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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.

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  • stelly blue style avatar for user Sahana Krishnaraj
    what is the point of setting g(x) = sin(x)?
    (2 votes)
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  • blobby green style avatar for user John Hunter
    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)) ?
    (1 vote)
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  • aqualine seed style avatar for user sergio
    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)
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  • aqualine ultimate style avatar for user Shimeko
    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)
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  • blobby green style avatar for user jakho nematov
    why doesn't F'(x)=h(sinx) or 2sinx-1?
    (2 votes)
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  • piceratops ultimate style avatar for user gingerseal8
    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)
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    • duskpin ultimate style avatar for user Ash_001
      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)
  • male robot hal style avatar for user Bill the Mather
    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)
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    • leaf green style avatar for user cossine
      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)
  • blobby green style avatar for user ariel
    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)
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    • cacteye blue style avatar for user Jerry Nilsson
      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)
  • blobby green style avatar for user ahmedaddous
    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)
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  • male robot johnny style avatar for user Mohamed Ibrahim
    In The integral function, do we regard the integrand as the input ?
    (1 vote)
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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.