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# Finding derivative with fundamental theorem of calculus: chain rule

AP.CALC:
FUN‑5 (EU)
,
FUN‑5.A (LO)
,
FUN‑5.A.1 (EK)
,
FUN‑5.A.2 (EK)

## Video transcript

- [Instructor] 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, 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. 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 I'll need to 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. And we could keep going. We could try to, we could try to simplify this a little bit or rewrite it in different ways, but there you have it.
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