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Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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? |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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? |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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? |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
2011 Calculus AB free response #3 (a & b) AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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? |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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? |
Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. |
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