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 1⁄2 power, so we can use the power rule, bring the 1 1⁄2 out front, so it becomes 1 1⁄2 x to the, and then take off one out of that exponent, so that's 1 1⁄2 minus one is negative 1 1⁄2 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?