be right over here. So it looks something like this. It looks something like that. And then when you come over here, it looks something like that."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this might be right over here. So it looks something like this. It looks something like that. And then when you come over here, it looks something like that. It's symmetric, so I'm trying my best to draw it reasonably. So there you go. That's the graph of g of x is equal to x squared."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then when you come over here, it looks something like that. It's symmetric, so I'm trying my best to draw it reasonably. So there you go. That's the graph of g of x is equal to x squared. Now let's graph g prime of x, or what the power rule is telling us that g prime of x is. So g prime of x is equal to 2x. So that's just a line that goes through the origin of slope 2."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the graph of g of x is equal to x squared. Now let's graph g prime of x, or what the power rule is telling us that g prime of x is. So g prime of x is equal to 2x. So that's just a line that goes through the origin of slope 2. So it looks something like that. When x is equal to 1, y is equal to 2. When x is equal to 2, y or g of x is equal to 4."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's just a line that goes through the origin of slope 2. So it looks something like that. When x is equal to 1, y is equal to 2. When x is equal to 2, y or g of x is equal to 4. So it looks something like this. Let me try my best to draw a straight line. It looks something like this."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is equal to 2, y or g of x is equal to 4. So it looks something like this. Let me try my best to draw a straight line. It looks something like this. Now, does this make sense? Well, if you just eyeball it really fast, if you look at this point right over here, and you want to think about the slope of the tangent line, the slope, I'm going to try my best to draw, let me do this in a color that pops out a little bit more. So the slope of the tangent line would look something like this."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks something like this. Now, does this make sense? Well, if you just eyeball it really fast, if you look at this point right over here, and you want to think about the slope of the tangent line, the slope, I'm going to try my best to draw, let me do this in a color that pops out a little bit more. So the slope of the tangent line would look something like this. So it looks like it has a reasonably high negative slope. It's definitely a negative slope, and it's a pretty steep negative slope. Well, for x is equal to 2, g prime of 2, or sorry, for x is equal to negative 2, g prime of negative 2 is equal to 2 times negative 2, which is equal to negative 4."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope of the tangent line would look something like this. So it looks like it has a reasonably high negative slope. It's definitely a negative slope, and it's a pretty steep negative slope. Well, for x is equal to 2, g prime of 2, or sorry, for x is equal to negative 2, g prime of negative 2 is equal to 2 times negative 2, which is equal to negative 4. So this is claiming that the slope at this point, so this right over here is negative 4, saying that the slope of this point is negative 4. m is equal to negative 4. That looks about right. It's a fairly steep negative slope."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, for x is equal to 2, g prime of 2, or sorry, for x is equal to negative 2, g prime of negative 2 is equal to 2 times negative 2, which is equal to negative 4. So this is claiming that the slope at this point, so this right over here is negative 4, saying that the slope of this point is negative 4. m is equal to negative 4. That looks about right. It's a fairly steep negative slope. Now, what happens if you go right over here when x is equal to 0? Well, our derivative, if you say g prime of 0, is telling us that the slope of our original function g at x is equal to 0 is 2 times 0 is 0. Well, does that make sense?"}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's a fairly steep negative slope. Now, what happens if you go right over here when x is equal to 0? Well, our derivative, if you say g prime of 0, is telling us that the slope of our original function g at x is equal to 0 is 2 times 0 is 0. Well, does that make sense? Well, if we go to our original parabola, it does indeed make sense. The slope of the tangent line looks something like this. We're at a minimum point."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, does that make sense? Well, if we go to our original parabola, it does indeed make sense. The slope of the tangent line looks something like this. We're at a minimum point. We're at the vertex. The slope really does look to be 0. And what if you go right over here to x equals 2?"}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're at a minimum point. We're at the vertex. The slope really does look to be 0. And what if you go right over here to x equals 2? The slope of the tangent line. Well, over here, the tangent line looks something like this. It looks like a fairly steep positive slope."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what if you go right over here to x equals 2? The slope of the tangent line. Well, over here, the tangent line looks something like this. It looks like a fairly steep positive slope. What is our derivative telling us based on the power rule? It's saying g prime. So this is essentially saying, hey, tell me what the slope of the tangent line for g is when x is equal to 2."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It looks like a fairly steep positive slope. What is our derivative telling us based on the power rule? It's saying g prime. So this is essentially saying, hey, tell me what the slope of the tangent line for g is when x is equal to 2. Well, we figured it out. It's going to be 2 times x. It's going to be 2 times 2, which is equal to 4."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is essentially saying, hey, tell me what the slope of the tangent line for g is when x is equal to 2. Well, we figured it out. It's going to be 2 times x. It's going to be 2 times 2, which is equal to 4. It's telling us that the slope over here is 4. That the slope, and I'm just using m. m is often the letter used to denote slope. They're saying that the slope of the tangent line there is 4, which seems completely, completely reasonable."}, {"video_title": "Justifying the power rule   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be 2 times 2, which is equal to 4. It's telling us that the slope over here is 4. That the slope, and I'm just using m. m is often the letter used to denote slope. They're saying that the slope of the tangent line there is 4, which seems completely, completely reasonable. So I encourage you to try this out yourself. I encourage you to try to estimate the slopes by calculating by taking points closer and closer around those points. And you'll see that the power rule really does give you results that actually make sense."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just get to it. And this is just one of many proofs of the chain rule. So the chain rule tells us that if y, y is a function of u, which is a function of x, and we want to figure out the derivative of this, so we want to differentiate this with respect to x. So we're gonna differentiate this with respect to x. We could write this as the derivative of y with respect to x, which is going to be equal to the derivative of y with respect to u times the derivative of u with respect to x. This is what the chain rule tells us. But how do we actually go about proving it?"}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna differentiate this with respect to x. We could write this as the derivative of y with respect to x, which is going to be equal to the derivative of y with respect to u times the derivative of u with respect to x. This is what the chain rule tells us. But how do we actually go about proving it? Well, we just have to remind ourselves that the derivative of y with respect to x, the derivative of y with respect to x, is equal to the limit as delta x approaches zero of change in y over change in x. Now we can do a little bit of algebraic manipulation here to introduce a change in u. So let's do that."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But how do we actually go about proving it? Well, we just have to remind ourselves that the derivative of y with respect to x, the derivative of y with respect to x, is equal to the limit as delta x approaches zero of change in y over change in x. Now we can do a little bit of algebraic manipulation here to introduce a change in u. So let's do that. So this is going to be the same thing as the limit as delta x approaches zero. And I'm gonna rewrite this part right over here. I'm gonna essentially divide and multiply by a change in u."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So this is going to be the same thing as the limit as delta x approaches zero. And I'm gonna rewrite this part right over here. I'm gonna essentially divide and multiply by a change in u. So I can rewrite this as delta y over delta u times delta u, whoops, times delta u over delta x. Change in y over change in u times change in u over change in x. And you can see, these are just going to be numbers here."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm gonna essentially divide and multiply by a change in u. So I can rewrite this as delta y over delta u times delta u, whoops, times delta u over delta x. Change in y over change in u times change in u over change in x. And you can see, these are just going to be numbers here. So our change in u, this would cancel with that, and you'd be left with change in y over change in x, which is exactly what we had here. So nothing earth-shattering just yet. But what's this going to be equal to?"}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you can see, these are just going to be numbers here. So our change in u, this would cancel with that, and you'd be left with change in y over change in x, which is exactly what we had here. So nothing earth-shattering just yet. But what's this going to be equal to? What's this going to be equal to? Well, the limit of the product is the same thing as the product of the limits. So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y over delta u times, maybe I'll put parentheses around it, times, times the limit, the limit as delta x approaches zero, delta x approaches zero of this business."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what's this going to be equal to? What's this going to be equal to? Well, the limit of the product is the same thing as the product of the limits. So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y over delta u times, maybe I'll put parentheses around it, times, times the limit, the limit as delta x approaches zero, delta x approaches zero of this business. So let me put some parentheses around it. Delta u over delta x. So what does this simplify to?"}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y over delta u times, maybe I'll put parentheses around it, times, times the limit, the limit as delta x approaches zero, delta x approaches zero of this business. So let me put some parentheses around it. Delta u over delta x. So what does this simplify to? Well, this right over here, this is the definition, and we're assuming, in order for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiable, are differentiable, are differentiable at x. And remember, also, if they're differentiable at x, that means they're continuous at x."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what does this simplify to? Well, this right over here, this is the definition, and we're assuming, in order for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiable, are differentiable, are differentiable at x. And remember, also, if they're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, and this is the derivative of, this is u prime of x, or du dx. So this right over here, we can rewrite as du dx. I think you see where this is going."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And remember, also, if they're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, and this is the derivative of, this is u prime of x, or du dx. So this right over here, we can rewrite as du dx. I think you see where this is going. Now, this right over here, just looking at it the way it's written right here, we can't quite yet call this dy du, because this is the limit as delta x approaches zero, not the limit as delta u approaches zero. But we just have to remind ourselves the results from probably the previous video, depending on how you're watching it, which is, if we have a function u that is continuous at a point that as delta x approaches zero, delta u approaches zero. So we can actually rewrite this."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I think you see where this is going. Now, this right over here, just looking at it the way it's written right here, we can't quite yet call this dy du, because this is the limit as delta x approaches zero, not the limit as delta u approaches zero. But we just have to remind ourselves the results from probably the previous video, depending on how you're watching it, which is, if we have a function u that is continuous at a point that as delta x approaches zero, delta u approaches zero. So we can actually rewrite this. We can rewrite this right over here. Instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta u is going to approach zero. As our change in x gets smaller and smaller and smaller, our change in u is going to get smaller and smaller and smaller."}, {"video_title": "Chain rule proof   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can actually rewrite this. We can rewrite this right over here. Instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta u is going to approach zero. As our change in x gets smaller and smaller and smaller, our change in u is going to get smaller and smaller and smaller. So we can rewrite this as our change in u approaches zero. And when we rewrite it like that, well, then this is just dy du. This is just dy, the derivative of y, with respect to u."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here I have this crazy function, 9x to the 7th minus 17x to the 6th plus 15 square roots of x, all of that over 3x to the 7th plus 1,000x to the 5th minus log base 2 of x. So what's going to happen as x approaches infinity? And the key here, like we've seen in other examples, is just to realize which terms will dominate. So for example, in the numerator, out of these three terms, the 9x to the 7th is going to grow much faster than any of these other terms. So this is the dominating term in the numerator. And the denominator, 3x to the 7th, is going to grow much faster than an x to the 5th term, definitely much faster than a log base 2 term. So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the 7th over 3x to the 7th."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, in the numerator, out of these three terms, the 9x to the 7th is going to grow much faster than any of these other terms. So this is the dominating term in the numerator. And the denominator, 3x to the 7th, is going to grow much faster than an x to the 5th term, definitely much faster than a log base 2 term. So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the 7th over 3x to the 7th. And so we can say, especially as we get larger and larger, as we get closer and closer to infinity, these two things are going to get closer and closer to each other. We can say this limit is going to be the same thing as this limit, which is going to be equal to the limit as x approaches infinity. Well, we can just cancel out the x to the 7th, so it's going to be 9 thirds or just 3, which is just going to be 3."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the 7th over 3x to the 7th. And so we can say, especially as we get larger and larger, as we get closer and closer to infinity, these two things are going to get closer and closer to each other. We can say this limit is going to be the same thing as this limit, which is going to be equal to the limit as x approaches infinity. Well, we can just cancel out the x to the 7th, so it's going to be 9 thirds or just 3, which is just going to be 3. So that is our limit as x approaches infinity of all of this craziness. Now let's do the same with this function over here. Once again, crazy function."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we can just cancel out the x to the 7th, so it's going to be 9 thirds or just 3, which is just going to be 3. So that is our limit as x approaches infinity of all of this craziness. Now let's do the same with this function over here. Once again, crazy function. We're going to negative infinity, but the same principles apply. Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude? Well, in the numerator, it's the 3x to the 3rd term."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Once again, crazy function. We're going to negative infinity, but the same principles apply. Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude? Well, in the numerator, it's the 3x to the 3rd term. In the denominator, it's the 6x to the 4th term. So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches negative infinity. And if we simplify this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, in the numerator, it's the 3x to the 3rd term. In the denominator, it's the 6x to the 4th term. So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches negative infinity. And if we simplify this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x. And what's this going to be? Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number, which is going to get us pretty darn close to 0, just as 1 over x as x approaches negative infinity gets us close to 0. So this right over here, the horizontal asymptote in this case is y is equal to 0."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we simplify this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x. And what's this going to be? Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number, which is going to get us pretty darn close to 0, just as 1 over x as x approaches negative infinity gets us close to 0. So this right over here, the horizontal asymptote in this case is y is equal to 0. And I encourage you to graph it or try it out with numbers to verify that for yourself. The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest. Now let's think about this one."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here, the horizontal asymptote in this case is y is equal to 0. And I encourage you to graph it or try it out with numbers to verify that for yourself. The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest. Now let's think about this one. What is the limit of this crazy function as x approaches infinity? Well, once again, what are the dominating terms? In the numerator, it's 4x to the 4th."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's think about this one. What is the limit of this crazy function as x approaches infinity? Well, once again, what are the dominating terms? In the numerator, it's 4x to the 4th. In the denominator, it's 250x to the 3rd. These are the highest degree terms. So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as the limit of 4."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "In the numerator, it's 4x to the 4th. In the denominator, it's 250x to the 3rd. These are the highest degree terms. So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as the limit of 4. Well, this is going to be the same thing as we could divide 200 and, well, I'll just leave it like this. It's going to be the limit of 4 over 250x to the 4th divided by x to the 3rd is just x times x as x approaches infinity. Or we could even say this is going to be 4 250ths times the limit as x approaches infinity of x."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as the limit of 4. Well, this is going to be the same thing as we could divide 200 and, well, I'll just leave it like this. It's going to be the limit of 4 over 250x to the 4th divided by x to the 3rd is just x times x as x approaches infinity. Or we could even say this is going to be 4 250ths times the limit as x approaches infinity of x. Now what's this? What's the limit of x as x approaches infinity? Well, it's just going to keep growing forever."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or we could even say this is going to be 4 250ths times the limit as x approaches infinity of x. Now what's this? What's the limit of x as x approaches infinity? Well, it's just going to keep growing forever. So this right over here is just going to be infinity. Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's just going to keep growing forever. So this right over here is just going to be infinity. Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kind of obvious way of seeing that right from the get-go is to realize that the numerator has a 4th degree term, while the highest degree term in the denominator is only a 3rd degree term. So the numerator is going to grow far faster than the denominator."}, {"video_title": "Limits at infinity of quotients (Part 2)   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kind of obvious way of seeing that right from the get-go is to realize that the numerator has a 4th degree term, while the highest degree term in the denominator is only a 3rd degree term. So the numerator is going to grow far faster than the denominator. So if the numerator is going far faster than the denominator, you're going to approach infinity in this case. If the numerator is going far slower than the denominator, if the denominator is going far faster than the numerator, like this case, you are then approaching 0. So hopefully you find that a little bit useful."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me see how well I can draw that. So it's moving radially outward. So that is the ripple that is formed from me dropping the rock into the water. So it's a circle centered at where the rock initially hit the water. And let's say right at this moment, the radius of this circle is equal to 3 centimeters. And we also know that the radius is increasing at a rate of 1 centimeter per second. So radius growing at rate of 1 centimeter per second."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's a circle centered at where the rock initially hit the water. And let's say right at this moment, the radius of this circle is equal to 3 centimeters. And we also know that the radius is increasing at a rate of 1 centimeter per second. So radius growing at rate of 1 centimeter per second. So given this, right now our circle, our ripple circle, has a radius of 3 centimeters. And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is the area of circle growing?"}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So radius growing at rate of 1 centimeter per second. So given this, right now our circle, our ripple circle, has a radius of 3 centimeters. And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is the area of circle growing? Area of circle growing. Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Given that, at what rate is the area of circle growing? Area of circle growing. Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal to 3 centimeters. We also know the rate at which r is changing with respect to time. We also know this information right over here."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal to 3 centimeters. We also know the rate at which r is changing with respect to time. We also know this information right over here. dr dt, the rate at which the radius is changing with respect to time, is 1 centimeter per second. Now what do we need to figure out? Well, they said, what rate is the area of the circle growing?"}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We also know this information right over here. dr dt, the rate at which the radius is changing with respect to time, is 1 centimeter per second. Now what do we need to figure out? Well, they said, what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle, where a is the area of the circle, at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle, and maybe take the derivative with respect to time."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, they said, what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle, where a is the area of the circle, at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle, and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule to do that. So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle, and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule to do that. So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now, what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time?"}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now, what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. Area is equal to pi r squared. Now I'm going to take the derivative of both sides with respect to time. So the derivative with respect to time."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pi r squared. Area is equal to pi r squared. Now I'm going to take the derivative of both sides with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r. I'm taking the derivative with respect to time. So on the left hand side right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative with respect to time. I'm not taking the derivative with respect to r. I'm taking the derivative with respect to time. So on the left hand side right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left hand side, And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left hand side, And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time, then area wouldn't be a function of time."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time, then area wouldn't be a function of time. So let me write, instead of just writing r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t, which we're squaring. We want to find the derivative of this with respect to time."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If r wasn't a function of time, then area wouldn't be a function of time. So let me write, instead of just writing r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t, which we're squaring. We want to find the derivative of this with respect to time. And here, we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "We want to find the derivative of this with respect to time. And here, we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. And then we're going to have to make it clear. So this is the derivative of r of t-squared with respect to r of t. The derivative of something squared with respect to that something. If it was the derivative of x squared with respect to x, we'd have 2x."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. And then we're going to have to make it clear. So this is the derivative of r of t-squared with respect to r of t. The derivative of something squared with respect to that something. If it was the derivative of x squared with respect to x, we'd have 2x. If it was the derivative of r of t squared with respect to r of t, it's 2r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. With the derivative, which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time. So the rate at which r of t changes with respect to time, well, we could just write that as dr dt."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "If it was the derivative of x squared with respect to x, we'd have 2x. If it was the derivative of r of t squared with respect to r of t, it's 2r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. With the derivative, which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time. So the rate at which r of t changes with respect to time, well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front. And I just want to emphasize, this is just the chain rule right over here."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the rate at which r of t changes with respect to time, well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front. And I just want to emphasize, this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something, so that's 2 times the something, times the derivative of that something with respect to time. I can't emphasize enough what we did right over here. This is the chain rule."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I just want to emphasize, this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something, so that's 2 times the something, times the derivative of that something with respect to time. I can't emphasize enough what we did right over here. This is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now, let me rewrite all of this again, just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times, actually, let me put that 2 out front, is equal to 2 times pi times, I can now switch back to just calling this r. We know that r is a function of t, so I'll just write 2 pi times r times dr dt."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now, let me rewrite all of this again, just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times, actually, let me put that 2 out front, is equal to 2 times pi times, I can now switch back to just calling this r. We know that r is a function of t, so I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue. 2 pi r dr dt. Now, what do we know?"}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have the derivative of our area with respect to time is equal to pi times, actually, let me put that 2 out front, is equal to 2 times pi times, I can now switch back to just calling this r. We know that r is a function of t, so I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue. 2 pi r dr dt. Now, what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now, r is 3 centimeters."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now, r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to?"}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right now, r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? What's going to be equal to, do that same green, 2 pi times 3 times 1 times 1 centimeter per second. And let's make sure we get the units right. So we have a centimeter times a centimeter."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's da dt going to be equal to? What's going to be equal to, do that same green, 2 pi times 3 times 1 times 1 centimeter per second. And let's make sure we get the units right. So we have a centimeter times a centimeter. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have a centimeter times a centimeter. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. The rate at which area is changing with respect to time is equal to 6 pi. So it's going to be a little bit over 18 centimeters squared per second, right at that moment."}, {"video_title": "Related rates intro   Applications of derivatives   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. The rate at which area is changing with respect to time is equal to 6 pi. So it's going to be a little bit over 18 centimeters squared per second, right at that moment. Yep, 3 times 2 pi. So 6 pi centimeters squared per second is how fast the area is changing. And we are done."}, {"video_title": "2015 AP Calculus AB 6a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Write an equation for the line tangent to the curve at the point negative one comma one. Alright, so we can figure out the equation for the line if we know the slope of the line and we know a point that it goes through, so that should be enough to figure out the equation of the line. So, the line's going to have a form y is equal to mx plus b. M is the slope and is going to be equal to dy dx at that point. And we know that that's going to be equal to, let's see, y is, when x is negative one, y is one, that sits on this curve. So y is one, so one, over three y squared. So three times one squared, which is three, minus x. When y is one, x is negative one, or when x is negative one, y is one."}, {"video_title": "2015 AP Calculus AB 6a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know that that's going to be equal to, let's see, y is, when x is negative one, y is one, that sits on this curve. So y is one, so one, over three y squared. So three times one squared, which is three, minus x. When y is one, x is negative one, or when x is negative one, y is one. So x is negative one here. And so this is the same thing as three plus positive one. And so this is equal to 1 4th."}, {"video_title": "2015 AP Calculus AB 6a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "When y is one, x is negative one, or when x is negative one, y is one. So x is negative one here. And so this is the same thing as three plus positive one. And so this is equal to 1 4th. And so the equation of our line is going to be y is equal to 1 4th x plus b. Now we need to solve for b. And we know that the point negative one comma one is on the line, so we can use that information to solve for b."}, {"video_title": "2015 AP Calculus AB 6a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is equal to 1 4th. And so the equation of our line is going to be y is equal to 1 4th x plus b. Now we need to solve for b. And we know that the point negative one comma one is on the line, so we can use that information to solve for b. This line is tangent to the curve, so it includes this point and only that point. That's what it has in common with the curve. And so y is equal to one when x is equal to negative one plus b."}, {"video_title": "2015 AP Calculus AB 6a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we know that the point negative one comma one is on the line, so we can use that information to solve for b. This line is tangent to the curve, so it includes this point and only that point. That's what it has in common with the curve. And so y is equal to one when x is equal to negative one plus b. And so we have one is equal to negative 1 4th plus b. You add 1 4th to both sides, you get b is equal to, we could either write it as 1 1 4th, which is equal to 5 4ths, which is equal to 1.25. We could write it any of those ways."}, {"video_title": "2015 AP Calculus AB 6a   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so y is equal to one when x is equal to negative one plus b. And so we have one is equal to negative 1 4th plus b. You add 1 4th to both sides, you get b is equal to, we could either write it as 1 1 4th, which is equal to 5 4ths, which is equal to 1.25. We could write it any of those ways. So the equation for the line tangent to the curve at this point is y is equal to, our slope is 1 4th, x plus, and I could write it in any of these ways. I'll write it as plus five over four. And we're done, at least with that part of the problem."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So an inflection point, an inflection point is a point where the sign of the second derivative, sign of second derivative, changes. Changes, so if you take the second derivative at that point, or as we go close to that point, or as we cross that point, it goes from positive to negative or negative to positive. And to think about that visually, you can think of some examples. So if you have a curve that looks something like this, if you have a curve that looks something like this, you'll notice that over here the slope is negative, but it's increasing. It's getting less negative, less negative, then it goes to zero, then it keeps increasing. Slope is increasing, increasing, all the way to there, and then it starts getting less positive. So it starts decreasing."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you have a curve that looks something like this, if you have a curve that looks something like this, you'll notice that over here the slope is negative, but it's increasing. It's getting less negative, less negative, then it goes to zero, then it keeps increasing. Slope is increasing, increasing, all the way to there, and then it starts getting less positive. So it starts decreasing. So it's increasing, the slope is increasing over this point right over here. So even though the slope is negative, it's getting less negative over here, so it's increasing. And then the slope keeps increasing, gets to getting more and more positive up to about this point, and then the slope is positive, but then it becomes less positive."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it starts decreasing. So it's increasing, the slope is increasing over this point right over here. So even though the slope is negative, it's getting less negative over here, so it's increasing. And then the slope keeps increasing, gets to getting more and more positive up to about this point, and then the slope is positive, but then it becomes less positive. So the slope begins decreasing after that. So then the slope begins decreasing after that. So this right over here is a point of inflection."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the slope keeps increasing, gets to getting more and more positive up to about this point, and then the slope is positive, but then it becomes less positive. So the slope begins decreasing after that. So then the slope begins decreasing after that. So this right over here is a point of inflection. The slope has gone from increasing to decreasing, and the other thing happened, if the slope went from decreasing to increasing, that would also be a point of inflection. So if this was maybe some type of a trigonometric curve, then you might see something like this. And so this also would be a point of inflection."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is a point of inflection. The slope has gone from increasing to decreasing, and the other thing happened, if the slope went from decreasing to increasing, that would also be a point of inflection. So if this was maybe some type of a trigonometric curve, then you might see something like this. And so this also would be a point of inflection. But for this, our g of x is kind of hard to visualize the way they've defined it right over here. So the best way to think about it is just figure out where its second derivative has a sign change. And to think about that, we have to find its second derivative."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this also would be a point of inflection. But for this, our g of x is kind of hard to visualize the way they've defined it right over here. So the best way to think about it is just figure out where its second derivative has a sign change. And to think about that, we have to find its second derivative. So let's write g of x over here. We know g of x is equal to two x plus the definite integral from zero to x of f of t dt. We've already taken its derivative, but we'll do it again."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And to think about that, we have to find its second derivative. So let's write g of x over here. We know g of x is equal to two x plus the definite integral from zero to x of f of t dt. We've already taken its derivative, but we'll do it again. G prime of x is equal to two plus fundamental theorem of calculus, the derivative of this right over here is just f of x. And if we have the second derivative of g, g prime prime of x, this is equal to derivative of two is just zero. And the derivative of f of x is f prime of x."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've already taken its derivative, but we'll do it again. G prime of x is equal to two plus fundamental theorem of calculus, the derivative of this right over here is just f of x. And if we have the second derivative of g, g prime prime of x, this is equal to derivative of two is just zero. And the derivative of f of x is f prime of x. So asking where this has a sign change, asking where our second derivative has a sign change, is equivalent to asking where does the first derivative of f have a sign change. And asking where the first derivative of f has a sign change is equivalent to saying where does the slope of f have a sign change? You can view this as the slope or the instantaneous slope of f. So we want to know when the slope of f has a sign change."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the derivative of f of x is f prime of x. So asking where this has a sign change, asking where our second derivative has a sign change, is equivalent to asking where does the first derivative of f have a sign change. And asking where the first derivative of f has a sign change is equivalent to saying where does the slope of f have a sign change? You can view this as the slope or the instantaneous slope of f. So we want to know when the slope of f has a sign change. So let's think about it. Over here, the slope is positive. It's going up."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can view this as the slope or the instantaneous slope of f. So we want to know when the slope of f has a sign change. So let's think about it. Over here, the slope is positive. It's going up. It's up. It's increasing, but it's positive. And that's what we care about."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going up. It's up. It's increasing, but it's positive. And that's what we care about. So let's write it. I'll do it in green. So the slope is positive this entire time."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's what we care about. So let's write it. I'll do it in green. So the slope is positive this entire time. It's increasing. It's increasing. It's positive."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope is positive this entire time. It's increasing. It's increasing. It's positive. It's getting less positive now. It's starting to decrease. But the slope is still positive."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's positive. It's getting less positive now. It's starting to decrease. But the slope is still positive. The slope is still positive all the way until we get right over there. It seems like it gets pretty close to 0. And then the slope gets a negative."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the slope is still positive. The slope is still positive all the way until we get right over there. It seems like it gets pretty close to 0. And then the slope gets a negative. And then right over here, the slope is negative. So this is interesting because even though f is actually not differentiable right here, so f is not differentiable at that point right over there. And you can see because the slope goes pretty close to 0, then it just jumps to negative 3."}, {"video_title": "2011 Calculus AB free response #4c   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the slope gets a negative. And then right over here, the slope is negative. So this is interesting because even though f is actually not differentiable right here, so f is not differentiable at that point right over there. And you can see because the slope goes pretty close to 0, then it just jumps to negative 3. So you have a discontinuity of the derivative right over there, but we do have a sign change. We go from having a positive slope on this part of the curve to having a negative slope over this part of the curve. So we experience a sign change right over here at x is equal to 0, a sign change in the first derivative of f, which is the same thing as saying a sign change in the second derivative of g. And a sign change in the second derivative of g tells us that when x is equal to 0, we have a point."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we know is that one of them is the function f, another is the first derivative of f, and then the third is the second derivative of f. And our goal is to figure out which function is which. Which one is f, which is the first derivative, and which is the second. Like always, pause this video and see if you can work through it on your own before we do it together. All right, now let's do this together. The way I'm going to tackle it is I'm going to try to sketch what we can about the derivatives of each of these graphs, or each of the functions represented by these graphs. So in this first graph here in this orange color, we can see that the slope is quite positive here, but then it becomes less and less and less positive up until this point where the slope is going to be zero, and then it becomes more and more and more and more negative. So the derivative of this curve right over here, or the function represented by this curve, it's gonna start off reasonably positive right over there."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, now let's do this together. The way I'm going to tackle it is I'm going to try to sketch what we can about the derivatives of each of these graphs, or each of the functions represented by these graphs. So in this first graph here in this orange color, we can see that the slope is quite positive here, but then it becomes less and less and less positive up until this point where the slope is going to be zero, and then it becomes more and more and more and more negative. So the derivative of this curve right over here, or the function represented by this curve, it's gonna start off reasonably positive right over there. At this point, the derivative is gonna cross zero because our derivative is zero there, slope of the tangent line. And then it's gonna get more and more negative, or at least over the interval that we see. So it might look, I don't know, something like this."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of this curve right over here, or the function represented by this curve, it's gonna start off reasonably positive right over there. At this point, the derivative is gonna cross zero because our derivative is zero there, slope of the tangent line. And then it's gonna get more and more negative, or at least over the interval that we see. So it might look, I don't know, something like this. I don't know if it's a line or not, it might be some type of a curve, but it would definitely have a trend something like that. Now we can immediately tell that this blue graph is not the derivative of this orange graph. Its trend is opposite."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it might look, I don't know, something like this. I don't know if it's a line or not, it might be some type of a curve, but it would definitely have a trend something like that. Now we can immediately tell that this blue graph is not the derivative of this orange graph. Its trend is opposite. Over that interval, it's going from being negative to positive as opposed to going from positive to negative. So we can rule out the blue graph as being the derivative of the orange graph. But what about this magenta graph?"}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "Its trend is opposite. Over that interval, it's going from being negative to positive as opposed to going from positive to negative. So we can rule out the blue graph as being the derivative of the orange graph. But what about this magenta graph? It does have, it does look like it has the right trend. In fact, it intersects the x-axis at the right place, right over there. And at least over this interval, it seems it's positive from here to here."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what about this magenta graph? It does have, it does look like it has the right trend. In fact, it intersects the x-axis at the right place, right over there. And at least over this interval, it seems it's positive from here to here. So it's positive. This graph is positive when the slope of the tangent line here is positive, and this graph is negative when the slope of the tangent line here is negative. Now one thing that might be causing some unease to immediately say that this last graph is the derivative of the first one is we're not used to situations where the derivative has more extreme points, more minima and maxima, than the original function."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "And at least over this interval, it seems it's positive from here to here. So it's positive. This graph is positive when the slope of the tangent line here is positive, and this graph is negative when the slope of the tangent line here is negative. Now one thing that might be causing some unease to immediately say that this last graph is the derivative of the first one is we're not used to situations where the derivative has more extreme points, more minima and maxima, than the original function. But in this case, it could just be because we don't see the entire original function. So for example, if this last graph is indeed the derivative of this first graph, then what we see is our derivative is our graph, our derivative is negative right over here, but then right around here, it starts becoming less negative. So if that point corresponds to roughly right over there, then over here, our slope will become less and less and less negative."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now one thing that might be causing some unease to immediately say that this last graph is the derivative of the first one is we're not used to situations where the derivative has more extreme points, more minima and maxima, than the original function. But in this case, it could just be because we don't see the entire original function. So for example, if this last graph is indeed the derivative of this first graph, then what we see is our derivative is our graph, our derivative is negative right over here, but then right around here, it starts becoming less negative. So if that point corresponds to roughly right over there, then over here, our slope will become less and less and less negative. And then at this point, our slope would become zero, which would be right around there. So for example, our graph might look something like this. We just didn't see it."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if that point corresponds to roughly right over there, then over here, our slope will become less and less and less negative. And then at this point, our slope would become zero, which would be right around there. So for example, our graph might look something like this. We just didn't see it. It fell off of the part of the graph that we actually showed. So I would actually say that this is a good candidate for being, the third function is a good candidate for being the derivative of the first function. So maybe we could say that this is f and that this is f prime."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "We just didn't see it. It fell off of the part of the graph that we actually showed. So I would actually say that this is a good candidate for being, the third function is a good candidate for being the derivative of the first function. So maybe we could say that this is f and that this is f prime. Now let's look at the second graph. What would its derivative look like? So over here, our slope is quite negative, and it becomes less and less and less negative until we go right over here where our slope is zero."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So maybe we could say that this is f and that this is f prime. Now let's look at the second graph. What would its derivative look like? So over here, our slope is quite negative, and it becomes less and less and less negative until we go right over here where our slope is zero. So our derivative would intersect the x-axis right over there. It would start out negative, and it would become less and less and less negative, and at this point, it crosses the x-axis, and then it becomes more and more positive. So we see here, our derivative becomes more and more positive, but then right around here, it seems like it's getting less positive again."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over here, our slope is quite negative, and it becomes less and less and less negative until we go right over here where our slope is zero. So our derivative would intersect the x-axis right over there. It would start out negative, and it would become less and less and less negative, and at this point, it crosses the x-axis, and then it becomes more and more positive. So we see here, our derivative becomes more and more positive, but then right around here, it seems like it's getting less positive again. So it might look something like this, where over here, it's becoming less positive again. Less positive, less positive, less positive. Right over here, our derivative would be zero."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we see here, our derivative becomes more and more positive, but then right around here, it seems like it's getting less positive again. So it might look something like this, where over here, it's becoming less positive again. Less positive, less positive, less positive. Right over here, our derivative would be zero. So our derivative would intersect the x-axis there, and then it just looks like it is, the slope is getting more and more and more negative. So our derivative is gonna get more and more and more negative. Well, what I very roughly just sketched out looks an awful lot like the brown graph right over here."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "Right over here, our derivative would be zero. So our derivative would intersect the x-axis there, and then it just looks like it is, the slope is getting more and more and more negative. So our derivative is gonna get more and more and more negative. Well, what I very roughly just sketched out looks an awful lot like the brown graph right over here. So this brown graph does indeed look like the derivative of this blue graph. So what I would say is that this is actually f, and then this would be f prime, and then if this is f prime, the derivative of that is going to be f prime prime. So that looks good."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what I very roughly just sketched out looks an awful lot like the brown graph right over here. So this brown graph does indeed look like the derivative of this blue graph. So what I would say is that this is actually f, and then this would be f prime, and then if this is f prime, the derivative of that is going to be f prime prime. So that looks good. I would actually go with this. And if you wanted, just for safe measure, you could try to sketch out what the derivative of this graph would be. Actually, let's just do that."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that looks good. I would actually go with this. And if you wanted, just for safe measure, you could try to sketch out what the derivative of this graph would be. Actually, let's just do that. So over here, the derivative of this, so right now, we have a positive. The slope of our tangent line is getting less and less positive. It hits zero right over there."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let's just do that. So over here, the derivative of this, so right now, we have a positive. The slope of our tangent line is getting less and less positive. It hits zero right over there. So the derivative might look something like this over that interval. Now, the slope of the tangent line is getting more and more and more and more negative, right, until about that point. So it's getting more and more and more negative until about that point."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "It hits zero right over there. So the derivative might look something like this over that interval. Now, the slope of the tangent line is getting more and more and more and more negative, right, until about that point. So it's getting more and more and more negative until about that point. And now it looks like it's getting less and less and less and less negative, all the way until the derivative goes back to being zero. And then it looks like it's getting more and more and more and more positive. So the derivative of this magenta curve looks like an upward-opening U."}, {"video_title": "Connecting f, f', and f'' graphically   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's getting more and more and more negative until about that point. And now it looks like it's getting less and less and less and less negative, all the way until the derivative goes back to being zero. And then it looks like it's getting more and more and more and more positive. So the derivative of this magenta curve looks like an upward-opening U. And we don't see that over here. So we could feel good that its derivative actually isn't depicted. So I feel good calling the middle graph F, calling the left graph F prime, and calling the right graph the second derivative."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm just gonna call that log base a of x. So how do we figure this out? Well, the key thing is, is what you might be familiar with from your algebra or your pre-calculus classes, which is having a change of base. So if I have some, I'll do it over here, log base a of b, and I want to change it to a different base. Let's say I want to change it to base c. This is the same thing as log base c of b divided by log base, log base c of a. Log base c of b divided by log base c of a. And this is a really useful thing. If you've never seen it before, you now have just seen it, this change of base, and we prove it in other videos on Khan Academy."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I have some, I'll do it over here, log base a of b, and I want to change it to a different base. Let's say I want to change it to base c. This is the same thing as log base c of b divided by log base, log base c of a. Log base c of b divided by log base c of a. And this is a really useful thing. If you've never seen it before, you now have just seen it, this change of base, and we prove it in other videos on Khan Academy. But it's really useful because, for example, your calculator has a log button, but the log on your calculator is log base 10. So if you do log, if you press 10 into your, or 100 into your calculator and press log, you will get a two there. So whenever you just see log of 100, it's implicitly base 10."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you've never seen it before, you now have just seen it, this change of base, and we prove it in other videos on Khan Academy. But it's really useful because, for example, your calculator has a log button, but the log on your calculator is log base 10. So if you do log, if you press 10 into your, or 100 into your calculator and press log, you will get a two there. So whenever you just see log of 100, it's implicitly base 10. And you also have a button for natural log, which is log base e. Natural log of x is equal to log base e of x. But sometimes you want to find all sorts of different base logarithms, and this is how you do it. So if you're using your calculator and you wanted to find, if you wanted to find what log base three of eight is, you would say, you would type in your calculator log of eight and log of three."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So whenever you just see log of 100, it's implicitly base 10. And you also have a button for natural log, which is log base e. Natural log of x is equal to log base e of x. But sometimes you want to find all sorts of different base logarithms, and this is how you do it. So if you're using your calculator and you wanted to find, if you wanted to find what log base three of eight is, you would say, you would type in your calculator log of eight and log of three. Or, let me write it this way, and log of three, where both of these are implicitly base 10. And you'd get the same value if you did natural log of eight divided by natural log of three, which you might also have on your calculator. And what we're gonna do in this video is leverage the natural log, because we know what the derivative of the natural log is."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you're using your calculator and you wanted to find, if you wanted to find what log base three of eight is, you would say, you would type in your calculator log of eight and log of three. Or, let me write it this way, and log of three, where both of these are implicitly base 10. And you'd get the same value if you did natural log of eight divided by natural log of three, which you might also have on your calculator. And what we're gonna do in this video is leverage the natural log, because we know what the derivative of the natural log is. So this derivative is the same thing as the derivative with respect to x of, well, log base a of x can be rewritten as natural log of x over natural log of a. And now, natural log of a, that's just a number. I could rewrite this, I could rewrite this as, write it this way, one over natural log of a times natural log of x."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we're gonna do in this video is leverage the natural log, because we know what the derivative of the natural log is. So this derivative is the same thing as the derivative with respect to x of, well, log base a of x can be rewritten as natural log of x over natural log of a. And now, natural log of a, that's just a number. I could rewrite this, I could rewrite this as, write it this way, one over natural log of a times natural log of x. And what's the derivative of that? Well, we could just take the constant out. One over natural log of a, that's just a number."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could rewrite this, I could rewrite this as, write it this way, one over natural log of a times natural log of x. And what's the derivative of that? Well, we could just take the constant out. One over natural log of a, that's just a number. So we're gonna get one over the natural log of a times the derivative, times the derivative with respect to x of natural log of x, of natural log of x, which we already know is one over x. So this thing right over here is one over x. So what we get is one over natural log of a times one over x, which we could write as one over natural log of a times x, times x, which is a really useful thing to know."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "One over natural log of a, that's just a number. So we're gonna get one over the natural log of a times the derivative, times the derivative with respect to x of natural log of x, of natural log of x, which we already know is one over x. So this thing right over here is one over x. So what we get is one over natural log of a times one over x, which we could write as one over natural log of a times x, times x, which is a really useful thing to know. So now we could take all sorts of derivatives. So if I were to tell you f of x is equal to log base seven of x, well, now we could say, well, f prime of x is going to be one over the natural log of seven times x. If we had a constant out front, if we had, for example, g of x, g of x is equal to negative three times log base, I don't know, log base pi."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we get is one over natural log of a times one over x, which we could write as one over natural log of a times x, times x, which is a really useful thing to know. So now we could take all sorts of derivatives. So if I were to tell you f of x is equal to log base seven of x, well, now we could say, well, f prime of x is going to be one over the natural log of seven times x. If we had a constant out front, if we had, for example, g of x, g of x is equal to negative three times log base, I don't know, log base pi. Pi is a number. Log base pi of x, well, g prime of x would be equal to one over the, oh, let me be careful. I have this constant out here, so it'd be negative three over, it's just that negative three, over the natural log of pi, so it's just the natural log of this number, times x."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're gonna base this argument based on a previous proof we made that the derivative with respect to x of sine of x is equal to cosine of x. So we're gonna assume this over here. I encourage you to watch that video. That's actually a fairly involved proof that proves this. But if we assume this, I'm gonna make a visual argument that this right over here is true, that the derivative with respect to x of cosine of x is negative sine of x. So right over here, we see sine of x in red and we see cosine of x in blue. And we're assuming that this blue graph is showing the derivative, the slope of the tangent line for any x value of the red graph."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's actually a fairly involved proof that proves this. But if we assume this, I'm gonna make a visual argument that this right over here is true, that the derivative with respect to x of cosine of x is negative sine of x. So right over here, we see sine of x in red and we see cosine of x in blue. And we're assuming that this blue graph is showing the derivative, the slope of the tangent line for any x value of the red graph. And we've gotten an intuition for that in previous videos. Now what I'm gonna do next is I'm gonna shift both of these graphs to the left by pi over two. Shift it to the left by pi over two."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're assuming that this blue graph is showing the derivative, the slope of the tangent line for any x value of the red graph. And we've gotten an intuition for that in previous videos. Now what I'm gonna do next is I'm gonna shift both of these graphs to the left by pi over two. Shift it to the left by pi over two. And I'm also gonna shift the blue graph to the left by pi over two. And so what am I going to get? Well, the blue graph is gonna look like this one right over here."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Shift it to the left by pi over two. And I'm also gonna shift the blue graph to the left by pi over two. And so what am I going to get? Well, the blue graph is gonna look like this one right over here. And if it was cosine of x up here, we can now say that this is equal to, y is equal to cosine of x plus pi over two. This is the blue graph, cosine of x, shifted to the left by pi over two. And this is y is equal to sine of x plus pi over two."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the blue graph is gonna look like this one right over here. And if it was cosine of x up here, we can now say that this is equal to, y is equal to cosine of x plus pi over two. This is the blue graph, cosine of x, shifted to the left by pi over two. And this is y is equal to sine of x plus pi over two. Now the visual argument is, all I did is I shifted both of these graphs to the left by pi over two. So it should still be the case that the derivative of the red graph is the blue graph. So we should still be able to say that the derivative with respect to x of the red graph, sine of x plus pi over two, that that is equal to the blue graph."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is y is equal to sine of x plus pi over two. Now the visual argument is, all I did is I shifted both of these graphs to the left by pi over two. So it should still be the case that the derivative of the red graph is the blue graph. So we should still be able to say that the derivative with respect to x of the red graph, sine of x plus pi over two, that that is equal to the blue graph. That that is equal to cosine of x plus pi over two. Now what is sine of x plus pi over two? Well, that's the same thing as cosine of x."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we should still be able to say that the derivative with respect to x of the red graph, sine of x plus pi over two, that that is equal to the blue graph. That that is equal to cosine of x plus pi over two. Now what is sine of x plus pi over two? Well, that's the same thing as cosine of x. So as you can see, this red graph is the same thing as cosine of x. We know that from our trig identities, and you can also see it intuitively or graphically just by looking at these graphs. And what is cosine of x plus pi over two?"}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's the same thing as cosine of x. So as you can see, this red graph is the same thing as cosine of x. We know that from our trig identities, and you can also see it intuitively or graphically just by looking at these graphs. And what is cosine of x plus pi over two? Well, once again, from our trig identities, we know that that is the exact same thing as negative sine of x. So there you have it, the visual argument. Just start with this knowledge, shift both of these graphs to the left by pi over two."}, {"video_title": "Proof of the derivative of cos(x)   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what is cosine of x plus pi over two? Well, once again, from our trig identities, we know that that is the exact same thing as negative sine of x. So there you have it, the visual argument. Just start with this knowledge, shift both of these graphs to the left by pi over two. It should still be true that the derivative with respect to x of sine of x plus pi over two is equal to cosine of x plus pi over two. And this is the same thing as saying what we have right over here. So now we should feel pretty good."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really simple idea. So let me draw a function here. Actually, let me define a function here. A kind of a simple function. So let's define f of x. Let's say that f of x is going to be x minus 1 over x minus 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Actually, let me define a function here. A kind of a simple function. So let's define f of x. Let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal, look, I have the same thing in the numerator and the denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1?"}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal, look, I have the same thing in the numerator and the denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true. The difference between f of x equals 1 and this thing right over here is that this thing is undefined when x is equal to 1. Because if you set, let me define it right, let me write it over here."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true. The difference between f of x equals 1 and this thing right over here is that this thing is undefined when x is equal to 1. Because if you set, let me define it right, let me write it over here. If you have f of, sorry, not f of 0, if you have f of 1, what happens? In the numerator, you get 1 minus 1, which is, let me just write it down. In the numerator, you get 0."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Because if you set, let me define it right, let me write it over here. If you have f of, sorry, not f of 0, if you have f of 1, what happens? In the numerator, you get 1 minus 1, which is, let me just write it down. In the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "In the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent. Both of these are going to be equal to 1 for all other x's other than 1, but at x equals 1, it becomes undefined."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So you can make the simplification. You can say that this is the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent. Both of these are going to be equal to 1 for all other x's other than 1, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function? So let me graph it."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Both of these are going to be equal to 1 for all other x's other than 1, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function? So let me graph it. So that is my y is equal to f of x-axis. y is equal to f of x-axis. And then this over here is my x-axis."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me graph it. So that is my y is equal to f of x-axis. y is equal to f of x-axis. And then this over here is my x-axis. x-axis. And then let's say that this is the point x is equal to 1. This over here would be x is equal to negative 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And then this over here is my x-axis. x-axis. And then let's say that this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1. Right up there, I could do negative 1, but that won't matter much relative to this function right over here. And let me graph it."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "This over here would be x is equal to negative 1. This is y is equal to 1. Right up there, I could do negative 1, but that won't matter much relative to this function right over here. And let me graph it. So it's essentially, for any x other than 1, f of x is going to be equal to 1. So it's going to look like this. It's going to look like this."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And let me graph it. So it's essentially, for any x other than 1, f of x is going to be equal to 1. So it's going to look like this. It's going to look like this. Except at 1. At 1, f of x is undefined. So I'm going to put a little bit of a gap right over here."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's going to look like this. Except at 1. At 1, f of x is undefined. So I'm going to put a little bit of a gap right over here. This circle to signify that this function is not defined. We don't know what this function equals at 1. We never defined it."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So I'm going to put a little bit of a gap right over here. This circle to signify that this function is not defined. We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us what to do with 1. It's literally undefined. Literally undefined when x is equal to 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "We never defined it. This definition of the function doesn't tell us what to do with 1. It's literally undefined. Literally undefined when x is equal to 1. So this is the function right over here. And so, once again, if someone were to ask you, what is f of 1, you'd go, and let's say that even if this was the function definition, you'd go, okay, x is equal to 1. Oh, wait, there's a gap in my function over here."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Literally undefined when x is equal to 1. So this is the function right over here. And so, once again, if someone were to ask you, what is f of 1, you'd go, and let's say that even if this was the function definition, you'd go, okay, x is equal to 1. Oh, wait, there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Oh, wait, there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it. f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1? Now, this is starting to touch on the idea of a limit."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's kind of redundant, but I'll rewrite it. f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1? Now, this is starting to touch on the idea of a limit. So as x gets closer and closer to 1, so as we get closer and closer x is to 1, as we get closer and closer x is to 1, what is the function approaching? Well, this entire time, the function, what's it getting closer and closer to? On the left-hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, this is starting to touch on the idea of a limit. So as x gets closer and closer to 1, so as we get closer and closer x is to 1, as we get closer and closer x is to 1, what is the function approaching? Well, this entire time, the function, what's it getting closer and closer to? On the left-hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here, from the right-hand side, you get the same thing. So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x, and lim, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1, and our function is going to be equal to 1. It's getting closer and closer to 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "On the left-hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here, from the right-hand side, you get the same thing. So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x, and lim, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1, and our function is going to be equal to 1. It's getting closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look, what is the function approaching as x gets closer and closer to 1?"}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's getting closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look, what is the function approaching as x gets closer and closer to 1? Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, and let me just, just for the sake of variety, let me call it g of x. Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared when x does not equal, I don't know, when x does not equal 2, and let's say that when x equals 2, when x equals 2, it is equal to 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So once again, it has very fancy notation, but it's just saying, look, what is the function approaching as x gets closer and closer to 1? Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, and let me just, just for the sake of variety, let me call it g of x. Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared when x does not equal, I don't know, when x does not equal 2, and let's say that when x equals 2, when x equals 2, it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous. It has a discontinuity. Let me graph it."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared when x does not equal, I don't know, when x does not equal 2, and let's say that when x equals 2, when x equals 2, it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous. It has a discontinuity. Let me graph it. So this is my y equals f of x axis. This is my x axis right over here. Let me draw x equals 2."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me graph it. So this is my y equals f of x axis. This is my x axis right over here. Let me draw x equals 2. Let's say this is x equals 1. This is x equals 2. This is negative 1."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me draw x equals 2. Let's say this is x equals 1. This is x equals 2. This is negative 1. This is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "This is negative 1. This is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. So it's going to be a parabola. It'll look something like this. It's going to look something, let me draw a better version of the parabola."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me draw it like this. So it's going to be a parabola. It'll look something like this. It's going to look something, let me draw a better version of the parabola. So it'll look something like this. It'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's going to look something, let me draw a better version of the parabola. So it'll look something like this. It'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric. Let me redraw it, because that's kind of ugly."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric. Let me redraw it, because that's kind of ugly. Okay, so let me, that's looking better. Okay, all right, there you go. All right, now, this would be the graph of just x squared, but it's not x squared when x is equal to 2."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me redraw it, because that's kind of ugly. Okay, so let me, that's looking better. Okay, all right, there you go. All right, now, this would be the graph of just x squared, but it's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2, the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that, so this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "All right, now, this would be the graph of just x squared, but it's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2, the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that, so this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that, so this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of x, of f of x is equal to x squared, except when you get to 2, it has this gap, because that's, you don't use the f of x is equal to x squared when x is equal to 2, you use f of x, or I should say g of x, you use g of x is equal to 1. I've been saying f of x, I apologize for that."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of x, of f of x is equal to x squared, except when you get to 2, it has this gap, because that's, you don't use the f of x is equal to x squared when x is equal to 2, you use f of x, or I should say g of x, you use g of x is equal to 1. I've been saying f of x, I apologize for that. You use g of x is equal to 1. So then at 2, just at 2, just exactly at 2, it drops down to 1, and then it keeps going along the function g of x is equal to, or I should say, along the function x squared. So my question to you, so there's a couple of things."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "I've been saying f of x, I apologize for that. You use g of x is equal to 1. So then at 2, just at 2, just exactly at 2, it drops down to 1, and then it keeps going along the function g of x is equal to, or I should say, along the function x squared. So my question to you, so there's a couple of things. If I were to just evaluate the function g of 2, well, you look at this definition. Okay, when x equals 2, I use this situation right over here, and it tells me it's going to be equal to 1. Let me ask a more interesting question, or perhaps a more interesting question."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So my question to you, so there's a couple of things. If I were to just evaluate the function g of 2, well, you look at this definition. Okay, when x equals 2, I use this situation right over here, and it tells me it's going to be equal to 1. Let me ask a more interesting question, or perhaps a more interesting question. What is the limit as x approaches 2 of g of x? Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me ask a more interesting question, or perhaps a more interesting question. What is the limit as x approaches 2 of g of x? Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9 and then 1.999 and then 1.999999 and then 1.999999, what is g of x approaching? Or if you were to go from the positive direction, if you were to say 2.1, what's g of 2.1?"}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9 and then 1.999 and then 1.999999 and then 1.999999, what is g of x approaching? Or if you were to go from the positive direction, if you were to say 2.1, what's g of 2.1? What's g of 2.01? What's g of 2.001? What is that approaching as we get closer and closer to it?"}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Or if you were to go from the positive direction, if you were to say 2.1, what's g of 2.1? What's g of 2.01? What's g of 2.001? What is that approaching as we get closer and closer to it? And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow it along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1, the limit of g of x as x approaches 2 is equal to 4."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "What is that approaching as we get closer and closer to it? And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow it along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1, the limit of g of x as x approaches 2 is equal to 4. And you can even do this numerically using a calculator. And let me do that, because I think that will be interesting. So let me get a calculator out."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Even though that's not where the function is, the function drops down to 1, the limit of g of x as x approaches 2 is equal to 4. And you can even do this numerically using a calculator. And let me do that, because I think that will be interesting. So let me get a calculator out. Let me get my trusty TI-85 out. Let me... So here is my calculator."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me get a calculator out. Let me get my trusty TI-85 out. Let me... So here is my calculator. And you can numerically say, okay, what's it going to approach as you approach x equals 2? So let's try 1.9. For x is equal to 1.9, you would use this top clause right over here."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So here is my calculator. And you can numerically say, okay, what's it going to approach as you approach x equals 2? So let's try 1.9. For x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. And so you'd get 3.61. Well, what if you get even closer to 2?"}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "For x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. And so you'd get 3.61. Well, what if you get even closer to 2? So 1.99. And once again, let me square that. Well, now I'm at 3.96."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, what if you get even closer to 2? So 1.99. And once again, let me square that. Well, now I'm at 3.96. Well, what if I do 1.999? And I square that. And I square that."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, now I'm at 3.96. Well, what if I do 1.999? And I square that. And I square that. I'm going to get 3.996. Notice I'm getting closer and closer and closer to our point. And if I got really close, 1.9999999999 squared, what am I going to get to?"}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And I square that. I'm going to get 3.996. Notice I'm getting closer and closer and closer to our point. And if I got really close, 1.9999999999 squared, what am I going to get to? It's not actually going to be exactly 4. This calculator just rounded things up. It's going to get to a number really, really, really, really, really, really, really close to 4."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And if I got really close, 1.9999999999 squared, what am I going to get to? It's not actually going to be exactly 4. This calculator just rounded things up. It's going to get to a number really, really, really, really, really, really, really close to 4. And we can do something from the positive direction, too. And it actually has to be the same number when we approach from the below what we're trying to approach and above what we're trying to approach. So if we try 2.1 squared, we get 4.4."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It's going to get to a number really, really, really, really, really, really, really close to 4. And we can do something from the positive direction, too. And it actually has to be the same number when we approach from the below what we're trying to approach and above what we're trying to approach. So if we try 2.1 squared, we get 4.4. If we do 2.0... Let's go a couple of steps ahead. 2.01. So this is much closer to 2 now squared."}, {"video_title": "Introduction to limits   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So if we try 2.1 squared, we get 4.4. If we do 2.0... Let's go a couple of steps ahead. 2.01. So this is much closer to 2 now squared. Now we're getting much closer to 4. So the closer we get to 2, the closer it seems like we're getting to 4. So once again, that's a numeric way of seeing that the limit as x approaches 2 from either direction of g of x, even though right at 2, the function is equal to 1 because it's discontinuous, the limit as we're approaching 2, we're getting closer and closer and closer to 4."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "If so, write a justification. So pause this video and see if you can think about this on your own before we do it together. Okay, well let's just visualize what's going on and visually think about the intermediate value theorem. So if that's my y-axis there, and then let's say that this is my x-axis right over here, we've been given some points over here. We know when x is equal to zero, f of x is equal to zero. Let me draw those. So we have that point."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if that's my y-axis there, and then let's say that this is my x-axis right over here, we've been given some points over here. We know when x is equal to zero, f of x is equal to zero. Let me draw those. So we have that point. When x is equal to two, y, or f of x, y equals f of x is going to be equal to negative two. So we have a negative two right over there. When x is equal to four, so three, four, f of x is equal to three, one, two, three."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have that point. When x is equal to two, y, or f of x, y equals f of x is going to be equal to negative two. So we have a negative two right over there. When x is equal to four, so three, four, f of x is equal to three, one, two, three. I'm doing them on a slightly different scale so that I can show everything. And when x is equal to six, so five, six, f of x is equal to seven, so three, four, five, six, seven. So right over here."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is equal to four, so three, four, f of x is equal to three, one, two, three. I'm doing them on a slightly different scale so that I can show everything. And when x is equal to six, so five, six, f of x is equal to seven, so three, four, five, six, seven. So right over here. Now they also tell us that our function is continuous. So one intuitive way of thinking about continuity is I can connect all of these dots without lifting my pencil. So the function might look, I'm just gonna make up some stuff."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So right over here. Now they also tell us that our function is continuous. So one intuitive way of thinking about continuity is I can connect all of these dots without lifting my pencil. So the function might look, I'm just gonna make up some stuff. It might look something, anything, like what I just drew just now. And it could have even wilder fluctuations. But that is what my f looks like."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the function might look, I'm just gonna make up some stuff. It might look something, anything, like what I just drew just now. And it could have even wilder fluctuations. But that is what my f looks like. Now the intermediate value theorem says, hey, pick a closed interval. And here we're picking the closed interval from four to six. So let me look at that."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that is what my f looks like. Now the intermediate value theorem says, hey, pick a closed interval. And here we're picking the closed interval from four to six. So let me look at that. So this is one, two, three, four here. This is six here. So we're gonna look at this closed interval."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me look at that. So this is one, two, three, four here. This is six here. So we're gonna look at this closed interval. And the intermediate value theorem tells us that look, if we're continuous over that closed interval, our function f is going to take on every value between f of four, f of four, which in this case, so this is f of four is equal to three, and f of six, and f of six, which is equal to seven. F of six, which is equal to seven. And so if someone said, hey, is there going to be a solution to f of x is equal to, say, five over this interval?"}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're gonna look at this closed interval. And the intermediate value theorem tells us that look, if we're continuous over that closed interval, our function f is going to take on every value between f of four, f of four, which in this case, so this is f of four is equal to three, and f of six, and f of six, which is equal to seven. F of six, which is equal to seven. And so if someone said, hey, is there going to be a solution to f of x is equal to, say, five over this interval? Yes, over this interval for some x, you're going to have f of x being equal to five. But they're not asking us for an f of x equaling something between these two values. They're asking us for an f of x equaling zero."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if someone said, hey, is there going to be a solution to f of x is equal to, say, five over this interval? Yes, over this interval for some x, you're going to have f of x being equal to five. But they're not asking us for an f of x equaling something between these two values. They're asking us for an f of x equaling zero. Zero isn't between f of four and f of six. And so we cannot use the intermediate value theorem here. And so if we wanted to write it out, we could say f is continuous, is continuous, but, but zero is not between, between f of four and f of six."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "They're asking us for an f of x equaling zero. Zero isn't between f of four and f of six. And so we cannot use the intermediate value theorem here. And so if we wanted to write it out, we could say f is continuous, is continuous, but, but zero is not between, between f of four and f of six. So the intermediate value theorem does not apply. All right, let's do the second one. So here they say, can we use the intermediate value theorem to say that there is a value c such that f of c equals zero and two is less than or equal to c is less than or equal to four?"}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if we wanted to write it out, we could say f is continuous, is continuous, but, but zero is not between, between f of four and f of six. So the intermediate value theorem does not apply. All right, let's do the second one. So here they say, can we use the intermediate value theorem to say that there is a value c such that f of c equals zero and two is less than or equal to c is less than or equal to four? If so, write a justification. So we are given that f is continuous. So let me write that down."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here they say, can we use the intermediate value theorem to say that there is a value c such that f of c equals zero and two is less than or equal to c is less than or equal to four? If so, write a justification. So we are given that f is continuous. So let me write that down. We are given that f is continuous. And if you want to be over that interval, but they're telling us it's continuous in general. And then we can just look at what is the value of the function at these endpoints."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write that down. We are given that f is continuous. And if you want to be over that interval, but they're telling us it's continuous in general. And then we can just look at what is the value of the function at these endpoints. So our interval goes from two to four. So we're talking about this closed interval right over here. We know that f of two, f of two is going to be equal to negative two, we see it in that table."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can just look at what is the value of the function at these endpoints. So our interval goes from two to four. So we're talking about this closed interval right over here. We know that f of two, f of two is going to be equal to negative two, we see it in that table. And what's f of four? F of four is equal to three. F of four is equal to three."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "We know that f of two, f of two is going to be equal to negative two, we see it in that table. And what's f of four? F of four is equal to three. F of four is equal to three. So zero is between f of two and f of four. And you can see it visually here. There's no way to draw between this point and that point without picking up your pen, without crossing the x-axis, without having a point where your function is equal to zero."}, {"video_title": "Justification with the intermediate value theorem table   AP Calculus AB   Khan Academy.mp3", "Sentence": "F of four is equal to three. So zero is between f of two and f of four. And you can see it visually here. There's no way to draw between this point and that point without picking up your pen, without crossing the x-axis, without having a point where your function is equal to zero. And so we can say, so according to the intermediate value theorem, there is a value c such that f of c is equal to zero and two is less than or equal to c is less than or equal to four. So all we're saying is, hey, there must be a value c, and the way I drew it here, that c value is right over here where c is between two and four, where f of c is equal to zero. And this seems all mathy and a little bit confusing sometimes, but it's saying something very intuitive."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "The graph of f consists of two quarter circles and one line segment. So this right here is one quarter circle, then we have another quarter circle, and then it has this line segment over here. As shown in the figure above, I put the figure on the side so that we have some screen real estate down here. Let g of x be equal to two x plus the definite integral from zero to x of f of t dt. All right, that's a little bit interesting. Now let's do part a. Part a, find g of negative three."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let g of x be equal to two x plus the definite integral from zero to x of f of t dt. All right, that's a little bit interesting. Now let's do part a. Part a, find g of negative three. So we wanna find g of negative three. So first of all, it tells us what g of x is. So g of negative three, every time we see an x, we just put a negative three there."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Part a, find g of negative three. So we wanna find g of negative three. So first of all, it tells us what g of x is. So g of negative three, every time we see an x, we just put a negative three there. So it's gonna be two times negative three, two times negative three, plus the definite integral from zero to negative three, f of t, f of t dt. So this first part is pretty straightforward. Two times negative three is negative six."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g of negative three, every time we see an x, we just put a negative three there. So it's gonna be two times negative three, two times negative three, plus the definite integral from zero to negative three, f of t, f of t dt. So this first part is pretty straightforward. Two times negative three is negative six. And then this part right over here is the definite integral from zero to negative three of f of t dt. So from zero to negative three, f of t dt. So this is essentially the area under f of t, or under f of x, if you wanna view it that way, from zero to three."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Two times negative three is negative six. And then this part right over here is the definite integral from zero to negative three of f of t dt. So from zero to negative three, f of t dt. So this is essentially the area under f of t, or under f of x, if you wanna view it that way, from zero to three. And this is pretty straightforward to figure out, but we have to be careful here. Because this area, this area right over here, this would be the integral, let me do this in a color that you can see, that area right over there would be the integral from negative three to zero, f of t, or we could either say f of t dt, or f of x dx, either way, dt. That would be this area right over here."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is essentially the area under f of t, or under f of x, if you wanna view it that way, from zero to three. And this is pretty straightforward to figure out, but we have to be careful here. Because this area, this area right over here, this would be the integral, let me do this in a color that you can see, that area right over there would be the integral from negative three to zero, f of t, or we could either say f of t dt, or f of x dx, either way, dt. That would be this area right over here. They've swapped it. They have the larger number as the lower bound. They have zero as the lower bound."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "That would be this area right over here. They've swapped it. They have the larger number as the lower bound. They have zero as the lower bound. So what we could do, we could rewrite this. This is the same thing. This is equal to negative six minus, if you swap the bounds of integration, you've swapped the sign on the integral."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "They have zero as the lower bound. So what we could do, we could rewrite this. This is the same thing. This is equal to negative six minus, if you swap the bounds of integration, you've swapped the sign on the integral. So we could say minus from negative three to zero of f of t dt. And now this thing right over here, this expression right over here, is the area under this quarter circle. And the area under this quarter circle, we can just use a little bit of geometry to figure that out."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is equal to negative six minus, if you swap the bounds of integration, you've swapped the sign on the integral. So we could say minus from negative three to zero of f of t dt. And now this thing right over here, this expression right over here, is the area under this quarter circle. And the area under this quarter circle, we can just use a little bit of geometry to figure that out. We know its radius. The radius here is three. Radius is equal to three."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the area under this quarter circle, we can just use a little bit of geometry to figure that out. We know its radius. The radius here is three. Radius is equal to three. The entire circle is pi r squared. So the entire circle, the area if this was an entire circle right over here, if this was an entire circle, that area would be pi r squared, so pi times three squared, so nine pi. So it would be nine pi, but this is only 1 1\u20444 of the entire circle, so we wanna divide it by four."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Radius is equal to three. The entire circle is pi r squared. So the entire circle, the area if this was an entire circle right over here, if this was an entire circle, that area would be pi r squared, so pi times three squared, so nine pi. So it would be nine pi, but this is only 1 1\u20444 of the entire circle, so we wanna divide it by four. So this right over here, we'll evaluate to nine pi over four. So part A, we get, it's equal to negative six, that's that right over there, minus nine pi, minus nine pi over four. And that's at least the first part of part A."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would be nine pi, but this is only 1 1\u20444 of the entire circle, so we wanna divide it by four. So this right over here, we'll evaluate to nine pi over four. So part A, we get, it's equal to negative six, that's that right over there, minus nine pi, minus nine pi over four. And that's at least the first part of part A. Then they say find g prime of x. So let's find g prime of x. So g prime of x."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's at least the first part of part A. Then they say find g prime of x. So let's find g prime of x. So g prime of x. We did the first part of part A right over there. Then we need to find g prime of x. G prime of x. Well, it's just going to be the derivative of g of x."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So g prime of x. We did the first part of part A right over there. Then we need to find g prime of x. G prime of x. Well, it's just going to be the derivative of g of x. The derivative of two x is just going to be two. And then the derivative, and this is the fundamental theorem of calculus right over here, the derivative of the definite integral from zero to x of f of t dt, that's just going to be f of x. That is just going to be f of x."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's just going to be the derivative of g of x. The derivative of two x is just going to be two. And then the derivative, and this is the fundamental theorem of calculus right over here, the derivative of the definite integral from zero to x of f of t dt, that's just going to be f of x. That is just going to be f of x. So that is g prime of x. We did this second part right over there. That is g prime of x."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is just going to be f of x. So that is g prime of x. We did this second part right over there. That is g prime of x. And then we need to evaluate it at negative three. So g prime of negative three, g prime of negative three is going to be equal to two plus f of negative three, which is equal to two plus, and let's look at our, let's see what f of negative three is. This is our function definition."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is g prime of x. And then we need to evaluate it at negative three. So g prime of negative three, g prime of negative three is going to be equal to two plus f of negative three, which is equal to two plus, and let's look at our, let's see what f of negative three is. This is our function definition. When x is equal to negative three, our function is at zero. So f of negative three is just zero. So it's just two plus zero, which is equal to two."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our function definition. When x is equal to negative three, our function is at zero. So f of negative three is just zero. So it's just two plus zero, which is equal to two. So g of negative three is equal to two. And we're done with part a. And the trickiest part of this is just realizing the bounds of integration."}, {"video_title": "2011 Calculus AB free response #4a   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's just two plus zero, which is equal to two. So g of negative three is equal to two. And we're done with part a. And the trickiest part of this is just realizing the bounds of integration. That you might be tempted to say, oh, the integral between zero and negative three, that's just this area over here, because we're going between zero and negative three. But you gotta realize that we actually have to swap them and make them negative if you really want to consider this area right over here. That this integral is going to be the negative of this area."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Find the volume of the solid. All right, so region S, we see it right over here. In the last part, we already said that this function is f and this function is g. You could try out some values and think about how they behave, or you could get a sense that, hey, look, a fourth degree quadratic's gonna have these ups and downs like that, while the, something, well, this is, it has an x term, but then it has an exponential term right here. That might have more of a, kind of an upward trend, especially as x increases, and then an upward trend also as x becomes more and more negative, because this exponent is going to become positive. So you can think about them in different ways like that. But what's important is region S. And there's a couple things we have to think about. We have to think about the bounds of region S. We know that the upper bound of region S, x is equal to, we know x is equal to two here, but we don't know this value right over here."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "That might have more of a, kind of an upward trend, especially as x increases, and then an upward trend also as x becomes more and more negative, because this exponent is going to become positive. So you can think about them in different ways like that. But what's important is region S. And there's a couple things we have to think about. We have to think about the bounds of region S. We know that the upper bound of region S, x is equal to, we know x is equal to two here, but we don't know this value right over here. We don't know, let's just call that a. I'll put a question mark over there. We don't know what this x value is, but we do know that at that x value, the functions are going to be equal to each other. So we know that f of a is going to be equal to g of a, or another way, in the last section, we talked about the absolute value of f of x minus g of x, which we've already inputted into our calculator and saved it."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have to think about the bounds of region S. We know that the upper bound of region S, x is equal to, we know x is equal to two here, but we don't know this value right over here. We don't know, let's just call that a. I'll put a question mark over there. We don't know what this x value is, but we do know that at that x value, the functions are going to be equal to each other. So we know that f of a is going to be equal to g of a, or another way, in the last section, we talked about the absolute value of f of x minus g of x, which we've already inputted into our calculator and saved it. We could also say that the absolute value of f of a minus g of a is equal to zero. And you don't need the absolute value there, because obviously the absolute value of zero is going to be zero, but we've already inputted this into our calculator, so it's going to be useful. So we can solve that thing we inputted into our calculator."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that f of a is going to be equal to g of a, or another way, in the last section, we talked about the absolute value of f of x minus g of x, which we've already inputted into our calculator and saved it. We could also say that the absolute value of f of a minus g of a is equal to zero. And you don't need the absolute value there, because obviously the absolute value of zero is going to be zero, but we've already inputted this into our calculator, so it's going to be useful. So we can solve that thing we inputted into our calculator. When does that equal zero? And actually, let's just do that right from the get-go. So in our calculator, we already inputted right over, let's make sure it's on."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can solve that thing we inputted into our calculator. When does that equal zero? And actually, let's just do that right from the get-go. So in our calculator, we already inputted right over, let's make sure it's on. In our calculator, we already inputted the absolute value of f of x minus g of x. And so we can use our solver now to figure out when does that equal zero. And it actually equals zero just in the, where we see it, it equals it one time, two times, and three times."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in our calculator, we already inputted right over, let's make sure it's on. In our calculator, we already inputted the absolute value of f of x minus g of x. And so we can use our solver now to figure out when does that equal zero. And it actually equals zero just in the, where we see it, it equals it one time, two times, and three times. So we want to find this place right over here. We want to figure out what x is, or what a is. So let's look at the math functions."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it actually equals zero just in the, where we see it, it equals it one time, two times, and three times. So we want to find this place right over here. We want to figure out what x is, or what a is. So let's look at the math functions. Actually, let's clear this. And actually, let me quit that mode. And so let's go to math."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at the math functions. Actually, let's clear this. And actually, let me quit that mode. And so let's go to math. Scroll down to the solver. The solver. And let's see, we want to redefine, okay, so we want to see, so we can go to the original screen of the solver."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's go to math. Scroll down to the solver. The solver. And let's see, we want to redefine, okay, so we want to see, so we can go to the original screen of the solver. And actually, I did this in the previous problem, although we redefined y sub one. But you say, okay, when does zero equal y sub one? Or when does y sub one, that's this here, equal zero?"}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's see, we want to redefine, okay, so we want to see, so we can go to the original screen of the solver. And actually, I did this in the previous problem, although we redefined y sub one. But you say, okay, when does zero equal y sub one? Or when does y sub one, that's this here, equal zero? And so you press enter. And then we can put in an initial guess. This looks someplace in between zero and two."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or when does y sub one, that's this here, equal zero? And so you press enter. And then we can put in an initial guess. This looks someplace in between zero and two. So I'll put an initial guess of one. And let's see what it gravitates to. Oh, whoops."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This looks someplace in between zero and two. So I'll put an initial guess of one. And let's see what it gravitates to. Oh, whoops. And then I could do alpha solve. Alpha solve. Let's let it munch on it a little bit."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Oh, whoops. And then I could do alpha solve. Alpha solve. Let's let it munch on it a little bit. Let's let it munch. It's going. And there you go, it got to 1.03, which is the one we want."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's let it munch on it a little bit. Let's let it munch. It's going. And there you go, it got to 1.03, which is the one we want. It didn't go to x equals zero or x equals two. It got, went to 1.03. So a is approximately equal to 1.03."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And there you go, it got to 1.03, which is the one we want. It didn't go to x equals zero or x equals two. It got, went to 1.03. So a is approximately equal to 1.03. So a is approximately equal to 1.03. And so what integral do we take? We don't want just the area of s. We're saying that s is the base."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So a is approximately equal to 1.03. So a is approximately equal to 1.03. And so what integral do we take? We don't want just the area of s. We're saying that s is the base. And if we take a cross section, we're going to have squares. So this is going to be a base of a square. So this length right over here is going to be the absolute value of f of x minus g of x."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't want just the area of s. We're saying that s is the base. And if we take a cross section, we're going to have squares. So this is going to be a base of a square. So this length right over here is going to be the absolute value of f of x minus g of x. That's the length of the base. But if we want the area of that cross section, you can imagine, let me draw it at a little bit of an angle here. So you have a cross section of a square there."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this length right over here is going to be the absolute value of f of x minus g of x. That's the length of the base. But if we want the area of that cross section, you can imagine, let me draw it at a little bit of an angle here. So you have a cross section of a square there. You have a cross section of a square here. And so the area of each of those cross sections is going to be the length of the base squared. And then you multiply that times the little dx's."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you have a cross section of a square there. You have a cross section of a square here. And so the area of each of those cross sections is going to be the length of the base squared. And then you multiply that times the little dx's. You get that volume of each of those little sections. And you sum them all up. And then you are going to get the volume of that figure."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you multiply that times the little dx's. You get that volume of each of those little sections. And you sum them all up. And then you are going to get the volume of that figure. So the volume of that figure is going to be equal to, so the volume is going to be equal to the integral from x equals a all the way, so that's approximately, you know, approximate, well I'll just write a over here, to x equals 2 of this business. This is the length of one side of the square. But we want to square it, because that's going to give us the area of that cross section."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you are going to get the volume of that figure. So the volume of that figure is going to be equal to, so the volume is going to be equal to the integral from x equals a all the way, so that's approximately, you know, approximate, well I'll just write a over here, to x equals 2 of this business. This is the length of one side of the square. But we want to square it, because that's going to give us the area of that cross section. And then, so let's do that. So times, I could say the absolute value of f of x minus g of x. I could just square that. Or if I want, I could just put parentheses there."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we want to square it, because that's going to give us the area of that cross section. And then, so let's do that. So times, I could say the absolute value of f of x minus g of x. I could just square that. Or if I want, I could just put parentheses there. I'll actually leave it as absolute value, since that's what I've already inputted into my calculator. And so that's going to give the area of each of those cross sections perpendicular to the x-axis. And you multiply it times dx."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or if I want, I could just put parentheses there. I'll actually leave it as absolute value, since that's what I've already inputted into my calculator. And so that's going to give the area of each of those cross sections perpendicular to the x-axis. And you multiply it times dx. You get the volume of that little section. Then you add them all up from x equals a to x equals 2. And luckily, we've already done a lot of the work here on our calculator."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you multiply it times dx. You get the volume of that little section. Then you add them all up from x equals a to x equals 2. And luckily, we've already done a lot of the work here on our calculator. Our value of a is stored in the variable x already. So let me quit this. And then, so let me go back to my math functions."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And luckily, we've already done a lot of the work here on our calculator. Our value of a is stored in the variable x already. So let me quit this. And then, so let me go back to my math functions. Let me go to definite integral, or function integral. Select, select that. And now, I'm not going to do y sub 1."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then, so let me go back to my math functions. Let me go to definite integral, or function integral. Select, select that. And now, I'm not going to do y sub 1. I'm going to do y sub 1 squared. So let me go variables, y variables. It's a function, y sub 1."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now, I'm not going to do y sub 1. I'm going to do y sub 1 squared. So let me go variables, y variables. It's a function, y sub 1. And I am going to, let me make sure, I'm going to square it. That's what I do right over here. My variable of integration is x. I'm going to do it from x."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's a function, y sub 1. And I am going to, let me make sure, I'm going to square it. That's what I do right over here. My variable of integration is x. I'm going to do it from x. The value of a is stored in the variable x in my calculator right now. It's a little confusing. This first x is a variable of integration."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "My variable of integration is x. I'm going to do it from x. The value of a is stored in the variable x in my calculator right now. It's a little confusing. This first x is a variable of integration. The next x is my lower bound of integration. And then, my upper bound is 2. And so, the most complex part here is some of the shortcuts I'm doing with the calculator."}, {"video_title": "2015 AP Calculus AP 2b   AP Calculus AB solved exams   AP Calculus AB   Khan Academy.mp3", "Sentence": "This first x is a variable of integration. The next x is my lower bound of integration. And then, my upper bound is 2. And so, the most complex part here is some of the shortcuts I'm doing with the calculator. And we let the calculator munch on it. And it's approximately 1.283. So it's approximately 1.283 cubic units."}, {"video_title": "Proof of the derivative of sin(x)   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you know that the derivative of sine of x with respect to x is cosine of x, and the derivative of cosine of x with respect to x is negative sine of x, that can empower you to do many more, far more complicated derivatives. But what we're gonna do in this video is dig a little bit deeper and actually prove this first derivative. I'm not gonna prove the second one. You can actually use it using the information we're going to do in this one. But it's just to make you feel good that someone's just not making this up, that there is a little bit of mathematical rigor behind it all. So let's try to calculate it. So the derivative with respect to x of sine of x."}, {"video_title": "Proof of the derivative of sin(x)   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can actually use it using the information we're going to do in this one. But it's just to make you feel good that someone's just not making this up, that there is a little bit of mathematical rigor behind it all. So let's try to calculate it. So the derivative with respect to x of sine of x. By definition, this is going to be the limit as delta x approaches zero of sine of x plus delta x minus sine of x, all of that over delta, all of that over delta x. This is really just the slope of the line between the point x comma sine of x and x plus delta x comma sine of x plus delta x. So how can we evaluate this?"}, {"video_title": "Proof of the derivative of sin(x)   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative with respect to x of sine of x. By definition, this is going to be the limit as delta x approaches zero of sine of x plus delta x minus sine of x, all of that over delta, all of that over delta x. This is really just the slope of the line between the point x comma sine of x and x plus delta x comma sine of x plus delta x. So how can we evaluate this? Well, we can rewrite sine of x plus delta x using our angle addition formulas that we learned during our trig identities. So this is going to be the same thing as the limit as delta x approaches zero. I'll rewrite this using our trig identity as cosine of x times sine of delta x plus sine of x times cosine of delta x."}, {"video_title": "Proof of the derivative of sin(x)   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how can we evaluate this? Well, we can rewrite sine of x plus delta x using our angle addition formulas that we learned during our trig identities. So this is going to be the same thing as the limit as delta x approaches zero. I'll rewrite this using our trig identity as cosine of x times sine of delta x plus sine of x times cosine of delta x. And then we're going to subtract this sine of x up here minus sine of x, all of that over, let me see if I can draw a relatively straight line, all of that over delta x. So this can be rewritten as being equal to the limit as delta x approaches zero of, let me write this part in red, so that would be cosine of x sine of delta x, all of that over delta x, and then that's going to be plus, I'll do all of this in orange, all I'm doing is I have the sum of things up here divided by delta x, I'm just breaking it up a little bit, plus sine of x cosine of delta x minus sine of x, all of that over delta x, and remember I'm taking the limit of this entire expression. Well the limit of a sum is equal to the sum of the limits, so this is going to be equal to, I'll do this first part in red, the limit as delta x approaches zero of, let's see I can rewrite this as cosine of x times sine of delta x over delta x plus the limit as delta x approaches zero of, and let's see I can factor out a sine of x here, so it's times sine of x, I factor that out and I'll be left with a cosine of delta x minus one, all of that over delta x, so that's this limit, and let's see if I can simplify this even more."}, {"video_title": "Proof of the derivative of sin(x)   Derivatives introduction   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll rewrite this using our trig identity as cosine of x times sine of delta x plus sine of x times cosine of delta x. And then we're going to subtract this sine of x up here minus sine of x, all of that over, let me see if I can draw a relatively straight line, all of that over delta x. So this can be rewritten as being equal to the limit as delta x approaches zero of, let me write this part in red, so that would be cosine of x sine of delta x, all of that over delta x, and then that's going to be plus, I'll do all of this in orange, all I'm doing is I have the sum of things up here divided by delta x, I'm just breaking it up a little bit, plus sine of x cosine of delta x minus sine of x, all of that over delta x, and remember I'm taking the limit of this entire expression. Well the limit of a sum is equal to the sum of the limits, so this is going to be equal to, I'll do this first part in red, the limit as delta x approaches zero of, let's see I can rewrite this as cosine of x times sine of delta x over delta x plus the limit as delta x approaches zero of, and let's see I can factor out a sine of x here, so it's times sine of x, I factor that out and I'll be left with a cosine of delta x minus one, all of that over delta x, so that's this limit, and let's see if I can simplify this even more. Let me scroll down a bit. So this left hand expression I can rewrite, this cosine of x has nothing to do with the limit as delta x approaches zero, so we can actually take that outside of the limit. So we have the cosine of x times the limit as delta x approaches zero of sine of delta x over delta x, and now we need to add this thing, and let's see how I could write this."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's take the derivative with respect to x of x to the n plus 1th power over n plus 1 plus some constant c. And we're going to assume here, because we want this expression to be defined, we're going to assume that n does not equal negative 1. If it equaled negative 1, we'd be dividing by 0, and we haven't defined what that means. So let's take the derivative here. So this is going to be equal to the derivative of x to the n plus 1 over n plus 1. We can just use the power rule over here. So our exponent is n plus 1. We can bring it out front."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to the derivative of x to the n plus 1 over n plus 1. We can just use the power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n plus 1 times x times x to the, I want to use that same color. Colors are the hard part. Times x to the, instead of n plus 1, we subtract 1 from the exponent."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can bring it out front. So it's going to be n plus 1 times x times x to the, I want to use that same color. Colors are the hard part. Times x to the, instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we have, we were dividing by this n plus 1. So we have divided by n plus 1."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Times x to the, instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we have, we were dividing by this n plus 1. So we have divided by n plus 1. And then we have plus c, the derivative of a constant with respect to x. A constant does not change as x changes. So it is just going to be 0."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have divided by n plus 1. And then we have plus c, the derivative of a constant with respect to x. A constant does not change as x changes. So it is just going to be 0. So plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it is just going to be 0. So plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing, and this is in very general terms, is equal to x to the n. So given that, what is the antiderivative, let me switch colors here, what is the antiderivative of x to the n, and remember this is just the kind of strange looking notation we use. It makes more sense when we start doing definite integrals. But what is the antiderivative of x to the n, and we could say the antiderivative with respect to x if we want to."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing, and this is in very general terms, is equal to x to the n. So given that, what is the antiderivative, let me switch colors here, what is the antiderivative of x to the n, and remember this is just the kind of strange looking notation we use. It makes more sense when we start doing definite integrals. But what is the antiderivative of x to the n, and we could say the antiderivative with respect to x if we want to. And another way of calling this is the indefinite integral. Indefinite integral. Well we know, this is saying x to the n is the derivative of what?"}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what is the antiderivative of x to the n, and we could say the antiderivative with respect to x if we want to. And another way of calling this is the indefinite integral. Indefinite integral. Well we know, this is saying x to the n is the derivative of what? Well we just figured out, it's the derivative of this thing. And we've written it in very general terms. We're actually encapsulating multiple constants here."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well we know, this is saying x to the n is the derivative of what? Well we just figured out, it's the derivative of this thing. And we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0 plus 1 plus 2 plus pi plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0 plus 1 plus 2 plus pi plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can kind of view this as the reverse power rule. And it applies for any n as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this. You could call it the reverse power rule if you want, or the anti-power rule."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this. You could call it the reverse power rule if you want, or the anti-power rule. So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well all we have to say is, well look, the 5 is equal to the n. We just have to increment the exponent by 1."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could call it the reverse power rule if you want, or the anti-power rule. So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well all we have to say is, well look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was, when you increment it by 1, we divide by that same value."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well all we have to say is, well look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was, when you increment it by 1, we divide by that same value. Divided by 5 plus 1. And of course we want to encapsulate all of the possible antiderivatives. So you put the c right over there."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Whatever the exponent was, when you increment it by 1, we divide by that same value. Divided by 5 plus 1. And of course we want to encapsulate all of the possible antiderivatives. So you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Now we'll do it in blue. Let's tie the antiderivative of, let's make it interesting, let's make it 5 times x to the negative 2 power dx. So how would we evaluate this?"}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's try another one. Now we'll do it in blue. Let's tie the antiderivative of, let's make it interesting, let's make it 5 times x to the negative 2 power dx. So how would we evaluate this? Well one simplification you can do, and I haven't rigorously proven it to you just yet, but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is indeed equal to 5 times the antiderivative of x to the negative 2 power dx. And now we can just use, I guess we could call it this anti-power rule."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how would we evaluate this? Well one simplification you can do, and I haven't rigorously proven it to you just yet, but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is indeed equal to 5 times the antiderivative of x to the negative 2 power dx. And now we can just use, I guess we could call it this anti-power rule. So this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1 plus some constant. And this is equal to 5 times negative x to the negative 1 plus some constant."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we can just use, I guess we could call it this anti-power rule. So this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1 plus some constant. And this is equal to 5 times negative x to the negative 1 plus some constant. And then if we want we can distribute the 5. So this is equal to negative 5 x to the negative 1. Now we could write plus 5 times some constant but this is just an arbitrary constant."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is equal to 5 times negative x to the negative 1 plus some constant. And then if we want we can distribute the 5. So this is equal to negative 5 x to the negative 1. Now we could write plus 5 times some constant but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could have written this, if you wanted to show this as different constants you could say this is c1 c1 c1. You multiply 5 times c1 you get another constant."}, {"video_title": "Reverse power rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now we could write plus 5 times some constant but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could have written this, if you wanted to show this as different constants you could say this is c1 c1 c1. You multiply 5 times c1 you get another constant. We could just call that c. Which is equal to 5 times c1. But there you have it. Negative 5 x to the negative 1 plus c. And once again all of these try to evaluate the derivative and you will see that you get this business right over there."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say that we are trying to take the derivative of the expression. So let's say we're taking the derivative of the expression, the natural log of sine of x. So the first key misconception or misunderstanding that many people have is when you're dealing with transcendental functions like this. And transcendental functions is just a fancy word for these functions like trigonometric functions, logarithmic functions that don't use standard algebraic operations. But when you see transcendental functions like this or compositions of them, many people confuse this with the product of functions. So at first when they look at this, they might see this as a product of functions. When they look at this, they might see this as being the same as the derivative with respect to x of natural log of x, natural log of x times sine of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And transcendental functions is just a fancy word for these functions like trigonometric functions, logarithmic functions that don't use standard algebraic operations. But when you see transcendental functions like this or compositions of them, many people confuse this with the product of functions. So at first when they look at this, they might see this as a product of functions. When they look at this, they might see this as being the same as the derivative with respect to x of natural log of x, natural log of x times sine of x. And you can see just the way that it's written, they look very similar. But this is the product of two functions. If you said natural log of x is f of x and sine of x is g of x, this is the product of sine and g of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "When they look at this, they might see this as being the same as the derivative with respect to x of natural log of x, natural log of x times sine of x. And you can see just the way that it's written, they look very similar. But this is the product of two functions. If you said natural log of x is f of x and sine of x is g of x, this is the product of sine and g of x. This is the product of f of x and g of x. And here you would use the product rule. So to actually compute this, you would use the product, the product rule."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you said natural log of x is f of x and sine of x is g of x, this is the product of sine and g of x. This is the product of f of x and g of x. And here you would use the product rule. So to actually compute this, you would use the product, the product rule. But this is a composition. Here you have f of g of x, not f of x times g of x. So here you have, that is our g of x, it equals sine of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So to actually compute this, you would use the product, the product rule. But this is a composition. Here you have f of g of x, not f of x times g of x. So here you have, that is our g of x, it equals sine of x. And then our f of g of x is the natural log of sine of x. So this is f of g of x. F of g of x, just like that. If someone asked you just what f of x was, well, that would be natural log of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here you have, that is our g of x, it equals sine of x. And then our f of g of x is the natural log of sine of x. So this is f of g of x. F of g of x, just like that. If someone asked you just what f of x was, well, that would be natural log of x. But f of g of x is natural log of our g of x, which is natural log of sine of x. So that's the key first thing. Always make sure whether you're gonna use, especially with these transcendental functions, that hey, if this is a composition, you've gotta use the chain rule, not the product rule."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "If someone asked you just what f of x was, well, that would be natural log of x. But f of g of x is natural log of our g of x, which is natural log of sine of x. So that's the key first thing. Always make sure whether you're gonna use, especially with these transcendental functions, that hey, if this is a composition, you've gotta use the chain rule, not the product rule. It's not the product. Now sometimes you have a combination. You have a product of compositions, and then things get a little bit more involved."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Always make sure whether you're gonna use, especially with these transcendental functions, that hey, if this is a composition, you've gotta use the chain rule, not the product rule. It's not the product. Now sometimes you have a combination. You have a product of compositions, and then things get a little bit more involved. But pay close attention to make sure that you're not dealing with a composition. Now the next misconception students have is even if they recognize, okay, I've gotta use the chain rule, sometimes it doesn't go fully to completion. So let's continue using this example."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have a product of compositions, and then things get a little bit more involved. But pay close attention to make sure that you're not dealing with a composition. Now the next misconception students have is even if they recognize, okay, I've gotta use the chain rule, sometimes it doesn't go fully to completion. So let's continue using this example. The chain rule here says, look, we have to take the derivative of the outer function with respect to the inner function. So if I were to say, in this case, f of x is natural log of x. f of g of x is this expression here. So if I wanna do this first part, f prime of g of x, f prime of g of x, well, the derivative of the natural log of x is one over x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's continue using this example. The chain rule here says, look, we have to take the derivative of the outer function with respect to the inner function. So if I were to say, in this case, f of x is natural log of x. f of g of x is this expression here. So if I wanna do this first part, f prime of g of x, f prime of g of x, well, the derivative of the natural log of x is one over x. So the derivative of natural log of x is one over x, but we don't want the derivative where the input is x. We want the derivative when the input is g of x. So instead of it being one over x, it's going to be one over g of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I wanna do this first part, f prime of g of x, f prime of g of x, well, the derivative of the natural log of x is one over x. So the derivative of natural log of x is one over x, but we don't want the derivative where the input is x. We want the derivative when the input is g of x. So instead of it being one over x, it's going to be one over g of x. One over g of x. And we know that g of x is equal to sine of x. That's equal to sine of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So instead of it being one over x, it's going to be one over g of x. One over g of x. And we know that g of x is equal to sine of x. That's equal to sine of x. Now one key misunderstanding that the folks at the College Board told us about is many students stop right there. They just do this first part and then they forget to multiply this second part. So here we are not done."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's equal to sine of x. Now one key misunderstanding that the folks at the College Board told us about is many students stop right there. They just do this first part and then they forget to multiply this second part. So here we are not done. We need to take this and multiply it times g prime of x. And let me write this down. G prime of x, what would that be?"}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here we are not done. We need to take this and multiply it times g prime of x. And let me write this down. G prime of x, what would that be? Well, the derivative of sine of x with respect to x, well, that's just going to be cosine of x. Cosine, cosine of x. Cosine of x. So in this example right over here, the derivative is going to be, let's see if I can squeeze it in over here, it's gonna be one over sine of x, which is this part, times cosine of x. So let me write it down."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "G prime of x, what would that be? Well, the derivative of sine of x with respect to x, well, that's just going to be cosine of x. Cosine, cosine of x. Cosine of x. So in this example right over here, the derivative is going to be, let's see if I can squeeze it in over here, it's gonna be one over sine of x, which is this part, times cosine of x. So let me write it down. It is going to be one over sine of x. Do that in that other color. One over sine of x and then times cosine of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me write it down. It is going to be one over sine of x. Do that in that other color. One over sine of x and then times cosine of x. So once again, just to make sure that you don't fall into one of these misconceptions, let me box this off so it's a little bit cleaner. So to just make sure that you don't fall into one of these misconceptions here, recognize the composition, that this is not the product of natural log of x and sine of x, it's natural log of sine of x. And then when you're actually applying the chain rule, derivative of the outside with respect to the inside, so the derivative of natural log of x is one over x, so that applied when the input is g of x is one over sine of x, and then multiply that times the derivative of the inner function."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "One over sine of x and then times cosine of x. So once again, just to make sure that you don't fall into one of these misconceptions, let me box this off so it's a little bit cleaner. So to just make sure that you don't fall into one of these misconceptions here, recognize the composition, that this is not the product of natural log of x and sine of x, it's natural log of sine of x. And then when you're actually applying the chain rule, derivative of the outside with respect to the inside, so the derivative of natural log of x is one over x, so that applied when the input is g of x is one over sine of x, and then multiply that times the derivative of the inner function. So don't forget to do this right over here. Now another misconception that students have is instead of doing what we just did, instead of applying the chain rule like this, they take the derivative of the outer function with respect to the derivative of the inner function. So for example, they would compute this, f prime of g prime of x. f prime of g prime of x."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then when you're actually applying the chain rule, derivative of the outside with respect to the inside, so the derivative of natural log of x is one over x, so that applied when the input is g of x is one over sine of x, and then multiply that times the derivative of the inner function. So don't forget to do this right over here. Now another misconception that students have is instead of doing what we just did, instead of applying the chain rule like this, they take the derivative of the outer function with respect to the derivative of the inner function. So for example, they would compute this, f prime of g prime of x. f prime of g prime of x. Which in this case, f prime of x is one over x, but if the input is g prime of x, g prime of x is cosine of x. So many students end up doing this, where they take the derivative of the outside and they apply, and the input into that, they use the derivative of the inside function. This, this is not right."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, they would compute this, f prime of g prime of x. f prime of g prime of x. Which in this case, f prime of x is one over x, but if the input is g prime of x, g prime of x is cosine of x. So many students end up doing this, where they take the derivative of the outside and they apply, and the input into that, they use the derivative of the inside function. This, this is not right. Be very careful that you're not doing that. You do the derivative of the outside function with respect to the inside function, not taking its derivative, and then multiply, don't forget to multiply, times the derivative of the inside function here. So hopefully that helps a little bit."}, {"video_title": "Common chain rule misunderstandings   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This, this is not right. Be very careful that you're not doing that. You do the derivative of the outside function with respect to the inside function, not taking its derivative, and then multiply, don't forget to multiply, times the derivative of the inside function here. So hopefully that helps a little bit. If all of this looks completely foreign to you, I encourage you to watch the whole series of chain rule introductory videos and worked examples we have. This is just a topping on top of that to make sure that you don't fall into these misconceptions of applying the product rule when you really need to be applying the chain rule, or forgetting to do part of the chain rule, multiplying by g prime of x, or evaluating f prime of g prime of x. So hopefully that helps."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And we see that f of x seems to be approaching this value, seems to be approaching this value right over here, it seems to be approaching five. And so the way we would denote that is the limit of f of x as x approaches two, and we're going to specify the direction, as x approaches two from the negative direction, we put the negative as a superscript after the two to denote the direction that we're approaching. This is not a negative two, we're approaching two from the negative direction. We're approaching two from values less than two, we're getting closer and closer to two, but from below, 1.9, 1.99, 1.999999, as x gets closer and closer to, from those values, what is f of x approaching? And we see here that it is approaching five. But what if we were asked a different question, or I guess the natural other question. What is the limit of f of x as x approaches two from values greater than two?"}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "We're approaching two from values less than two, we're getting closer and closer to two, but from below, 1.9, 1.99, 1.999999, as x gets closer and closer to, from those values, what is f of x approaching? And we see here that it is approaching five. But what if we were asked a different question, or I guess the natural other question. What is the limit of f of x as x approaches two from values greater than two? So this is a little superscript positive right over here. So now we're going to approach x equals two, but we're going to approach it from this direction. x equals three, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001, and we're going to get closer and closer to, but we're coming from values that are larger than two."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "What is the limit of f of x as x approaches two from values greater than two? So this is a little superscript positive right over here. So now we're going to approach x equals two, but we're going to approach it from this direction. x equals three, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001, and we're going to get closer and closer to, but we're coming from values that are larger than two. So here, when x equals three, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "x equals three, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001, and we're going to get closer and closer to, but we're coming from values that are larger than two. So here, when x equals three, f of x is here. When x equals 2.5, f of x is here. When x equals 2.01, f of x looks like it's right over here. So in this situation, we're getting closer and closer to f of x equaling one. It never does quite equal that. It actually then just has a jump discontinuity, but it seems to be approaching, this seems to be the limiting value when we approach x from values greater than, or when we approach two from values greater than two."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "When x equals 2.01, f of x looks like it's right over here. So in this situation, we're getting closer and closer to f of x equaling one. It never does quite equal that. It actually then just has a jump discontinuity, but it seems to be approaching, this seems to be the limiting value when we approach x from values greater than, or when we approach two from values greater than two. So this right over here is equal to one. And so when we think about limits in general, the only way that a limit at two will actually exist is if both of these limits, both of these one-sided limits are actually the same thing. In this situation, they aren't."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It actually then just has a jump discontinuity, but it seems to be approaching, this seems to be the limiting value when we approach x from values greater than, or when we approach two from values greater than two. So this right over here is equal to one. And so when we think about limits in general, the only way that a limit at two will actually exist is if both of these limits, both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach two from values below two, the function seems to be approaching five, and as we approach two from values above two, the function seems to be approaching one. So in this case, the limit, let me write this down, the limit of f of x as x approaches two from the negative direction does not equal the limit of f of x as x approaches two from the positive direction. And since this is the case that they're not equal, the limit does not exist."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "In this situation, they aren't. As we approach two from values below two, the function seems to be approaching five, and as we approach two from values above two, the function seems to be approaching one. So in this case, the limit, let me write this down, the limit of f of x as x approaches two from the negative direction does not equal the limit of f of x as x approaches two from the positive direction. And since this is the case that they're not equal, the limit does not exist. The limit as x approaches two in general of f of x, or the limit of f of x as x approaches two does not exist. In order for it to have existed, these two things would have had to been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches four, well, then we could think about the two one-sided limits, the one-sided limit from below and the one-sided limit from above."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And since this is the case that they're not equal, the limit does not exist. The limit as x approaches two in general of f of x, or the limit of f of x as x approaches two does not exist. In order for it to have existed, these two things would have had to been equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches four, well, then we could think about the two one-sided limits, the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x as x approaches four from below. So let me draw that."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "For example, if someone were to say, what is the limit of f of x as x approaches four, well, then we could think about the two one-sided limits, the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x as x approaches four from below. So let me draw that. So we care about x equals four. As x equals four from below, so when x equals three, we're here where f of three is negative two. f of 3.5 seems to be right over here."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me draw that. So we care about x equals four. As x equals four from below, so when x equals three, we're here where f of three is negative two. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999, we're getting closer and closer to our function equaling negative five. So this, the limit as we approach four from below, this one-sided limit from the left, we could say, this is going to be equal to negative five."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999, we're getting closer and closer to our function equaling negative five. So this, the limit as we approach four from below, this one-sided limit from the left, we could say, this is going to be equal to negative five. And if we were to ask ourselves the limit of f of x as x approaches four from the right, from values larger than four, x approaches four from the right, well, same exercise. f of five gets us here. f of 4.5 seems right around here."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So this, the limit as we approach four from below, this one-sided limit from the left, we could say, this is going to be equal to negative five. And if we were to ask ourselves the limit of f of x as x approaches four from the right, from values larger than four, x approaches four from the right, well, same exercise. f of five gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of four is actually defined, but we're getting closer and closer to it."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of four is actually defined, but we're getting closer and closer to it. And we see once again we are approaching five. Even if f of four was not defined, on either side we would be approaching negative five. So this is also approaching negative five."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "And even f of four is actually defined, but we're getting closer and closer to it. And we see once again we are approaching five. Even if f of four was not defined, on either side we would be approaching negative five. So this is also approaching negative five. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say, so these two things are equal. And because these two things are equal, we know that the limit of f of x as x approaches four is equal to five. Let's look at a few more examples."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is also approaching negative five. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say, so these two things are equal. And because these two things are equal, we know that the limit of f of x as x approaches four is equal to five. Let's look at a few more examples. So let's ask ourselves the limit of f of x, now this is our new f of x depicted here, as x approaches eight. And let's approach eight from the left. As x approaches eight from values less than eight."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's look at a few more examples. So let's ask ourselves the limit of f of x, now this is our new f of x depicted here, as x approaches eight. And let's approach eight from the left. As x approaches eight from values less than eight. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to eight."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "As x approaches eight from values less than eight. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to eight. So if x is seven, f of seven is here. If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to three."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So x is getting closer and closer to eight. So if x is seven, f of seven is here. If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to three. So it looks like the limit of f of x as x approaches eight from the negative side is equal to three. What about from the positive side? What about the limit of f of x as x approaches eight from the positive direction or from the right side?"}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So it looks like our value of f of x is getting closer and closer and closer to three. So it looks like the limit of f of x as x approaches eight from the negative side is equal to three. What about from the positive side? What about the limit of f of x as x approaches eight from the positive direction or from the right side? Well, here we see as x is nine, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling one."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "What about the limit of f of x as x approaches eight from the positive direction or from the right side? Well, here we see as x is nine, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling one. So notice, these two limits are different. So the limit actually does, the non-one-sided limit or the two-sided limit does not exist at f of x or as we approach eight. So let me write that down."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "It seems like we're approaching f of x equaling one. So notice, these two limits are different. So the limit actually does, the non-one-sided limit or the two-sided limit does not exist at f of x or as we approach eight. So let me write that down. The limit of f of x as x approaches eight, because these two things are not the same value, this does not exist. Let's do one more example, and here they're actually asking us a question. The function f is graphed below."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So let me write that down. The limit of f of x as x approaches eight, because these two things are not the same value, this does not exist. Let's do one more example, and here they're actually asking us a question. The function f is graphed below. What appears to be the value of the one-sided limit, the limit of f of x, this is f of x, as x approaches negative two from the negative direction? So this is the negative two from the negative direction. So we care what happens as x approaches negative two."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "The function f is graphed below. What appears to be the value of the one-sided limit, the limit of f of x, this is f of x, as x approaches negative two from the negative direction? So this is the negative two from the negative direction. So we care what happens as x approaches negative two. We see f of x is actually undefined right over there. But let's see what happens as we approach from the negative direction or as we approach from values less than negative two or as we approach from the left. As we approach from the left, f of negative four is right over here."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So we care what happens as x approaches negative two. We see f of x is actually undefined right over there. But let's see what happens as we approach from the negative direction or as we approach from values less than negative two or as we approach from the left. As we approach from the left, f of negative four is right over here. So this is f of negative four. F of negative three is right over here. Negative three is right over there."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "As we approach from the left, f of negative four is right over here. So this is f of negative four. F of negative three is right over here. Negative three is right over there. F of negative 2.5 seems to be right over here. We seem to be getting closer and closer to f of x being equal to four, at least visually. So I would say that it looks, at least graphically, the limit of f of x as x approaches two from the negative direction is equal to four."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "Negative three is right over there. F of negative 2.5 seems to be right over here. We seem to be getting closer and closer to f of x being equal to four, at least visually. So I would say that it looks, at least graphically, the limit of f of x as x approaches two from the negative direction is equal to four. Now, if we also ask ourselves the limit of f of x as x approaches negative two from the positive direction, we would get a similar result. Now we're going to approach from when x is zero, f of x seems to be right over here. When x is one, f of x is right over here."}, {"video_title": "One-sided limits from graphs   Limits   Differential Calculus   Khan Academy.mp3", "Sentence": "So I would say that it looks, at least graphically, the limit of f of x as x approaches two from the negative direction is equal to four. Now, if we also ask ourselves the limit of f of x as x approaches negative two from the positive direction, we would get a similar result. Now we're going to approach from when x is zero, f of x seems to be right over here. When x is one, f of x is right over here. When x is negative 1.99, f of x is there. When x is negative 1.9, f of x seems to be right over here. So once again, we seem to be getting closer and closer to four."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we want to figure out what does it obtain its maximum acceleration? So let's just review what they gave us. They gave us velocity as a function of time. So let's just remind ourselves. If we have, let's say our position is a function of time, so let's say x of t is position as a function of time, then if we were to take the derivative of that, so x prime of t, well, that's going to be the rate of change of position with respect to time or the velocity as a function of time. And if we were to take the derivative of our velocity, then that's going to be the rate of change of velocity with respect to time. Well, that's going to be acceleration as a function of time."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just remind ourselves. If we have, let's say our position is a function of time, so let's say x of t is position as a function of time, then if we were to take the derivative of that, so x prime of t, well, that's going to be the rate of change of position with respect to time or the velocity as a function of time. And if we were to take the derivative of our velocity, then that's going to be the rate of change of velocity with respect to time. Well, that's going to be acceleration as a function of time. So they give us velocity, so from velocity we can figure out acceleration. So let me just rewrite that. So we know that v of t is equal to negative t to the third power plus six t squared plus two t, and so from that we can figure out the acceleration as a function of time, which is just going to be the derivative with respect to t of the velocity."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's going to be acceleration as a function of time. So they give us velocity, so from velocity we can figure out acceleration. So let me just rewrite that. So we know that v of t is equal to negative t to the third power plus six t squared plus two t, and so from that we can figure out the acceleration as a function of time, which is just going to be the derivative with respect to t of the velocity. So I'll just use the power rule a bunch. So that's going to be, this is a third power right there, so negative three t squared plus two times six is 12 t to the first plus two, so that's our acceleration as a function of time. And we want to figure out when we obtain our maximum acceleration."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that v of t is equal to negative t to the third power plus six t squared plus two t, and so from that we can figure out the acceleration as a function of time, which is just going to be the derivative with respect to t of the velocity. So I'll just use the power rule a bunch. So that's going to be, this is a third power right there, so negative three t squared plus two times six is 12 t to the first plus two, so that's our acceleration as a function of time. And we want to figure out when we obtain our maximum acceleration. And just inspecting this acceleration function here, we see it's a quadratic, it has a second degree polynomial, so we have a negative coefficient out in front of the highest degree term, in front of the second degree term, so it is going to be a downward opening parabola. So it is going to be a downward opening, let me draw it in the same color. So it is going to have that general shape, and so it will indeed take on, it will indeed take on a maximum value."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we want to figure out when we obtain our maximum acceleration. And just inspecting this acceleration function here, we see it's a quadratic, it has a second degree polynomial, so we have a negative coefficient out in front of the highest degree term, in front of the second degree term, so it is going to be a downward opening parabola. So it is going to be a downward opening, let me draw it in the same color. So it is going to have that general shape, and so it will indeed take on, it will indeed take on a maximum value. But how do we figure out that maximum value? Well, that maximum value is going to happen when the acceleration value, when the slope of its tangent line is equal to, when the slope of its tangent line is equal to zero. And we could also verify that it is concave downwards at that point using the second derivative test, by showing that the second derivative is negative there."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it is going to have that general shape, and so it will indeed take on, it will indeed take on a maximum value. But how do we figure out that maximum value? Well, that maximum value is going to happen when the acceleration value, when the slope of its tangent line is equal to, when the slope of its tangent line is equal to zero. And we could also verify that it is concave downwards at that point using the second derivative test, by showing that the second derivative is negative there. So let's do that, let's look at the first and second derivatives of our acceleration, of our acceleration function. So, and I'll switch colors, that one's actually a little bit hard to see. So, the first derivative, the rate of change of acceleration is going to be equal to, so this is negative six t plus 12."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could also verify that it is concave downwards at that point using the second derivative test, by showing that the second derivative is negative there. So let's do that, let's look at the first and second derivatives of our acceleration, of our acceleration function. So, and I'll switch colors, that one's actually a little bit hard to see. So, the first derivative, the rate of change of acceleration is going to be equal to, so this is negative six t plus 12. Now let's think about when does this thing equal zero? Well, if we subtract 12 from both sides, we get negative six t is equal to negative 12. Divide both sides by negative six, you get t is equal to two."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, the first derivative, the rate of change of acceleration is going to be equal to, so this is negative six t plus 12. Now let's think about when does this thing equal zero? Well, if we subtract 12 from both sides, we get negative six t is equal to negative 12. Divide both sides by negative six, you get t is equal to two. So, a couple of things, you could just say, alright, look, I know that this is a downward opening quadra, parabola right over here, I have a negative coefficient on my second degree term. I know that the slope of the tangent line here is zero at t equals two, so that's gonna be my maximum point. Or you could go a little bit further, you could take the second derivative, you, let's do that, just for kicks."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Divide both sides by negative six, you get t is equal to two. So, a couple of things, you could just say, alright, look, I know that this is a downward opening quadra, parabola right over here, I have a negative coefficient on my second degree term. I know that the slope of the tangent line here is zero at t equals two, so that's gonna be my maximum point. Or you could go a little bit further, you could take the second derivative, you, let's do that, just for kicks. So we could take the second derivative of our acceleration function. So this is going to be equal to negative six, right? So the derivative of negative six t is negative six, the derivative of constant is just zero."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or you could go a little bit further, you could take the second derivative, you, let's do that, just for kicks. So we could take the second derivative of our acceleration function. So this is going to be equal to negative six, right? So the derivative of negative six t is negative six, the derivative of constant is just zero. So this thing, the second derivative is always negative. So we are always, always concave, concave downward, downward. And so by the second derivative test, at t equals two, well at t equals two, our second derivative of our acceleration function is going to be negative."}, {"video_title": "Motion problems finding the maximum acceleration   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of negative six t is negative six, the derivative of constant is just zero. So this thing, the second derivative is always negative. So we are always, always concave, concave downward, downward. And so by the second derivative test, at t equals two, well at t equals two, our second derivative of our acceleration function is going to be negative. And so we know that this is our maximum value, or our max at t is equal to two. So at what value of t does the particle obtain its maximum acceleration? At t is equal to, at t is equal to two."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Some of them are one-sided, and some of them are regular limits or two-sided limits. All right, let's start with this first one. The limit as x approaches four from values larger than equaling four. So that's what that plus tells us. And so when x is greater than four, our f of x is equal to square root of x. So as we are approaching four from the right, we are really thinking about this part of the function. And so this is going to be equal to the square root of four."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's what that plus tells us. And so when x is greater than four, our f of x is equal to square root of x. So as we are approaching four from the right, we are really thinking about this part of the function. And so this is going to be equal to the square root of four. Even though right at four, our f of x is equal to this, we are approaching from values greater than four. We're approaching from the right. So we would use this part of our function definition."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to the square root of four. Even though right at four, our f of x is equal to this, we are approaching from values greater than four. We're approaching from the right. So we would use this part of our function definition. And so this is going to be equal to two. Now what about our limit of f of x as we approach four from the left? Well, then we would use this part of our function definition."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would use this part of our function definition. And so this is going to be equal to two. Now what about our limit of f of x as we approach four from the left? Well, then we would use this part of our function definition. And so this is going to be equal to four plus two over four minus one, which is equal to six over three, which is equal to two. And so if we wanna say what is the limit of f of x as x approaches four? Well, this is a good scenario here because from both the left and the right, as we approach x equals four, we're approaching the same value."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, then we would use this part of our function definition. And so this is going to be equal to four plus two over four minus one, which is equal to six over three, which is equal to two. And so if we wanna say what is the limit of f of x as x approaches four? Well, this is a good scenario here because from both the left and the right, as we approach x equals four, we're approaching the same value. And we know that in order for the two-sided limit to have a limit, you have to be approaching the same thing from the right and the left. And we are, and so this is going to be equal to two. Now what's the limit as x approaches two of f of x?"}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, this is a good scenario here because from both the left and the right, as we approach x equals four, we're approaching the same value. And we know that in order for the two-sided limit to have a limit, you have to be approaching the same thing from the right and the left. And we are, and so this is going to be equal to two. Now what's the limit as x approaches two of f of x? Well, as x approaches two, we are going to be completely in this scenario right over here. Now interesting things do happen at x equals one here. Our denominator goes to zero."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what's the limit as x approaches two of f of x? Well, as x approaches two, we are going to be completely in this scenario right over here. Now interesting things do happen at x equals one here. Our denominator goes to zero. But at x equals two, this part of the curve is gonna be continuous. So we can just substitute the value. It's going to be two plus two over two minus one, which is four over one, which is equal to four."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Our denominator goes to zero. But at x equals two, this part of the curve is gonna be continuous. So we can just substitute the value. It's going to be two plus two over two minus one, which is four over one, which is equal to four. Let's do another example. So we have another piecewise function. And so let's pause our video and figure out these things."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be two plus two over two minus one, which is four over one, which is equal to four. Let's do another example. So we have another piecewise function. And so let's pause our video and figure out these things. All right, now let's do this together. So what's the limit as x approaches negative one from the right? So if we're approaching from the right, when we are greater than or equal to negative one, we are in this part of our piecewise function."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let's pause our video and figure out these things. All right, now let's do this together. So what's the limit as x approaches negative one from the right? So if we're approaching from the right, when we are greater than or equal to negative one, we are in this part of our piecewise function. And so we would say this is going to approach, this is gonna be two to the negative one power, which is equal to 1 1\u20442. What about if we're approaching from the left? Well, if we're approaching from the left, we're in this scenario right over here."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we're approaching from the right, when we are greater than or equal to negative one, we are in this part of our piecewise function. And so we would say this is going to approach, this is gonna be two to the negative one power, which is equal to 1 1\u20442. What about if we're approaching from the left? Well, if we're approaching from the left, we're in this scenario right over here. We're to the left of x equals negative one. And so this is going to be equal to the sine, because we're in this case for our piecewise function, of negative one plus one, which is a sine of zero, which is equal to zero. Now what's the two-sided limit as x approaches negative one of g of x?"}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, if we're approaching from the left, we're in this scenario right over here. We're to the left of x equals negative one. And so this is going to be equal to the sine, because we're in this case for our piecewise function, of negative one plus one, which is a sine of zero, which is equal to zero. Now what's the two-sided limit as x approaches negative one of g of x? Well, we're approaching two different values as we approach from the right and as we approach from the left. And if our one-sided limits aren't approaching the same value, well, then this limit does not exist. Does not exist."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what's the two-sided limit as x approaches negative one of g of x? Well, we're approaching two different values as we approach from the right and as we approach from the left. And if our one-sided limits aren't approaching the same value, well, then this limit does not exist. Does not exist. And what's the limit of g of x as x approaches zero from the right? Well, if we're talking about approaching zero from the right, we are going to be in this case right over here. Zero is definitely in this interval."}, {"video_title": "Limits of piecewise functions   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Does not exist. And what's the limit of g of x as x approaches zero from the right? Well, if we're talking about approaching zero from the right, we are going to be in this case right over here. Zero is definitely in this interval. And over this interval, this right over here is going to be continuous. And so we can just substitute x equals zero there. So it's gonna be two to the zero, which is indeed equal to one."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what I'm concerned with is finding the area under the curve y is equal to f of x, so that's my y axis. This is my x axis. And let me draw my function. My function looks like this, at least in the first quadrant. That's where I'll graph it for now. I could also graph it, obviously, in the second quadrant. But what I care about is the area under this curve and above the positive x axis between x equals 1 and x equals 4."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "My function looks like this, at least in the first quadrant. That's where I'll graph it for now. I could also graph it, obviously, in the second quadrant. But what I care about is the area under this curve and above the positive x axis between x equals 1 and x equals 4. And I'm tired of approximating areas. I want to find the exact area under this curve, above the x axis. And the way we denote the exact area under the curve, this little brown shaded area, is using the definite integral."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what I care about is the area under this curve and above the positive x axis between x equals 1 and x equals 4. And I'm tired of approximating areas. I want to find the exact area under this curve, above the x axis. And the way we denote the exact area under the curve, this little brown shaded area, is using the definite integral. It's a definite integral from 1 to 4 of f of x dx. And the way I conceptualize where this notation comes from is we imagine an infinite number of infinitely thin rectangles that we sum up to find this area. And let me draw one of those infinitely thin rectangles, maybe not so infinitely thin."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the way we denote the exact area under the curve, this little brown shaded area, is using the definite integral. It's a definite integral from 1 to 4 of f of x dx. And the way I conceptualize where this notation comes from is we imagine an infinite number of infinitely thin rectangles that we sum up to find this area. And let me draw one of those infinitely thin rectangles, maybe not so infinitely thin. So let me draw it like this. So that would be one of the rectangles. That would be another rectangle."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let me draw one of those infinitely thin rectangles, maybe not so infinitely thin. So let me draw it like this. So that would be one of the rectangles. That would be another rectangle. This should be reminiscent of a Riemann sum. In fact, that's where the Riemann integral comes from, taking a Riemann sum where you have an infinite number of these rectangles where the width of each of the rectangles, this is how I conceptualize it, is dx. And the height of this rectangle is the function evaluated at an x that's within this interval right over here."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "That would be another rectangle. This should be reminiscent of a Riemann sum. In fact, that's where the Riemann integral comes from, taking a Riemann sum where you have an infinite number of these rectangles where the width of each of the rectangles, this is how I conceptualize it, is dx. And the height of this rectangle is the function evaluated at an x that's within this interval right over here. And so this part right over here is the area of one of those rectangles and we're summing them all up. And this is kind of an elongated s reminiscent of a sigma for summing. We're summing up the infinite number of those infinitely thin rectangles or the areas of those infinitely thin rectangles between 1 and 4."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the height of this rectangle is the function evaluated at an x that's within this interval right over here. And so this part right over here is the area of one of those rectangles and we're summing them all up. And this is kind of an elongated s reminiscent of a sigma for summing. We're summing up the infinite number of those infinitely thin rectangles or the areas of those infinitely thin rectangles between 1 and 4. So that's where the notation of the definite integral comes from. But we still haven't done anything. We've just written some notation that says the exact area between 1 and 4 under the curve f of x and above the x-axis."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're summing up the infinite number of those infinitely thin rectangles or the areas of those infinitely thin rectangles between 1 and 4. So that's where the notation of the definite integral comes from. But we still haven't done anything. We've just written some notation that says the exact area between 1 and 4 under the curve f of x and above the x-axis. In order to actually do anything really productive with this, we have to turn to the second fundamental theorem of calculus, sometimes called part 2 of the fundamental theorem of calculus, which tells us that if f has an antiderivative, so if we have the antiderivative of f, so f of x is derivative of some function capital F of x, or another way of saying it is capital F of x is the antiderivative of lowercase f of x, then I can evaluate this thing, and we do a whole video on conceptually understanding why this makes sense, we can evaluate this by evaluating the antiderivative of f or an antiderivative of f at 4 and from that subtract the antiderivative evaluated at 1. So let's do it for this particular case right over here. So we are taking, I'll just rewrite this statement, instead of writing f of x I'll write x squared, so the definite integral from 1 to 4 of x squared dx, well, we're just going to have to figure out what the antiderivative is."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've just written some notation that says the exact area between 1 and 4 under the curve f of x and above the x-axis. In order to actually do anything really productive with this, we have to turn to the second fundamental theorem of calculus, sometimes called part 2 of the fundamental theorem of calculus, which tells us that if f has an antiderivative, so if we have the antiderivative of f, so f of x is derivative of some function capital F of x, or another way of saying it is capital F of x is the antiderivative of lowercase f of x, then I can evaluate this thing, and we do a whole video on conceptually understanding why this makes sense, we can evaluate this by evaluating the antiderivative of f or an antiderivative of f at 4 and from that subtract the antiderivative evaluated at 1. So let's do it for this particular case right over here. So we are taking, I'll just rewrite this statement, instead of writing f of x I'll write x squared, so the definite integral from 1 to 4 of x squared dx, well, we're just going to have to figure out what the antiderivative is. So if f of x is equal to x squared, what is capital F of x equal to? What is the antiderivative? Well, you might remember from your power rules that if you take the derivative with respect to x of x to the third, you are going to get 3x squared, which is pretty darn close to x squared except for this factor of 3, so let's divide both sides by 3."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we are taking, I'll just rewrite this statement, instead of writing f of x I'll write x squared, so the definite integral from 1 to 4 of x squared dx, well, we're just going to have to figure out what the antiderivative is. So if f of x is equal to x squared, what is capital F of x equal to? What is the antiderivative? Well, you might remember from your power rules that if you take the derivative with respect to x of x to the third, you are going to get 3x squared, which is pretty darn close to x squared except for this factor of 3, so let's divide both sides by 3. And you get the derivative of x to the third divided by 3 is indeed x squared, or you could say that this is the same thing as the derivative with respect to x of x to the third over 3. So what is the derivative of this? It will be 3 times 1 third, and then you'll decrement the power, it will just be x squared."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you might remember from your power rules that if you take the derivative with respect to x of x to the third, you are going to get 3x squared, which is pretty darn close to x squared except for this factor of 3, so let's divide both sides by 3. And you get the derivative of x to the third divided by 3 is indeed x squared, or you could say that this is the same thing as the derivative with respect to x of x to the third over 3. So what is the derivative of this? It will be 3 times 1 third, and then you'll decrement the power, it will just be x squared. So this right over here, once again, is x squared. It's just equal to x squared. So in this case, our capital F of x, our antiderivative, is x to the third over 3."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "It will be 3 times 1 third, and then you'll decrement the power, it will just be x squared. So this right over here, once again, is x squared. It's just equal to x squared. So in this case, our capital F of x, our antiderivative, is x to the third over 3. And so we just have to evaluate that at 4 and at 1. And sometimes the notation we'll use is, we'll say that the antiderivative is x to the third over 3, and we're going to evaluate it, the one I always like to just write the numbers up here, at 4 and from that subtract it, evaluate it at 1. Sometimes you'll see people write a little line here too, where they'll say we're going to evaluate it at 4 and then at 1, but I'll just do it without the line."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in this case, our capital F of x, our antiderivative, is x to the third over 3. And so we just have to evaluate that at 4 and at 1. And sometimes the notation we'll use is, we'll say that the antiderivative is x to the third over 3, and we're going to evaluate it, the one I always like to just write the numbers up here, at 4 and from that subtract it, evaluate it at 1. Sometimes you'll see people write a little line here too, where they'll say we're going to evaluate it at 4 and then at 1, but I'll just do it without the line. So we're going to evaluate this thing at 4, and from that subtract it, evaluate it at 1. So this is going to be equal to 4 to the third power of 64, so it's going to be 64 over 3. Let me color code it."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sometimes you'll see people write a little line here too, where they'll say we're going to evaluate it at 4 and then at 1, but I'll just do it without the line. So we're going to evaluate this thing at 4, and from that subtract it, evaluate it at 1. So this is going to be equal to 4 to the third power of 64, so it's going to be 64 over 3. Let me color code it. This right over here is this right over there, and then from that we're going to subtract this business evaluated at 1. Well, when you evaluate it at 1, you get 1 to the third is 1 over 3. You get 1 third."}, {"video_title": "Area between a curve and the x-axis   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me color code it. This right over here is this right over there, and then from that we're going to subtract this business evaluated at 1. Well, when you evaluate it at 1, you get 1 to the third is 1 over 3. You get 1 third. So just to be clear, this is this right over there. And then we are ready to just subtract these fractions, 64 over 3 minus 1 third is equal to 63 over 3, and 3 goes into 63 exactly 21 times. So whatever the units are, the area of this brown area is 21 square units."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the way I'll do it is by setting up 5 trapezoids of equal width. So this will be the left boundary of the first trapezoid. This will be its right boundary, which will also be the left boundary of the second trapezoid. This will be the right boundary of the second trapezoid. This is the right boundary of the third trapezoid. This will be the right boundary of the fourth trapezoid. fourth trapezoid, and then finally this will be the route boundary of the fifth trapezoid."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This will be the right boundary of the second trapezoid. This is the right boundary of the third trapezoid. This will be the right boundary of the fourth trapezoid. fourth trapezoid, and then finally this will be the route boundary of the fifth trapezoid. And since we're traveling, we're going from one to six, so we're traveling six minus one in the x direction, and I want to split it into five sections, the width of each trapezoid is just going to be equal to one. And so if we say that the width of a trapezoid is delta x, we just can now say that delta x is equal to one. So let's set up our trapezoids."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "fourth trapezoid, and then finally this will be the route boundary of the fifth trapezoid. And since we're traveling, we're going from one to six, so we're traveling six minus one in the x direction, and I want to split it into five sections, the width of each trapezoid is just going to be equal to one. And so if we say that the width of a trapezoid is delta x, we just can now say that delta x is equal to one. So let's set up our trapezoids. So the first trapezoid is going to look like that. Actually, it's going to be a triangle, not really a trapezoid. Then the second trapezoid is going to look like this."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's set up our trapezoids. So the first trapezoid is going to look like that. Actually, it's going to be a triangle, not really a trapezoid. Then the second trapezoid is going to look like this. I guess you could say a trapezoid where one of the sides has length zero turns into a triangle. Then the third trapezoid is going to look like this. And then the fourth trapezoid is going to look like that."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then the second trapezoid is going to look like this. I guess you could say a trapezoid where one of the sides has length zero turns into a triangle. Then the third trapezoid is going to look like this. And then the fourth trapezoid is going to look like that. And then finally you have the fifth trapezoid. So let's calculate the area of each of these, and then we will have our approximation for the area under the curve. So let's do trapezoid, or I really should say triangle, this first shape, whatever you want to call it."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the fourth trapezoid is going to look like that. And then finally you have the fifth trapezoid. So let's calculate the area of each of these, and then we will have our approximation for the area under the curve. So let's do trapezoid, or I really should say triangle, this first shape, whatever you want to call it. What is the area of that going to be? Well, the area of a trapezoid, and you'll see this will just turn into the area of a triangle, it's the average of the heights of the two sides of the trapezoid, the way we've looked at it, or you could say the average of the heights of the two parallel sides, I guess is the best way to say it. So f of 1, that's the height here, plus f of 2, all of that over 2, and then we're going to multiply it times our delta x."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do trapezoid, or I really should say triangle, this first shape, whatever you want to call it. What is the area of that going to be? Well, the area of a trapezoid, and you'll see this will just turn into the area of a triangle, it's the average of the heights of the two sides of the trapezoid, the way we've looked at it, or you could say the average of the heights of the two parallel sides, I guess is the best way to say it. So f of 1, that's the height here, plus f of 2, all of that over 2, and then we're going to multiply it times our delta x. Actually, let me do that in that same red color to show you that this is the area of that first trapezoid. So times delta x. And as you see right over here, if you look at it, the f of 1 is just going to be 0, so you're going to have f of 2 times, so it's going to be this height times this base times 1 half, which is just the area of a triangle."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of 1, that's the height here, plus f of 2, all of that over 2, and then we're going to multiply it times our delta x. Actually, let me do that in that same red color to show you that this is the area of that first trapezoid. So times delta x. And as you see right over here, if you look at it, the f of 1 is just going to be 0, so you're going to have f of 2 times, so it's going to be this height times this base times 1 half, which is just the area of a triangle. But let's look at the second trapezoid, trapezoid 2. Right over here. What is its area going to be?"}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as you see right over here, if you look at it, the f of 1 is just going to be 0, so you're going to have f of 2 times, so it's going to be this height times this base times 1 half, which is just the area of a triangle. But let's look at the second trapezoid, trapezoid 2. Right over here. What is its area going to be? Well, it's going to be f of 2 plus f of 3. f of 2 is this height, f of 3 is this height, so we're taking the average of those two heights, divided by 2, that's the average of those two heights, times the base, times delta x. And then let's do trapezoid 3. I think you're getting the idea here."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is its area going to be? Well, it's going to be f of 2 plus f of 3. f of 2 is this height, f of 3 is this height, so we're taking the average of those two heights, divided by 2, that's the average of those two heights, times the base, times delta x. And then let's do trapezoid 3. I think you're getting the idea here. Trapezoid 3 is going to be f of 3 plus f of 4, divided by 2 times delta x. And then, let's see, I'm running out of colors. This is trapezoid 4 right over here."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I think you're getting the idea here. Trapezoid 3 is going to be f of 3 plus f of 4, divided by 2 times delta x. And then, let's see, I'm running out of colors. This is trapezoid 4 right over here. So plus f of 4 plus f of 5, all of that over 2 times delta x. And then we have our last trapezoid, which I will do in yellow. So this is trapezoid number 5."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is trapezoid 4 right over here. So plus f of 4 plus f of 5, all of that over 2 times delta x. And then we have our last trapezoid, which I will do in yellow. So this is trapezoid number 5. Scroll down a little bit, get some more real estate. So it's going to be plus, I'll just write the plus over here, plus f of 5 plus f of 6 over 2 times our delta x. So let's see how we can simplify this a little bit."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is trapezoid number 5. Scroll down a little bit, get some more real estate. So it's going to be plus, I'll just write the plus over here, plus f of 5 plus f of 6 over 2 times our delta x. So let's see how we can simplify this a little bit. All of these terms, we have a 1 half delta x, so let's actually factor that out. So remember, this is our approximation of our area. So area is approximately, remember this is just a rough approximation, it's very clear actually."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see how we can simplify this a little bit. All of these terms, we have a 1 half delta x, so let's actually factor that out. So remember, this is our approximation of our area. So area is approximately, remember this is just a rough approximation, it's very clear actually. You might say this is pretty good using the trapezoids, but it is pretty clear that we are letting go of some of the area. We're letting go of that area, we're letting go of some of this right over here, you can barely see it, some of this right over here, you can barely see it. But we are, this is going to be, it looks like an underestimate, but it is a decent approximation."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So area is approximately, remember this is just a rough approximation, it's very clear actually. You might say this is pretty good using the trapezoids, but it is pretty clear that we are letting go of some of the area. We're letting go of that area, we're letting go of some of this right over here, you can barely see it, some of this right over here, you can barely see it. But we are, this is going to be, it looks like an underestimate, but it is a decent approximation. Let's see if we can simplify this expression. So it's approximately going to be equal to, I'm going to factor out a delta x over 2. And then what I'm left with, and I will switch to a neutral color, if I factor out a delta x over 2, then I have just an f of 1, and then I have 2 f of 2's, so plus 2 times f of 2, and I'm doing this because you might see a formula that looks something like this in your calculus book, and it's not some mysterious thing, they just really summed up the areas of the trapezoids."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we are, this is going to be, it looks like an underestimate, but it is a decent approximation. Let's see if we can simplify this expression. So it's approximately going to be equal to, I'm going to factor out a delta x over 2. And then what I'm left with, and I will switch to a neutral color, if I factor out a delta x over 2, then I have just an f of 1, and then I have 2 f of 2's, so plus 2 times f of 2, and I'm doing this because you might see a formula that looks something like this in your calculus book, and it's not some mysterious thing, they just really summed up the areas of the trapezoids. And then we're going to have 2 f of 3's, plus 2 times f of 3's, plus we're going to have 2 f of 4's, plus 2 times f of 4, and then we're going to have 2 f of 5's, plus 2 times f of 5, and then finally we're going to have 1 f of 6, plus f of 6. If you were to generalize it, you have one of the first endpoint, the function evaluated at the first endpoint, one of it at the very last endpoint, and then two of all of the rest of them. But this is just the area of trapezoids."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then what I'm left with, and I will switch to a neutral color, if I factor out a delta x over 2, then I have just an f of 1, and then I have 2 f of 2's, so plus 2 times f of 2, and I'm doing this because you might see a formula that looks something like this in your calculus book, and it's not some mysterious thing, they just really summed up the areas of the trapezoids. And then we're going to have 2 f of 3's, plus 2 times f of 3's, plus we're going to have 2 f of 4's, plus 2 times f of 4, and then we're going to have 2 f of 5's, plus 2 times f of 5, and then finally we're going to have 1 f of 6, plus f of 6. If you were to generalize it, you have one of the first endpoint, the function evaluated at the first endpoint, one of it at the very last endpoint, and then two of all of the rest of them. But this is just the area of trapezoids. I'm not actually a big fan when textbooks write this, because when you see this, it's hard to visualize the trapezoids. When you see this, it's much clearer how you might visualize that. But with that out of the way, let's actually evaluate this."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But this is just the area of trapezoids. I'm not actually a big fan when textbooks write this, because when you see this, it's hard to visualize the trapezoids. When you see this, it's much clearer how you might visualize that. But with that out of the way, let's actually evaluate this. Lucky for us, the math is simple. Our delta x is just 1. And then we just have to evaluate all of this business."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But with that out of the way, let's actually evaluate this. Lucky for us, the math is simple. Our delta x is just 1. And then we just have to evaluate all of this business. f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1, so that is just going to be 0."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we just have to evaluate all of this business. f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1, so that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1, so this is just going to be 2. Actually, let me do it in that same... Well, I'm now using the purple for a different purpose than just the first trapezoid."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of 1 is the square root of 1 minus 1, so that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1, so this is just going to be 2. Actually, let me do it in that same... Well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully you realize that. I was just sticking with that pen color. Then f of 3, 3 minus 1 is 2, square root of 2."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Actually, let me do it in that same... Well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully you realize that. I was just sticking with that pen color. Then f of 3, 3 minus 1 is 2, square root of 2. So the function evaluated at 3 is the square root of 2, so this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate at 4, you get the square root of 3, so this is going to be 2 times the square root of 3."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then f of 3, 3 minus 1 is 2, square root of 2. So the function evaluated at 3 is the square root of 2, so this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate at 4, you get the square root of 3, so this is going to be 2 times the square root of 3. Then you get 2 times the square root of 4. 5 minus 1 is 4. 2 times the square root of 4 is just 4."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you evaluate at 4, you get the square root of 3, so this is going to be 2 times the square root of 3. Then you get 2 times the square root of 4. 5 minus 1 is 4. 2 times the square root of 4 is just 4. Then finally, you get f of 6 is square root of 6 minus 1 is the square root of 5. I think we are now ready to evaluate. Let me get my handy TI-85 out and calculate this."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "2 times the square root of 4 is just 4. Then finally, you get f of 6 is square root of 6 minus 1 is the square root of 5. I think we are now ready to evaluate. Let me get my handy TI-85 out and calculate this. It's going to be... I'm just going to calculate... Well, I'll just multiply. 0.5 times, open parentheses, well, it's a 0."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me get my handy TI-85 out and calculate this. It's going to be... I'm just going to calculate... Well, I'll just multiply. 0.5 times, open parentheses, well, it's a 0. I'll just write it just so you know what I'm doing. 0 plus 2... Whoops, lost my calculator. Plus 2 times the square root of 2 plus 2 times the square root of 3 plus 4, I'm almost done, plus the square root of 5."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "0.5 times, open parentheses, well, it's a 0. I'll just write it just so you know what I'm doing. 0 plus 2... Whoops, lost my calculator. Plus 2 times the square root of 2 plus 2 times the square root of 3 plus 4, I'm almost done, plus the square root of 5. Let me write that. Plus the square root of 5 gives me... Now we are ready for our drum roll. It gives me, and I'll just round it, 7.26."}, {"video_title": "Trapezoidal sums   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Plus 2 times the square root of 2 plus 2 times the square root of 3 plus 4, I'm almost done, plus the square root of 5. Let me write that. Plus the square root of 5 gives me... Now we are ready for our drum roll. It gives me, and I'll just round it, 7.26. The area is approximately equal to 7.26 under the curve. y is equal to the square root of x minus 1 between x equals 1 and x equals 6. We did this using trapezoids."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "f of e is 1. The natural log of e is 1. And I've drawn the slope of the tangent line, or I've drawn the tangent line, and we need to figure out what the slope of it is, or at least come up with an expression for it. And I'm going to come up with an expression using both the formal definition and the alternate definition, and that will allow us to compare them a little bit. So let's think about first the formal definition. So the formal definition wants us to find an expression for the derivative of our function at any x. So let's say that this is some arbitrary x right over here."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "And I'm going to come up with an expression using both the formal definition and the alternate definition, and that will allow us to compare them a little bit. So let's think about first the formal definition. So the formal definition wants us to find an expression for the derivative of our function at any x. So let's say that this is some arbitrary x right over here. This would be the point x, f of x. And let's say that this is, let's call this x plus h. So this distance right over here is going to be h. This right over here is going to be the point x plus h, f of x plus h. Now the whole underlying idea of the formal definition of limits is to find the slope of the secant line between these two points, and then take the limit as h approaches 0. As h gets closer and closer, this blue point is going to get closer and closer and closer to x, and this point is going to approach it on the curve, and the secant line is going to become a better and better and better approximation of the tangent line at x."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's say that this is some arbitrary x right over here. This would be the point x, f of x. And let's say that this is, let's call this x plus h. So this distance right over here is going to be h. This right over here is going to be the point x plus h, f of x plus h. Now the whole underlying idea of the formal definition of limits is to find the slope of the secant line between these two points, and then take the limit as h approaches 0. As h gets closer and closer, this blue point is going to get closer and closer and closer to x, and this point is going to approach it on the curve, and the secant line is going to become a better and better and better approximation of the tangent line at x. So let's actually do that. So what's the slope of the secant line? Well, it's the change in your vertical axis, which is going to be f of x plus h minus f of x over the change in your horizontal axis."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "As h gets closer and closer, this blue point is going to get closer and closer and closer to x, and this point is going to approach it on the curve, and the secant line is going to become a better and better and better approximation of the tangent line at x. So let's actually do that. So what's the slope of the secant line? Well, it's the change in your vertical axis, which is going to be f of x plus h minus f of x over the change in your horizontal axis. And that's x plus h minus x, and we see here the difference is just h over h. And we're going to take the limit of that as h approaches 0. So in the case when f of x is the natural log of x, this will reduce to the limit as h approaches 0, f of x plus h is the natural log of x plus h minus the natural log of x, all of that over h. So this right over here, for our particular f of x, this is equal to f prime of x. So if we wanted to evaluate this, when x is equal to e, then everywhere we see an x, we just have to replace it with an e. This is essentially expressing our derivative as a function of x."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, it's the change in your vertical axis, which is going to be f of x plus h minus f of x over the change in your horizontal axis. And that's x plus h minus x, and we see here the difference is just h over h. And we're going to take the limit of that as h approaches 0. So in the case when f of x is the natural log of x, this will reduce to the limit as h approaches 0, f of x plus h is the natural log of x plus h minus the natural log of x, all of that over h. So this right over here, for our particular f of x, this is equal to f prime of x. So if we wanted to evaluate this, when x is equal to e, then everywhere we see an x, we just have to replace it with an e. This is essentially expressing our derivative as a function of x. It's kind of a crazy-looking function of x. You have a limit here and all of that, but every place you see an x, like any function definition, you can replace it now with an e. So we can, let me just do that. Whoops, I lost my screen."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "So if we wanted to evaluate this, when x is equal to e, then everywhere we see an x, we just have to replace it with an e. This is essentially expressing our derivative as a function of x. It's kind of a crazy-looking function of x. You have a limit here and all of that, but every place you see an x, like any function definition, you can replace it now with an e. So we can, let me just do that. Whoops, I lost my screen. Here we go. So we could write f prime of e is equal to the limit as h approaches 0 of natural log, I'll do the same color so we can keep track of things, natural log of e plus h, I'll just leave that blank for now, minus the natural log of e, all of that over h. So just like that, this right over here, if we evaluate this limit, if we're able to, and we actually can, if we're able to evaluate this limit, this would give us the slope of the tangent line when x equals e. This is doing the formal definition. Now let's do the alternate definition."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "Whoops, I lost my screen. Here we go. So we could write f prime of e is equal to the limit as h approaches 0 of natural log, I'll do the same color so we can keep track of things, natural log of e plus h, I'll just leave that blank for now, minus the natural log of e, all of that over h. So just like that, this right over here, if we evaluate this limit, if we're able to, and we actually can, if we're able to evaluate this limit, this would give us the slope of the tangent line when x equals e. This is doing the formal definition. Now let's do the alternate definition. The alternate definition, if you don't want to find a general derivative expressed as a function of x like this, and you just want to find the slope at a particular point, the alternate definition kind of just gets straight to the point there. So what they say is, hey, look, let's imagine some other x value here. So let's imagine some other x value."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "Now let's do the alternate definition. The alternate definition, if you don't want to find a general derivative expressed as a function of x like this, and you just want to find the slope at a particular point, the alternate definition kind of just gets straight to the point there. So what they say is, hey, look, let's imagine some other x value here. So let's imagine some other x value. This right over here is the point x, well, we could say f of x or we could even say the natural log of x. What is the slope of the secant line between those two points? Well, it's going to be your change in y values, so it's going to be natural log of x minus 1 over your change in x values."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's imagine some other x value. This right over here is the point x, well, we could say f of x or we could even say the natural log of x. What is the slope of the secant line between those two points? Well, it's going to be your change in y values, so it's going to be natural log of x minus 1 over your change in x values. That's x minus e. So that's the slope of the secant line between those two points. Well, what if you want to get the tangent line? Well, let's just take the limit as x approaches e. As x gets closer and closer and closer, these points are going to get closer and closer and closer, and the secant line is going to better approximate the tangent line."}, {"video_title": "Formal and alternate form of the derivative for ln x   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, it's going to be your change in y values, so it's going to be natural log of x minus 1 over your change in x values. That's x minus e. So that's the slope of the secant line between those two points. Well, what if you want to get the tangent line? Well, let's just take the limit as x approaches e. As x gets closer and closer and closer, these points are going to get closer and closer and closer, and the secant line is going to better approximate the tangent line. So we're just going to take the limit as x approaches e. So either one of this, this is using the formal definition of a limit. Let me make clear that that h does not belong part of it. So we can either do it using the formal definition or the alternate definition of the derivative."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe under the curve and above the x-axis? And let's say between two boundaries. Let's say between x is equal to a and x is equal to b. So let me draw these boundaries right over here. That's our left boundary. This is our right boundary and we wanna think about this area right over here. Well, without calculus, you could actually get better and better approximations for it."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me draw these boundaries right over here. That's our left boundary. This is our right boundary and we wanna think about this area right over here. Well, without calculus, you could actually get better and better approximations for it. How would you do it? Well, you could divide this section into a bunch of delta x's that go from a to b. They could be equal sections or not, but let's just say for the sake of visualizations, I'm gonna draw roughly equal sections here."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, without calculus, you could actually get better and better approximations for it. How would you do it? Well, you could divide this section into a bunch of delta x's that go from a to b. They could be equal sections or not, but let's just say for the sake of visualizations, I'm gonna draw roughly equal sections here. So that's the first, that's the second, this is the third, this is the fourth, this is the fifth, and then we have the sixth right over here. And so each of these, this is delta x, let's just call that delta x one, this is delta x two, this width right over here, this is delta x three, all the way to delta x n. I'll try to be general here. And so what we could do is let's try to sum up the area of the rectangles defined here."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "They could be equal sections or not, but let's just say for the sake of visualizations, I'm gonna draw roughly equal sections here. So that's the first, that's the second, this is the third, this is the fourth, this is the fifth, and then we have the sixth right over here. And so each of these, this is delta x, let's just call that delta x one, this is delta x two, this width right over here, this is delta x three, all the way to delta x n. I'll try to be general here. And so what we could do is let's try to sum up the area of the rectangles defined here. And we could make the height, maybe we make the height based on the value of the function at the right bound. It doesn't have to be, it could be the value of the function someplace in this delta x. But that's one solution."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what we could do is let's try to sum up the area of the rectangles defined here. And we could make the height, maybe we make the height based on the value of the function at the right bound. It doesn't have to be, it could be the value of the function someplace in this delta x. But that's one solution. We're gonna go into a lot more depth into it in future videos. And so we do that. And so now we have an approximation."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that's one solution. We're gonna go into a lot more depth into it in future videos. And so we do that. And so now we have an approximation. Or we could say, look, the area of each of these rectangles are going to be f of x sub i, or maybe x sub i is the right boundary the way I've drawn it, times delta x i. That's each of these rectangles. And then we can sum them up."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so now we have an approximation. Or we could say, look, the area of each of these rectangles are going to be f of x sub i, or maybe x sub i is the right boundary the way I've drawn it, times delta x i. That's each of these rectangles. And then we can sum them up. And that would give us an approximation for the area. But as long as we use a finite number, we might say, well, we can always get better by making our delta x's smaller, and then by having more of these rectangles. Or get to a situation, here we're going from i is equal to one to i is equal to n. But what happens is delta x gets thinner and thinner and thinner, and we have, and n gets larger and larger and larger."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we can sum them up. And that would give us an approximation for the area. But as long as we use a finite number, we might say, well, we can always get better by making our delta x's smaller, and then by having more of these rectangles. Or get to a situation, here we're going from i is equal to one to i is equal to n. But what happens is delta x gets thinner and thinner and thinner, and we have, and n gets larger and larger and larger. As delta x gets infinitesimally small, and then as n approaches infinity. And so you're probably sensing something. Then maybe we could think about the limit as we could say as n approaches infinity, or the limit as delta x becomes very, very, very, very small."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or get to a situation, here we're going from i is equal to one to i is equal to n. But what happens is delta x gets thinner and thinner and thinner, and we have, and n gets larger and larger and larger. As delta x gets infinitesimally small, and then as n approaches infinity. And so you're probably sensing something. Then maybe we could think about the limit as we could say as n approaches infinity, or the limit as delta x becomes very, very, very, very small. And this notion of getting better and better approximations as we take the limit as n approaches infinity, this is the core idea of integral calculus. And it's called integral calculus because the central operation we use, the summing up of an infinite number of infinitesimally thin things is one way to visualize it, is the integral. That this is going to be the integral, in this case, from a to b."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then maybe we could think about the limit as we could say as n approaches infinity, or the limit as delta x becomes very, very, very, very small. And this notion of getting better and better approximations as we take the limit as n approaches infinity, this is the core idea of integral calculus. And it's called integral calculus because the central operation we use, the summing up of an infinite number of infinitesimally thin things is one way to visualize it, is the integral. That this is going to be the integral, in this case, from a to b. And we're gonna learn a lot more depth. In this case, it is a definite integral of f of x, f of x dx. But you can already see the parallels here."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "That this is going to be the integral, in this case, from a to b. And we're gonna learn a lot more depth. In this case, it is a definite integral of f of x, f of x dx. But you can already see the parallels here. You can view the integral sign as like a sigma notation, as a summation sign, but instead of taking the sum of a discrete number of things, you're taking the sum of an infinite number infinitely thin things. Instead of delta x, you now have dx, infinitesimally small things. And this is a notion of an integral."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But you can already see the parallels here. You can view the integral sign as like a sigma notation, as a summation sign, but instead of taking the sum of a discrete number of things, you're taking the sum of an infinite number infinitely thin things. Instead of delta x, you now have dx, infinitesimally small things. And this is a notion of an integral. So this right over here is an integral. Now what makes it interesting to calculus, it is using this notion of a limit. But what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is a notion of an integral. So this right over here is an integral. Now what makes it interesting to calculus, it is using this notion of a limit. But what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics. As we will see in the fundamental theorem of calculus, that integration, the notion of an integral, is closely, tied closely to the notion of a derivative. In fact, the notion of an antiderivative. In differential calculus, we looked at the problem of, hey, if I have some function, I can take its derivative and I can get the derivative of the function."}, {"video_title": "Introduction to integral calculus   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics. As we will see in the fundamental theorem of calculus, that integration, the notion of an integral, is closely, tied closely to the notion of a derivative. In fact, the notion of an antiderivative. In differential calculus, we looked at the problem of, hey, if I have some function, I can take its derivative and I can get the derivative of the function. Integral calculus, we're going to be doing a lot of, well, what if we start with a derivative, can we figure out through integration, can we figure out its antiderivative, or the function whose derivative it is? As we will see, all of these are related. The idea of the area under a curve, the idea of a limit of summing an infinite number of infinitely thin things, and the notion of an antiderivative, they all come together in our journey in integral calculus."}, {"video_title": "Definite integral over a single point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's think about a definite integral of f of x dx. So it's the area under the curve f of x, but instead of it being between two different x values, say a and b, like we've seen multiple times, let's say it's between the same one. Let's say it's between c and c, or let's say c is right over here. What do you think this thing right over here is going to be equal to? What does this represent? What does this equal to? And I encourage you to pause the video and try to think about it."}, {"video_title": "Definite integral over a single point   AP Calculus AB   Khan Academy.mp3", "Sentence": "What do you think this thing right over here is going to be equal to? What does this represent? What does this equal to? And I encourage you to pause the video and try to think about it. Well, if you try to visualize it, you're thinking, okay, the area under the curve f of x above the x-axis from x equals c to x equals c. So this region, I guess we could call it, that we think about it, does have a height. The height here is f of c. But what's the width? Well, there is no width."}, {"video_title": "Definite integral over a single point   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to think about it. Well, if you try to visualize it, you're thinking, okay, the area under the curve f of x above the x-axis from x equals c to x equals c. So this region, I guess we could call it, that we think about it, does have a height. The height here is f of c. But what's the width? Well, there is no width. We're just at a single point. We're not going from c to c plus some delta x, or c plus some even very small change in x, or some other, or c plus some other very small value. We're just saying at the point c. So we really, when we're thinking about area, we're thinking about a, we're thinking about things, we're thinking about how much two-dimensional space you're taking up."}, {"video_title": "Definite integral over a single point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, there is no width. We're just at a single point. We're not going from c to c plus some delta x, or c plus some even very small change in x, or some other, or c plus some other very small value. We're just saying at the point c. So we really, when we're thinking about area, we're thinking about a, we're thinking about things, we're thinking about how much two-dimensional space you're taking up. But this idea, this is just a one-dimensional, this is just a one-dimensional, I guess you could think of it as a line segment. What's the area of a line segment? Well, a line segment has no area."}, {"video_title": "Definite integral over a single point   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're just saying at the point c. So we really, when we're thinking about area, we're thinking about a, we're thinking about things, we're thinking about how much two-dimensional space you're taking up. But this idea, this is just a one-dimensional, this is just a one-dimensional, I guess you could think of it as a line segment. What's the area of a line segment? Well, a line segment has no area. So this thing right over here is going to be equal to zero. Now, you might say, okay, I get that. I see why that could make sense, why that makes intuitive sense."}, {"video_title": "Definite integral over a single point   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, a line segment has no area. So this thing right over here is going to be equal to zero. Now, you might say, okay, I get that. I see why that could make sense, why that makes intuitive sense. I have, I'm trying to find the area of a rectangle where I know its height, but it has, its width is zero, so that area is going to be zero, is one way to think about it. But Sal, why are you even pointing this out to me? And as we'll see, especially when we do more complex, definite integration problems and solving things, sometimes, sometimes, recognizing this will help you simplify an integral, an integration problem dramatically."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is clearly a nonlinear function. F of x is equal to one over x minus one. This is its graph, or at least part of its graph right over here. But where you want to approximate it with a linear function, especially around a certain value. And so what we're going to do is we want to find an approximation. Let me write this down. I want to find an approximation for, and actually let me be clear."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "But where you want to approximate it with a linear function, especially around a certain value. And so what we're going to do is we want to find an approximation. Let me write this down. I want to find an approximation for, and actually let me be clear. I want to find a linear approximation. So I'm going to approximate it with a line. I want to find a linear approximation, approximation of f, of f around, and you need to know where you're going to be approximating it, around x equals negative one."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I want to find an approximation for, and actually let me be clear. I want to find a linear approximation. So I'm going to approximate it with a line. I want to find a linear approximation, approximation of f, of f around, and you need to know where you're going to be approximating it, around x equals negative one. So what do we mean by that? Well, let's look at this graph over here. On this curve, when x is equal to negative one, f of negative one is negative, is negative one half, which sticks us right over there."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "I want to find a linear approximation, approximation of f, of f around, and you need to know where you're going to be approximating it, around x equals negative one. So what do we mean by that? Well, let's look at this graph over here. On this curve, when x is equal to negative one, f of negative one is negative, is negative one half, which sticks us right over there. Let me do this in a better color. So it's right over there. And what we want to do is approximate it with a line around that."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "On this curve, when x is equal to negative one, f of negative one is negative, is negative one half, which sticks us right over there. Let me do this in a better color. So it's right over there. And what we want to do is approximate it with a line around that. And what we're essentially going to do is we're going to approximate it with the equation of the tangent line. The tangent line is going to look something, something like that. And as we can see, as we get further and further from x equals negative one, the approximation gets worse and worse."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we want to do is approximate it with a line around that. And what we're essentially going to do is we're going to approximate it with the equation of the tangent line. The tangent line is going to look something, something like that. And as we can see, as we get further and further from x equals negative one, the approximation gets worse and worse. But if we stay around x equals negative one, well, it's a decent, it is as good as you can get for a linear approximation, or at least in this example, it is a very good linear approximation. So when people say, hey, find a linear approximation of f around x equals negative one, or if they say, what is the following is the best approximation, and all of your choices are lines, well, essentially they're asking you to find the equation of the tangent line at x equals negative one. So let's do that."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as we can see, as we get further and further from x equals negative one, the approximation gets worse and worse. But if we stay around x equals negative one, well, it's a decent, it is as good as you can get for a linear approximation, or at least in this example, it is a very good linear approximation. So when people say, hey, find a linear approximation of f around x equals negative one, or if they say, what is the following is the best approximation, and all of your choices are lines, well, essentially they're asking you to find the equation of the tangent line at x equals negative one. So let's do that. So in order to find the equation of the tangent line, the equation of a line is y is equal to mx plus b, where m is the slope and b is the y-intercept. There's other ways that you could think about it. You could think about it in terms of point slope, where you could say y minus some y that sits on that line is equal to the slope times x minus the corresponding x one."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. So in order to find the equation of the tangent line, the equation of a line is y is equal to mx plus b, where m is the slope and b is the y-intercept. There's other ways that you could think about it. You could think about it in terms of point slope, where you could say y minus some y that sits on that line is equal to the slope times x minus the corresponding x one. So x one comma y one sits on that line someplace. Actually, I like to write this point slope form like this sometimes. Y minus y one over x minus x one is equal to b."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could think about it in terms of point slope, where you could say y minus some y that sits on that line is equal to the slope times x minus the corresponding x one. So x one comma y one sits on that line someplace. Actually, I like to write this point slope form like this sometimes. Y minus y one over x minus x one is equal to b. Because this comes straight out of the idea of, look, if x one and y one are on the line, the slope between any other point on the line and that point is going to be your slope of your line. So we could think about it any of these ways. So let's first find the slope of the tangent line, and that's where the derivative is useful."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Y minus y one over x minus x one is equal to b. Because this comes straight out of the idea of, look, if x one and y one are on the line, the slope between any other point on the line and that point is going to be your slope of your line. So we could think about it any of these ways. So let's first find the slope of the tangent line, and that's where the derivative is useful. So f, well, actually, let me just write f of x again. So I'm gonna write it as x minus one to the negative one power, because that makes it a little bit clearer that we can use the power rule and a little bit of the chain rule. So the derivative of f with respect to x is equal to, so the derivative of x minus one to the negative one with respect to x minus one, well, that's just going to be, we're just going to use the power rule here."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's first find the slope of the tangent line, and that's where the derivative is useful. So f, well, actually, let me just write f of x again. So I'm gonna write it as x minus one to the negative one power, because that makes it a little bit clearer that we can use the power rule and a little bit of the chain rule. So the derivative of f with respect to x is equal to, so the derivative of x minus one to the negative one with respect to x minus one, well, that's just going to be, we're just going to use the power rule here. It's gonna be negative one times x minus one to the negative two. And then we're gonna multiply that times the derivative of x minus one with respect to x. Well, that's just going to be one, right?"}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the derivative of f with respect to x is equal to, so the derivative of x minus one to the negative one with respect to x minus one, well, that's just going to be, we're just going to use the power rule here. It's gonna be negative one times x minus one to the negative two. And then we're gonna multiply that times the derivative of x minus one with respect to x. Well, that's just going to be one, right? The derivative of x with respect to x is one. The derivative of negative one with respect to x is zero. So we could say times one here if we like, or we could just not write that because it doesn't change the value."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's just going to be one, right? The derivative of x with respect to x is one. The derivative of negative one with respect to x is zero. So we could say times one here if we like, or we could just not write that because it doesn't change the value. And so let's evaluate that when x is equal to negative one. So f prime of negative one is equal to, I could just write this as negative, all right, like this way, negative one over negative one minus one squared. And so this is going to be negative two down here."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we could say times one here if we like, or we could just not write that because it doesn't change the value. And so let's evaluate that when x is equal to negative one. So f prime of negative one is equal to, I could just write this as negative, all right, like this way, negative one over negative one minus one squared. And so this is going to be negative two down here. So this is equal to negative negative 1 4th. So the slope of our tangent line is, so I could write it this way, m is equal to negative one negative 1 4th. Negative 1 4th, and so now we just have to write its equation down."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be negative two down here. So this is equal to negative negative 1 4th. So the slope of our tangent line is, so I could write it this way, m is equal to negative one negative 1 4th. Negative 1 4th, and so now we just have to write its equation down. So we already know an x one and a y one that sits on the line. In fact, we want to use the point when x equals negative one. So we know that the point negative one comma, we could just input it right over here, f of negative one is negative 1 1.5, one over negative one minus one, negative 1 1.5."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative 1 4th, and so now we just have to write its equation down. So we already know an x one and a y one that sits on the line. In fact, we want to use the point when x equals negative one. So we know that the point negative one comma, we could just input it right over here, f of negative one is negative 1 1.5, one over negative one minus one, negative 1 1.5. So we know that this negative one comma negative 1 1.5, that that is on our curve and it is on our line, that's the point at which the tangent and the curve actually intersect. And so we can use any of these to now write the equation of our line. We could say y, I'll do it right here, y minus y one, so minus negative 1 1.5 is going to be equal to, is going to be equal to our slope, negative 1 4th, I'm just using the point slope version of our equation, is equal to our slope times x minus x one."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we know that the point negative one comma, we could just input it right over here, f of negative one is negative 1 1.5, one over negative one minus one, negative 1 1.5. So we know that this negative one comma negative 1 1.5, that that is on our curve and it is on our line, that's the point at which the tangent and the curve actually intersect. And so we can use any of these to now write the equation of our line. We could say y, I'll do it right here, y minus y one, so minus negative 1 1.5 is going to be equal to, is going to be equal to our slope, negative 1 4th, I'm just using the point slope version of our equation, is equal to our slope times x minus x one. So x minus our x coordinate that we know sits on this. So minus negative one. And so let me now write all of this in a neutral color."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could say y, I'll do it right here, y minus y one, so minus negative 1 1.5 is going to be equal to, is going to be equal to our slope, negative 1 4th, I'm just using the point slope version of our equation, is equal to our slope times x minus x one. So x minus our x coordinate that we know sits on this. So minus negative one. And so let me now write all of this in a neutral color. This will be y plus 1 1.5 is equal to, and I can, so this is going to be plus one right over there. So I can distribute the negative 1 4th. So it's negative 1 4th x minus 1 4th, minus 1 4th, and then I can subtract 1 1.5 from both sides."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so let me now write all of this in a neutral color. This will be y plus 1 1.5 is equal to, and I can, so this is going to be plus one right over there. So I can distribute the negative 1 4th. So it's negative 1 4th x minus 1 4th, minus 1 4th, and then I can subtract 1 1.5 from both sides. So I'm going to get y is equal to negative 1 4th x, and then if I already am subtracting 1 4th and I subtract another half, that's going to be negative 3 4ths. So minus three, minus 3 4ths. So, and that's actually pretty close to what I drew up here."}, {"video_title": "Linear approximation of a rational function   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's negative 1 4th x minus 1 4th, minus 1 4th, and then I can subtract 1 1.5 from both sides. So I'm going to get y is equal to negative 1 4th x, and then if I already am subtracting 1 4th and I subtract another half, that's going to be negative 3 4ths. So minus three, minus 3 4ths. So, and that's actually pretty close to what I drew up here. This should be intersecting the y axis at negative 3 4ths. So there you have it. This line, or you could even say this equation, is going to be a very good linear approximation, about as good as you can get for a linear approximation, for that nonlinear function around x equals negative one."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The continuous function g is graphed. We're interested in the area under the curve between x equals negative seven and x equals seven. And we're considering using Riemann sums to approximate it. So this is the area that we're thinking about in this light blue color. Order the areas from least on top to greatest on bottom. So this is a screenshot from a Khan Academy exercise where you would be expected to actually click and drag these around. But it's just a screenshot."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the area that we're thinking about in this light blue color. Order the areas from least on top to greatest on bottom. So this is a screenshot from a Khan Academy exercise where you would be expected to actually click and drag these around. But it's just a screenshot. So what I'm gonna do instead of dragging them around, I'm just gonna write numbers ordering them from least to greatest, where one would be the least and then three would be the greatest. So pause this video and try to think about these. Which of these is the least, which is in the middle, and which is the greatest?"}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it's just a screenshot. So what I'm gonna do instead of dragging them around, I'm just gonna write numbers ordering them from least to greatest, where one would be the least and then three would be the greatest. So pause this video and try to think about these. Which of these is the least, which is in the middle, and which is the greatest? So let's just draw out what a left Riemann sum, a right Riemann sum would actually look like and compare it to the actual area. And we could do an arbitrary number of subdivisions. I would encourage us to do fewer because we're just trying to get a general sense of things."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Which of these is the least, which is in the middle, and which is the greatest? So let's just draw out what a left Riemann sum, a right Riemann sum would actually look like and compare it to the actual area. And we could do an arbitrary number of subdivisions. I would encourage us to do fewer because we're just trying to get a general sense of things. And they don't even have to be equal subdivisions. So let's start with a left Riemann sum. So we wanna start at x equals negative seven."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "I would encourage us to do fewer because we're just trying to get a general sense of things. And they don't even have to be equal subdivisions. So let's start with a left Riemann sum. So we wanna start at x equals negative seven. And we wanna go to x equals seven. Well, let's say that this is the first rectangle right over here. So this is our first subdivision."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we wanna start at x equals negative seven. And we wanna go to x equals seven. Well, let's say that this is the first rectangle right over here. So this is our first subdivision. And it's a left Riemann sum. So we would use the value of the function at the left end of that subdivision, which is negative seven, x equals negative seven. The value of the function there is 12."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is our first subdivision. And it's a left Riemann sum. So we would use the value of the function at the left end of that subdivision, which is negative seven, x equals negative seven. The value of the function there is 12. And so this would be our first rectangle. You already get a sense that this is gonna be an overestimate relative to the actual area. And so the next subdivision would start here."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "The value of the function there is 12. And so this would be our first rectangle. You already get a sense that this is gonna be an overestimate relative to the actual area. And so the next subdivision would start here. So this would be our height of our rectangle. And once again, they don't have to be equal subdivisions. They often are, but I'm gonna show you unequal subdivisions just to show you that this is still a valid Riemann sum."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so the next subdivision would start here. So this would be our height of our rectangle. And once again, they don't have to be equal subdivisions. They often are, but I'm gonna show you unequal subdivisions just to show you that this is still a valid Riemann sum. And once again, this is an overestimate where the actual area that we're trying to approximate is smaller than the area of this rectangle. And then let's say this third subdivision right over here starts right over there at x equals three. And we use the left end of the subdivision, the value of the function there to define the height of the rectangle."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "They often are, but I'm gonna show you unequal subdivisions just to show you that this is still a valid Riemann sum. And once again, this is an overestimate where the actual area that we're trying to approximate is smaller than the area of this rectangle. And then let's say this third subdivision right over here starts right over there at x equals three. And we use the left end of the subdivision, the value of the function there to define the height of the rectangle. And once again, you see it is an overestimate. So the left Riemann sum is clearly an overestimate. And it's pretty clear why."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we use the left end of the subdivision, the value of the function there to define the height of the rectangle. And once again, you see it is an overestimate. So the left Riemann sum is clearly an overestimate. And it's pretty clear why. This function is, this function never increases. It's either decreasing or it looks like it stays flat at certain points. And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value, than any other value the function takes on over that interval for the subdivision."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's pretty clear why. This function is, this function never increases. It's either decreasing or it looks like it stays flat at certain points. And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value, than any other value the function takes on over that interval for the subdivision. And so you get left with all of this extra area that is part of the overestimate or this area that is larger than the actual area that you're trying to approximate. Now let's think about a right Riemann sum. And I'll do different subdivisions."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value, than any other value the function takes on over that interval for the subdivision. And so you get left with all of this extra area that is part of the overestimate or this area that is larger than the actual area that you're trying to approximate. Now let's think about a right Riemann sum. And I'll do different subdivisions. Let's say the first subdivision goes from negative seven to negative five. And here we would use the right edge to define the height. So f of negative five or g of negative five, I should say."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'll do different subdivisions. Let's say the first subdivision goes from negative seven to negative five. And here we would use the right edge to define the height. So f of negative five or g of negative five, I should say. So that's right over there. That's our first rectangle. Maybe our next rectangle, the right edge is zero."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So f of negative five or g of negative five, I should say. So that's right over there. That's our first rectangle. Maybe our next rectangle, the right edge is zero. So this would be it right over there. And then maybe we'll do four rectangles. Maybe our third subdivision, the right edge, is at x is equal to three."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe our next rectangle, the right edge is zero. So this would be it right over there. And then maybe we'll do four rectangles. Maybe our third subdivision, the right edge, is at x is equal to three. So it would be right over there. And then our fourth subdivision, let's just do it at x equals seven. And we're using the right edge of the subdivisions."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe our third subdivision, the right edge, is at x is equal to three. So it would be right over there. And then our fourth subdivision, let's just do it at x equals seven. And we're using the right edge of the subdivisions. Remember, this is a right Riemann sum. So we use the right edge, the value of the function there is just like that. And now you can see for any one of these subdivisions, our rectangles are underestimates of the area under the curve."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're using the right edge of the subdivisions. Remember, this is a right Riemann sum. So we use the right edge, the value of the function there is just like that. And now you can see for any one of these subdivisions, our rectangles are underestimates of the area under the curve. Under, underestimate. And that's because, once again, in this particular case, the function never increases. It's either decreasing or staying flat."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now you can see for any one of these subdivisions, our rectangles are underestimates of the area under the curve. Under, underestimate. And that's because, once again, in this particular case, the function never increases. It's either decreasing or staying flat. So if you use the value of the function at the right edge, it's going to be smaller, it's never going to be larger than the value that the function takes on in the rest of that subdivision. And so we are continuously underestimating. We're missing all of this area right over there is not being included, so we have an underestimate."}, {"video_title": "Worked example over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's either decreasing or staying flat. So if you use the value of the function at the right edge, it's going to be smaller, it's never going to be larger than the value that the function takes on in the rest of that subdivision. And so we are continuously underestimating. We're missing all of this area right over there is not being included, so we have an underestimate. So if we want to rank these from least to greatest, well, the right Riemann sum is the least. It is underestimating it. Then you have the actual area of the curve, which is just the area of the curve."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Looks like it's between 0 and some positive value. And I want to think about the maximum and minimum points on this. So we've already talked a little bit about absolute maximum and absolute minimum points on an interval. And those are pretty obvious. We hit a maximum point right over here, right at the beginning of our interval. Looks like when x is equal to 0, this is the absolute maximum point for the interval. And the absolute minimum point for the interval happens at the other end point."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And those are pretty obvious. We hit a maximum point right over here, right at the beginning of our interval. Looks like when x is equal to 0, this is the absolute maximum point for the interval. And the absolute minimum point for the interval happens at the other end point. So if this is a, this is b, the absolute minimum point is f of b. And the absolute maximum point is f of a. And it looks like a is equal to 0."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And the absolute minimum point for the interval happens at the other end point. So if this is a, this is b, the absolute minimum point is f of b. And the absolute maximum point is f of a. And it looks like a is equal to 0. But you're probably thinking, hey, there are other kind of interesting points right over here. This point right over here, it isn't the largest. We're not taking on this value right over here."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And it looks like a is equal to 0. But you're probably thinking, hey, there are other kind of interesting points right over here. This point right over here, it isn't the largest. We're not taking on this value right over here. It is definitely not the largest value that the function takes on in that interval. But relative to the other values around it, it seems like a little bit of a hill. It's larger than the other ones."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "We're not taking on this value right over here. It is definitely not the largest value that the function takes on in that interval. But relative to the other values around it, it seems like a little bit of a hill. It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called, this value right over here, let's call this, let's say this right over here is c. This is c, so this is f of c. We would call f of c is a relative maximum value. And we're saying relative because it's obviously the function takes on other values that are larger than it."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called, this value right over here, let's call this, let's say this right over here is c. This is c, so this is f of c. We would call f of c is a relative maximum value. And we're saying relative because it's obviously the function takes on other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly, I can never say that word, if this point right over here is d, f of d looks like a relative minimum point or relative minimum value. f of d is a relative minimum or a local minimum value."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And we're saying relative because it's obviously the function takes on other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly, I can never say that word, if this point right over here is d, f of d looks like a relative minimum point or relative minimum value. f of d is a relative minimum or a local minimum value. Once again, over the whole interval, there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "f of d is a relative minimum or a local minimum value. Once again, over the whole interval, there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here, I'll just give you the definition that really is just a more formal way of saying what we just said."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here, I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x. We could just say kind of in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c?"}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "So here, I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x. We could just say kind of in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in, and you just have to find one open interval. There might be many open intervals where this is true."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in, and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that looks something like that. So this value right over here is c plus h. That value right over here is c minus h. And you see that over that interval, the function at c, f of c, is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "There might be many open intervals where this is true. But if we construct an open interval that looks something like that. So this value right over here is c plus h. That value right over here is c minus h. And you see that over that interval, the function at c, f of c, is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write, let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here."}, {"video_title": "Introduction to minimum and maximum points   Functions   Algebra I   Khan Academy.mp3", "Sentence": "And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write, let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always less than or equal to any of the other values. The f is of all of these other x's in that interval. And that's why we say that it's a relative minimum point."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I have y equals x, and y is equal to x squared minus 2x right over here. And we're going to rotate the region in between these two functions. So that's this region right over here. And we're not going to rotate it just around the x-axis. We're going to rotate it around the horizontal line y equals 4. So we're going to rotate it around this. And if we do that, we'll get a shape that looks like this."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we're not going to rotate it just around the x-axis. We're going to rotate it around the horizontal line y equals 4. So we're going to rotate it around this. And if we do that, we'll get a shape that looks like this. I drew it ahead of time just so I could draw it nicely. And as you can see, it looks like some type of a, I don't know, a vase with a hole at the bottom. And so what we're going to do is attempt to do this using, I guess you could call it the washer method, which is a variant of the disk method."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we do that, we'll get a shape that looks like this. I drew it ahead of time just so I could draw it nicely. And as you can see, it looks like some type of a, I don't know, a vase with a hole at the bottom. And so what we're going to do is attempt to do this using, I guess you could call it the washer method, which is a variant of the disk method. So let's construct a washer. So let's look at a given x. So let's say an x right over here."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what we're going to do is attempt to do this using, I guess you could call it the washer method, which is a variant of the disk method. So let's construct a washer. So let's look at a given x. So let's say an x right over here. So let's say that we're at an x right over there. And what we're going to do is we're going to rotate this region. We're going to give it some depth, dx."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's say an x right over here. So let's say that we're at an x right over there. And what we're going to do is we're going to rotate this region. We're going to give it some depth, dx. So that is dx. And we're going to rotate this around the line y is equal to 4. So if you were to visualize it over here, you have some depth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to give it some depth, dx. So that is dx. And we're going to rotate this around the line y is equal to 4. So if you were to visualize it over here, you have some depth. And when you rotate it around, the inner radius is going to look like the inner radius of our washer is going to look something like that. And then the outer radius of our washer is going to contour around x squared minus 2x. So it's going to look something like that."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you were to visualize it over here, you have some depth. And when you rotate it around, the inner radius is going to look like the inner radius of our washer is going to look something like that. And then the outer radius of our washer is going to contour around x squared minus 2x. So it's going to look something like that. And of course, our washer is going to have some depth. So let me draw the depth. So it's going to have some depth, dx."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to look something like that. And of course, our washer is going to have some depth. So let me draw the depth. So it's going to have some depth, dx. So this is my best attempt at drawing some of that to depth. So this is the depth of our washer. And then just to make the face of the washer a little bit clearer, let me do it in this green color."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to have some depth, dx. So this is my best attempt at drawing some of that to depth. So this is the depth of our washer. And then just to make the face of the washer a little bit clearer, let me do it in this green color. So the face of the washer is going to be all of this business. So if we can figure out the volume of one of these washers for a given x, then we just have to sum up all of the washers for all of the x's in our interval. So let's see if we can set up the integral."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then just to make the face of the washer a little bit clearer, let me do it in this green color. So the face of the washer is going to be all of this business. So if we can figure out the volume of one of these washers for a given x, then we just have to sum up all of the washers for all of the x's in our interval. So let's see if we can set up the integral. And maybe then in the next video, we'll just forge ahead and actually evaluate the integral. So let's think about the volume of the washer. To think about the volume of the washer, we really just have to think about the area of the face of the washer."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if we can set up the integral. And maybe then in the next video, we'll just forge ahead and actually evaluate the integral. So let's think about the volume of the washer. To think about the volume of the washer, we really just have to think about the area of the face of the washer. So area of face is going to be equal to what? Well, it would be the area of the washer, if it wasn't a washer, if it was just a coin, and then subtract out the area of the part that you're cutting out. So the area of the washer, if it didn't have a hole in the middle, would just be pi times the outer radius squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "To think about the volume of the washer, we really just have to think about the area of the face of the washer. So area of face is going to be equal to what? Well, it would be the area of the washer, if it wasn't a washer, if it was just a coin, and then subtract out the area of the part that you're cutting out. So the area of the washer, if it didn't have a hole in the middle, would just be pi times the outer radius squared. It would be pi times this radius squared. That we could call the outer radius. And since it's a washer, we need to subtract out the area of this inner circle."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area of the washer, if it didn't have a hole in the middle, would just be pi times the outer radius squared. It would be pi times this radius squared. That we could call the outer radius. And since it's a washer, we need to subtract out the area of this inner circle. So minus pi times inner radius squared. So we really just have to figure out what the outer and inner radius, or radii, I should say, are. So let's think about it."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "And since it's a washer, we need to subtract out the area of this inner circle. So minus pi times inner radius squared. So we really just have to figure out what the outer and inner radius, or radii, I should say, are. So let's think about it. So our outer radius is going to be equal to what? Well, we can visualize it over here. This is our outer radius, which is also going to be equal to that right over there."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about it. So our outer radius is going to be equal to what? Well, we can visualize it over here. This is our outer radius, which is also going to be equal to that right over there. So that's the distance between y equals 4 and the function that's defining our outside. The distance between y equals 4 and the function that is defining our outside. So this is essentially, this height right over here is going to be equal to 4 minus x squared minus 2x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is our outer radius, which is also going to be equal to that right over there. So that's the distance between y equals 4 and the function that's defining our outside. The distance between y equals 4 and the function that is defining our outside. So this is essentially, this height right over here is going to be equal to 4 minus x squared minus 2x. I'm just finding the distance, or the height, between these two functions. So the outer radius is going to be 4 minus this. Minus x squared minus 2x, which is just 4 minus x squared plus 2x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is essentially, this height right over here is going to be equal to 4 minus x squared minus 2x. I'm just finding the distance, or the height, between these two functions. So the outer radius is going to be 4 minus this. Minus x squared minus 2x, which is just 4 minus x squared plus 2x. Now what is the inner radius? Inner radius. What is that going to be?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "Minus x squared minus 2x, which is just 4 minus x squared plus 2x. Now what is the inner radius? Inner radius. What is that going to be? Well, that's just going to be this distance. That's just going to be the distance between y equals 4 and y equals x. So this is going to be 4 minus x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is that going to be? Well, that's just going to be this distance. That's just going to be the distance between y equals 4 and y equals x. So this is going to be 4 minus x. It's just going to be 4 minus x. So if we wanted to find the area of the face of one of these washers for a given x, it's going to be, and we can factor out this pi, it's going to be pi times the outer radius squared, which is all of this business squared. So it's going to be 4 minus x squared plus 2x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be 4 minus x. It's just going to be 4 minus x. So if we wanted to find the area of the face of one of these washers for a given x, it's going to be, and we can factor out this pi, it's going to be pi times the outer radius squared, which is all of this business squared. So it's going to be 4 minus x squared plus 2x squared. Minus pi times the inner radius, although we factored out the pi. So minus the inner radius squared. So minus 4 minus x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be 4 minus x squared plus 2x squared. Minus pi times the inner radius, although we factored out the pi. So minus the inner radius squared. So minus 4 minus x squared. So this will give us the surface, the surface there, or the area of the surface or the face of one of these washers. If we want the volume of one of those washers, we then just have to multiply times the depth dx. We just multiply times the depth dx."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So minus 4 minus x squared. So this will give us the surface, the surface there, or the area of the surface or the face of one of these washers. If we want the volume of one of those washers, we then just have to multiply times the depth dx. We just multiply times the depth dx. And then if we want to actually find the volume of this entire figure, then we just have to sum up all of these washers for each of our x's. So let's do that. So we're going to sum up the washers for each of our x's and take the limit as they approach 0."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "We just multiply times the depth dx. And then if we want to actually find the volume of this entire figure, then we just have to sum up all of these washers for each of our x's. So let's do that. So we're going to sum up the washers for each of our x's and take the limit as they approach 0. But we have to make sure we got our interval right. So what are these? We care about the entire region between the points where they intersect."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're going to sum up the washers for each of our x's and take the limit as they approach 0. But we have to make sure we got our interval right. So what are these? We care about the entire region between the points where they intersect. So let's make sure we get our interval. So to figure out our interval, we just say, well, when does y equal x intersect y equal x squared minus 2x? So we just have to say, when does x, let me do this in a different color."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "We care about the entire region between the points where they intersect. So let's make sure we get our interval. So to figure out our interval, we just say, well, when does y equal x intersect y equal x squared minus 2x? So we just have to say, when does x, let me do this in a different color. We just have to think about when does x equal x squared minus 2x? Equal x squared minus 2x. When are our two functions equal to each other?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we just have to say, when does x, let me do this in a different color. We just have to think about when does x equal x squared minus 2x? Equal x squared minus 2x. When are our two functions equal to each other? Which is equivalent to, if we just subtract x from both sides, we get, when does x squared minus 3x equal 0? We can factor out an x on the right-hand side. So this is going to be, when does x times x minus 3 equal 0?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "When are our two functions equal to each other? Which is equivalent to, if we just subtract x from both sides, we get, when does x squared minus 3x equal 0? We can factor out an x on the right-hand side. So this is going to be, when does x times x minus 3 equal 0? Well, if the product is equal to 0, at least one of these need to be equal to 0. So x could be equal to 0, or x minus 3 is equal to 0. So x is equal to 0, or x is equal to 3."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be, when does x times x minus 3 equal 0? Well, if the product is equal to 0, at least one of these need to be equal to 0. So x could be equal to 0, or x minus 3 is equal to 0. So x is equal to 0, or x is equal to 3. So this is x is 0, and this right over here is x is equal to 3. So that gives us our interval. We're going to go from x equals 0 to x equals 3 to get our volume."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And like always, pause this video and see if you can figure it out. Well, whenever we're trying to find limits at either positive or negative infinity of rational expressions like this, it's useful to look at, well, what is the highest degree term in the numerator or in the denominator, or actually in the numerator and the denominator, and then divide the numerator and the denominator by that highest degree, by x to that degree. Because if we do that, then we're going to end up with some constants and some other things that will go to zero as we approach positive or negative infinity, and we should be able to find this limit. So what I'm talking about, let's divide the numerator by one over x squared, and let's divide the denominator by one over x squared. Now you might be saying, wait, wait, I see an x to the fourth here, that's a higher degree. Remember, it's under the radical here. So if you want to look at it at a very high level, you're saying, okay, well, x to the fourth, but it's under, you're gonna take the square root of this entire expression, so you could really view this as a second degree term."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what I'm talking about, let's divide the numerator by one over x squared, and let's divide the denominator by one over x squared. Now you might be saying, wait, wait, I see an x to the fourth here, that's a higher degree. Remember, it's under the radical here. So if you want to look at it at a very high level, you're saying, okay, well, x to the fourth, but it's under, you're gonna take the square root of this entire expression, so you could really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of an aside here. So if I have one over x squared, or let me write it, let me, one over x squared times the square root of four x to the fourth minus x, like we have in the numerator here."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if you want to look at it at a very high level, you're saying, okay, well, x to the fourth, but it's under, you're gonna take the square root of this entire expression, so you could really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of an aside here. So if I have one over x squared, or let me write it, let me, one over x squared times the square root of four x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of four x to the fourth minus x. And so this is equal to the square root of four x to the fourth minus x over x to the fourth, which is equal to the square root of, and all I did is I brought the radical in here. This is, you could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of four x to the fourth minus x over x to the fourth."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I have one over x squared, or let me write it, let me, one over x squared times the square root of four x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of four x to the fourth minus x. And so this is equal to the square root of four x to the fourth minus x over x to the fourth, which is equal to the square root of, and all I did is I brought the radical in here. This is, you could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of four x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator is going to be the square root of four minus one x to the third power. And then the denominator is going to be equal to, well you divide two x squared by x squared."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is, you could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of four x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator is going to be the square root of four minus one x to the third power. And then the denominator is going to be equal to, well you divide two x squared by x squared. You're just going to be left with two. And then three divided by x squared is going to be three over x squared. Now let's think about the limit as we approach negative infinity."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the denominator is going to be equal to, well you divide two x squared by x squared. You're just going to be left with two. And then three divided by x squared is going to be three over x squared. Now let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero. One divided by things that are becoming more and more and more and more and more negative, their magnitude is getting larger, so this is going to approach zero. This over here is also going to be, this thing is also going to be approaching zero."}, {"video_title": "Limits at infinity of quotients with square roots (even power)   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero. One divided by things that are becoming more and more and more and more and more negative, their magnitude is getting larger, so this is going to approach zero. This over here is also going to be, this thing is also going to be approaching zero. We're dividing by larger and larger and larger values. And so what this is going to result in is the square root of four, the principal root of four over two, which is the same thing as two over two, which is equal to one. And we are done."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's review what we're doing when we take just a regular definite integral. So if we take the definite integral between say 0 and 2 of x squared dx, what does that represent? Let's look at our endpoints. So this is x is equal to 0. Let's say that this right over here is x is equal to 2. What we're doing is for each x we're finding a little dx around it. So this right over here is a little dx."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is x is equal to 0. Let's say that this right over here is x is equal to 2. What we're doing is for each x we're finding a little dx around it. So this right over here is a little dx. And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height. Times this height right over here."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this right over here is a little dx. And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height. Times this height right over here. The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Times this height right over here. The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles. All of these rectangles for all of the x's between x is equal to 0 and x is equal to 2. But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, would not being equal to 0, and we have an infinite number of them. That's the whole power of the definite integral."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the integral sign is literally the sum of all of these rectangles. All of these rectangles for all of the x's between x is equal to 0 and x is equal to 2. But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, would not being equal to 0, and we have an infinite number of them. That's the whole power of the definite integral. And so you can imagine as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we're getting a better and better approximation of the area under the curve until at the limit we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x axis. So this is going to stretch our powers of visualization here."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's the whole power of the definite integral. And so you can imagine as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we're getting a better and better approximation of the area under the curve until at the limit we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x axis. So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x axis. So let's figure to rotate it, and I'll look at, say we're looking at a little bit from the right. So we get kind of a base that looks something like this."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x axis. So let's figure to rotate it, and I'll look at, say we're looking at a little bit from the right. So we get kind of a base that looks something like this. So this is my best attempt to draw it. So you have a base that looks something like that, and then the rest of the function. So it looks kind of like, if we just think about it between 0 and 2, it looks like one of those pieces from, I don't know if you ever played the game Sorry, or it looks like a little bit of a kind of a weird hat."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we get kind of a base that looks something like this. So this is my best attempt to draw it. So you have a base that looks something like that, and then the rest of the function. So it looks kind of like, if we just think about it between 0 and 2, it looks like one of those pieces from, I don't know if you ever played the game Sorry, or it looks like a little bit of a kind of a weird hat. So it looks like this, and let me shade it in a little bit. So it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated, and we care about the entire volume of the thing, let me draw it from a few different angles."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it looks kind of like, if we just think about it between 0 and 2, it looks like one of those pieces from, I don't know if you ever played the game Sorry, or it looks like a little bit of a kind of a weird hat. So it looks like this, and let me shade it in a little bit. So it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated, and we care about the entire volume of the thing, let me draw it from a few different angles. So if I drew it from the top, it would look something like this. It would become a little more obvious. It would look something like a hat."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just so that we're making sure we can visualize this thing that's being rotated, and we care about the entire volume of the thing, let me draw it from a few different angles. So if I drew it from the top, it would look something like this. It would become a little more obvious. It would look something like a hat. Point up like this, and it goes down like that. It would look something like that. So we're not seeing, in this angle, we're not seeing the bottom of it."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would look something like a hat. Point up like this, and it goes down like that. It would look something like that. So we're not seeing, in this angle, we're not seeing the bottom of it. And if you were to just orient yourself, the axes in this case look like this. So this is the y-axis, and then the x-axis goes right inside of this thing, and then pops out the other side. And if this thing was transparent, then you could see the back side."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we're not seeing, in this angle, we're not seeing the bottom of it. And if you were to just orient yourself, the axes in this case look like this. So this is the y-axis, and then the x-axis goes right inside of this thing, and then pops out the other side. And if this thing was transparent, then you could see the back side. It would look something like that. The x-axis, if you could see through it, would pop the base right over there, would go right through the base right over there, and then come out on the other side. So this is one orientation for the same thing."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if this thing was transparent, then you could see the back side. It would look something like that. The x-axis, if you could see through it, would pop the base right over there, would go right through the base right over there, and then come out on the other side. So this is one orientation for the same thing. You could visualize it from different angles. So let's think about how we can take the volume of it. Well, instead of thinking about the area of each of these rectangles, what happens if we rotate each of these rectangles around the x-axis?"}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is one orientation for the same thing. You could visualize it from different angles. So let's think about how we can take the volume of it. Well, instead of thinking about the area of each of these rectangles, what happens if we rotate each of these rectangles around the x-axis? So let's do it. So let's take each of these. Let's say you have this dx right over here, and you rotate it around the x-axis."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, instead of thinking about the area of each of these rectangles, what happens if we rotate each of these rectangles around the x-axis? So let's do it. So let's take each of these. Let's say you have this dx right over here, and you rotate it around the x-axis. So if you were to rotate this thing around the x-axis, so I'm trying my best to, around the x-axis, you rotate it, what do you end up with? What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say you have this dx right over here, and you rotate it around the x-axis. So if you were to rotate this thing around the x-axis, so I'm trying my best to, around the x-axis, you rotate it, what do you end up with? What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. Let me draw it out here. So if you were to, that same disk out here would look something like this. It would look something like this, and it has a depth of dx."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. Let me draw it out here. So if you were to, that same disk out here would look something like this. It would look something like this, and it has a depth of dx. So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It would look something like this, and it has a depth of dx. So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly. So this is my x-axis. My disk looks something like this. My best attempt."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's really important to visualize this stuff properly. So this is my x-axis. My disk looks something like this. My best attempt. The x-axis hits it right over there, comes out of the center, and then this is the surface of my disk, and then this right over here is my depth, dx. So that looks pretty good. And then let me just shade it in a little bit to give you a little bit of the depth."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "My best attempt. The x-axis hits it right over there, comes out of the center, and then this is the surface of my disk, and then this right over here is my depth, dx. So that looks pretty good. And then let me just shade it in a little bit to give you a little bit of the depth. So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this face?"}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then let me just shade it in a little bit to give you a little bit of the depth. So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this face? Well, we know that the area of a circle is equal to pi r squared. If we know the radius of this face, we can figure out the area of the face. Well, what's the radius?"}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what's the area of this face? Well, we know that the area of a circle is equal to pi r squared. If we know the radius of this face, we can figure out the area of the face. Well, what's the radius? Well, the radius is just the height of that original rectangle, and for any x, the height over here is going to be equal to f of x, and in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, what's the radius? Well, the radius is just the height of that original rectangle, and for any x, the height over here is going to be equal to f of x, and in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared. Now what's our volume going to be? Well, our volume is going to be our area times the depth here."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared. Now what's our volume going to be? Well, our volume is going to be our area times the depth here. It's going to be that times the depth times dx. So the volume of this thing right over here, so the volume just of this coin, I guess you could call it, is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi x squared squared is x to the fourth."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, our volume is going to be our area times the depth here. It's going to be that times the depth times dx. So the volume of this thing right over here, so the volume just of this coin, I guess you could call it, is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi x squared squared is x to the fourth. Pi x to the fourth dx. Now this expression right over here, this gave us the volume just of one of those disks, but what we want is the volume of this entire hat or this entire kind of bugle or cone-looking, or I guess you could say the front of a trumpet-looking thing. So how could we do that?"}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's equal to pi x squared squared is x to the fourth. Pi x to the fourth dx. Now this expression right over here, this gave us the volume just of one of those disks, but what we want is the volume of this entire hat or this entire kind of bugle or cone-looking, or I guess you could say the front of a trumpet-looking thing. So how could we do that? Well, the exact same technique. What happens if we were to take the sum of all of these things? So let's do that."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how could we do that? Well, the exact same technique. What happens if we were to take the sum of all of these things? So let's do that. Take the sum of all of these things, and I'll switch to one color, pi times x to the fourth dx. We're going to take the sum of all of these things from x is equal to 0 to 2. Those are the boundaries that we started off with."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. Take the sum of all of these things, and I'll switch to one color, pi times x to the fourth dx. We're going to take the sum of all of these things from x is equal to 0 to 2. Those are the boundaries that we started off with. I just defined them arbitrarily. We could do this really for any two x values. Between x is equal to 0 and x equals 2, and we're going to take the sum of all of these, the volumes of all of these coins, but the limit as the depths get smaller and smaller and smaller and we have more and more and more coins, at the limit we're actually going to get the volume of our cone or our bugle or whatever we want to call it."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Those are the boundaries that we started off with. I just defined them arbitrarily. We could do this really for any two x values. Between x is equal to 0 and x equals 2, and we're going to take the sum of all of these, the volumes of all of these coins, but the limit as the depths get smaller and smaller and smaller and we have more and more and more coins, at the limit we're actually going to get the volume of our cone or our bugle or whatever we want to call it. So if we just evaluate this definite integral, we have our volume. So let's see if we can do that. So now this is just taking a standard definite integral."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Between x is equal to 0 and x equals 2, and we're going to take the sum of all of these, the volumes of all of these coins, but the limit as the depths get smaller and smaller and smaller and we have more and more and more coins, at the limit we're actually going to get the volume of our cone or our bugle or whatever we want to call it. So if we just evaluate this definite integral, we have our volume. So let's see if we can do that. So now this is just taking a standard definite integral. So this is going to be equal to, and I encourage you to try it out before I do it. So we could take the pi out. So it's going to be equal to pi times the integral from 0 to 2 of x to the fourth dx."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now this is just taking a standard definite integral. So this is going to be equal to, and I encourage you to try it out before I do it. So we could take the pi out. So it's going to be equal to pi times the integral from 0 to 2 of x to the fourth dx. I don't like that color. Now the antiderivative of x to the fourth is x to the fifth over 5. So this is going to be equal to pi times x to the fifth over 5."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be equal to pi times the integral from 0 to 2 of x to the fourth dx. I don't like that color. Now the antiderivative of x to the fourth is x to the fifth over 5. So this is going to be equal to pi times x to the fifth over 5. And we're going to go from 0 to 2. So this is going to be equal to pi times, this thing evaluated at 2. Let's see, 2 to the third is 8, 2 to the fourth is 16, 2 to the fifth is, let me actually just write it down, 2 to the fifth over 5 minus 0 to the fifth over 5."}, {"video_title": "Disc method around x-axis   Applications of definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to pi times x to the fifth over 5. And we're going to go from 0 to 2. So this is going to be equal to pi times, this thing evaluated at 2. Let's see, 2 to the third is 8, 2 to the fourth is 16, 2 to the fifth is, let me actually just write it down, 2 to the fifth over 5 minus 0 to the fifth over 5. And this is going to be equal to, 2 to the fifth is 32. So it's going to be equal to pi times 32 over 5 minus, well this is just 0, minus 0, which is equal to 32 pi over 5. And we're done."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now in this video, we can actually evaluate the definite integral. So what we need to do is really just expand out this expression, the square root of y plus 1 plus 2. So let's do that. So this is going to be equal to pi times the definite integral from y is equal to negative 1 to y is equal to 3. If you expand this out, you get square root of y plus 1 squared, which is just going to be y plus 1. And then you're going to have 2 times the product of both of these terms. 2 times 2 times square root of y plus 1 is going to be plus 4 times the square root of y plus 1."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be equal to pi times the definite integral from y is equal to negative 1 to y is equal to 3. If you expand this out, you get square root of y plus 1 squared, which is just going to be y plus 1. And then you're going to have 2 times the product of both of these terms. 2 times 2 times square root of y plus 1 is going to be plus 4 times the square root of y plus 1. And then you have 2 squared, so plus 4. So you have this whole thing times dy. We can simplify a little bit."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "2 times 2 times square root of y plus 1 is going to be plus 4 times the square root of y plus 1. And then you have 2 squared, so plus 4. So you have this whole thing times dy. We can simplify a little bit. You have a 1 plus a 4. We can add the 1 to the 4 and get a 5. And now we're ready to take the antiderivative."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We can simplify a little bit. You have a 1 plus a 4. We can add the 1 to the 4 and get a 5. And now we're ready to take the antiderivative. This is going to be equal to pi times, let's take the antiderivative of all of this business, and I'll color code it. The antiderivative of y is just y squared over 2. The antiderivative of 4 times the square root of y plus 1, you just really have to think of it as 4 times y plus 1 to the 1 half power."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we're ready to take the antiderivative. This is going to be equal to pi times, let's take the antiderivative of all of this business, and I'll color code it. The antiderivative of y is just y squared over 2. The antiderivative of 4 times the square root of y plus 1, you just really have to think of it as 4 times y plus 1 to the 1 half power. We could use u substitution explicitly, but you probably are pretty practiced and this can do this in your head. y plus 1 raised to the 1 half power. Derivative of y plus 1 is just 1, which is essentially out here."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of 4 times the square root of y plus 1, you just really have to think of it as 4 times y plus 1 to the 1 half power. We could use u substitution explicitly, but you probably are pretty practiced and this can do this in your head. y plus 1 raised to the 1 half power. Derivative of y plus 1 is just 1, which is essentially out here. So you can really just, if you did u substitution, you'd say u is equal to y plus 1. But this antiderivative is going to be equal to, well, if you increment this exponent, you get 3 halves multiplied by the reciprocal 2 thirds. 2 thirds times 4 is 8 thirds."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative of y plus 1 is just 1, which is essentially out here. So you can really just, if you did u substitution, you'd say u is equal to y plus 1. But this antiderivative is going to be equal to, well, if you increment this exponent, you get 3 halves multiplied by the reciprocal 2 thirds. 2 thirds times 4 is 8 thirds. So it's plus 8 thirds times y plus 1 to the 3 halves. And you can verify, if you take the derivative here, you will get this expression right over here. 3 halves times 8 thirds is 4."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "2 thirds times 4 is 8 thirds. So it's plus 8 thirds times y plus 1 to the 3 halves. And you can verify, if you take the derivative here, you will get this expression right over here. 3 halves times 8 thirds is 4. Decrement it, you have y plus 1 to the 1 half power. And then finally, you have this 5. The antiderivative of 5 is just 5y."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "3 halves times 8 thirds is 4. Decrement it, you have y plus 1 to the 1 half power. And then finally, you have this 5. The antiderivative of 5 is just 5y. And we are going to evaluate it from at 3 and at negative 1. y is equals 3 and y equals negative 1. So this is going to be equal to pi. So let's evaluate all this business at 3."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "The antiderivative of 5 is just 5y. And we are going to evaluate it from at 3 and at negative 1. y is equals 3 and y equals negative 1. So this is going to be equal to pi. So let's evaluate all this business at 3. So 3 squared over 2 is 9 halves. 3 plus 1 is 4 to the 3 halves. Well, that's, let's see, square root of 4 is 2 to the third power is 8."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's evaluate all this business at 3. So 3 squared over 2 is 9 halves. 3 plus 1 is 4 to the 3 halves. Well, that's, let's see, square root of 4 is 2 to the third power is 8. 8 times 8 thirds is 64 over 3. So plus 64 over 3. You have 5 times 3."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, that's, let's see, square root of 4 is 2 to the third power is 8. 8 times 8 thirds is 64 over 3. So plus 64 over 3. You have 5 times 3. Well, that's going to be 15. Plus 15. And from that, we're going to subtract all this business evaluated at negative 1."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "You have 5 times 3. Well, that's going to be 15. Plus 15. And from that, we're going to subtract all this business evaluated at negative 1. So you have negative 1 squared over 2. Well, that's just 1 half. Negative 1 plus 1 is 0 to the 3 half power."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And from that, we're going to subtract all this business evaluated at negative 1. So you have negative 1 squared over 2. Well, that's just 1 half. Negative 1 plus 1 is 0 to the 3 half power. That's going to be 0 times 8 thirds. This is all going to be 0, so we don't have to even write it. And then finally, you have negative 1 times 5."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Negative 1 plus 1 is 0 to the 3 half power. That's going to be 0 times 8 thirds. This is all going to be 0, so we don't have to even write it. And then finally, you have negative 1 times 5. Well, this is going to be negative 5. And we are in the home stretch. We really just have to do a little bit of arithmetic, add some hairy fractions right over here."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then finally, you have negative 1 times 5. Well, this is going to be negative 5. And we are in the home stretch. We really just have to do a little bit of arithmetic, add some hairy fractions right over here. So let's do it. So this whole thing is going to simplify to pi times. And it looks like, let's see, our least common multiple of all of these denominators is going to be 6."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We really just have to do a little bit of arithmetic, add some hairy fractions right over here. So let's do it. So this whole thing is going to simplify to pi times. And it looks like, let's see, our least common multiple of all of these denominators is going to be 6. So let's put everything over a denominator of 6. So 9 halves is the same thing as 27 over 6. 64 thirds is the same thing as 128 over 6."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it looks like, let's see, our least common multiple of all of these denominators is going to be 6. So let's put everything over a denominator of 6. So 9 halves is the same thing as 27 over 6. 64 thirds is the same thing as 128 over 6. 15 is the same thing as 90 over 6. 1 half is the same thing as 3 over 6. So this is, we distribute the negative signs."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "64 thirds is the same thing as 128 over 6. 15 is the same thing as 90 over 6. 1 half is the same thing as 3 over 6. So this is, we distribute the negative signs. This is negative 3 6. And negative times negative is positive. 5 is the same thing as 30 over 6."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is, we distribute the negative signs. This is negative 3 6. And negative times negative is positive. 5 is the same thing as 30 over 6. So plus 30 over 6. And so this is going to give us, our denominator is going to be over 6. We're going to multiply something times pi."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "5 is the same thing as 30 over 6. So plus 30 over 6. And so this is going to give us, our denominator is going to be over 6. We're going to multiply something times pi. We have this pi over here. And then we just have to figure out what the numerator is. So let's see if I can do this in my head."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're going to multiply something times pi. We have this pi over here. And then we just have to figure out what the numerator is. So let's see if I can do this in my head. So 27 plus 128 is going to be, let me see, that's going to be 155. Is that right? 155."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's see if I can do this in my head. So 27 plus 128 is going to be, let me see, that's going to be 155. Is that right? 155. Let's see, if we get to 48 plus another 7. Yeah, 155 plus 90 gets us to 245. Is that right?"}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "155. Let's see, if we get to 48 plus another 7. Yeah, 155 plus 90 gets us to 245. Is that right? Yeah, plus 90 gets us to 245. Minus 3, you subtract 3 from that, you get to 242. And then you add 30 to that, you get to 272."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "Is that right? Yeah, plus 90 gets us to 245. Minus 3, you subtract 3 from that, you get to 242. And then you add 30 to that, you get to 272. So we're left with 272 pi over 6. But then we can, let's see, 272 and 6 are both divisible by 2. So this is equal to, see, 272 divided by 2 is going to be 136 pi over, and if you divide this denominator right over here, by 2 over 3."}, {"video_title": "Calculating integral disc around vertical line   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you add 30 to that, you get to 272. So we're left with 272 pi over 6. But then we can, let's see, 272 and 6 are both divisible by 2. So this is equal to, see, 272 divided by 2 is going to be 136 pi over, and if you divide this denominator right over here, by 2 over 3. Is that right? Yeah, 136 pi over 3. And 136 is not divisible by 3."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Imagine we're asked to approximate the area between the x-axis and the graph of f from x equals one to x equals 10 using a right Riemann sum with three equal subdivisions. To do that, we are given a table of values for f. So I encourage you to pause the video and see if you can come up with an approximation for the area between the x-axis and the graph from x equals one to x equals 10 using a right Riemann sum with three equal subdivisions. So I'm assuming you've had a go at it. So now let's try to do that together. And this is interesting because we don't have a graph of the entire function, but we just have the value of the function at certain points. But as we'll see, this is all we need in order to get an approximation for the area. We don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation or an approximation using a right Riemann sum."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So now let's try to do that together. And this is interesting because we don't have a graph of the entire function, but we just have the value of the function at certain points. But as we'll see, this is all we need in order to get an approximation for the area. We don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation or an approximation using a right Riemann sum. So let me just draw some axes here because whenever I do Riemann sums, you can do them without graphs, but it helps to think about what's going on if you can visualize it graphically. So let's see. We are going from x equals one to x equals 10."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation or an approximation using a right Riemann sum. So let me just draw some axes here because whenever I do Riemann sums, you can do them without graphs, but it helps to think about what's going on if you can visualize it graphically. So let's see. We are going from x equals one to x equals 10. So this is one, two, three, four, five, six, seven, eight, nine, 10. And so they give us the value of f of x when x equals one, when x equals two, three, four, when x equals seven, five, six, seven, eight, nine, 10, and x equals 10. And they tell us that when x is one, we're at six, and we go to eight, to three, five."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are going from x equals one to x equals 10. So this is one, two, three, four, five, six, seven, eight, nine, 10. And so they give us the value of f of x when x equals one, when x equals two, three, four, when x equals seven, five, six, seven, eight, nine, 10, and x equals 10. And they tell us that when x is one, we're at six, and we go to eight, to three, five. So let me mark these off. So we're gonna go up to eight. So one, two, three, four, five, six, seven, eight."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And they tell us that when x is one, we're at six, and we go to eight, to three, five. So let me mark these off. So we're gonna go up to eight. So one, two, three, four, five, six, seven, eight. And so what we know, when x is equal to one, f of one is six. So this is one, two, three, four, five, six, seven, eight. So this point right over here is f of one."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one, two, three, four, five, six, seven, eight. And so what we know, when x is equal to one, f of one is six. So this is one, two, three, four, five, six, seven, eight. So this point right over here is f of one. This is the point one comma six. And then we have the point four comma eight. Four comma eight will put us right about there."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this point right over here is f of one. This is the point one comma six. And then we have the point four comma eight. Four comma eight will put us right about there. And then we have seven comma three is on our graph, y equals f of x. So seven comma three would put us right over there. And then we have 10 comma five."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Four comma eight will put us right about there. And then we have seven comma three is on our graph, y equals f of x. So seven comma three would put us right over there. And then we have 10 comma five. So 10 comma five would put us right over there. That's all we know about the function. We don't know exactly what it looks like."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then we have 10 comma five. So 10 comma five would put us right over there. That's all we know about the function. We don't know exactly what it looks like. Our function might look like this. It might do something like this. Whoops, I drew a part that didn't look like a function."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know exactly what it looks like. Our function might look like this. It might do something like this. Whoops, I drew a part that didn't look like a function. It might do something like this and oscillate really quickly. It might be nice and smooth and just kind of go and do something just like that, kind of connect the dots. We don't know."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Whoops, I drew a part that didn't look like a function. It might do something like this and oscillate really quickly. It might be nice and smooth and just kind of go and do something just like that, kind of connect the dots. We don't know. But we can still do the approximation using a right Riemann sum with three equal subdivisions. How do we do that? Well, we're thinking about the area from x equals one to x equals 10."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know. But we can still do the approximation using a right Riemann sum with three equal subdivisions. How do we do that? Well, we're thinking about the area from x equals one to x equals 10. So let me make those boundaries clear. So this is from x equals one to x equals 10. And what we want to do is have three equal subdivisions."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're thinking about the area from x equals one to x equals 10. So let me make those boundaries clear. So this is from x equals one to x equals 10. And what we want to do is have three equal subdivisions. And there's three very natural subdivisions here. If we make each of our subdivisions three wide, so this could be a subdivision. And then this is another subdivision."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And what we want to do is have three equal subdivisions. And there's three very natural subdivisions here. If we make each of our subdivisions three wide, so this could be a subdivision. And then this is another subdivision. And when you do Riemann sums, you don't have to have equal subdivisions, although that's what you'll often see. So we've just divided going from one to 10 into three equal sections that are three wide. So that's three, this is three, and this is three."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then this is another subdivision. And when you do Riemann sums, you don't have to have equal subdivisions, although that's what you'll often see. So we've just divided going from one to 10 into three equal sections that are three wide. So that's three, this is three, and this is three. And so the question is, how do we define the height of these subdivisions which are going to end up being rectangles? And that's where the right Riemann sum applies. If we were doing a left Riemann sum, we would use the left boundary of each of the subdivisions and the value of the function there to define the height of the rectangle."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's three, this is three, and this is three. And so the question is, how do we define the height of these subdivisions which are going to end up being rectangles? And that's where the right Riemann sum applies. If we were doing a left Riemann sum, we would use the left boundary of each of the subdivisions and the value of the function there to define the height of the rectangle. So this would be doing a left Riemann sum. But we're doing a right Riemann sum. So we use the right boundary of each of these subdivisions to define the height."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we were doing a left Riemann sum, we would use the left boundary of each of the subdivisions and the value of the function there to define the height of the rectangle. So this would be doing a left Riemann sum. But we're doing a right Riemann sum. So we use the right boundary of each of these subdivisions to define the height. So our right boundary is when x equals four for this first section. What is f of four? It's eight."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we use the right boundary of each of these subdivisions to define the height. So our right boundary is when x equals four for this first section. What is f of four? It's eight. So we're gonna use that as the height of this first rectangle that's approximating the area for this part of the curve. Similarly, for this second one, since we're using a right Riemann sum, we use the value of the function at the right boundary, the right boundary seven. So the value of the function is three."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's eight. So we're gonna use that as the height of this first rectangle that's approximating the area for this part of the curve. Similarly, for this second one, since we're using a right Riemann sum, we use the value of the function at the right boundary, the right boundary seven. So the value of the function is three. So this would be our second rectangle, our second division, I guess, used to approximate the area. And then last but not least, we would use the right boundary of this third subdivision when x equals 10, f of 10 is five. So just like that."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the value of the function is three. So this would be our second rectangle, our second division, I guess, used to approximate the area. And then last but not least, we would use the right boundary of this third subdivision when x equals 10, f of 10 is five. So just like that. And so then our right Riemann approximation using our right Riemann sum with three equal subdivisions to approximate the area, we would just add the area of these rectangles. So this first rectangle, let's see, it is three wide. And how high is it?"}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So just like that. And so then our right Riemann approximation using our right Riemann sum with three equal subdivisions to approximate the area, we would just add the area of these rectangles. So this first rectangle, let's see, it is three wide. And how high is it? Well, the height here is f of four, which is eight. So this is going to be 24 square units, whatever the units happen to be. This is going to be three times, the height here is three, f of seven is three."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "And how high is it? Well, the height here is f of four, which is eight. So this is going to be 24 square units, whatever the units happen to be. This is going to be three times, the height here is three, f of seven is three. So that is nine square units. And then here, this is three, the width is three times the height, f of 10 is five. So three times five, which gets us 15."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is going to be three times, the height here is three, f of seven is three. So that is nine square units. And then here, this is three, the width is three times the height, f of 10 is five. So three times five, which gets us 15. And so our approximation of the area would be summing these three values up. So this would be 24 plus, let's see, nine plus 15 would give us another 24. So it's 24 plus 24, it gets us to 48."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So three times five, which gets us 15. And so our approximation of the area would be summing these three values up. So this would be 24 plus, let's see, nine plus 15 would give us another 24. So it's 24 plus 24, it gets us to 48. There you go. We, just using that table of values, we've been able to find an approximation. Now once again, we don't know how good of an approximation is, it depends on what the function is doing."}, {"video_title": "Worked example finding a Riemann sum using a table   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's 24 plus 24, it gets us to 48. There you go. We, just using that table of values, we've been able to find an approximation. Now once again, we don't know how good of an approximation is, it depends on what the function is doing. There's a world where it could be a very good approximation. Maybe the function does something like this. Maybe the function just happens, let me make it a little bit, maybe the function does something like this, where in this case, what we just did would be a very good approximation."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have here f of x being equal to the natural log of the square root of x. And what we want to do in this video is find the derivative of f. And the key here is to recognize that f can actually be viewed as a composition of two functions. And we can diagram that out. What's going on here? Well, if you input an x into our function f, what's the first thing that you do? Well, you take the square root of it. So if we start off with some x, you input it, the first thing that you do is you take the square root of it."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "What's going on here? Well, if you input an x into our function f, what's the first thing that you do? Well, you take the square root of it. So if we start off with some x, you input it, the first thing that you do is you take the square root of it. So you are going to take the square root of the input to produce the square root of x. And then what do you do? Well, you take the square root, and then you take the natural log of that."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if we start off with some x, you input it, the first thing that you do is you take the square root of it. So you are going to take the square root of the input to produce the square root of x. And then what do you do? Well, you take the square root, and then you take the natural log of that. And so then you take the natural log of that. So you could view that as inputting it into another function that takes the natural log of whatever is inputted in. I'm making these little squares to show what you do with the input."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, you take the square root, and then you take the natural log of that. And so then you take the natural log of that. So you could view that as inputting it into another function that takes the natural log of whatever is inputted in. I'm making these little squares to show what you do with the input. And then what do you produce? Well, you produce the natural log of the square root of x. Natural log of the square root of x, which is equal to f of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm making these little squares to show what you do with the input. And then what do you produce? Well, you produce the natural log of the square root of x. Natural log of the square root of x, which is equal to f of x. So you could view f of x as this entire, as this entire set, or this entire, I guess you could say, this combination of functions right over there. And that is f of x, which is essentially a composition of two functions. You're inputting into one function, then taking that output and inputting it into another."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Natural log of the square root of x, which is equal to f of x. So you could view f of x as this entire, as this entire set, or this entire, I guess you could say, this combination of functions right over there. And that is f of x, which is essentially a composition of two functions. You're inputting into one function, then taking that output and inputting it into another. So you could have a function u here, which takes the square root of whatever it's input in, is. So u of x is equal to the square root of x. And then you take another, and then you take that output and input it into another function that we could call v. And what does v do?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're inputting into one function, then taking that output and inputting it into another. So you could have a function u here, which takes the square root of whatever it's input in, is. So u of x is equal to the square root of x. And then you take another, and then you take that output and input it into another function that we could call v. And what does v do? Well, it takes the natural log of whatever the input is. In this case, in the case of f, or in the case of how I've just diagrammed it, v is taking the natural log, the input happens to be square root of x. So it outputs the natural log of the square root of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you take another, and then you take that output and input it into another function that we could call v. And what does v do? Well, it takes the natural log of whatever the input is. In this case, in the case of f, or in the case of how I've just diagrammed it, v is taking the natural log, the input happens to be square root of x. So it outputs the natural log of the square root of x. If we wanted to write v with x as an input, we would just say, well, that's just the natural log. That is just the natural log of x. And as you can see here, f of x, and I color-coded it ahead of time, is equal to, f of x, f of x is equal to the natural log of the square root of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it outputs the natural log of the square root of x. If we wanted to write v with x as an input, we would just say, well, that's just the natural log. That is just the natural log of x. And as you can see here, f of x, and I color-coded it ahead of time, is equal to, f of x, f of x is equal to the natural log of the square root of x. So that is v of the square root of x, or v of u of x. So it is a composition, which tells you that, okay, if I'm trying to find the derivative here, the chain rule is going to be very, very, very, very useful. And the chain rule tells us that f prime of x is going to be equal to the derivative of, you can view it as the outside function, with respect to this inside function."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And as you can see here, f of x, and I color-coded it ahead of time, is equal to, f of x, f of x is equal to the natural log of the square root of x. So that is v of the square root of x, or v of u of x. So it is a composition, which tells you that, okay, if I'm trying to find the derivative here, the chain rule is going to be very, very, very, very useful. And the chain rule tells us that f prime of x is going to be equal to the derivative of, you can view it as the outside function, with respect to this inside function. So it's going to be v prime of u of x, v prime of u of x, times the derivative of this inside function with respect to x. So that's just u prime, u prime of x. So how do we evaluate these things?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the chain rule tells us that f prime of x is going to be equal to the derivative of, you can view it as the outside function, with respect to this inside function. So it's going to be v prime of u of x, v prime of u of x, times the derivative of this inside function with respect to x. So that's just u prime, u prime of x. So how do we evaluate these things? Well, we know how to take the derivative of u of x and v of x. u prime of x here is going to be equal to, well, remember, square root of x is just the same thing as x to the 1 1\u20442 power, so we can use the power rule, bring the 1 1\u20442 out front, so it becomes 1 1\u20442 x to the, and then take off one out of that exponent, so that's 1 1\u20442 minus one is negative 1 1\u20442 power. And what is v of x, sorry, what is v prime of x? Well, the derivative of the natural log of x is one over x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So how do we evaluate these things? Well, we know how to take the derivative of u of x and v of x. u prime of x here is going to be equal to, well, remember, square root of x is just the same thing as x to the 1 1\u20442 power, so we can use the power rule, bring the 1 1\u20442 out front, so it becomes 1 1\u20442 x to the, and then take off one out of that exponent, so that's 1 1\u20442 minus one is negative 1 1\u20442 power. And what is v of x, sorry, what is v prime of x? Well, the derivative of the natural log of x is one over x. We show that in other videos. And so we now know what u prime of x is. What is, we know what v prime of x is, but what is v prime of u of x?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the derivative of the natural log of x is one over x. We show that in other videos. And so we now know what u prime of x is. What is, we know what v prime of x is, but what is v prime of u of x? Well, v prime of u of x, wherever we see the x, we replace it, let me write that a little bit neater, we replace that with a u of x. So v prime of u of x is going to be equal to, is going to be equal to one over u of x, one over u of x, u of x, which is equal to, which is equal to one over, u of x is just the square root of x, one over the square root of x. So this thing right over here, we have figured out is one over the square root of x, and this thing, u prime of x, we figured out is 1 1\u20442 times x to the negative 1 1\u20442, and x to the negative 1 1\u20442, I could rewrite that as 1 1\u20442 times one over x to the 1 1\u20442, which is the same thing as 1 1\u20442 times one over the square root of x, or I could write that as one over two square roots of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "What is, we know what v prime of x is, but what is v prime of u of x? Well, v prime of u of x, wherever we see the x, we replace it, let me write that a little bit neater, we replace that with a u of x. So v prime of u of x is going to be equal to, is going to be equal to one over u of x, one over u of x, u of x, which is equal to, which is equal to one over, u of x is just the square root of x, one over the square root of x. So this thing right over here, we have figured out is one over the square root of x, and this thing, u prime of x, we figured out is 1 1\u20442 times x to the negative 1 1\u20442, and x to the negative 1 1\u20442, I could rewrite that as 1 1\u20442 times one over x to the 1 1\u20442, which is the same thing as 1 1\u20442 times one over the square root of x, or I could write that as one over two square roots of x. So what is this thing going to be? Well, this is going to be equal to, in green, v prime of u of x is one over the square root of x times, times u prime of x is one over two times the square root of x. Now what is this going to be equal to?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this thing right over here, we have figured out is one over the square root of x, and this thing, u prime of x, we figured out is 1 1\u20442 times x to the negative 1 1\u20442, and x to the negative 1 1\u20442, I could rewrite that as 1 1\u20442 times one over x to the 1 1\u20442, which is the same thing as 1 1\u20442 times one over the square root of x, or I could write that as one over two square roots of x. So what is this thing going to be? Well, this is going to be equal to, in green, v prime of u of x is one over the square root of x times, times u prime of x is one over two times the square root of x. Now what is this going to be equal to? Well, this is going to be equal to, this is just algebra at this point, one over, we have our two, and square root of x times square root of x is just x. So it just simplifies to one over two x. So hopefully this made sense, and I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions, and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is this going to be equal to? Well, this is going to be equal to, this is just algebra at this point, one over, we have our two, and square root of x times square root of x is just x. So it just simplifies to one over two x. So hopefully this made sense, and I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions, and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook. But as you get more practice, you'll be able to do it essentially without having to write out all of this. You'll say, okay, look, I have a composition. This is the natural log of the square root of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So hopefully this made sense, and I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions, and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook. But as you get more practice, you'll be able to do it essentially without having to write out all of this. You'll say, okay, look, I have a composition. This is the natural log of the square root of x. This is v of u of x. So what I want to do is I want to take the derivative of this outside function with respect to this inside function. So, the derivative of natural log of something with respect to that something is one over that something."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the natural log of the square root of x. This is v of u of x. So what I want to do is I want to take the derivative of this outside function with respect to this inside function. So, the derivative of natural log of something with respect to that something is one over that something. So it is one over that something. The derivative of natural log of something with respect to that something is one over that something. So that's what we just did here."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule   AP Calculus AB   Khan Academy.mp3", "Sentence": "So, the derivative of natural log of something with respect to that something is one over that something. So it is one over that something. The derivative of natural log of something with respect to that something is one over that something. So that's what we just did here. One way to think about it, what would natural log of x be? Well that'd be one over x, but it's not natural log of x, it's one over square root of x. So, it's going to be one over the square root of x."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "is the function given below, continuous slash differentiable at x equals one, and they define the function g piecewise right over here, and then they give us a bunch of choices, continuous but not differentiable, differentiable but not continuous, both continuous and differentiable, neither continuous nor differentiable. And like always, pause this video and see if you could figure this out. So let's do step by step. So first let's think about continuity. So for continuity, for g to be continuous at x equals one, that means that g of one, that means that g of one must be equal to the limit as x approaches one of g, of g of x. Well g of one, what is that going to be? G of one, we're gonna fall into this case, one minus one squared is going to be zero."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "So first let's think about continuity. So for continuity, for g to be continuous at x equals one, that means that g of one, that means that g of one must be equal to the limit as x approaches one of g, of g of x. Well g of one, what is that going to be? G of one, we're gonna fall into this case, one minus one squared is going to be zero. So if we can show that the limit of g of x as x approaches one is the same as g of one, is equal to zero, then we know we're continuous there. Let's do the left and right-handed limits here. So if we do the left-handed limit, and that's especially useful because we're in these different clauses here as we approach from the left and the right-hand side."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "G of one, we're gonna fall into this case, one minus one squared is going to be zero. So if we can show that the limit of g of x as x approaches one is the same as g of one, is equal to zero, then we know we're continuous there. Let's do the left and right-handed limits here. So if we do the left-handed limit, and that's especially useful because we're in these different clauses here as we approach from the left and the right-hand side. So as x approaches one from the left-hand side of g of x, well we're gonna be falling into this situation here as we approach from the left, as x is less than one. So this is going to be the same thing as that. That's what g of x is equal to when we are less than one as we're approaching from the left."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "So if we do the left-handed limit, and that's especially useful because we're in these different clauses here as we approach from the left and the right-hand side. So as x approaches one from the left-hand side of g of x, well we're gonna be falling into this situation here as we approach from the left, as x is less than one. So this is going to be the same thing as that. That's what g of x is equal to when we are less than one as we're approaching from the left. Well this thing is defined and it's continuous for all real numbers, so we can just substitute one in for x and we get this is equal to zero. So so far so good, let's do one more of these. Let's approach from the right-hand side as x approaches one from the right-hand side of g of x."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "That's what g of x is equal to when we are less than one as we're approaching from the left. Well this thing is defined and it's continuous for all real numbers, so we can just substitute one in for x and we get this is equal to zero. So so far so good, let's do one more of these. Let's approach from the right-hand side as x approaches one from the right-hand side of g of x. Well now we're falling into this case. So g of x, if we're to the right of one, if our value's greater than or equal to one, it's gonna be x minus one squared. Well once again, x minus one squared, that is defined for all real numbers, it's continuous for all real numbers, so we could just pop that one in there, you get one minus one squared, well that's just zero again."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "Let's approach from the right-hand side as x approaches one from the right-hand side of g of x. Well now we're falling into this case. So g of x, if we're to the right of one, if our value's greater than or equal to one, it's gonna be x minus one squared. Well once again, x minus one squared, that is defined for all real numbers, it's continuous for all real numbers, so we could just pop that one in there, you get one minus one squared, well that's just zero again. So the left-hand limit, the right-hand limit are both equal zero, which means that the limit is equal, the limit of g of x as x approaches one is equal to zero, which is the same thing as g of one, so we are good with continuity. So we can rule out all the ones that are saying that it's not continuous. So we could rule out that one, and we can rule out that one right over there."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "Well once again, x minus one squared, that is defined for all real numbers, it's continuous for all real numbers, so we could just pop that one in there, you get one minus one squared, well that's just zero again. So the left-hand limit, the right-hand limit are both equal zero, which means that the limit is equal, the limit of g of x as x approaches one is equal to zero, which is the same thing as g of one, so we are good with continuity. So we can rule out all the ones that are saying that it's not continuous. So we could rule out that one, and we can rule out that one right over there. So now let's think about whether it is differentiable. So differentiability, so differentiability, I'll write differentiability, ability. Did I, let's see, that's a long word, differentiability."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "So we could rule out that one, and we can rule out that one right over there. So now let's think about whether it is differentiable. So differentiability, so differentiability, I'll write differentiability, ability. Did I, let's see, that's a long word, differentiability. All right, differentiability, what needs to be true here? Well we have to have a defined limit as x approaches one for f of x minus f of one over, oh let me be careful, it's not f, it's g. So we need to have a defined limit for g of x minus g of one over x minus one. And so let's just try to evaluate this limit from the left and right-hand sides."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "Did I, let's see, that's a long word, differentiability. All right, differentiability, what needs to be true here? Well we have to have a defined limit as x approaches one for f of x minus f of one over, oh let me be careful, it's not f, it's g. So we need to have a defined limit for g of x minus g of one over x minus one. And so let's just try to evaluate this limit from the left and right-hand sides. And we could simplify it, we already know that g of one is zero, so that's just going to be zero. So we just need to find the limit as x approaches one of g of x over x minus one, or see if we can find a limit. So let's first think about the limit as we approach from the left-hand side of g of x over x minus, g of x over x minus one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "And so let's just try to evaluate this limit from the left and right-hand sides. And we could simplify it, we already know that g of one is zero, so that's just going to be zero. So we just need to find the limit as x approaches one of g of x over x minus one, or see if we can find a limit. So let's first think about the limit as we approach from the left-hand side of g of x over x minus, g of x over x minus one. Well as we approach from the left-hand side, g of x is that right over there. So we could write this, instead of writing g of x, we could write this as x minus one, x minus one over x minus one. And as long as we aren't equal to one, this thing is going to be equal, as long as x does not equal one, x minus one over x minus one is just going to be one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "So let's first think about the limit as we approach from the left-hand side of g of x over x minus, g of x over x minus one. Well as we approach from the left-hand side, g of x is that right over there. So we could write this, instead of writing g of x, we could write this as x minus one, x minus one over x minus one. And as long as we aren't equal to one, this thing is going to be equal, as long as x does not equal one, x minus one over x minus one is just going to be one. So this limit is going to be one. So that one worked out. Now let's think about the limit as x approaches one from the right-hand side."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "And as long as we aren't equal to one, this thing is going to be equal, as long as x does not equal one, x minus one over x minus one is just going to be one. So this limit is going to be one. So that one worked out. Now let's think about the limit as x approaches one from the right-hand side. Of, once again, I could write g of x minus g of one, but g of one is just zero, so I'll just write g of x over x minus one. Well what's g of x now? Well it's x minus one squared."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "Now let's think about the limit as x approaches one from the right-hand side. Of, once again, I could write g of x minus g of one, but g of one is just zero, so I'll just write g of x over x minus one. Well what's g of x now? Well it's x minus one squared. So instead of writing g of x, I could write this as x minus one squared over x minus one. And so as long as x does not equal one, and we're just doing the limit, we're saying as we approach one from the right-hand side, well this expression right over here, you have x minus one squared divided by x minus one, well that's just going to give us x minus one. X minus one squared divided by x minus one is just going to be x minus one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "Well it's x minus one squared. So instead of writing g of x, I could write this as x minus one squared over x minus one. And so as long as x does not equal one, and we're just doing the limit, we're saying as we approach one from the right-hand side, well this expression right over here, you have x minus one squared divided by x minus one, well that's just going to give us x minus one. X minus one squared divided by x minus one is just going to be x minus one. And this limit, well this expression right over here is going to be continuous and defined for sure all x's that are not equaling one. Actually let me, well it was before it was this, x minus one squared over x minus one. This thing right over here, as I said, it's not defined for x equals one, but it is defined for anything not, for x does not equal one, and we're just approaching one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "X minus one squared divided by x minus one is just going to be x minus one. And this limit, well this expression right over here is going to be continuous and defined for sure all x's that are not equaling one. Actually let me, well it was before it was this, x minus one squared over x minus one. This thing right over here, as I said, it's not defined for x equals one, but it is defined for anything not, for x does not equal one, and we're just approaching one. And if we wanted to simplify this expression, it would get, this would just be, I think I just did this, but I'm just making sure I'm doing it right. This is going to be the same thing as that for x not being equal to one. Well this is just going to be zero."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "This thing right over here, as I said, it's not defined for x equals one, but it is defined for anything not, for x does not equal one, and we're just approaching one. And if we wanted to simplify this expression, it would get, this would just be, I think I just did this, but I'm just making sure I'm doing it right. This is going to be the same thing as that for x not being equal to one. Well this is just going to be zero. You could just evaluate when x is equal to one here, this is going to be equal to zero. And so notice, you get a different limit for this definition of derivative as we approach from the left-hand side or the right-hand side. And that makes sense."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "Well this is just going to be zero. You could just evaluate when x is equal to one here, this is going to be equal to zero. And so notice, you get a different limit for this definition of derivative as we approach from the left-hand side or the right-hand side. And that makes sense. This graph is going to look something like, we have a slope of one, so it's going to look something like this. And then right when x is equal to one and the value of our function is zero, it looks something like this. It looks something like this."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable)   Khan Academy.mp3", "Sentence": "And that makes sense. This graph is going to look something like, we have a slope of one, so it's going to look something like this. And then right when x is equal to one and the value of our function is zero, it looks something like this. It looks something like this. And so the graph is continuous. The graph for sure is continuous, but our slope coming into that point is one and our slope right when we leave that point is zero. So it is not differentiable over there."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's start with a little bit of a geometric or trigonometric construction that I have here. So this white circle, this is a unit circle. Let me label it as such. So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function?"}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle?"}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about."}, {"video_title": "Limit of sin(x) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta? Well, cosine of zero is just one, and it's a continuous function, so this is just going to be one. So let's see."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "What we want to do in this video is figure out what the limit as x approaches zero of one minus cosine of x over x is equal to. And we're going to assume we know one thing ahead of time. We're going to assume we know that the limit as x approaches zero of sine of x over x, that this is equal to one. And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x?"}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0   Derivative rules   AP Calculus AB   Khan Academy (2).mp3", "Sentence": "Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done. Using that fact and a little bit of trig identities and a little bit of algebraic manipulation, we were able to show that our original limit, the limit as x approaches zero of one minus cosine of x over x is equal to zero."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "What we're going to do in this video is get some practice applying u substitution to definite integrals. So let's say we have the integral. So we're gonna go from x equals one to x equals two, and the integral is two x times x squared plus one to the third power dx. So I already told you that we're gonna apply u substitution, but it's interesting to be able to recognize when to use it. And the key giveaway here is, well I have this x squared plus one business to the third power, but then I also have the derivative of x squared plus one, which is two x right over here. So I could do the substitution. I could say u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero, or just two x. I could write that in differential form."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I already told you that we're gonna apply u substitution, but it's interesting to be able to recognize when to use it. And the key giveaway here is, well I have this x squared plus one business to the third power, but then I also have the derivative of x squared plus one, which is two x right over here. So I could do the substitution. I could say u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero, or just two x. I could write that in differential form. A mathematical hand-wavy way of thinking about it is multiplying both sides by dx. And so you get du is equal to two x dx. And so at least this part of the integral I can rewrite."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could say u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero, or just two x. I could write that in differential form. A mathematical hand-wavy way of thinking about it is multiplying both sides by dx. And so you get du is equal to two x dx. And so at least this part of the integral I can rewrite. So let me at least write, so this is going to be, I'll write the integral. We're gonna think about the bounds in a second. So we have u to the third power."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so at least this part of the integral I can rewrite. So let me at least write, so this is going to be, I'll write the integral. We're gonna think about the bounds in a second. So we have u to the third power. U, do the same orange color. U to the third power. That's this stuff right over here."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have u to the third power. U, do the same orange color. U to the third power. That's this stuff right over here. And then two x times dx. Remember, you could just view this as two x times x squared plus one to the third power times dx. So two x times dx, well two x times dx, that is du."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's this stuff right over here. And then two x times dx. Remember, you could just view this as two x times x squared plus one to the third power times dx. So two x times dx, well two x times dx, that is du. So that and that together, that is du. Now an interesting question, because this isn't an indefinite integral. We're not just trying to find the antiderivative."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So two x times dx, well two x times dx, that is du. So that and that together, that is du. Now an interesting question, because this isn't an indefinite integral. We're not just trying to find the antiderivative. This is a definite integral. So what happens to our bounds of integration? Well, there's two ways that you can approach this."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're not just trying to find the antiderivative. This is a definite integral. So what happens to our bounds of integration? Well, there's two ways that you can approach this. You can change your bounds of integration, because this one is x equals one to x equals two. But now we're integrating with respect to u. So one way, if you want to keep your bounds of integration, or if you want to keep this a definite integral, I guess you could say, you would change your bounds from u is equal to something to u is equal to something else."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, there's two ways that you can approach this. You can change your bounds of integration, because this one is x equals one to x equals two. But now we're integrating with respect to u. So one way, if you want to keep your bounds of integration, or if you want to keep this a definite integral, I guess you could say, you would change your bounds from u is equal to something to u is equal to something else. And so your bounds of integration, let's see, when x is equal to one, what is u? Well, when x is equal to one, you have one squared plus one, so you have two. U is equal to two in that situation."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one way, if you want to keep your bounds of integration, or if you want to keep this a definite integral, I guess you could say, you would change your bounds from u is equal to something to u is equal to something else. And so your bounds of integration, let's see, when x is equal to one, what is u? Well, when x is equal to one, you have one squared plus one, so you have two. U is equal to two in that situation. When x is equal to two, what is u? Well, you have two squared, which is four, plus one, which is five. So u is equal to five."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "U is equal to two in that situation. When x is equal to two, what is u? Well, you have two squared, which is four, plus one, which is five. So u is equal to five. And you won't typically see someone writing u equal to or u equals five. It's often just from two to five, because we're integrating with respect to u. You assume it's u equals two to u equals five."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "So u is equal to five. And you won't typically see someone writing u equal to or u equals five. It's often just from two to five, because we're integrating with respect to u. You assume it's u equals two to u equals five. And so we could just rewrite this as being equal to the integral from two to five of u to the third du. But it's really important to realize why we changed our bounds. We are now integrating with respect to u."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "You assume it's u equals two to u equals five. And so we could just rewrite this as being equal to the integral from two to five of u to the third du. But it's really important to realize why we changed our bounds. We are now integrating with respect to u. And the way we did it is we used our substitution right over here. When x equals one, u is equal to two. When x is equal to two, two squared plus one, u is equal to five."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are now integrating with respect to u. And the way we did it is we used our substitution right over here. When x equals one, u is equal to two. When x is equal to two, two squared plus one, u is equal to five. And then we can just evaluate this right over here. Let's see, this is going to be equal to the antiderivative of u to the third power is u to the fourth over four. We're gonna evaluate that at five and two."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "When x is equal to two, two squared plus one, u is equal to five. And then we can just evaluate this right over here. Let's see, this is going to be equal to the antiderivative of u to the third power is u to the fourth over four. We're gonna evaluate that at five and two. And so this is going to be five to the fourth over four minus two to the fourth over four. And then we could simplify this if we like. But we've just evaluated this definite integral."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna evaluate that at five and two. And so this is going to be five to the fourth over four minus two to the fourth over four. And then we could simplify this if we like. But we've just evaluated this definite integral. Now another way to do it is to think about the, is to try to solve the indefinite integral in terms of x and use u substitution as an intermediate. So one way to think about this is to say, well let's just try to evaluate what the indefinite integral of two x times x squared plus one to the third power dx is. And then whatever this expression ends up being algebraically I'm going to evaluate it at x equals two and at x equals one."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "But we've just evaluated this definite integral. Now another way to do it is to think about the, is to try to solve the indefinite integral in terms of x and use u substitution as an intermediate. So one way to think about this is to say, well let's just try to evaluate what the indefinite integral of two x times x squared plus one to the third power dx is. And then whatever this expression ends up being algebraically I'm going to evaluate it at x equals two and at x equals one. And so then you use the u substitution right over here and you would get, this would simplify using the same substitution as the integral of u, u to the third power du, u to the third power du. And once again you're going to evaluate this whole thing from x equals two and then subtract from that, it evaluated at x equals one. And so this is going to be equal to, well let's see, this is u to the fourth power over four."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then whatever this expression ends up being algebraically I'm going to evaluate it at x equals two and at x equals one. And so then you use the u substitution right over here and you would get, this would simplify using the same substitution as the integral of u, u to the third power du, u to the third power du. And once again you're going to evaluate this whole thing from x equals two and then subtract from that, it evaluated at x equals one. And so this is going to be equal to, well let's see, this is u to the fourth power over four. And once again evaluating it at x equals two and then subtracting from that at x equals one. And then now we can just back substitute. We could say hey look, u is equal to x squared plus one."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this is going to be equal to, well let's see, this is u to the fourth power over four. And once again evaluating it at x equals two and then subtracting from that at x equals one. And then now we can just back substitute. We could say hey look, u is equal to x squared plus one. So this is the same thing as x squared plus one to the fourth power over four. And we're now going to evaluate that at x equals two and at x equals one. And you will notice that you will get the exact same thing."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "We could say hey look, u is equal to x squared plus one. So this is the same thing as x squared plus one to the fourth power over four. And we're now going to evaluate that at x equals two and at x equals one. And you will notice that you will get the exact same thing. When you put x equals two here, you get two squared plus one which is five to the fourth power over four, that right over there. And then minus one squared plus one is two to the fourth power over four, that right over there. So either way you'll get the same result."}, {"video_title": "_-substitution definite integrals   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you will notice that you will get the exact same thing. When you put x equals two here, you get two squared plus one which is five to the fourth power over four, that right over there. And then minus one squared plus one is two to the fourth power over four, that right over there. So either way you'll get the same result. You can either keep it a definite integral and then change your bounds of integration and express them in terms of u, that's one way to do it. The other way is to try to evaluate the indefinite integral. Use u substitution as an intermediary step."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And before I even get into the nitty gritty of it, I really just want to get an intuitive feel for what the second derivative test is telling us. So let me just draw some axes here. So let's say that's my y-axis. Let's say this is my x-axis. And let's say I have a function that has a relative maximum value at x equals c. So let's say we have a situation that looks something like that, and x equals c is right over, so that's the point c, f of c. So, I can draw a straighter dotted line. So that is x being equal to c. And we visually see that we have a local maximum point there, and we can use our calculus tools to think about what's going on there. Well, one thing that we know, we know that the slope of the tangent line, at least the way I've drawn it right over here, is equal to zero."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say this is my x-axis. And let's say I have a function that has a relative maximum value at x equals c. So let's say we have a situation that looks something like that, and x equals c is right over, so that's the point c, f of c. So, I can draw a straighter dotted line. So that is x being equal to c. And we visually see that we have a local maximum point there, and we can use our calculus tools to think about what's going on there. Well, one thing that we know, we know that the slope of the tangent line, at least the way I've drawn it right over here, is equal to zero. So we can say f prime of c is equal to zero. And the other thing we can see is that we are concave downward in the neighborhood around x equals c. So notice our slope is constantly decreasing. And since our slope is, notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative, and even more negative."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, one thing that we know, we know that the slope of the tangent line, at least the way I've drawn it right over here, is equal to zero. So we can say f prime of c is equal to zero. And the other thing we can see is that we are concave downward in the neighborhood around x equals c. So notice our slope is constantly decreasing. And since our slope is, notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative, and even more negative. So we know that f prime prime, we know that f prime prime of c is less than zero. And so, I haven't done any deep mathematical proof here, but if I have a critical point where f prime, where our critical point at x equals c, so f prime of c is equal to zero, and we also see that the second derivative there is less than zero, well, intuitively, this makes sense that we are at a maximum value. And we could go the other way."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And since our slope is, notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative, and even more negative. So we know that f prime prime, we know that f prime prime of c is less than zero. And so, I haven't done any deep mathematical proof here, but if I have a critical point where f prime, where our critical point at x equals c, so f prime of c is equal to zero, and we also see that the second derivative there is less than zero, well, intuitively, this makes sense that we are at a maximum value. And we could go the other way. If we are at a local minimum point at x equals c, or relative minimum point, so our first derivative should still be equal to zero, because our slope of a tangent line right over there is still zero, so f prime of c is equal to zero. But in this second situation, we are concave upwards. The slope is constantly increasing."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we could go the other way. If we are at a local minimum point at x equals c, or relative minimum point, so our first derivative should still be equal to zero, because our slope of a tangent line right over there is still zero, so f prime of c is equal to zero. But in this second situation, we are concave upwards. The slope is constantly increasing. We have an upward opening bowl. And so here, we have a relative minimum value. Or we could say our second derivative is greater than zero."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope is constantly increasing. We have an upward opening bowl. And so here, we have a relative minimum value. Or we could say our second derivative is greater than zero. Visually, we see it's a relative minimum value, and we can tell just looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero, and we are concave upwards. Second derivative is greater than zero. And so this intuition that we hopefully just built up is what the second derivative test tells us."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Or we could say our second derivative is greater than zero. Visually, we see it's a relative minimum value, and we can tell just looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero, and we are concave upwards. Second derivative is greater than zero. And so this intuition that we hopefully just built up is what the second derivative test tells us. So it says, hey, look, if we're dealing with some function f, let's say it's a twice-differentiable function, so that means that over some interval, so that means that you could find its first and second derivatives are defined. And so let's say there's some point, x equals c, where its first derivative is equal to zero. So the slope of the tangent line is equal to zero."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so this intuition that we hopefully just built up is what the second derivative test tells us. So it says, hey, look, if we're dealing with some function f, let's say it's a twice-differentiable function, so that means that over some interval, so that means that you could find its first and second derivatives are defined. And so let's say there's some point, x equals c, where its first derivative is equal to zero. So the slope of the tangent line is equal to zero. And the derivative exists in a neighborhood around c, and most of the functions we deal with, if it's differentiable at c, it tends to be differentiable in the neighborhood around c. And then we also assume that the second derivative exists, it's twice-differentiable. Well then, we might be dealing with a maximum point, we might be dealing with a minimum point, or we might not know what we're dealing with. And it might be neither a minimum or a maximum point."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope of the tangent line is equal to zero. And the derivative exists in a neighborhood around c, and most of the functions we deal with, if it's differentiable at c, it tends to be differentiable in the neighborhood around c. And then we also assume that the second derivative exists, it's twice-differentiable. Well then, we might be dealing with a maximum point, we might be dealing with a minimum point, or we might not know what we're dealing with. And it might be neither a minimum or a maximum point. But using the second derivative test, if we take the second derivative, and if we see that the second derivative is indeed less than zero, then we have a relative maximum point. So this is a situation that we started with right up there. If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it might be neither a minimum or a maximum point. But using the second derivative test, if we take the second derivative, and if we see that the second derivative is indeed less than zero, then we have a relative maximum point. So this is a situation that we started with right up there. If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards. Where the slope is zero, that's the bottom of the bowl, we have a relative minimum point. And if our second derivative is zero, it's inconclusive. We don't know what is actually going on at that point, so we can't make any strong statement."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards. Where the slope is zero, that's the bottom of the bowl, we have a relative minimum point. And if our second derivative is zero, it's inconclusive. We don't know what is actually going on at that point, so we can't make any strong statement. So with that out of the way, let's just do a quick example, just to see if this has gelled. Let's say that I have some twice-differentiable function h. And let's say that I tell you that h of eight is equal to five. I tell you that h prime of eight is equal to zero."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "We don't know what is actually going on at that point, so we can't make any strong statement. So with that out of the way, let's just do a quick example, just to see if this has gelled. Let's say that I have some twice-differentiable function h. And let's say that I tell you that h of eight is equal to five. I tell you that h prime of eight is equal to zero. And I tell you that the second derivative at x equals eight is equal to negative four. So given this, can you tell me whether the point eight comma five, so the point eight comma five, is it a relative, is it a relative minimum, relative minimum maximum point, or not enough info? Not enough info or inconclusive."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "I tell you that h prime of eight is equal to zero. And I tell you that the second derivative at x equals eight is equal to negative four. So given this, can you tell me whether the point eight comma five, so the point eight comma five, is it a relative, is it a relative minimum, relative minimum maximum point, or not enough info? Not enough info or inconclusive. And like always, pause the video and see if you can figure it out. Well, we're assuming it's twice-differentiable. I think it's safe to assume that it's, and we'll, for the sake of this problem, we're gonna assume that the derivative exists in a neighborhood around x equals eight."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Not enough info or inconclusive. And like always, pause the video and see if you can figure it out. Well, we're assuming it's twice-differentiable. I think it's safe to assume that it's, and we'll, for the sake of this problem, we're gonna assume that the derivative exists in a neighborhood around x equals eight. So this example, c is eight, so the point eight five is definitely on the curve. The derivative is equal to zero, so we're dealing potentially with one of these scenarios. And our second derivative is less than zero."}, {"video_title": "Second derivative test   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "I think it's safe to assume that it's, and we'll, for the sake of this problem, we're gonna assume that the derivative exists in a neighborhood around x equals eight. So this example, c is eight, so the point eight five is definitely on the curve. The derivative is equal to zero, so we're dealing potentially with one of these scenarios. And our second derivative is less than zero. Second derivative is less than zero. So this threw us, so the fact that the second derivative, so h prime prime of eight is less than zero, tells us that we fall into this situation right over here. So just with the information they've given us, we can say that at the point eight comma five, we have a relative maximum value, or that this is a relative maximum point for this."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we have listed here are two significant properties of indefinite integrals, and we will see in the future that they are very, very powerful. All this is saying is the indefinite integral of the sum of two different functions is equal to the sum of the indefinite integral of each of those functions. This one right over here says the indefinite integral of a constant, that's not going to be a function of x, of a constant times f of x is the same thing as the constant times the indefinite integral of f of x. So one way to think about it is we took the constant out of the integral, which we'll see in the future. Both of these are very useful techniques. Now, if you're satisfied with them as they are written, then that's fine, you can move on. If you want a little bit of a proof, what I'm going to do here to give an argument for why this is true is use the derivative properties."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "So one way to think about it is we took the constant out of the integral, which we'll see in the future. Both of these are very useful techniques. Now, if you're satisfied with them as they are written, then that's fine, you can move on. If you want a little bit of a proof, what I'm going to do here to give an argument for why this is true is use the derivative properties. Take the derivative of both sides and see that the equality holds once we get rid of the integrals. So let's do that. Let's take the derivative with respect to x of both sides of this, derivative with respect to x."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you want a little bit of a proof, what I'm going to do here to give an argument for why this is true is use the derivative properties. Take the derivative of both sides and see that the equality holds once we get rid of the integrals. So let's do that. Let's take the derivative with respect to x of both sides of this, derivative with respect to x. The left side here, well, this will just become whatever's inside of the indefinite integral. This will just become f of x plus g of x, plus g of x. Now, what would this become?"}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's take the derivative with respect to x of both sides of this, derivative with respect to x. The left side here, well, this will just become whatever's inside of the indefinite integral. This will just become f of x plus g of x, plus g of x. Now, what would this become? Well, we could just go to our derivative properties. The derivative of the sum of two things, that's just the same thing as the sum of the derivatives. So this will be a little bit lengthy."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, what would this become? Well, we could just go to our derivative properties. The derivative of the sum of two things, that's just the same thing as the sum of the derivatives. So this will be a little bit lengthy. So this is going to be the derivative with respect to x of this first part plus the derivative with respect to x of this second part. And so this first part is the integral of f of x dx. We're gonna add it."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this will be a little bit lengthy. So this is going to be the derivative with respect to x of this first part plus the derivative with respect to x of this second part. And so this first part is the integral of f of x dx. We're gonna add it. And then this is the integral of g of x dx. So let me write it down. This is f of x."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're gonna add it. And then this is the integral of g of x dx. So let me write it down. This is f of x. And then this is g of x. Now, what are these things? Well, these things, let me just write this equal sign right over here."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is f of x. And then this is g of x. Now, what are these things? Well, these things, let me just write this equal sign right over here. So in the end, this is going to be equal to, the derivative of this with respect to x is just going to be f of x. And then the derivative with respect to here is just going to be g of x. And this is obviously true."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, these things, let me just write this equal sign right over here. So in the end, this is going to be equal to, the derivative of this with respect to x is just going to be f of x. And then the derivative with respect to here is just going to be g of x. And this is obviously true. So now let's tackle this. Well, let's just do the same thing. Let's take the derivative of both sides."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this is obviously true. So now let's tackle this. Well, let's just do the same thing. Let's take the derivative of both sides. So the derivative with respect to x of that and the derivative with respect to x of that. So the left-hand side will clearly become c times f of x. The right-hand side is going to become, well, we know from our derivative properties, the derivative of a constant times something is the same thing as the constant times the derivative of that something."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's take the derivative of both sides. So the derivative with respect to x of that and the derivative with respect to x of that. So the left-hand side will clearly become c times f of x. The right-hand side is going to become, well, we know from our derivative properties, the derivative of a constant times something is the same thing as the constant times the derivative of that something. So then we have the integral, indefinite integral of f of x dx. And then this thing is just going to be f of x. So this is all going to be equal to c times f of x."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "The right-hand side is going to become, well, we know from our derivative properties, the derivative of a constant times something is the same thing as the constant times the derivative of that something. So then we have the integral, indefinite integral of f of x dx. And then this thing is just going to be f of x. So this is all going to be equal to c times f of x. So once again, you can see that the equality clearly holds. So hopefully this makes you feel good that those properties are true. But the more important thing is that you know when to use it."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is all going to be equal to c times f of x. So once again, you can see that the equality clearly holds. So hopefully this makes you feel good that those properties are true. But the more important thing is that you know when to use it. So for example, if I were to take the integral of, let's say, x squared plus cosine of x, the indefinite integral of that, we now know, and it's going to be useful in the future, say, well, this is the same thing as the integral of x squared dx plus the integral of cosine of x dx. So this is the same thing as that plus that. And then you can separately evaluate them."}, {"video_title": "Indefinite integrals sums & multiples   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the more important thing is that you know when to use it. So for example, if I were to take the integral of, let's say, x squared plus cosine of x, the indefinite integral of that, we now know, and it's going to be useful in the future, say, well, this is the same thing as the integral of x squared dx plus the integral of cosine of x dx. So this is the same thing as that plus that. And then you can separately evaluate them. And this is helpful because we know that if we are trying to figure out the integral of, let's say, pi times sine of x dx, that we can take this constant out. Pi is in no way dependent on x. It's just going to stay being equal to pi."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Alright, let's see if we can find the limit of one over square root of two sine of theta over cosine of two theta as theta approaches negative pi over four. And like always, try to give it a shot before we go through it together. Well, one take on it is, well, let's just say that this is going to be the same thing as the limit as theta approaches negative pi over four of one plus square root of two sine theta theta over the limit as theta approaches negative pi over four, make sure we can see that negative there, of cosine of two theta. And both of these expressions are, if these were function definitions, or if we were to graph y equals one plus square root of two times sine theta, or y equals cosine of two theta, we would get continuous functions, especially at theta is equal to negative pi over four, so we could just substitute in. We say, well, this is going to be equal to, this expression evaluated at negative pi over four, so one plus square root of two times sine of negative pi over four over cosine of two times negative pi over four. Now, negative pi over four, sine of negative pi over four is going to be negative square root of two over two. So this is negative square root of two over two."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And both of these expressions are, if these were function definitions, or if we were to graph y equals one plus square root of two times sine theta, or y equals cosine of two theta, we would get continuous functions, especially at theta is equal to negative pi over four, so we could just substitute in. We say, well, this is going to be equal to, this expression evaluated at negative pi over four, so one plus square root of two times sine of negative pi over four over cosine of two times negative pi over four. Now, negative pi over four, sine of negative pi over four is going to be negative square root of two over two. So this is negative square root of two over two. We're assuming this is in radians. If we're thinking in degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know. And so if you have one, so let's see."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is negative square root of two over two. We're assuming this is in radians. If we're thinking in degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know. And so if you have one, so let's see. Well, actually, let me just rewrite it. So this is going to be equal to one plus square root of two times that is going to be negative two over two. So this is going to be minus one."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so if you have one, so let's see. Well, actually, let me just rewrite it. So this is going to be equal to one plus square root of two times that is going to be negative two over two. So this is going to be minus one. That's the numerator over here. All of this stuff simplifies to negative one over, this is going to be cosine of negative pi over two. All right, this is negative pi over two."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is going to be minus one. That's the numerator over here. All of this stuff simplifies to negative one over, this is going to be cosine of negative pi over two. All right, this is negative pi over two. Cosine of negative pi over two, if you thought in degrees, that's gonna be negative 90 degrees. Well, cosine of that is just going to be zero. So what we end up with is equal to zero over zero."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, this is negative pi over two. Cosine of negative pi over two, if you thought in degrees, that's gonna be negative 90 degrees. Well, cosine of that is just going to be zero. So what we end up with is equal to zero over zero. And as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up. But when we have this indeterminate form, it does not mean that the limit does not exist."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what we end up with is equal to zero over zero. And as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up. But when we have this indeterminate form, it does not mean that the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or is not an indeterminate form, at theta is equal to negative pi over four, and I will see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But when we have this indeterminate form, it does not mean that the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or is not an indeterminate form, at theta is equal to negative pi over four, and I will see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities. And in particular, cosine of two theta seems interesting. Let me write some trig identities involving cosine of two theta. I'll write it over here."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities. And in particular, cosine of two theta seems interesting. Let me write some trig identities involving cosine of two theta. I'll write it over here. So we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta, which is equal to one minus two sine squared of theta, which is equal to two cosine squared theta minus one. And you can go from this one to this one to this one just using the Pythagorean identity, and we proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful?"}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll write it over here. So we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta, which is equal to one minus two sine squared of theta, which is equal to two cosine squared theta minus one. And you can go from this one to this one to this one just using the Pythagorean identity, and we proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful? Well, all of these three are gonna be differences of squares so we can factor them in interesting ways. And remember, our goal at the end of the day is maybe cancel things out that are making us get this zero over zero. And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, do any of these look useful? Well, all of these three are gonna be differences of squares so we can factor them in interesting ways. And remember, our goal at the end of the day is maybe cancel things out that are making us get this zero over zero. And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business. And it looks like this right over here, that can be factored as one plus square root of two sine theta times one minus square root of two sine theta. So let me use this. Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business. And it looks like this right over here, that can be factored as one plus square root of two sine theta times one minus square root of two sine theta. So let me use this. Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares. We can rewrite that as, if this is a squared minus b squared, this is a plus b times a minus b. So I can just replace this with one plus square root of two sine theta times one minus square root of two sine theta. And now we have some nice canceling, or potential canceling that can occur."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares. We can rewrite that as, if this is a squared minus b squared, this is a plus b times a minus b. So I can just replace this with one plus square root of two sine theta times one minus square root of two sine theta. And now we have some nice canceling, or potential canceling that can occur. So we could say that cancels with that. And we could say that that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have one, in the denominator we just are left with one minus square root of two sine theta. And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we have some nice canceling, or potential canceling that can occur. So we could say that cancels with that. And we could say that that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have one, in the denominator we just are left with one minus square root of two sine theta. And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain. So this one right over here, this one we already saw is not defined at theta is equal to negative pi over four. And so this one, in order for these to be equivalent, we have to say that this one is also not. And actually other places, but let's just say theta does not equal negative pi over four."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain. So this one right over here, this one we already saw is not defined at theta is equal to negative pi over four. And so this one, in order for these to be equivalent, we have to say that this one is also not. And actually other places, but let's just say theta does not equal negative pi over four. And we could think about all of this happening in some type of an open interval around negative pi over four if we wanted to get very precise. But if we, for this particular case, well let's just say everything we're doing is in the open interval. So in open interval, in open interval between theta, or say negative one and one."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And actually other places, but let's just say theta does not equal negative pi over four. And we could think about all of this happening in some type of an open interval around negative pi over four if we wanted to get very precise. But if we, for this particular case, well let's just say everything we're doing is in the open interval. So in open interval, in open interval between theta, or say negative one and one. And I think that covers it, because if we have pi over four, that is not going to get us the zero over zero form. And pi over four would make this denominator equal to zero, and it also makes, let's see, pi over four also will make this denominator equal to zero. Because we would get one minus one."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So in open interval, in open interval between theta, or say negative one and one. And I think that covers it, because if we have pi over four, that is not going to get us the zero over zero form. And pi over four would make this denominator equal to zero, and it also makes, let's see, pi over four also will make this denominator equal to zero. Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four. So this is just going to be equal to one over one minus the square root of two times sine of negative pi over four. Sine of negative pi over four. Sine of negative pi over four, which we've already seen is negative square root of two over two."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four. So this is just going to be equal to one over one minus the square root of two times sine of negative pi over four. Sine of negative pi over four. Sine of negative pi over four, which we've already seen is negative square root of two over two. And so this is going to be equal to one over one minus square root of two times the negative square root of two over two. So negative negative, you get a positive. Square root of two times square root of two is two over two is gonna be one."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Sine of negative pi over four, which we've already seen is negative square root of two over two. And so this is going to be equal to one over one minus square root of two times the negative square root of two over two. So negative negative, you get a positive. Square root of two times square root of two is two over two is gonna be one. So this is going to be equal to 1 1 2. And so, I wanna be very clear. This expression is not the same thing as this expression."}, {"video_title": "Trig limit using double angle identity   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Square root of two times square root of two is two over two is gonna be one. So this is going to be equal to 1 1 2. And so, I wanna be very clear. This expression is not the same thing as this expression. Where the same thing at all values of theta, especially if we're dealing in this open interval, except at theta equals negative pi over four, this one is not defined, and this one is defined. But as we've seen multiple times before, if we find a function that is equal to our original, or an expression that's equal to our original expression, at all values of theta, except where the original one was not defined at a certain point, but this new one is defined and is continuous there, well then these two limits are going to be equal. So if this limit is 1 1 2, then this limit is going to be 1 1 2."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Consider the left and right Riemann sums that would approximate the area under y is equal to g of x between x equals two and x equals eight. So we want to approximate this light blue area right over here. Are the approximations overestimations or underestimations? So let's just think about each of them. Let's consider the left and the right Riemann sums. So first the left, and I'm just gonna write left for short, but I'm talking about the left Riemann sum. So they don't tell us how many subdivisions to make for our approximation, so that's up to us to decide."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just think about each of them. Let's consider the left and the right Riemann sums. So first the left, and I'm just gonna write left for short, but I'm talking about the left Riemann sum. So they don't tell us how many subdivisions to make for our approximation, so that's up to us to decide. Let's say we went with three subdivisions. Let's say we wanted to make them equal. They don't have to be, but let's say we do."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So they don't tell us how many subdivisions to make for our approximation, so that's up to us to decide. Let's say we went with three subdivisions. Let's say we wanted to make them equal. They don't have to be, but let's say we do. So the first one would go from two to four, the next one would go from four to six, and the next one would go from six to eight. If we do a left Riemann sum, you use the left side of each of these subdivisions in order to find the height. You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "They don't have to be, but let's say we do. So the first one would go from two to four, the next one would go from four to six, and the next one would go from six to eight. If we do a left Riemann sum, you use the left side of each of these subdivisions in order to find the height. You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles. So we would use g of two to approximate for, or to set the height of our first approximating rectangle, just like that, and then we would use g of four for the next rectangle. So we would be right over there, and then you'd use g of six to represent the height of our third and our final rectangle right over there. Now, when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles. So we would use g of two to approximate for, or to set the height of our first approximating rectangle, just like that, and then we would use g of four for the next rectangle. So we would be right over there, and then you'd use g of six to represent the height of our third and our final rectangle right over there. Now, when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation. Why do we know that? Because these rectangles, the area that they're trying to approximate are always contained in the rectangles, and these rectangles have this surplus area, so they're always going to be larger than the areas that they're trying to approximate. And in general, if you have a function that's decreasing over the interval that we care about right over here, and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate, because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation. Why do we know that? Because these rectangles, the area that they're trying to approximate are always contained in the rectangles, and these rectangles have this surplus area, so they're always going to be larger than the areas that they're trying to approximate. And in general, if you have a function that's decreasing over the interval that we care about right over here, and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate, because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision. And so that's why, for a decreasing function, the left Riemann sum is going to be an overestimation. Now, let's think about the right Riemann sum, and you might already guess it's going to be the opposite, but let's visualize that. So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And in general, if you have a function that's decreasing over the interval that we care about right over here, and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate, because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision. And so that's why, for a decreasing function, the left Riemann sum is going to be an overestimation. Now, let's think about the right Riemann sum, and you might already guess it's going to be the opposite, but let's visualize that. So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height. So for this first rectangle, the height is going to be defined by g of four, so that's right over there. And then for the second one, it's going to be g of six, so that is right over there. And for the third one, it's going to be g of eight."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height. So for this first rectangle, the height is going to be defined by g of four, so that's right over there. And then for the second one, it's going to be g of six, so that is right over there. And for the third one, it's going to be g of eight. And so let me shade these in to make it clear which rectangles we're talking about. This would be the right Riemann sum to approximate the area. And it's very clear here that this is going to be an underestimate, underestimate, because we see in each of these intervals, the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum is a subset of the area that it's trying to estimate."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And for the third one, it's going to be g of eight. And so let me shade these in to make it clear which rectangles we're talking about. This would be the right Riemann sum to approximate the area. And it's very clear here that this is going to be an underestimate, underestimate, because we see in each of these intervals, the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum is a subset of the area that it's trying to estimate. We're not able to, it doesn't capture this extra area right over here. And once again, that is because this is a strictly decreasing function. So if you use the right endpoint of any one of these, or the right side of any of these subdivisions in order to define the height, that right value of g is going to be the lowest value of g in that subdivision."}, {"video_title": "Over- and under-estimation of Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "And it's very clear here that this is going to be an underestimate, underestimate, because we see in each of these intervals, the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum is a subset of the area that it's trying to estimate. We're not able to, it doesn't capture this extra area right over here. And once again, that is because this is a strictly decreasing function. So if you use the right endpoint of any one of these, or the right side of any of these subdivisions in order to define the height, that right value of g is going to be the lowest value of g in that subdivision. And so it's going to be a lower height than what you could even say is the average height of the value of the function over that interval. So you're going to have an underestimate in this situation. Now, if your function was strictly increasing, then these two things would be swapped around."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we have a question here from the 2015 AP Calculus A-B test, and it says, consider the curve given by the equation y to the third minus xy is equal to two. It can be shown that the first derivative of y with respect to x is equal to that, so they solved that for us. And then part c of it, I skipped parts a and b for the sake of this video, evaluate the second derivative of y with respect to x at the point on the curve where x equals negative one and y is equal to one. So pause this video and see if you can do that. All right, now let's do it together. And so let me just first write down the first derivative. So dy, derivative of y with respect to x is equal to y over three y squared minus x."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So pause this video and see if you can do that. All right, now let's do it together. And so let me just first write down the first derivative. So dy, derivative of y with respect to x is equal to y over three y squared minus x. Well, if we're concerning ourselves with the second derivative, well, then we wanna take the derivative with respect to x of both sides of this. So let's just do that, do the derivative operator on both sides right over here. Now on the left-hand side, we of course are going to get the second derivative of y with respect to x."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So dy, derivative of y with respect to x is equal to y over three y squared minus x. Well, if we're concerning ourselves with the second derivative, well, then we wanna take the derivative with respect to x of both sides of this. So let's just do that, do the derivative operator on both sides right over here. Now on the left-hand side, we of course are going to get the second derivative of y with respect to x. But what do we get on the right-hand side? And there's multiple ways to approach this, but for something like this, the quotient rule probably is the best way to tackle it. I sometimes complain about the quotient rule, saying, hey, it's just a variation of the product rule, but it's actually quite useful in something like this."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now on the left-hand side, we of course are going to get the second derivative of y with respect to x. But what do we get on the right-hand side? And there's multiple ways to approach this, but for something like this, the quotient rule probably is the best way to tackle it. I sometimes complain about the quotient rule, saying, hey, it's just a variation of the product rule, but it's actually quite useful in something like this. We just have to remind ourselves that this is going to be equal to the derivative of the numerator with respect to x, and so that's just going to be derivative of y with respect to x times the denominator, three y squared minus x, minus the numerator, y, times the derivative of the denominator with respect to x. Well, what's the derivative of this denominator with respect to x? Well, the derivative of three y squared with respect to x, that's going to be the derivative of three y squared with respect to y, which is just going to be six y, I'm just using the power rule there, times the derivative of y with respect to x."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "I sometimes complain about the quotient rule, saying, hey, it's just a variation of the product rule, but it's actually quite useful in something like this. We just have to remind ourselves that this is going to be equal to the derivative of the numerator with respect to x, and so that's just going to be derivative of y with respect to x times the denominator, three y squared minus x, minus the numerator, y, times the derivative of the denominator with respect to x. Well, what's the derivative of this denominator with respect to x? Well, the derivative of three y squared with respect to x, that's going to be the derivative of three y squared with respect to y, which is just going to be six y, I'm just using the power rule there, times the derivative of y with respect to x. All I did just now is I took the derivative of that with respect to x, which is the derivative of that with respect to y times the derivative of y with respect to x, comes straight out of the chain rule, minus the derivative of this with respect to x, which is just going to be equal to one. All of that over, remember, we're in the middle of the quotient rule right over here. All of that over the denominator squared."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the derivative of three y squared with respect to x, that's going to be the derivative of three y squared with respect to y, which is just going to be six y, I'm just using the power rule there, times the derivative of y with respect to x. All I did just now is I took the derivative of that with respect to x, which is the derivative of that with respect to y times the derivative of y with respect to x, comes straight out of the chain rule, minus the derivative of this with respect to x, which is just going to be equal to one. All of that over, remember, we're in the middle of the quotient rule right over here. All of that over the denominator squared. All of that over three y squared minus x squared. Now lucky for us, they want us to evaluate this at a point as opposed to have to do a bunch of algebraic simplification here. So we can say when, let me do it over here."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that over the denominator squared. All of that over three y squared minus x squared. Now lucky for us, they want us to evaluate this at a point as opposed to have to do a bunch of algebraic simplification here. So we can say when, let me do it over here. So when, I'll do it right here. When x is equal to negative one and y is equal to one, well, first of all, what's dy dx going to be? The derivative of y with respect to x."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can say when, let me do it over here. So when, I'll do it right here. When x is equal to negative one and y is equal to one, well, first of all, what's dy dx going to be? The derivative of y with respect to x. Let me scroll down a little bit so we have a little bit more space. The derivative of y with respect to x is going to be equal to one over three times one squared, which is just three, minus negative one. So that's just going to be plus one."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of y with respect to x. Let me scroll down a little bit so we have a little bit more space. The derivative of y with respect to x is going to be equal to one over three times one squared, which is just three, minus negative one. So that's just going to be plus one. It's going to be equal to 1 1\u2074. And so this whole expression over here, so I can write the second derivative of y with respect to x is going to be equal to, well, we know that that's going to be equal to 1 1\u2074. 1 1\u2074 times three times one squared, which is just three, minus negative one, so plus one, minus one, so I'll just leave that minus out there, times six times one, times 1\u2074."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's just going to be plus one. It's going to be equal to 1 1\u2074. And so this whole expression over here, so I can write the second derivative of y with respect to x is going to be equal to, well, we know that that's going to be equal to 1 1\u2074. 1 1\u2074 times three times one squared, which is just three, minus negative one, so plus one, minus one, so I'll just leave that minus out there, times six times one, times 1\u2074. Let me just write it out. Six, six times one, times 1\u2074, minus one. All of that over, let's see, this is going to be three times y squared."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "1 1\u2074 times three times one squared, which is just three, minus negative one, so plus one, minus one, so I'll just leave that minus out there, times six times one, times 1\u2074. Let me just write it out. Six, six times one, times 1\u2074, minus one. All of that over, let's see, this is going to be three times y squared. Y is one, so this is going to be three, three minus negative one, so plus one, squared. Now what is this going to be? And this is just simplifying something here."}, {"video_title": "Second derivatives (implicit equations) evaluate derivative   AP Calculus AB   Khan Academy.mp3", "Sentence": "All of that over, let's see, this is going to be three times y squared. Y is one, so this is going to be three, three minus negative one, so plus one, squared. Now what is this going to be? And this is just simplifying something here. 1\u2074 times four, that's going to simplify to one. And let's see, this is going to be 1 1\u20442 minus one, so that's going to be 1\u20442. And then we're going to have all of that over 16."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "which slope field is generated by the differential equation the derivative of y with respect to x is equal to x minus y. And like always, pause this video and see if you can figure it out on your own. Well, the easiest way to think about a slope field, if I needed to plot this slope field by hand, I would sample a bunch of x and y points and then I would figure out what the derivative would have to be at that point. And so what we can do here, since they've already drawn some candidate slope fields for us, is figure out what we think the slope field should be at some point and see which of these diagrams, these graphs or these slope fields, actually show that. So let's, let me make a little table here. So I'm gonna have, I'm gonna have x, y, and then the derivative of y with respect to x. And we can do it at a bunch of values, so let's think about it."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so what we can do here, since they've already drawn some candidate slope fields for us, is figure out what we think the slope field should be at some point and see which of these diagrams, these graphs or these slope fields, actually show that. So let's, let me make a little table here. So I'm gonna have, I'm gonna have x, y, and then the derivative of y with respect to x. And we can do it at a bunch of values, so let's think about it. Let's think about when, we're at this point right over here, when x is two and y is two. When x is two and y is two, the derivative of y with respect to x is going to be two minus two. It's going to be equal to zero."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we can do it at a bunch of values, so let's think about it. Let's think about when, we're at this point right over here, when x is two and y is two. When x is two and y is two, the derivative of y with respect to x is going to be two minus two. It's going to be equal to zero. And just with that, let's see, here, this slope on this slope field does not look like it's zero. This looks like it's negative one. So already I could rule this one out."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's going to be equal to zero. And just with that, let's see, here, this slope on this slope field does not look like it's zero. This looks like it's negative one. So already I could rule this one out. This slope right over here looks like it's positive one, so I'll rule that out. It's definitely not zero. This slope also looks like positive one, so I can rule that one out."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So already I could rule this one out. This slope right over here looks like it's positive one, so I'll rule that out. It's definitely not zero. This slope also looks like positive one, so I can rule that one out. This slope at two comma two actually does look like zero, so I'm liking this one right over here. This slope at two comma two looks larger than one, so I could rule that out. So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "This slope also looks like positive one, so I can rule that one out. This slope at two comma two actually does look like zero, so I'm liking this one right over here. This slope at two comma two looks larger than one, so I could rule that out. So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one. But just for kicks, we could keep going to verify that this is indeed the slope field. So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here. When you're at four, four, derivative equals zero."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one. But just for kicks, we could keep going to verify that this is indeed the slope field. So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here. When you're at four, four, derivative equals zero. When it's six, six, derivative equals zero. At negative two, negative two, derivative equals zero. So that feels good that this is the right slope field."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "When you're at four, four, derivative equals zero. When it's six, six, derivative equals zero. At negative two, negative two, derivative equals zero. So that feels good that this is the right slope field. And then we could pick other arbitrary points. Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two. So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that feels good that this is the right slope field. And then we could pick other arbitrary points. Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two. So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here. And if it was the other way around, when x is, when x is, let's say, x is negative four, and y is negative two. So negative four, negative two. Well, negative four minus negative two is going to be negative two."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here. And if it was the other way around, when x is, when x is, let's say, x is negative four, and y is negative two. So negative four, negative two. Well, negative four minus negative two is going to be negative two. And you can see that right over here. Negative four, negative two. You can see the slope right over here."}, {"video_title": "Worked example slope field from equation   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, negative four minus negative two is going to be negative two. And you can see that right over here. Negative four, negative two. You can see the slope right over here. It's a little harder to see. Looks like negative two. So once again, using even just this first two comma two coordinates, we were able to deduce that this was the choice."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "The following graph gives the object's position relative to its starting point over time. For each point on the graph, is the object moving forward, backward, or neither? So pause this video and try to figure that out. All right, so we see we have position in meters versus time. So for example, this point right over here tells us that after one second, we are four meters ahead of our starting point. Or for example, this point right over here says that after four seconds, we are almost, it seems, almost four meters behind our starting point. So let's look at each of these points and think about whether we're moving forward, backward, or neither."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "All right, so we see we have position in meters versus time. So for example, this point right over here tells us that after one second, we are four meters ahead of our starting point. Or for example, this point right over here says that after four seconds, we are almost, it seems, almost four meters behind our starting point. So let's look at each of these points and think about whether we're moving forward, backward, or neither. So at this point right over here, at that moment, we're about two and a half meters in front of our starting point. We're at a positive position of two and a half meters. But as time goes on, we are moving backwards closer and closer to the starting point."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's look at each of these points and think about whether we're moving forward, backward, or neither. So at this point right over here, at that moment, we're about two and a half meters in front of our starting point. We're at a positive position of two and a half meters. But as time goes on, we are moving backwards closer and closer to the starting point. So this is, we are moving backward. One way to think about it, at this time, we're at two and a half meters. If you go forward about half a second, we are then back at our starting point."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "But as time goes on, we are moving backwards closer and closer to the starting point. So this is, we are moving backward. One way to think about it, at this time, we're at two and a half meters. If you go forward about half a second, we are then back at our starting point. So we have to go backwards. And if we look at this point right over here, it looks like we were going backwards this entire time while our curve is downward sloping. But at this point right over here, when we are about, it looks like, five meters behind our starting point, we start going forward again."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "If you go forward about half a second, we are then back at our starting point. So we have to go backwards. And if we look at this point right over here, it looks like we were going backwards this entire time while our curve is downward sloping. But at this point right over here, when we are about, it looks like, five meters behind our starting point, we start going forward again. But right at that moment, we are going neither forward nor backwards. It's right at that moment where we just finished going backwards and we're about to go forward. And one way to think about it is, what would be the slope of the tangent line at that point?"}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "But at this point right over here, when we are about, it looks like, five meters behind our starting point, we start going forward again. But right at that moment, we are going neither forward nor backwards. It's right at that moment where we just finished going backwards and we're about to go forward. And one way to think about it is, what would be the slope of the tangent line at that point? And the slope of the tangent line at that point would be horizontal. And so this is neither. So we can use that same technique to think about this point."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "And one way to think about it is, what would be the slope of the tangent line at that point? And the slope of the tangent line at that point would be horizontal. And so this is neither. So we can use that same technique to think about this point. The slope is positive. And we see that, all right, right at that moment, it looks like we are at the starting point. But if you fast forward even a few, even a fraction of a second, we are now in front of our starting point."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we can use that same technique to think about this point. The slope is positive. And we see that, all right, right at that moment, it looks like we are at the starting point. But if you fast forward even a few, even a fraction of a second, we are now in front of our starting point. So we are moving forward. We are moving forward right over here. And at this point, we are at our starting point."}, {"video_title": "Interpreting direction of motion from position-time graph   AP Calculus AB   Khan Academy.mp3", "Sentence": "But if you fast forward even a few, even a fraction of a second, we are now in front of our starting point. So we are moving forward. We are moving forward right over here. And at this point, we are at our starting point. But if you think about what's going to happen a moment later, a moment later, we're going to be a little bit behind our starting point. And so here we are moving backward. And we're done."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "We'd have to resort to numeric techniques to estimate the solutions. But let's go to what I would argue is the simplest form of differential equation to solve, and that's what's called a separable, separable differential equation. And we will see in a second why it is called a separable differential equation. So let's say that we have the derivative of y with respect to x is equal to negative x over y e to the x squared. So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one. And I encourage you to pause this video, and I'll give you a hint. If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So let's say that we have the derivative of y with respect to x is equal to negative x over y e to the x squared. So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one. And I encourage you to pause this video, and I'll give you a hint. If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it, don't worry, because we're about to work through it. So as I said, let's use a little bit of algebra to get all the y's and d y's on one side, and all the x's and d x's on the other side. So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it, don't worry, because we're about to work through it. So as I said, let's use a little bit of algebra to get all the y's and d y's on one side, and all the x's and d x's on the other side. So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side. Well I can multiply both sides times y. So I can multiply both sides times y. That has the effect of putting the y's on the left hand side."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side. Well I can multiply both sides times y. So I can multiply both sides times y. That has the effect of putting the y's on the left hand side. And then I can multiply both sides times d x. I can multiply both sides times d x. And we kind of treat, you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables. And so this will cancel with that."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "That has the effect of putting the y's on the left hand side. And then I can multiply both sides times d x. I can multiply both sides times d x. And we kind of treat, you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables. And so this will cancel with that. And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way. Let me write it as negative x e, actually I might want a little more space. So negative x e to the negative x squared d x."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "And so this will cancel with that. And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way. Let me write it as negative x e, actually I might want a little more space. So negative x e to the negative x squared d x. D x. Now why is this interesting? Because we can integrate both sides."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So negative x e to the negative x squared d x. D x. Now why is this interesting? Because we can integrate both sides. And now this also highlights why we call this separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Because we can integrate both sides. And now this also highlights why we call this separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side. But this one we were able to. And so that's why this is called a separable differential equation. Differential, differential equation."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side. But this one we were able to. And so that's why this is called a separable differential equation. Differential, differential equation. And it's usually the first technique that you should try. Hey, can I separate the y's and the x's? And as I said, this is not going to be true of many, if not most differential equations."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Differential, differential equation. And it's usually the first technique that you should try. Hey, can I separate the y's and the x's? And as I said, this is not going to be true of many, if not most differential equations. But now that we did this, we can integrate both sides. So let's do that. So I'll find a nice color to integrate with."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "And as I said, this is not going to be true of many, if not most differential equations. But now that we did this, we can integrate both sides. So let's do that. So I'll find a nice color to integrate with. So I'm going to integrate, integrate both sides. Now if you integrate the left-hand side, what do you get? You get, and remember, we're integrating with respect to y here."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So I'll find a nice color to integrate with. So I'm going to integrate, integrate both sides. Now if you integrate the left-hand side, what do you get? You get, and remember, we're integrating with respect to y here. So this is going to be y squared over two. And we could put some constant there. I could call that plus c one."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "You get, and remember, we're integrating with respect to y here. So this is going to be y squared over two. And we could put some constant there. I could call that plus c one. And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x. And let's see, you could do u substitution, or you could recognize that look, the derivative of negative x squared is going to be negative two x. So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "I could call that plus c one. And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x. And let's see, you could do u substitution, or you could recognize that look, the derivative of negative x squared is going to be negative two x. So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there. So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point. So I have something and its derivative, so I really could just integrate with respect to that something, with respect to that u. So this is going to be 1 1 2, this 1 1 2 right over here."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there. So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point. So I have something and its derivative, so I really could just integrate with respect to that something, with respect to that u. So this is going to be 1 1 2, this 1 1 2 right over here. The antiderivative of this is e to the negative x squared, and then of course I might have some other constant. I'll just call that c two. And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So this is going to be 1 1 2, this 1 1 2 right over here. The antiderivative of this is e to the negative x squared, and then of course I might have some other constant. I'll just call that c two. And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece. Now, what can I do here? Well I have a constant on the left-hand side, it's an arbitrary constant, we don't know what it is. I haven't used this initial condition yet, we could call it."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece. Now, what can I do here? Well I have a constant on the left-hand side, it's an arbitrary constant, we don't know what it is. I haven't used this initial condition yet, we could call it. So let me just subtract c one from both sides. So if I just subtract c one from both sides, I have an arbitrary, so this is going to cancel, and I have c two, sorry, let me, so this is c one, so these are going to cancel, and c two minus c one, these are both constants, arbitrary constants, we don't know what they are yet, and so we could just rewrite this as, on the left-hand side we have y squared over two is equal to, on the right-hand side, I'll write one half e, let me write that in blue, just because I wrote it in blue before, one half, one half e to the negative x squared, and I'll just say c two minus c one, let's just call that c. So if you take the sum of those two things, let's just call that c. And so now, this is kind of a general solution, we don't know what this constant is, and we haven't explicitly solved for y yet, but even in this form, we can now find a particular solution using this initial condition. Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "I haven't used this initial condition yet, we could call it. So let me just subtract c one from both sides. So if I just subtract c one from both sides, I have an arbitrary, so this is going to cancel, and I have c two, sorry, let me, so this is c one, so these are going to cancel, and c two minus c one, these are both constants, arbitrary constants, we don't know what they are yet, and so we could just rewrite this as, on the left-hand side we have y squared over two is equal to, on the right-hand side, I'll write one half e, let me write that in blue, just because I wrote it in blue before, one half, one half e to the negative x squared, and I'll just say c two minus c one, let's just call that c. So if you take the sum of those two things, let's just call that c. And so now, this is kind of a general solution, we don't know what this constant is, and we haven't explicitly solved for y yet, but even in this form, we can now find a particular solution using this initial condition. Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition. So it tells us when x is zero, y needs to be equal to one. So we would have one squared, which is just one, over two is equal to one half, e to the negative zero squared, well that's just going to be, e to the zero is just one, so it's going to be one half plus c, and just like that, we're able to figure out if you subtract one half from both sides, c is equal to zero. So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition. So it tells us when x is zero, y needs to be equal to one. So we would have one squared, which is just one, over two is equal to one half, e to the negative zero squared, well that's just going to be, e to the zero is just one, so it's going to be one half plus c, and just like that, we're able to figure out if you subtract one half from both sides, c is equal to zero. So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two. Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared. Now we can take the square root of both sides, and you could say, well look, y squared is equal to this, so y could be equal to the plus or minus square root of e to the negative x squared, of e to the negative x squared, but they gave us an initial condition where y is actually positive. So we're finding the particular solution that goes through this point, that means y is going to be the positive square root."}, {"video_title": "Separable differential equations introduction   First order differential equations   Khan Academy.mp3", "Sentence": "So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two. Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared. Now we can take the square root of both sides, and you could say, well look, y squared is equal to this, so y could be equal to the plus or minus square root of e to the negative x squared, of e to the negative x squared, but they gave us an initial condition where y is actually positive. So we're finding the particular solution that goes through this point, that means y is going to be the positive square root. If this was the point zero, negative one, then we would say y is the negative square root, but we know that y is the positive square root, it's the principal root right over there. Actually let me do that a little bit neater, so we can get rid of, whoops, I thought I was writing in black. So we can get rid of this right over here."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "What I have here in yellow is the graph of y equals f of x. Then here in this mauve color, I've graphed y is equal to the derivative of f, is f prime of x. And then here in blue, I've graphed y is equal to the second derivative of our function. So this is the derivative of this, of the first derivative right over there. And we've already seen examples of how can we identify minimum and maximum points. Obviously, if we have the graph in front of us, it's not hard for a human brain to identify this as a local maximum point. The function might take on higher values later on."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is the derivative of this, of the first derivative right over there. And we've already seen examples of how can we identify minimum and maximum points. Obviously, if we have the graph in front of us, it's not hard for a human brain to identify this as a local maximum point. The function might take on higher values later on. And to identify this as a local minimum point, the function might take on lower values later on. But we saw, even if we don't have the graph in front of us, if we're able to take the derivative of the function, we might, or even if we're not able to take the derivative of the function, we might be able to identify these points as minimum or maximum. The way that we did it, we said, OK, what are the critical points for this function?"}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The function might take on higher values later on. And to identify this as a local minimum point, the function might take on lower values later on. But we saw, even if we don't have the graph in front of us, if we're able to take the derivative of the function, we might, or even if we're not able to take the derivative of the function, we might be able to identify these points as minimum or maximum. The way that we did it, we said, OK, what are the critical points for this function? Well, critical points are where the function's derivative is either undefined or 0. This is the function's derivative. It is 0 here and here."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The way that we did it, we said, OK, what are the critical points for this function? Well, critical points are where the function's derivative is either undefined or 0. This is the function's derivative. It is 0 here and here. So we would call those critical points. I don't see any undefined, any points at which the derivative is undefined just yet. So we would call here and here critical points."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is 0 here and here. So we would call those critical points. I don't see any undefined, any points at which the derivative is undefined just yet. So we would call here and here critical points. So these are candidate minimum, these are candidate points at which our function might take on a minimum or a maximum value. And the way that we figured out whether there was a minimum or maximum value is to look at the behavior of the derivative around that point. And over here, we saw the derivative is positive as we approach that point."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we would call here and here critical points. So these are candidate minimum, these are candidate points at which our function might take on a minimum or a maximum value. And the way that we figured out whether there was a minimum or maximum value is to look at the behavior of the derivative around that point. And over here, we saw the derivative is positive as we approach that point. And then it becomes negative. It goes from being positive to negative as we cross that point, which means that the function was increasing. If the derivative is positive, that means that the function was increasing as we approached that point, and then decreasing as we leave that point, which is a pretty good way to think about this being a maximum point."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And over here, we saw the derivative is positive as we approach that point. And then it becomes negative. It goes from being positive to negative as we cross that point, which means that the function was increasing. If the derivative is positive, that means that the function was increasing as we approached that point, and then decreasing as we leave that point, which is a pretty good way to think about this being a maximum point. If we're increasing as we approach it and decreasing as we leave it, then this is definitely going to be a maximum point. Similarly, right over here, we see that the derivative is negative as we approach the point, which means that the function is decreasing. And we see that the derivative is positive as we exit that point."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "If the derivative is positive, that means that the function was increasing as we approached that point, and then decreasing as we leave that point, which is a pretty good way to think about this being a maximum point. If we're increasing as we approach it and decreasing as we leave it, then this is definitely going to be a maximum point. Similarly, right over here, we see that the derivative is negative as we approach the point, which means that the function is decreasing. And we see that the derivative is positive as we exit that point. We go from having a negative derivative to a positive derivative, which means the function goes from decreasing to increasing right around that point, which is a pretty good indication, or that is the indication, that this critical point is a point at which the function takes on a minimum value. What I want to do now is extend things by using the idea of concavity. And I know I'm mispronouncing it."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we see that the derivative is positive as we exit that point. We go from having a negative derivative to a positive derivative, which means the function goes from decreasing to increasing right around that point, which is a pretty good indication, or that is the indication, that this critical point is a point at which the function takes on a minimum value. What I want to do now is extend things by using the idea of concavity. And I know I'm mispronouncing it. Maybe it's con-cavity. But thinking about concavity, we can start to look at the second derivative rather than seeing just this transition to think about whether this is a minimum or a maximum point. So let's think about what's happening in this first region, this part of the curve up here where it looks like a arc, where it's opening downward, where it looks like an A without the cross beam or an upside down U."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I know I'm mispronouncing it. Maybe it's con-cavity. But thinking about concavity, we can start to look at the second derivative rather than seeing just this transition to think about whether this is a minimum or a maximum point. So let's think about what's happening in this first region, this part of the curve up here where it looks like a arc, where it's opening downward, where it looks like an A without the cross beam or an upside down U. And then we'll think about what's happening in this upward opening U part of the curve. So over this first interval right over here, if we start over here, the slope is very positive. Slope is very positive."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think about what's happening in this first region, this part of the curve up here where it looks like a arc, where it's opening downward, where it looks like an A without the cross beam or an upside down U. And then we'll think about what's happening in this upward opening U part of the curve. So over this first interval right over here, if we start over here, the slope is very positive. Slope is very positive. Then it becomes less positive. Then it becomes even less positive. It eventually gets to 0."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Slope is very positive. Then it becomes less positive. Then it becomes even less positive. It eventually gets to 0. Then it keeps decreasing. Now it becomes slightly negative. Then it becomes even more negative."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "It eventually gets to 0. Then it keeps decreasing. Now it becomes slightly negative. Then it becomes even more negative. Then it becomes even more negative. And then it stops decreasing right around there. So the slope stops decreasing right around there."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then it becomes even more negative. Then it becomes even more negative. And then it stops decreasing right around there. So the slope stops decreasing right around there. And you see that in the derivative. The slope is decreasing, decreasing, decreasing, decreasing until that point. And then it starts to increase."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the slope stops decreasing right around there. And you see that in the derivative. The slope is decreasing, decreasing, decreasing, decreasing until that point. And then it starts to increase. So this entire section right over here, the slope is decreasing. And you see it right over here when we take the derivative. The derivative right over here, over this entire interval, is decreasing."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then it starts to increase. So this entire section right over here, the slope is decreasing. And you see it right over here when we take the derivative. The derivative right over here, over this entire interval, is decreasing. And we also see that when we take the second derivative. If the derivative is decreasing, that means that the second derivative, the derivative of the derivative is negative. And we see that that is indeed the case."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative right over here, over this entire interval, is decreasing. And we also see that when we take the second derivative. If the derivative is decreasing, that means that the second derivative, the derivative of the derivative is negative. And we see that that is indeed the case. Over this entire interval, the second derivative is indeed negative. Now what happens as we start to transition to this upward opening u part of the curve? Well, here the derivative is reasonably negative."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And we see that that is indeed the case. Over this entire interval, the second derivative is indeed negative. Now what happens as we start to transition to this upward opening u part of the curve? Well, here the derivative is reasonably negative. It's reasonably negative right there. But then it starts to get, it's still negative, but it becomes less negative and less negative and less negative, less negative and less negative and less negative. Then it becomes 0 right over here."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, here the derivative is reasonably negative. It's reasonably negative right there. But then it starts to get, it's still negative, but it becomes less negative and less negative and less negative, less negative and less negative and less negative. Then it becomes 0 right over here. And then it becomes more and more and more positive. And you see that right over here. So over this entire interval, the slope or the derivative is increasing."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Then it becomes 0 right over here. And then it becomes more and more and more positive. And you see that right over here. So over this entire interval, the slope or the derivative is increasing. So the slope is in. The slope is increasing. And you see this over here."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So over this entire interval, the slope or the derivative is increasing. So the slope is in. The slope is increasing. And you see this over here. Over here, the slope is 0. The slope of the derivative is 0. The slope, the derivative itself, isn't changing right at this moment."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you see this over here. Over here, the slope is 0. The slope of the derivative is 0. The slope, the derivative itself, isn't changing right at this moment. And then you see that the slope is increasing. And once again, we can visualize that on the second derivative. The derivative of the derivative, if the derivative is increasing, that means the derivative of that must be positive."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The slope, the derivative itself, isn't changing right at this moment. And then you see that the slope is increasing. And once again, we can visualize that on the second derivative. The derivative of the derivative, if the derivative is increasing, that means the derivative of that must be positive. And it is indeed the case that the derivative is positive. And we have a word for this downward opening u and this upward opening u. We call this concave downwards."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "The derivative of the derivative, if the derivative is increasing, that means the derivative of that must be positive. And it is indeed the case that the derivative is positive. And we have a word for this downward opening u and this upward opening u. We call this concave downwards. Let me make this clear. Concave downwards, and we call this concave upwards. So let's review how we can identify concave downward intervals and concave upwards intervals."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "We call this concave downwards. Let me make this clear. Concave downwards, and we call this concave upwards. So let's review how we can identify concave downward intervals and concave upwards intervals. So if we're talking about concave downwards, we see several things. We see that the slope is decreasing, which is another way of saying that f prime of x is decreasing, which is another way of saying that the second derivative must be negative. If the first derivative is decreasing, the second derivative must be negative, which is another way of saying that the second derivative over that interval must be negative."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's review how we can identify concave downward intervals and concave upwards intervals. So if we're talking about concave downwards, we see several things. We see that the slope is decreasing, which is another way of saying that f prime of x is decreasing, which is another way of saying that the second derivative must be negative. If the first derivative is decreasing, the second derivative must be negative, which is another way of saying that the second derivative over that interval must be negative. So if you have a negative second derivative, then you are in a concave downward interval. Similarly, I have trouble saying that word. Let's think about concave upwards, where you have an upward opening u. Concave upwards, in these intervals, the slope is increasing."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "If the first derivative is decreasing, the second derivative must be negative, which is another way of saying that the second derivative over that interval must be negative. So if you have a negative second derivative, then you are in a concave downward interval. Similarly, I have trouble saying that word. Let's think about concave upwards, where you have an upward opening u. Concave upwards, in these intervals, the slope is increasing. We have a negative slope, less negative, less negative, 0, positive, more positive, more positive, even more positive. So slope is increasing, which means that the derivative of the function is increasing. And you see that right over here."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's think about concave upwards, where you have an upward opening u. Concave upwards, in these intervals, the slope is increasing. We have a negative slope, less negative, less negative, 0, positive, more positive, more positive, even more positive. So slope is increasing, which means that the derivative of the function is increasing. And you see that right over here. This derivative is increasing in value, which means that the second derivative over an interval where we are concave upwards must be greater than 0. If the second derivative is greater than 0, that means that the first derivative is increasing, which means that the slope is increasing. We are in a concave upward interval."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you see that right over here. This derivative is increasing in value, which means that the second derivative over an interval where we are concave upwards must be greater than 0. If the second derivative is greater than 0, that means that the first derivative is increasing, which means that the slope is increasing. We are in a concave upward interval. Now, given all of these definitions that we've just given for concave downwards and concave upwards, can we come up with another way of identifying whether a critical point is a minimum point or a maximum point? Well, if you have a maximum point, if you have a critical point where the function is concave downwards, then you're going to be at a maximum point. Concave downwards, let's just be clear here, means that it's opening down like this."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "We are in a concave upward interval. Now, given all of these definitions that we've just given for concave downwards and concave upwards, can we come up with another way of identifying whether a critical point is a minimum point or a maximum point? Well, if you have a maximum point, if you have a critical point where the function is concave downwards, then you're going to be at a maximum point. Concave downwards, let's just be clear here, means that it's opening down like this. And when we're talking about a critical point, if we're assuming it's concave downwards over here, we're assuming differentiability over this interval. And so the critical point is going to be one where the slope is 0. So it's going to be that point right over there."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "Concave downwards, let's just be clear here, means that it's opening down like this. And when we're talking about a critical point, if we're assuming it's concave downwards over here, we're assuming differentiability over this interval. And so the critical point is going to be one where the slope is 0. So it's going to be that point right over there. So if you're concave downwards and you have a point where f prime of, let's say, a is equal to 0, then we have a maximum point at a. We have a maximum point at a. And similarly, if we're concave upwards, that means that our function looks something like this."}, {"video_title": "Concavity introduction   Using derivatives to analyze functions   AP Calculus AB   Khan Academy.mp3", "Sentence": "So it's going to be that point right over there. So if you're concave downwards and you have a point where f prime of, let's say, a is equal to 0, then we have a maximum point at a. We have a maximum point at a. And similarly, if we're concave upwards, that means that our function looks something like this. And if we found a point, obviously a critical point could also be where the function is not defined. But if we're assuming that our first derivative and second derivative is defined here, then the critical point is going to be one where the first derivative is going to be 0. So f prime of a is equal to 0."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So infinity is kind of a strange number. You can't just plug in infinity and see what happens. But if you wanted to evaluate this limit, what you might try to do is just evaluate, if you want to find the limit as this numerator approaches infinity, you put in really large numbers there, you're going to see that it approaches infinity, that the numerator approaches infinity as x approaches infinity. And if you put really large numbers in the denominator, you're going to see that that also, well, not quite infinity, 3x squared will approach infinity, but we're subtracting it. So if you subtract infinity from some non-infinite number, this is going to be negative infinity. So if you were to just kind of evaluate it at infinity, the numerator, you would get positive infinity, the denominator, you would get negative infinity. And that's one of the indeterminate forms that L'Hopital's Rule can be applied to."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And if you put really large numbers in the denominator, you're going to see that that also, well, not quite infinity, 3x squared will approach infinity, but we're subtracting it. So if you subtract infinity from some non-infinite number, this is going to be negative infinity. So if you were to just kind of evaluate it at infinity, the numerator, you would get positive infinity, the denominator, you would get negative infinity. And that's one of the indeterminate forms that L'Hopital's Rule can be applied to. And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule? I know how to do this without L'Hopital's Rule, and you probably do, or you should, and we'll do that in a second. But I just want to show you that L'Hopital's Rule also works for this type of problems, and I really just want to show you an example that had an infinity over a negative or positive infinity indeterminate form."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And that's one of the indeterminate forms that L'Hopital's Rule can be applied to. And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule? I know how to do this without L'Hopital's Rule, and you probably do, or you should, and we'll do that in a second. But I just want to show you that L'Hopital's Rule also works for this type of problems, and I really just want to show you an example that had an infinity over a negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exists, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "But I just want to show you that L'Hopital's Rule also works for this type of problems, and I really just want to show you an example that had an infinity over a negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exists, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x. And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x. And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist, or the rational function of the derivative of this guy divided by the derivative of that guy, if that exists, then this limit's going to be equal to the limit as x approaches infinity of arbitrary switch colors. Derivative of 8x minus 5 is just 8."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist, or the rational function of the derivative of this guy divided by the derivative of that guy, if that exists, then this limit's going to be equal to the limit as x approaches infinity of arbitrary switch colors. Derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be, I mean, this is just a constant here, so it doesn't matter what limit you're approaching. This is just going to equal this value, which is what?"}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be, I mean, this is just a constant here, so it doesn't matter what limit you're approaching. This is just going to equal this value, which is what? If we put it in lowest common form, simplified form, it's negative 4 over 3. So this limit exists. This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This is just going to equal this value, which is what? If we put it in lowest common form, simplified form, it's negative 4 over 3. So this limit exists. This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. So this limit must also equal negative 4 over 3. And by that same argument, that limit also must be equal to negative 4 over 3. And for those of you who say, hey, we already knew how to do this."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. So this limit must also equal negative 4 over 3. And by that same argument, that limit also must be equal to negative 4 over 3. And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right, and I'll show you that right here. Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right, and I'll show you that right here. Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit, so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here to show you that this is equal to that, not to this thing over here."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit, so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Let me draw a little line here to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. Divided by, let's factor an x squared out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. Divided by, let's factor an x squared out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then these x squareds cancel out. This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument, this right here is going to approach 0."}, {"video_title": "L'H\u00f4pital's rule example 2   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument, this right here is going to approach 0. So all you're left with is the 4 and the negative 3. So this is going to be equal to negative 4 over a negative 3 or negative 4 thirds. So you didn't have to use L'Hospital's Rule for this problem."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "For each point on the graph, is the object speeding up, slowing down, or neither? So pause this video and see if you can figure that out. All right, now let's do it together. And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed?"}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed?"}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing."}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed?"}, {"video_title": "Interpreting change in speed from velocity-time graph   Differential Calculus   Khan Academy.mp3", "Sentence": "And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed? Well, the speed here is two meters per second. Remember, it would be the absolute value of the velocity. And the absolute value is actually going down if we forward in time a little bit."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that we have the indefinite integral and the function is three x squared plus two x times e to the x to the third plus x squared dx. So how would we go about solving this? So first when you look at it, it seems like a really complicated integral. We have this polynomial right over here being multiplied by this exponential expression and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "We have this polynomial right over here being multiplied by this exponential expression and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x?"}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. But it is often useful to kind of pretend that it is a fraction."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? You could multiply both sides times dx."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here?"}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared. U is x to the third plus x squared, x to the third plus x squared."}, {"video_title": "_-substitution intro   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared. U is x to the third plus x squared, x to the third plus x squared. And then we have our plus c. And we are done. We have found the antiderivative. And I encourage you to take the derivative of this, and I think you will find yourself using the chain rule and getting right back to what we had over here."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "So let's say we wanted to figure out the limit as x approaches zero of f of g of x, f of g of x. First of all, pause this video and think about whether this theorem even applies. Well, the first thing to think about is what is the limit as x approaches zero of g of x to see if we meet this first condition. So if we look at g of x right over here, as x approaches zero from the left, it looks like g is approaching two. As x approaches zero from the right, it looks like g is approaching two. And so it looks like this is going to be equal to two. So that's a check."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "So if we look at g of x right over here, as x approaches zero from the left, it looks like g is approaching two. As x approaches zero from the right, it looks like g is approaching two. And so it looks like this is going to be equal to two. So that's a check. Now let's see the second condition. Is f continuous at that limit, at two? So when x is equal to two, it does not look like f is continuous."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "So that's a check. Now let's see the second condition. Is f continuous at that limit, at two? So when x is equal to two, it does not look like f is continuous. So we do not meet this second condition right over here. So we can't just directly apply this theorem. But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "So when x is equal to two, it does not look like f is continuous. So we do not meet this second condition right over here. So we can't just directly apply this theorem. But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. For example, in this situation, the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above. And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. For example, in this situation, the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above. And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. And then we can go the other way. If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below. Now, if we approach two from below, it looks like the value of f is approaching zero."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. And then we can go the other way. If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below. Now, if we approach two from below, it looks like the value of f is approaching zero. So in both of these scenarios, our value of our function f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "Now, if we approach two from below, it looks like the value of f is approaching zero. So in both of these scenarios, our value of our function f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example. Let's say we wanted to figure out the limit as x approaches two of f of g of x. Pause this video, and we'll first see if this theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "Now let me give you another example. Let's say we wanted to figure out the limit as x approaches two of f of g of x. Pause this video, and we'll first see if this theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching two from the left, it looks like g is approaching negative two. When we approach x equals two from the right, it looks like g is approaching zero. So our right and left-hand limits are not the same here, so this thing does not exist."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met   AP Calculus   Khan Academy.mp3", "Sentence": "Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching two from the left, it looks like g is approaching negative two. When we approach x equals two from the right, it looks like g is approaching zero. So our right and left-hand limits are not the same here, so this thing does not exist. Does not exist. And so we don't meet this condition right over here, so we can't apply the theorem. But as we've already seen, just because you can't apply the theorem does not mean that the limit does not exist."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "So what I've drawn in red. At the point x equals a. And we've already seen this with the definition of the derivative. We could try to find a general function that gives us the slope of the tangent line at any point. So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "We could try to find a general function that gives us the slope of the tangent line at any point. So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical. Which would be f of x plus h minus f of x over the change in the horizontal."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical. Which would be f of x plus h minus f of x over the change in the horizontal. Which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "Which would be f of x plus h minus f of x over the change in the horizontal. Which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x. This is a function of x. You give me an x, you give me an arbitrary x where the derivative is defined."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x. This is a function of x. You give me an x, you give me an arbitrary x where the derivative is defined. I'm going to plug it into this. Whatever this ends up being, it might be some nice clean algebraic expression. And then I'm going to give you a number."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "You give me an x, you give me an arbitrary x where the derivative is defined. I'm going to plug it into this. Whatever this ends up being, it might be some nice clean algebraic expression. And then I'm going to give you a number. So for example, if you wanted to find, you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "And then I'm going to give you a number. So for example, if you wanted to find, you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now. Blank plus h minus f of blank. All of that over h. I left those blanks so I could write the a in red."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now. Blank plus h minus f of blank. All of that over h. I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "All of that over h. I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way, and this is often used as the alternate form of the derivative, would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point, some here, some place."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "So this is one way to find the slope of the tangent line when x equals a. Another way, and this is often used as the alternate form of the derivative, would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point, some here, some place. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points?"}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "Let's just take another arbitrary point, some here, some place. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. So let me do that in a purple color, over x minus a. Now, how can we get a better and better approximation for the slope of the tangent line here?"}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "And so what's the slope of the secant line between these two points? It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. So let me do that in a purple color, over x minus a. Now, how can we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "Now, how can we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. Either way, we're doing a very similar, we're doing the exact same thing. We're finding, we're taking, we have an expression for the slope of a secant line, and then we're bringing those x values of those points closer and closer together, closer and closer together, so the slopes of those secant lines better and better and better approximate that slope of the tangent line, and at the limit, it does become the slope of the tangent line. That is the derivative of, or that's the definition of the derivative."}, {"video_title": "Formal and alternate form of the derivative   Differential Calculus   Khan Academy.mp3", "Sentence": "So we would want to take the limit as x approaches a here. Either way, we're doing a very similar, we're doing the exact same thing. We're finding, we're taking, we have an expression for the slope of a secant line, and then we're bringing those x values of those points closer and closer together, closer and closer together, so the slopes of those secant lines better and better and better approximate that slope of the tangent line, and at the limit, it does become the slope of the tangent line. That is the derivative of, or that's the definition of the derivative. So this is the kind of the more standard definition of a derivative. It would give you your derivative as a function of x, and then you can then input your x that you want, the particular value of x, or you could use the alternate form of the derivative. If you know that, hey, look, I'm just looking to find the derivative exactly at a, I need a general function of f, then you could do this, but they're doing the same thing."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what I want to do in this video is introduce a third value, c, that is in between a and b. And it could be equal to a or it could be equal to b. So let me just introduce it. Write it just like that. And I could write that a is less than or equal to c, which is less than or equal to b. And what I want to think about is, how does this definite integral relate to the definite integral from a to c and the definite integral from c to b? So let's think through that."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Write it just like that. And I could write that a is less than or equal to c, which is less than or equal to b. And what I want to think about is, how does this definite integral relate to the definite integral from a to c and the definite integral from c to b? So let's think through that. So we have the definite integral from a to c of f of x. Actually, I've already used that purple color for the function itself. So I'm going to use green."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let's think through that. So we have the definite integral from a to c of f of x. Actually, I've already used that purple color for the function itself. So I'm going to use green. So we have the integral from a to c of f of x dx. And that, of course, is going to represent this area right over here, from a to c under the curve f of x, above the x axis. So that's that."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm going to use green. So we have the integral from a to c of f of x dx. And that, of course, is going to represent this area right over here, from a to c under the curve f of x, above the x axis. So that's that. And then we could have the integral from c to b of f of x dx. And that, of course, is going to represent this area right over here. Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that's that. And then we could have the integral from c to b of f of x dx. And that, of course, is going to represent this area right over here. Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. So this is just equal to that plus that over there. And once again, you might say, why is this integration property useful? That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way?"}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. So this is just equal to that plus that over there. And once again, you might say, why is this integration property useful? That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? Well, as you'll see, this is really useful. It can be very useful when you're looking at functions that have discontinuities. If they have step functions, you can break up the larger integral into smaller integrals."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? Well, as you'll see, this is really useful. It can be very useful when you're looking at functions that have discontinuities. If they have step functions, you can break up the larger integral into smaller integrals. You'll also see that this is useful when we prove the fundamental theorem of calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an integral where it might be very useful to utilize that property."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "If they have step functions, you can break up the larger integral into smaller integrals. You'll also see that this is useful when we prove the fundamental theorem of calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an integral where it might be very useful to utilize that property. So if this is a, this is b, and let's say the function, I'm just going to make it constant over an interval. So it's constant from there to there, and then it drops down from there to there. Let's say the function looked like this."}, {"video_title": "Breaking up integral interval   Accumulation and Riemann sums   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me actually draw an integral where it might be very useful to utilize that property. So if this is a, this is b, and let's say the function, I'm just going to make it constant over an interval. So it's constant from there to there, and then it drops down from there to there. Let's say the function looked like this. Well, you could say that the larger integral, which would be the area under the curve, it would be all of this. Let's just say it's a gap right there, or it jumps down there. So this entire area, you can break up into two smaller areas."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "What I encourage you in this video is to pause it and try to do the same type of proof for the derivative of the inverse cosine of x. So what is, so our goal here is to figure out, I want the derivative with respect to x of the inverse cosine of x. What is this going to be equal to? So assuming you've had a go at it, let's work through it. So just like last time, we could write, let's just set y being equal to this. Y is equal to the inverse cosine of x, which means the same thing as saying that x is equal to the cosine of y. Now let's take the derivative of both sides with respect to x."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So assuming you've had a go at it, let's work through it. So just like last time, we could write, let's just set y being equal to this. Y is equal to the inverse cosine of x, which means the same thing as saying that x is equal to the cosine of y. Now let's take the derivative of both sides with respect to x. On the left-hand side, you're just going to have a one. We're just going to have a one. And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Now let's take the derivative of both sides with respect to x. On the left-hand side, you're just going to have a one. We're just going to have a one. And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x. And we know that x is cosine of y. So let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine of y. Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "We want it in terms of x. And we know that x is cosine of y. So let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine of y. Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. We know that sine of y is equal to the square root of one minus cosine squared of y. So this is equal to negative one. This is just a manipulation of the Pythagorean trig identity."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. We know that sine of y is equal to the square root of one minus cosine squared of y. So this is equal to negative one. This is just a manipulation of the Pythagorean trig identity. This is equal to one minus cosine. I could write it like this, cosine squared of y, but I'll write it like this, because it'll make it a little bit clearer. And what is cosine of y?"}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This is just a manipulation of the Pythagorean trig identity. This is equal to one minus cosine. I could write it like this, cosine squared of y, but I'll write it like this, because it'll make it a little bit clearer. And what is cosine of y? Well, of course, that is x. So this is equal to negative one over the square root of one minus, instead of writing cosine y, instead of writing cosine y, I'm trying to switch colors, instead of writing cosine y, we could write one minus x, one minus x squared. So there you have it."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "And what is cosine of y? Well, of course, that is x. So this is equal to negative one over the square root of one minus, instead of writing cosine y, instead of writing cosine y, I'm trying to switch colors, instead of writing cosine y, we could write one minus x, one minus x squared. So there you have it. The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared. So this is a neat thing. This right over here is a neat thing to know."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "So there you have it. The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared. So this is a neat thing. This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sine. So let me copy and paste that."}, {"video_title": "Derivative of inverse cosine   Taking derivatives   Differential Calculus   Khan Academy.mp3", "Sentence": "This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sine. So let me copy and paste that. So copy and paste it. I'm gonna paste it down here. And now let's look at them side by side."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So we'll now think about the extreme value theorem, which we'll see is a bit of common sense. But in all of these theorems, it's always fun to think about the edge cases. Why is it laid out the way it is? And that might give us a little bit of more intuition about it. So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval's from a to b. And when we say a closed interval, that means we include the endpoints a and b. That's why we have these brackets here instead of parentheses."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And that might give us a little bit of more intuition about it. So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval's from a to b. And when we say a closed interval, that means we include the endpoints a and b. That's why we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit and this probably is pretty intuitive for you. You're probably saying, well, why do they even have to write a theorem here?"}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "That's why we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit and this probably is pretty intuitive for you. You're probably saying, well, why do they even have to write a theorem here? And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're probably saying, well, why do they even have to write a theorem here? And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis. That's my y-axis. And let's draw the interval. So the interval's from a to b."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is my x-axis. That's my y-axis. And let's draw the interval. So the interval's from a to b. So let's say that this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So the interval's from a to b. So let's say that this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a. And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So that is f of a. And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this. Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's say the function does something like this. Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval. So there are members of this set that are in the interval such that, and I'm just using the logical notation here, such that f of c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval. So there are members of this set that are in the interval such that, and I'm just using the logical notation here, such that f of c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values. Now, one thing, and we could draw other continuous functions. And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Just like that. So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values. Now, one thing, and we could draw other continuous functions. And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. And just you could draw a bunch of functions here that are continuous over this closed interval. Here, our maximum point happens right when we hit b, and our minimum point happens at a. For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. And just you could draw a bunch of functions here that are continuous over this closed interval. Here, our maximum point happens right when we hit b, and our minimum point happens at a. For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval. So first, let's think about why does f need to be continuous. Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval. So first, let's think about why does f need to be continuous. Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually, pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult, or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well, let's see."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually, pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult, or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well, let's see. Let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, let's see. Let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a. That's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that's a. That's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value. Let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here, you could say, well, look."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say our function did something right where you would have expected to have a maximum value. Let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here, you could say, well, look. The function is clearly approaching. As x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maximum value because the function never gets to that."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so right over here, you could say, well, look. The function is clearly approaching. As x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maximum value because the function never gets to that. So you could say, well, let's get a little closer here. Maybe this number right over here is 5. So you could say maybe the maximum value is 4.9."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "But that limit can't be the maximum value because the function never gets to that. So you could say, well, let's get a little closer here. Maybe this number right over here is 5. So you could say maybe the maximum value is 4.9. But then you could get your x even closer to this value and say your y be 4.99 or 4.999. You could keep adding another 9. So there is no maximum value."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So you could say maybe the maximum value is 4.9. But then you could get your x even closer to this value and say your y be 4.99 or 4.999. You could keep adding another 9. So there is no maximum value. Similar here on the minimum. Let me draw it a little bit so it looks more like a minimum. You can get closer and closer to it, but there's no minimum."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "So there is no maximum value. Similar here on the minimum. Let me draw it a little bit so it looks more like a minimum. You can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1. So you could get to 1.1 or 1.01 or 1.0001. And so you could keep drawing some 0's between the two 1's, but there is no absolute minimum value there."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "You can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1. So you could get to 1.1 or 1.01 or 1.0001. And so you could keep drawing some 0's between the two 1's, but there is no absolute minimum value there. Now let's think about why it being a closed interval matters, why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well, let's imagine that it was an open interval. Let's imagine an open interval."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And so you could keep drawing some 0's between the two 1's, but there is no absolute minimum value there. Now let's think about why it being a closed interval matters, why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well, let's imagine that it was an open interval. Let's imagine an open interval. And sometimes if we want to be particular, we could make this is the closed interval right over here, brackets. And if we want to do an open interval right over here, that's a, that's b. And let's just pick a very simple function."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's imagine an open interval. And sometimes if we want to be particular, we could make this is the closed interval right over here, brackets. And if we want to do an open interval right over here, that's a, that's b. And let's just pick a very simple function. Let's say a function like this. So right over here, if a were in our interval, it looks like we hit our minimum value at a. f of a would have been our minimum value. And f of b looks like it would have been our maximum value."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And let's just pick a very simple function. Let's say a function like this. So right over here, if a were in our interval, it looks like we hit our minimum value at a. f of a would have been our minimum value. And f of b looks like it would have been our maximum value. So f of a would have been our maximum value. But we're not including a and b in the interval. This is an open interval."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "And f of b looks like it would have been our maximum value. So f of a would have been our maximum value. But we're not including a and b in the interval. This is an open interval. So you can keep getting closer and closer and closer to b, and keep getting higher and higher and higher values without ever quite getting to b. Because once again, we're not including the point b. Similarly, you could get closer and closer and closer to a and get smaller and smaller values, but a is not included in your set, in your consideration."}, {"video_title": "Extreme value theorem   Existence theorems   AP Calculus AB   Khan Academy.mp3", "Sentence": "This is an open interval. So you can keep getting closer and closer and closer to b, and keep getting higher and higher and higher values without ever quite getting to b. Because once again, we're not including the point b. Similarly, you could get closer and closer and closer to a and get smaller and smaller values, but a is not included in your set, in your consideration. So f of a cannot be your minimum value. So that's, on one level, it's kind of a very intuitive, almost obvious theorem. But on the other hand, it is nice to know why they had to say continuous and why they had to say a closed interval like this."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So let's think about what they're saying. So we're looking for a place where the first derivative is greater than zero. That means that the slope of the tangent line is positive. That means that the function is increasing over that interval. So if we just think about it here, over this whole region right over here, the function is clearly decreasing. Then the slope becomes zero right over here. And then the function starts increasing again, all the way until this point right over here."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "That means that the function is increasing over that interval. So if we just think about it here, over this whole region right over here, the function is clearly decreasing. Then the slope becomes zero right over here. And then the function starts increasing again, all the way until this point right over here. It hits zero, and then it goes, and the function starts decreasing. So we're going to, just this first constraint right over here, tells us it's going to be something in this interval right over there. And then they say where the second derivative is less than zero."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then the function starts increasing again, all the way until this point right over here. It hits zero, and then it goes, and the function starts decreasing. So we're going to, just this first constraint right over here, tells us it's going to be something in this interval right over there. And then they say where the second derivative is less than zero. So this means that the slope itself, whether it's positive or negative, that it's actually decreasing. We are going to be concave downwards right over here. The slope itself, it could be positive, but it'll be coming less and less and less positive."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "And then they say where the second derivative is less than zero. So this means that the slope itself, whether it's positive or negative, that it's actually decreasing. We are going to be concave downwards right over here. The slope itself, it could be positive, but it'll be coming less and less and less positive. And so we're looking for a place where the slope is positive, but it's becoming less and less and less positive. If you look right over here, the slope is positive, but the slope is increasing. It's getting steeper and steeper and steeper as we go."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "The slope itself, it could be positive, but it'll be coming less and less and less positive. And so we're looking for a place where the slope is positive, but it's becoming less and less and less positive. If you look right over here, the slope is positive, but the slope is increasing. It's getting steeper and steeper and steeper as we go. And then all of a sudden, it starts getting less steep, less steep, less steep, less steep, all the way to when the slope gets back to zero. So if we want to select an interval, it would be this interval right over here. Our slope is positive."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "It's getting steeper and steeper and steeper as we go. And then all of a sudden, it starts getting less steep, less steep, less steep, less steep, all the way to when the slope gets back to zero. So if we want to select an interval, it would be this interval right over here. Our slope is positive. Our function is clearly increasing. But it is increasing at a lower and lower rate. So I will select that right over there."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Our slope is positive. Our function is clearly increasing. But it is increasing at a lower and lower rate. So I will select that right over there. Let's do one more example. A function f of x is plotted below. Highlight an interval where f prime of x is greater than zero."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "So I will select that right over there. Let's do one more example. A function f of x is plotted below. Highlight an interval where f prime of x is greater than zero. So same thing, where our function is increasing, but it's increasing at a slower and slower rate. So our function is increasing. So our function is increasing in this whole region right over here."}, {"video_title": "Recognizing concavity exercise   Derivative applications   Differential Calculus   Khan Academy.mp3", "Sentence": "Highlight an interval where f prime of x is greater than zero. So same thing, where our function is increasing, but it's increasing at a slower and slower rate. So our function is increasing. So our function is increasing in this whole region right over here. And we see it's really steep here. Then it's getting less steep and less steep. And it's getting closer and closer to zero, the slope of the tangent line or the rate of increase of the function."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "And what I would like you to do is pause this video and figure out what the average acceleration is of this particle over the interval, the closed interval from t is equal to one to t is equal to two. What is this going to be equal to? So assuming you've given a go at it, and the first thing you might have realized is we're trying to take the average value of a function that we don't know explicitly yet. We know the position function, but not the acceleration function. But luckily, we also know that the acceleration function is the derivative with respect to time of the velocity function, which is the derivative with respect to time of the position function. So the acceleration function is the second derivative of this, and then we have to just find its average value over this interval. So let's do that."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "We know the position function, but not the acceleration function. But luckily, we also know that the acceleration function is the derivative with respect to time of the velocity function, which is the derivative with respect to time of the position function. So the acceleration function is the second derivative of this, and then we have to just find its average value over this interval. So let's do that. Let's take the derivative of this twice. And before we do it, let me just even rewrite this so it's going to be a little bit easier to differentiate it. So if we just take each of these two terms of the numerator and divide them by t squared, we're going to get t to the third divided by t squared is just t, and then two divided by t squared, we could write that as plus two t to the negative two power."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So let's do that. Let's take the derivative of this twice. And before we do it, let me just even rewrite this so it's going to be a little bit easier to differentiate it. So if we just take each of these two terms of the numerator and divide them by t squared, we're going to get t to the third divided by t squared is just t, and then two divided by t squared, we could write that as plus two t to the negative two power. And now let's take the derivative. So the velocity function, the velocity as a function of time, just the derivative of this with respect to time. So it's going to be derivative of t with respect to t is one."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So if we just take each of these two terms of the numerator and divide them by t squared, we're going to get t to the third divided by t squared is just t, and then two divided by t squared, we could write that as plus two t to the negative two power. And now let's take the derivative. So the velocity function, the velocity as a function of time, just the derivative of this with respect to time. So it's going to be derivative of t with respect to t is one. Derivative of two t to the negative two, let's see, negative two times positive two is negative four t to the, and we just decremented the exponent here, t to the negative negative three power. Now, to find acceleration as a function of time, we just take the derivative of this with respect to time. So acceleration as a function of time is equal to, actually since I've already used that color for the average, let me do a different color now."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So it's going to be derivative of t with respect to t is one. Derivative of two t to the negative two, let's see, negative two times positive two is negative four t to the, and we just decremented the exponent here, t to the negative negative three power. Now, to find acceleration as a function of time, we just take the derivative of this with respect to time. So acceleration as a function of time is equal to, actually since I've already used that color for the average, let me do a different color now. So acceleration as a function of time is just the derivative of this with respect to t. So derivative of a constant with respect to time, well, it's not changing, so it's a zero, and then over here, negative three times negative four is positive 12 times t to the, let's decrement that exponent, to the negative four power. Now, to find the average value, all we have to do now, average value, is essentially take the definite integral of this over the interval and divide that by the width of the interval. So, or we could say, we could take, we could divide by the width of the interval, one over two minus one, and this all simplifies to one, times the definite integral over the interval."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So acceleration as a function of time is equal to, actually since I've already used that color for the average, let me do a different color now. So acceleration as a function of time is just the derivative of this with respect to t. So derivative of a constant with respect to time, well, it's not changing, so it's a zero, and then over here, negative three times negative four is positive 12 times t to the, let's decrement that exponent, to the negative four power. Now, to find the average value, all we have to do now, average value, is essentially take the definite integral of this over the interval and divide that by the width of the interval. So, or we could say, we could take, we could divide by the width of the interval, one over two minus one, and this all simplifies to one, times the definite integral over the interval. So one to two of a of t, which is, so this can be 12t to the negative four power dt. So what does this simplify to? Once again, this is one over one."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So, or we could say, we could take, we could divide by the width of the interval, one over two minus one, and this all simplifies to one, times the definite integral over the interval. So one to two of a of t, which is, so this can be 12t to the negative four power dt. So what does this simplify to? Once again, this is one over one. That's just going to be one. If we take the antiderivative of this, and then we actually, well, let me just, so this is going to be equal to, the antiderivative of this is, so we're gonna go t to the negative three power, but then we divide by negative three. So an antiderivative of this is going to be, if we don't take the, well, an antiderivative is going to be negative four t to the negative three power, and we saw that over here."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "Once again, this is one over one. That's just going to be one. If we take the antiderivative of this, and then we actually, well, let me just, so this is going to be equal to, the antiderivative of this is, so we're gonna go t to the negative three power, but then we divide by negative three. So an antiderivative of this is going to be, if we don't take the, well, an antiderivative is going to be negative four t to the negative three power, and we saw that over here. Obviously, if you were really just taking an indefinite integral, we would have to put some constant here, but in the definite integral, even if we put a constant here, it would get, assuming the same constant, it would get canceled out when you actually do the calculation. But the antiderivative of this, we increment the exponent, and then we divide by that new exponent. So 12 divided by negative three is negative four, and we're going to evaluate that from at two and at one."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So an antiderivative of this is going to be, if we don't take the, well, an antiderivative is going to be negative four t to the negative three power, and we saw that over here. Obviously, if you were really just taking an indefinite integral, we would have to put some constant here, but in the definite integral, even if we put a constant here, it would get, assuming the same constant, it would get canceled out when you actually do the calculation. But the antiderivative of this, we increment the exponent, and then we divide by that new exponent. So 12 divided by negative three is negative four, and we're going to evaluate that from at two and at one. And so this is going to be equal to, when we evaluate it at two, at the upper bound of our interval, it's gonna be negative four times two to the negative three power. So it's negative four times, what is that, two, that's one over two to the third, times 1 1 8th is one way to think about that. And then we're gonna have minus this evaluated at one."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "So 12 divided by negative three is negative four, and we're going to evaluate that from at two and at one. And so this is going to be equal to, when we evaluate it at two, at the upper bound of our interval, it's gonna be negative four times two to the negative three power. So it's negative four times, what is that, two, that's one over two to the third, times 1 1 8th is one way to think about that. And then we're gonna have minus this evaluated at one. So minus negative four times t to the negative three, or one to the negative three is just one. So it's just gonna be negative four times one. And this is going to be equal to, we're really in the home stretch now, this is equal to, this part right over here is negative 1 1 2."}, {"video_title": "Average acceleration over interval   AP Calculus BC   Khan Academy.mp3", "Sentence": "And then we're gonna have minus this evaluated at one. So minus negative four times t to the negative three, or one to the negative three is just one. So it's just gonna be negative four times one. And this is going to be equal to, we're really in the home stretch now, this is equal to, this part right over here is negative 1 1 2. So this is negative 1 1 2. And this part right over here is positive four. So positive four minus 1 1 2, we could either write that as three and a half, or if we wanted to write it as an improper fraction, we could write this as seven halves."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "We've done several videos already where we're approximating the area under a curve by breaking up that area into rectangles and then finding the sum of the areas of those rectangles as an approximation. This was actually the first example that we looked at where each of the rectangles had an equal width. So we equally partitioned the interval between our two boundaries, between a and b. And the height of the rectangle was the function evaluated at the left endpoint of each rectangle. And we wanted to generalize it and write it in sigma notation. It looked something like this. And this was one case."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "And the height of the rectangle was the function evaluated at the left endpoint of each rectangle. And we wanted to generalize it and write it in sigma notation. It looked something like this. And this was one case. Later on we looked at a situation where you define the height by the function evaluated at the right endpoint or at the midpoint. And then we even constructed trapezoids. And these are all particular instances of Riemann sums."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "And this was one case. Later on we looked at a situation where you define the height by the function evaluated at the right endpoint or at the midpoint. And then we even constructed trapezoids. And these are all particular instances of Riemann sums. So this right over here is a Riemann sum. And when people talk about Riemann sums, they're talking about the more general notion. You don't have to just do it this way."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "And these are all particular instances of Riemann sums. So this right over here is a Riemann sum. And when people talk about Riemann sums, they're talking about the more general notion. You don't have to just do it this way. You could use trapezoids. You don't even have to have equally spaced partitions. I used equally spaced partitions because it made things a little bit conceptually simpler."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "You don't have to just do it this way. You could use trapezoids. You don't even have to have equally spaced partitions. I used equally spaced partitions because it made things a little bit conceptually simpler. And this right here is a picture of the person that Riemann sums was named after. This is Bernard Riemann. And he made many contributions to mathematics."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "I used equally spaced partitions because it made things a little bit conceptually simpler. And this right here is a picture of the person that Riemann sums was named after. This is Bernard Riemann. And he made many contributions to mathematics. But what he's most known for, at least if you're taking a first year calculus course, is the Riemann sum and how this is used to define the Riemann integral. Both Newton and Leibniz had come up with the idea of the integral when they had formulated calculus. But the Riemann integral is kind of the most mainstream formal or I would say rigorous definition of what an integral is."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "And he made many contributions to mathematics. But what he's most known for, at least if you're taking a first year calculus course, is the Riemann sum and how this is used to define the Riemann integral. Both Newton and Leibniz had come up with the idea of the integral when they had formulated calculus. But the Riemann integral is kind of the most mainstream formal or I would say rigorous definition of what an integral is. So as you can imagine, this is one instance of a Riemann sum. We have n right over here. The larger n is, the better an approximation it's going to be."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "But the Riemann integral is kind of the most mainstream formal or I would say rigorous definition of what an integral is. So as you can imagine, this is one instance of a Riemann sum. We have n right over here. The larger n is, the better an approximation it's going to be. So his definition of an integral, which is the actual area under the curve, or his definition of a definite integral, which is the actual area under a curve between a and b, is to take this Riemann sum, it doesn't have to be this one, take any Riemann sum, and take the limit as n approaches infinity. So just to be clear, what's happening when n approaches infinity? Let me draw another diagram here."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "The larger n is, the better an approximation it's going to be. So his definition of an integral, which is the actual area under the curve, or his definition of a definite integral, which is the actual area under a curve between a and b, is to take this Riemann sum, it doesn't have to be this one, take any Riemann sum, and take the limit as n approaches infinity. So just to be clear, what's happening when n approaches infinity? Let me draw another diagram here. So let's say that's my y-axis, this is my x-axis, this is my function. As n approaches infinity, so this is a, this is b, you're just going to have a ton of rectangles. You're just going to get a ton of rectangles over there and they're going to become better and better approximations for the actual area."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let me draw another diagram here. So let's say that's my y-axis, this is my x-axis, this is my function. As n approaches infinity, so this is a, this is b, you're just going to have a ton of rectangles. You're just going to get a ton of rectangles over there and they're going to become better and better approximations for the actual area. And the actual area under the curve is denoted by the integral from a to b of f of x times dx. And just to make it, you see where this is coming from, or how these notations are close, or at least in my brain, how they're connected, delta x was the distance for each of these, was the width for each of these sections. This right here is delta x, so that is a delta x."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "You're just going to get a ton of rectangles over there and they're going to become better and better approximations for the actual area. And the actual area under the curve is denoted by the integral from a to b of f of x times dx. And just to make it, you see where this is coming from, or how these notations are close, or at least in my brain, how they're connected, delta x was the distance for each of these, was the width for each of these sections. This right here is delta x, so that is a delta x. This is another delta x, this is another delta x. A reasonable way to conceptualize what dx is, or what a differential is, is what delta x approaches if it becomes infinitely small. So you can view this, you can conceptualize this, and it's not a very rigorous way of thinking about it, it's an infinitely small delta x."}, {"video_title": "Definite integral as the limit of a Riemann sum   AP Calculus AB   Khan Academy.mp3", "Sentence": "This right here is delta x, so that is a delta x. This is another delta x, this is another delta x. A reasonable way to conceptualize what dx is, or what a differential is, is what delta x approaches if it becomes infinitely small. So you can view this, you can conceptualize this, and it's not a very rigorous way of thinking about it, it's an infinitely small delta x. It's one way that you can conceptualize this. So once again, you have your function times a little small change in delta x, and you are summing, although you're summing an infinite number of these things from a to b. So I'm going to leave you there, just so you see the connection, you know the name for these things, and once again, this one over here, this isn't the only Riemann sum, in fact this is often called the left Riemann sum, if you're using it with rectangles, you can do a right Riemann sum, you could use the midpoint, you could use a trapezoid, but if you take the limit of any of those Riemann sums, as n approaches infinity, then that you get as a Riemann definition of the integral."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Now what I want to explore in this video, and it'll come up with kind of an answer that you probably could have guessed on your own, but at least get an intuition for it, is I want to start thinking about the area under the curve that's a scaled version of f of x. Let's say it's y is equal to c times f of x. Y is equal to some number times f of x, so it's scaling f of x. And so I want this to be kind of some arbitrary number, but just to help me visualize, you know, I have to draw something. So I'm just gonna kind of in my head, let's just pretend like the c is a three, just for visualization purposes. So it's gonna be three times this all, instead of one, instead of this far right over here, it's gonna be about this far far right over here. Instead of this far right over here, it's gonna be that, and another right over there. And then instead of, it's gonna be about there, and then instead of it being like that, it's going to be one, two, and then three right around there."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So I'm just gonna kind of in my head, let's just pretend like the c is a three, just for visualization purposes. So it's gonna be three times this all, instead of one, instead of this far right over here, it's gonna be about this far far right over here. Instead of this far right over here, it's gonna be that, and another right over there. And then instead of, it's gonna be about there, and then instead of it being like that, it's going to be one, two, and then three right around there. So I'm starting to get a sense of what this curve is going to look like. It's going to, a scaled version of f of x, and for at least what I'm drawing, it's pretty close to three times f of x, but just to give you an idea. It's gonna look something like, and over here, let's see, at this distance, do a second one, a third one, it's gonna be up here."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then instead of, it's gonna be about there, and then instead of it being like that, it's going to be one, two, and then three right around there. So I'm starting to get a sense of what this curve is going to look like. It's going to, a scaled version of f of x, and for at least what I'm drawing, it's pretty close to three times f of x, but just to give you an idea. It's gonna look something like, and over here, let's see, at this distance, do a second one, a third one, it's gonna be up here. It's gonna look something, something like this. It's gonna look something like that. So this is a scaled version."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's gonna look something like, and over here, let's see, at this distance, do a second one, a third one, it's gonna be up here. It's gonna look something, something like this. It's gonna look something like that. So this is a scaled version. And the scale I did right here, I assumed a positive, I assumed a positive c greater than zero, but this is just for visualization purposes. Now, what do we think the area under this curve is going to be between a and b? So what do we think this, what do we think, what do we think this area, this area right over here is going to be?"}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is a scaled version. And the scale I did right here, I assumed a positive, I assumed a positive c greater than zero, but this is just for visualization purposes. Now, what do we think the area under this curve is going to be between a and b? So what do we think this, what do we think, what do we think this area, this area right over here is going to be? And we already know how we can denote it. That area right over there is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question a little bit clearer, how does this relate to this?"}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "So what do we think this, what do we think, what do we think this area, this area right over here is going to be? And we already know how we can denote it. That area right over there is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question a little bit clearer, how does this relate to this? How does this green area relate to this yellow area? Well, one way to think about it is we just scaled the vertical dimension up by c. So one way that you could reason it is, well, if I'm finding the area of something, if I have the area of a rectangle, and I have the vertical dimension is, let's say the vertical dimension is, I don't want to use those same letters over and over again. Well, I'll just, well, let's say the vertical dimension is alpha and the horizontal dimension is beta."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "I guess to make the question a little bit clearer, how does this relate to this? How does this green area relate to this yellow area? Well, one way to think about it is we just scaled the vertical dimension up by c. So one way that you could reason it is, well, if I'm finding the area of something, if I have the area of a rectangle, and I have the vertical dimension is, let's say the vertical dimension is, I don't want to use those same letters over and over again. Well, I'll just, well, let's say the vertical dimension is alpha and the horizontal dimension is beta. We know that the area, we know that the area is going to be alpha times beta. Now, if I scale up the vertical dimension by c, so instead of alpha, this is c times alpha, and this is the width is beta. If I scale up the vertical dimension by c, so this is, I have essentially, this is now c times alpha, what's the area going to be?"}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, I'll just, well, let's say the vertical dimension is alpha and the horizontal dimension is beta. We know that the area, we know that the area is going to be alpha times beta. Now, if I scale up the vertical dimension by c, so instead of alpha, this is c times alpha, and this is the width is beta. If I scale up the vertical dimension by c, so this is, I have essentially, this is now c times alpha, what's the area going to be? Well, it's going to be c alpha times beta. Or another way to think of it, I have just taken, when I scale one of the dimensions by c, I take my old area and I scale up my old area by c. And that's what we're doing. We're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "If I scale up the vertical dimension by c, so this is, I have essentially, this is now c times alpha, what's the area going to be? Well, it's going to be c alpha times beta. Or another way to think of it, I have just taken, when I scale one of the dimensions by c, I take my old area and I scale up my old area by c. And that's what we're doing. We're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now, obviously, that changes as our x changes, but when you think back to the Riemann sums, the f of x is what gave us the height of our rectangles. We're now scaling up the height, or scaling, I should say, because we might be scaling down, depending on the c. We're scaling it, we're scaling one dimension by c. If you scale one dimension by c, you're going to scale the area by c. So this right over here, the integral, let me just rewrite it, the integral from a to b of c f of x dx, that's just going to be the scaled, we're just going to take the area of f of x, so let me do that in that same color. We're going to take the area under the curve f of x from a to b f of x dx, and we're just going to scale it up, we're going to scale it up by this c. We're just going to scale it up by this c. So you might say, okay, maybe I could have felt that that was, you know, if it was a c inside the integral, now I can take the c out of the integral."}, {"video_title": "Integrating scaled version of function   AP Calculus AB   Khan Academy.mp3", "Sentence": "We're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now, obviously, that changes as our x changes, but when you think back to the Riemann sums, the f of x is what gave us the height of our rectangles. We're now scaling up the height, or scaling, I should say, because we might be scaling down, depending on the c. We're scaling it, we're scaling one dimension by c. If you scale one dimension by c, you're going to scale the area by c. So this right over here, the integral, let me just rewrite it, the integral from a to b of c f of x dx, that's just going to be the scaled, we're just going to take the area of f of x, so let me do that in that same color. We're going to take the area under the curve f of x from a to b f of x dx, and we're just going to scale it up, we're going to scale it up by this c. We're just going to scale it up by this c. So you might say, okay, maybe I could have felt that that was, you know, if it was a c inside the integral, now I can take the c out of the integral. Once again, this is not a rigorous proof based on the definition of the definite integral, but it hopefully gives you a little bit of intuition why you can do this. If you scale up the function, you're essentially scaling up the vertical dimension, so the area under this is going to just be a scaled up version of the area under the original function f of x. And once again, really, really, really useful property, really, really useful property of definite integrals that's going to help us solve a bunch of definite integrals and kind of clarify what we're even doing with them."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So here on the left, you see that this curve looks just like y equals x squared until we get to x equals three, and instead of it being three squared, at this point, you have this opening, and instead, the function at three is defined at four, but then it keeps going, and it looks just like y equals x squared. This is known as a point or a removable discontinuity, and it's called that for obvious reasons. You're discontinuous at that point. You might imagine of defining or redefining the function at that point, so it is continuous, so the discontinuity is removable, but then how does this relate to our definition of continuity? Well, let's remind ourselves our definition of continuity. We say f is continuous, continuous, if and only if, or let me write f continuous at x equals c, if and only if the limit as x approaches c of f of x is equal to the actual value of the function when x is equal to c. So why does this one fail? Well, the two-sided limit actually exists."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "You might imagine of defining or redefining the function at that point, so it is continuous, so the discontinuity is removable, but then how does this relate to our definition of continuity? Well, let's remind ourselves our definition of continuity. We say f is continuous, continuous, if and only if, or let me write f continuous at x equals c, if and only if the limit as x approaches c of f of x is equal to the actual value of the function when x is equal to c. So why does this one fail? Well, the two-sided limit actually exists. You could find, if we say c in this case is three, the limit as x approaches three of f of x, it looks like, if you graphically inspect this, and I actually know this is the graph of y equals x squared except at that discontinuity right over there, this is equal to nine, but the issue is is the way this graph has been depicted, this is not the same thing as the value of the function. This function, f of three, the way it's been graphed, f of three is equal to four. So this is a situation where this two-sided limit exists, but it's not equal to the value of that function."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, the two-sided limit actually exists. You could find, if we say c in this case is three, the limit as x approaches three of f of x, it looks like, if you graphically inspect this, and I actually know this is the graph of y equals x squared except at that discontinuity right over there, this is equal to nine, but the issue is is the way this graph has been depicted, this is not the same thing as the value of the function. This function, f of three, the way it's been graphed, f of three is equal to four. So this is a situation where this two-sided limit exists, but it's not equal to the value of that function. You might see other circumstances where the function isn't even defined there, so that isn't even there, and so once again, the limit might exist, but the function might not be defined there, so in either case, you aren't going to meet this criteria for continuity, and so that's how a point or removable discontinuity, why it is discontinuous, with regards to our limit definition of continuity. So now let's look at this second example. If we looked at our intuitive continuity test, if we were just trying to trace this thing, we see that once we get to x equals two, I have to pick up my pencil to keep tracing it, and so that's a pretty good sign that we are discontinuous."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So this is a situation where this two-sided limit exists, but it's not equal to the value of that function. You might see other circumstances where the function isn't even defined there, so that isn't even there, and so once again, the limit might exist, but the function might not be defined there, so in either case, you aren't going to meet this criteria for continuity, and so that's how a point or removable discontinuity, why it is discontinuous, with regards to our limit definition of continuity. So now let's look at this second example. If we looked at our intuitive continuity test, if we were just trying to trace this thing, we see that once we get to x equals two, I have to pick up my pencil to keep tracing it, and so that's a pretty good sign that we are discontinuous. We see that over here as well. If I'm tracing this function, I gotta pick up my pencil to, I can't go through that point or I have to jump down here and then keep going right over there, so in either case, I have to pick up my pencil, and so intuitively, it is discontinuous, but this particular type of discontinuity, where I'm making a jump from one point and then I'm making a jump down here to continue, it is intuitively called a jump discontinuity. Discontinuity, and this is, of course, a point or removable discontinuity."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "If we looked at our intuitive continuity test, if we were just trying to trace this thing, we see that once we get to x equals two, I have to pick up my pencil to keep tracing it, and so that's a pretty good sign that we are discontinuous. We see that over here as well. If I'm tracing this function, I gotta pick up my pencil to, I can't go through that point or I have to jump down here and then keep going right over there, so in either case, I have to pick up my pencil, and so intuitively, it is discontinuous, but this particular type of discontinuity, where I'm making a jump from one point and then I'm making a jump down here to continue, it is intuitively called a jump discontinuity. Discontinuity, and this is, of course, a point or removable discontinuity. And so how does this relate to limits? Well, here, the left and right-handed limits exist, but they're not the same thing, so you don't have a two-sided limit. So for example, for this one in particular, but for all the x values up to and including x equals two, this is the graph of y equals x squared, and then for x greater than two, it's the graph of square root of x."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Discontinuity, and this is, of course, a point or removable discontinuity. And so how does this relate to limits? Well, here, the left and right-handed limits exist, but they're not the same thing, so you don't have a two-sided limit. So for example, for this one in particular, but for all the x values up to and including x equals two, this is the graph of y equals x squared, and then for x greater than two, it's the graph of square root of x. So in this scenario, if you were to take the limit of f of x as x approaches two from the left, from the left, this is going to be equal to four. You're approaching this value, and that actually is the value of the function, but if you were to take the limit as x approaches two from the right of f of x, what is that going to be equal to? Well, we're approaching from the right, this is actually the square root of x, so it's approaching the square root of two."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "So for example, for this one in particular, but for all the x values up to and including x equals two, this is the graph of y equals x squared, and then for x greater than two, it's the graph of square root of x. So in this scenario, if you were to take the limit of f of x as x approaches two from the left, from the left, this is going to be equal to four. You're approaching this value, and that actually is the value of the function, but if you were to take the limit as x approaches two from the right of f of x, what is that going to be equal to? Well, we're approaching from the right, this is actually the square root of x, so it's approaching the square root of two. You can't, you wouldn't know it's the square root of two just by looking at this. I know that just because when I went onto Desmos and defined the function, that's the function that I used. But it's clear, even visually, that you're approaching two different values when you approach from the left than when you approach from the right."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Well, we're approaching from the right, this is actually the square root of x, so it's approaching the square root of two. You can't, you wouldn't know it's the square root of two just by looking at this. I know that just because when I went onto Desmos and defined the function, that's the function that I used. But it's clear, even visually, that you're approaching two different values when you approach from the left than when you approach from the right. So even though the one-sided limits exist, they're not approaching the same thing, so the two-sided limit doesn't exist, and if the two-sided limit doesn't exist, it for sure cannot be equal to the value of the function there, even if the function is defined. So that's why the jump discontinuity is failing this test. And once again, it's intuitive."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "But it's clear, even visually, that you're approaching two different values when you approach from the left than when you approach from the right. So even though the one-sided limits exist, they're not approaching the same thing, so the two-sided limit doesn't exist, and if the two-sided limit doesn't exist, it for sure cannot be equal to the value of the function there, even if the function is defined. So that's why the jump discontinuity is failing this test. And once again, it's intuitive. You're seeing that, hey, I gotta jump, I gotta pick up my pencil. These two things are not connected to each other. Finally, what you see here is, when you learned precalculus, often known as an asymptotic discontinuity, asymptotic, asymptotic discontinuity, discontinuity, and intuitively, you have an asymptote here."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And once again, it's intuitive. You're seeing that, hey, I gotta jump, I gotta pick up my pencil. These two things are not connected to each other. Finally, what you see here is, when you learned precalculus, often known as an asymptotic discontinuity, asymptotic, asymptotic discontinuity, discontinuity, and intuitively, you have an asymptote here. If you, it's a vertical asymptote at x equals two. If I were to try to trace the graph from the left, I'd be, I would just keep on going. In fact, I would be doing it forever because it's, it would be, it would be infinitely, it would be unbounded as I get closer and closer to x equals two from the left."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "Finally, what you see here is, when you learned precalculus, often known as an asymptotic discontinuity, asymptotic, asymptotic discontinuity, discontinuity, and intuitively, you have an asymptote here. If you, it's a vertical asymptote at x equals two. If I were to try to trace the graph from the left, I'd be, I would just keep on going. In fact, I would be doing it forever because it's, it would be, it would be infinitely, it would be unbounded as I get closer and closer to x equals two from the left. And if I get to, try to get to x equals two from the right, once again, I get unbounded up. But even if I could, and when I say it's unbounded, it goes to infinity, so it's actually impossible in a mortal's lifespan to try to trace the whole thing, but you get the sense that, hey, there's no way that I could draw from here to here without picking up my pencil. And if you wanna relate it to our notion of limits, it's that both the left and right-handed limits are unbounded, so they officially don't exist."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "In fact, I would be doing it forever because it's, it would be, it would be infinitely, it would be unbounded as I get closer and closer to x equals two from the left. And if I get to, try to get to x equals two from the right, once again, I get unbounded up. But even if I could, and when I say it's unbounded, it goes to infinity, so it's actually impossible in a mortal's lifespan to try to trace the whole thing, but you get the sense that, hey, there's no way that I could draw from here to here without picking up my pencil. And if you wanna relate it to our notion of limits, it's that both the left and right-handed limits are unbounded, so they officially don't exist. So if they don't exist, then we can't meet these conditions. So if I were to say the limit as x approaches two from the left-hand side of f of x, we can see that it goes unbounded in the negative direction. You might sometimes see someone write something like this, negative infinity, but that's a little hand-wavy with the math."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "And if you wanna relate it to our notion of limits, it's that both the left and right-handed limits are unbounded, so they officially don't exist. So if they don't exist, then we can't meet these conditions. So if I were to say the limit as x approaches two from the left-hand side of f of x, we can see that it goes unbounded in the negative direction. You might sometimes see someone write something like this, negative infinity, but that's a little hand-wavy with the math. The more correct way to say it, it's just unbounded. Unbounded. And likewise, if we thought about the limit as x approaches two from the right of f of x, it is now unbounded towards positive infinity."}, {"video_title": "Types of discontinuities   Limits and continuity   AP Calculus AB   Khan Academy.mp3", "Sentence": "You might sometimes see someone write something like this, negative infinity, but that's a little hand-wavy with the math. The more correct way to say it, it's just unbounded. Unbounded. And likewise, if we thought about the limit as x approaches two from the right of f of x, it is now unbounded towards positive infinity. So this, once again, this is also unbounded. And because it's unbounded and this limit does not exist, it can't meet these conditions, and so we are going to be discontinuous. So this is a point or removable discontinuity, jump discontinuity, I'm jumping, and then we have these asymptotes, a vertical asymptote."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "I'll put that in quotes right over here because it's kind of a little loosey-goosey is how close is that. But as close as you want by getting x sufficiently close to c. So another way of saying this is if you tell me, hey, I want to get my f of x to be within 0.5 of this limit, then you're telling me if this limit is actually true, you should be able to hand me a value around c that if x is within that range, that f of x is definitely going to be as close to L as I desire. So let me draw that out to make it a little bit clearer. And I'm going to zoom in. I'll draw another diagram. So let's say that this right over here is my y-axis. And I'm going to zoom in."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm going to zoom in. I'll draw another diagram. So let's say that this right over here is my y-axis. And I'm going to zoom in. I'm going to draw a slightly different function just so we can really focus on what's going on around here, the ranges around c and the ranges around L. So that's x. This right over here is y. Let's say that this is c. And let's just zoom in on our function."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "And I'm going to zoom in. I'm going to draw a slightly different function just so we can really focus on what's going on around here, the ranges around c and the ranges around L. So that's x. This right over here is y. Let's say that this is c. And let's just zoom in on our function. So let's say our function is doing something like, let's say it does something like, let's see, I don't want it to be defined at c, at least just for the, it could be. You can always find a limit even where it is defined. But let's say our function looks something like that."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "Let's say that this is c. And let's just zoom in on our function. So let's say our function is doing something like, let's say it does something like, let's see, I don't want it to be defined at c, at least just for the, it could be. You can always find a limit even where it is defined. But let's say our function looks something like that. And it can even have a little kink in it the way I drew it. So it looks something like this. It's undefined."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "But let's say our function looks something like that. And it can even have a little kink in it the way I drew it. So it looks something like this. It's undefined. Let me draw it a little bit different. So it is undefined when x is equal to c. So this is the point where there is a hole. It is undefined when x is equal to c. And it even has a little kink in it just like that."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "It's undefined. Let me draw it a little bit different. So it is undefined when x is equal to c. So this is the point where there is a hole. It is undefined when x is equal to c. And it even has a little kink in it just like that. And what we want to do is prove that the limit of f of x, and let me make it clear, this is the graph of y is equal to f of x. We want to get an idea for what this definition is saying if we're claiming that the limit of f of x as x approaches c is L. So conceptually, we get the gist already. We already get the gist that this right over here is L. But what is this definition saying?"}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "It is undefined when x is equal to c. And it even has a little kink in it just like that. And what we want to do is prove that the limit of f of x, and let me make it clear, this is the graph of y is equal to f of x. We want to get an idea for what this definition is saying if we're claiming that the limit of f of x as x approaches c is L. So conceptually, we get the gist already. We already get the gist that this right over here is L. But what is this definition saying? What's saying that you can get f of x as close to L as you want. So if you tell someone, I want to get f of x within a certain range of L, then if this limit is actually true, if the limit of f of x as x approaches c really is equal to L, then they should be able to find a range around c that as long as x is around that range, your f of x is going to be in the range that you want. So let me actually go through that exercise."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "We already get the gist that this right over here is L. But what is this definition saying? What's saying that you can get f of x as close to L as you want. So if you tell someone, I want to get f of x within a certain range of L, then if this limit is actually true, if the limit of f of x as x approaches c really is equal to L, then they should be able to find a range around c that as long as x is around that range, your f of x is going to be in the range that you want. So let me actually go through that exercise. It really is a little bit like a game. So someone comes up to you and says, well, OK, I don't necessarily believe that you're claiming that the limit of f of x as x approaches c is equal to L. I'm not really sure if that's the case. But I agree with this definition."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "So let me actually go through that exercise. It really is a little bit like a game. So someone comes up to you and says, well, OK, I don't necessarily believe that you're claiming that the limit of f of x as x approaches c is equal to L. I'm not really sure if that's the case. But I agree with this definition. So I want to get f of x within 0.5 of L. So this right over here would be L plus 0.5. And this right over here is L minus 0.5. And then you say, fine, I'm going to give you a range around c that if you take any x within that range, your f of x is always going to fall in this range that you care about."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "But I agree with this definition. So I want to get f of x within 0.5 of L. So this right over here would be L plus 0.5. And this right over here is L minus 0.5. And then you say, fine, I'm going to give you a range around c that if you take any x within that range, your f of x is always going to fall in this range that you care about. And so you look at this. And obviously, we haven't explicitly defined this function. But you can even eyeball it the way this function is defined."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "And then you say, fine, I'm going to give you a range around c that if you take any x within that range, your f of x is always going to fall in this range that you care about. And so you look at this. And obviously, we haven't explicitly defined this function. But you can even eyeball it the way this function is defined. It won't be that easy for all functions. But you look at it like this. And you say that this value, just the way it's drawn right over here, let's say that this is c minus 0.25."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "But you can even eyeball it the way this function is defined. It won't be that easy for all functions. But you look at it like this. And you say that this value, just the way it's drawn right over here, let's say that this is c minus 0.25. And let's say that this value right over here is c plus 0.25. And so you tell them, look, as long as you get x within 0.25 of c, so as long as your x's are sitting someplace over here, the corresponding f of x is going to sit in the range that you care about. And you say, OK, fine."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you say that this value, just the way it's drawn right over here, let's say that this is c minus 0.25. And let's say that this value right over here is c plus 0.25. And so you tell them, look, as long as you get x within 0.25 of c, so as long as your x's are sitting someplace over here, the corresponding f of x is going to sit in the range that you care about. And you say, OK, fine. You won that round. Let me make it even tighter. Maybe instead of saying within 0.5, I want to get within 0.05."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "And you say, OK, fine. You won that round. Let me make it even tighter. Maybe instead of saying within 0.5, I want to get within 0.05. And then you'd have to do this exercise again and find another range. And in order for this to be true, you have to be able to do this for any range that they give you. For any range around L that they give you, you have to be able to get f of x within that range by finding a range around c that as long as x is that range around c, f of x is going to sit within that range."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "Maybe instead of saying within 0.5, I want to get within 0.05. And then you'd have to do this exercise again and find another range. And in order for this to be true, you have to be able to do this for any range that they give you. For any range around L that they give you, you have to be able to get f of x within that range by finding a range around c that as long as x is that range around c, f of x is going to sit within that range. I'll let you think about that a little bit. There's a lot to think about. But hopefully this made sense."}, {"video_title": "Formal definition of limits Part 2 building the idea   AP Calculus AB   Khan Academy.mp3", "Sentence": "For any range around L that they give you, you have to be able to get f of x within that range by finding a range around c that as long as x is that range around c, f of x is going to sit within that range. I'll let you think about that a little bit. There's a lot to think about. But hopefully this made sense. We did it for the particular example of someone hands you the 0.5. I want f of x within 0.5 of L. And you say, well, as long as x is within 0.25 of c, you're going to match it. You need to be able to do that for any range they give you around L. And then this limit will definitely be true."}, {"video_title": "Product rule example.mp3", "Sentence": "So let's see if we can find the derivative with respect to x of e to the x times cosine of x. And like always, pause this video and give it a go on your own before we work through it. So when you look at this, you might say, well I know how to find the derivative with respect of e to the x. That's in fact just e to the x. And let me write this down. We know a few things. We know the derivative with respect to x of e to the x. E to the x is e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "That's in fact just e to the x. And let me write this down. We know a few things. We know the derivative with respect to x of e to the x. E to the x is e to the x. We know how to find the derivative of cosine of x. The derivative with respect to x of cosine of x is equal to negative sine of x. But how do we find the derivative of their product?"}, {"video_title": "Product rule example.mp3", "Sentence": "We know the derivative with respect to x of e to the x. E to the x is e to the x. We know how to find the derivative of cosine of x. The derivative with respect to x of cosine of x is equal to negative sine of x. But how do we find the derivative of their product? Well as you can imagine, this might involve the product rule. And let me just write down the product rule generally first. So if we would take the derivative with respect to x of the first expression in terms of x."}, {"video_title": "Product rule example.mp3", "Sentence": "But how do we find the derivative of their product? Well as you can imagine, this might involve the product rule. And let me just write down the product rule generally first. So if we would take the derivative with respect to x of the first expression in terms of x. So this is, we could call this u of x u of x times another expression that involves x. So u times v of x. This is going to be equal to, and I'm color coding it so we can really keep track of things."}, {"video_title": "Product rule example.mp3", "Sentence": "So if we would take the derivative with respect to x of the first expression in terms of x. So this is, we could call this u of x u of x times another expression that involves x. So u times v of x. This is going to be equal to, and I'm color coding it so we can really keep track of things. This is going to be equal to the derivative of the first expression. So I could write that as u prime of x times just the second expression, not the derivative of it, just the second expression. So times v of x. V of x."}, {"video_title": "Product rule example.mp3", "Sentence": "This is going to be equal to, and I'm color coding it so we can really keep track of things. This is going to be equal to the derivative of the first expression. So I could write that as u prime of x times just the second expression, not the derivative of it, just the second expression. So times v of x. V of x. And then we have plus, plus the first expression, not its derivative, just the first expression, u of x times the derivative of the second expression. Times the derivative of the second expression. So the way you remember it is you have these two things here."}, {"video_title": "Product rule example.mp3", "Sentence": "So times v of x. V of x. And then we have plus, plus the first expression, not its derivative, just the first expression, u of x times the derivative of the second expression. Times the derivative of the second expression. So the way you remember it is you have these two things here. You're going to end up with two different terms. And each of them, you're going to take the derivative of one of them but not the other one. And then the other one, you'll take the derivative of the other one but not the first one."}, {"video_title": "Product rule example.mp3", "Sentence": "So the way you remember it is you have these two things here. You're going to end up with two different terms. And each of them, you're going to take the derivative of one of them but not the other one. And then the other one, you'll take the derivative of the other one but not the first one. So derivative of u times v is u prime times v plus u times v prime. Now when you just look at it like that, it seems a little bit abstract and that might even be a little confusing but that's why we have a tangible example here. And I color coded it intentionally so we can say that u of x is equal to e to the x and v of x is equal to cosine of x."}, {"video_title": "Product rule example.mp3", "Sentence": "And then the other one, you'll take the derivative of the other one but not the first one. So derivative of u times v is u prime times v plus u times v prime. Now when you just look at it like that, it seems a little bit abstract and that might even be a little confusing but that's why we have a tangible example here. And I color coded it intentionally so we can say that u of x is equal to e to the x and v of x is equal to cosine of x. So v of x is equal to cosine of x. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. That's one of the most magical things in mathematics."}, {"video_title": "Product rule example.mp3", "Sentence": "And I color coded it intentionally so we can say that u of x is equal to e to the x and v of x is equal to cosine of x. So v of x is equal to cosine of x. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. That's one of the most magical things in mathematics. One of the things that makes e so special. So u prime of x is still equal to e to the x. And v prime of x, we know is negative sine of x."}, {"video_title": "Product rule example.mp3", "Sentence": "That's one of the most magical things in mathematics. One of the things that makes e so special. So u prime of x is still equal to e to the x. And v prime of x, we know is negative sine of x. And so what's this going to be equal to? Well this is going to be equal to the derivative of the first expression. So the derivative of e to the x, which is just e to the x, times the second expression, not taking its derivative, so times cosine of x plus the first expression, not taking its derivative."}, {"video_title": "Product rule example.mp3", "Sentence": "And v prime of x, we know is negative sine of x. And so what's this going to be equal to? Well this is going to be equal to the derivative of the first expression. So the derivative of e to the x, which is just e to the x, times the second expression, not taking its derivative, so times cosine of x plus the first expression, not taking its derivative. So e to the x times the derivative of the second expression. So times the derivative of cosine of x, is negative sine, negative sine of x. And it might be a little bit confusing because e to the x is its own derivative, but this right over here, you can view this as this was the derivative of e to the x, which happens to be e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "So the derivative of e to the x, which is just e to the x, times the second expression, not taking its derivative, so times cosine of x plus the first expression, not taking its derivative. So e to the x times the derivative of the second expression. So times the derivative of cosine of x, is negative sine, negative sine of x. And it might be a little bit confusing because e to the x is its own derivative, but this right over here, you can view this as this was the derivative of e to the x, which happens to be e to the x. That's what's exciting about that expression or that function. And then this is just e to the x without taking its derivative. They're of course the same thing."}, {"video_title": "Product rule example.mp3", "Sentence": "And it might be a little bit confusing because e to the x is its own derivative, but this right over here, you can view this as this was the derivative of e to the x, which happens to be e to the x. That's what's exciting about that expression or that function. And then this is just e to the x without taking its derivative. They're of course the same thing. But anyway, now we can just simplify it. This is going to be equal to, this is going to be equal to, we could write this either as e to the x times cosine of x, times cosine of x, minus e to the x, e to the x, times sine of x, times sine of x. Or if you want, you could factor out an e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "They're of course the same thing. But anyway, now we can just simplify it. This is going to be equal to, this is going to be equal to, we could write this either as e to the x times cosine of x, times cosine of x, minus e to the x, e to the x, times sine of x, times sine of x. Or if you want, you could factor out an e to the x. This is the same thing as e to the x times cosine of x minus sine of x, cosine of x minus sine of x. So hopefully this makes the product rule a little bit more tangible. And once you have this in your tool belt, there's a whole broader class of functions and expressions that we can start to differentiate."}, {"video_title": "Product rule example.mp3", "Sentence": "Or if you want, you could factor out an e to the x. This is the same thing as e to the x times cosine of x minus sine of x, cosine of x minus sine of x. So hopefully this makes the product rule a little bit more tangible. And once you have this in your tool belt, there's a whole broader class of functions and expressions that we can start to differentiate."}, {"video_title": "Product rule example.mp3", "Sentence": "And once you have this in your tool belt, there's a whole broader class of functions and expressions that we can start to differentiate."}]