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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

# 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)?
• To magnify that sin(x) is a function of x, so the chain rule applies
• 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)) ?
• ∫(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
• 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?
• 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?
• why doesn't F'(x)=h(sinx) or 2sinx-1?
• 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)
• At , 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.
• 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.
• 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