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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 6

Lesson 4: The fundamental theorem of calculus and accumulation functions- The fundamental theorem of calculus and accumulation functions
- Functions defined by definite integrals (accumulation functions)
- Functions defined by definite integrals (accumulation functions)
- Finding derivative with fundamental theorem of calculus
- Finding derivative with fundamental theorem of calculus
- Finding derivative with fundamental theorem of calculus: chain rule
- Finding derivative with fundamental theorem of calculus: chain rule

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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)

The Fundamental Theorem of Calculus tells us how to find the derivative of the integral from 𝘢 to 𝘹 of a certain function. But what if instead of 𝘹 we have a function of 𝘹, for example sin(𝘹)? Then we need to also use the chain rule.

## Want to join the conversation?

- what is the point of setting g(x) = sin(x)?(2 votes)
- To magnify that sin(x) is a function of x, so the chain rule applies(8 votes)

- Sorry, I'm not getting this. Why can we write F(x)=h(g(x)) ? FTIC2 surely only leads us to F'(x) = h(g(x)) ?(2 votes)
- ∫(2𝑡 − 1)𝑑𝑡 = 𝑡² − 𝑡 + 𝐶

FToC2 gives us

𝐹(𝑥) = sin²𝑥 − sin 𝑥 + 𝐶 − (1² − 1 + 𝐶) = sin²𝑥 − sin 𝑥

Thereby,

𝐹 '(𝑥) = 2 sin 𝑥 cos 𝑥 − cos 𝑥 = (2 sin 𝑥 − 1)cos 𝑥(4 votes)

- If F(x) is a function of x and you can find its derrivative, then does that mean that you can graph F, where F(x) is the area that different x and x is the different upper bounds?(2 votes)
- How would you solve it if sin(x) was the a of the integral notation (the value below the integral symbol). Would the methods be the same?(2 votes)
- why doesn't F'(x)=h(sinx) or 2sinx-1?(2 votes)
- because it’s F(x) which is equal to h(sin x) so you need to derivate it to get the h prime. and you can’t say that (F(x)=2sin(x)-1) because F(x) is a function of( sin (x)) not in x so that's why you need to do it in that way.(1 vote)

- At1:34, Sal says that F(x)=h(g(x)). Why isn't that what F'(x) is equal to? The integral sign seems to be missing in h(x).(1 vote)
- It's h(g(x)) because the integral (on the upper bound) approaches sin(x) and not x, and this makes it a composite function because h(x) = the integral but with x as the upper bound rather than sin(x) and g(x) = sin(x) which makes F(x) = h(g(x)) and F'(x) = h'(g(x)) * g'(x) by the chain rule. This might seem weird but if the bound is not x, and is something else like a function of x then we need to represent it as a composite function like this problem and solve from there. I hope this helped, if this seems confusing or unclear, please let me know.(2 votes)

- I'm not exactly sure what is the intent behind changing the upper bound from x to sin(x) or any other complex type of function. Also, why does chain rule take upper bounds into effect when computing area?(1 vote)
- I would say so long you know what the chain rule is and what the fundamental theorem of calculus is you are good to go.

This is from the video description.

The Fundamental Theorem of Calculus tells us how to find the derivative of the integral from 𝘢 to 𝘹 of a certain function. But what if instead of 𝘹 we have a function of 𝘹, for example sin(𝘹)? Then we need to also use the chain rule.(2 votes)

- Still can't wrap my head around this, so basically Sal defines sine(x) as another function because it's a composite function? What about anything other than just x? such as In(x) or even a number multiplied by x like 2x? Do you also define it as another function?(1 vote)
- Yes, and that's what we do every time we use the chain rule.

For example when finding the derivative of sin(ln 𝑥),

we can define 𝑔(𝑥) = ln 𝑥

and 𝑓(𝑥) = sin 𝑥 ⇒ 𝑓(𝑔(𝑥)) = sin(𝑔(𝑥)) = sin(ln(𝑥))

The chain rule gives us

𝑑∕𝑑𝑥[sin(ln 𝑥)] = 𝑑∕𝑑𝑥[𝑓(𝑔(𝑥))] = 𝑓 '(𝑔(𝑥))⋅𝑔'(𝑥)

𝑓 '(𝑥) = cos 𝑥 ⇒ 𝑓 '(𝑔(𝑥)) = cos(𝑔(𝑥)) = cos(ln 𝑥)

𝑔'(𝑥) = 1∕𝑥

Thus, 𝑑∕𝑑𝑥[sin(ln 𝑥)] = cos(ln 𝑥)⋅1∕𝑥(2 votes)

- How could you generalize this reasoning so that we could find the derivative of a function defined as a definite integral with BOTH limits as functions?(1 vote)
- Isn't one of the criteria for applying the fundamental theorem is that the function's continuous over the lower and upper bounds? So why can we just apply it here willy-nilly if sinx isn't even an x-coordinate?(1 vote)
- Integrating with functions as bounds will get pretty common if you take multivariable calculus. Essentially, when we say the bounds of integration are f(x,y) and g(x,y), we mean that while integrating, we go from the surface given by f(x,y) to the surface given by g(x,y). Here too, you can think of going between two functions while integrating, where both functions are continuous.(1 vote)

## 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.