You know that there is chain rule in derivative problems, but don't forget to apply chain rule as well in integral problems when the upper bound has a variable! They basically are telling you that y = integral of some function f(x) and want you to find y '. In order to do this the fundamental theorem of calculus says that the top bound on the integral needs to be the variable with which you are taking the derivative with respect to. Therefore if you have something like y = integral from 1 to x^2 of f(x) and you want to find dy/dx, you need to subsitute the x^2 with a u and use chain rule then to find dy/dx = dy/du times du/dx, and then apply the fundamental theorem of calculus on the dy/du part.

Let's do an example:

`f(x)=int_(1)^(x^2)dt` , find f'(x).

First of all, as what I have mentioned, we substitube the x^2 with u: u=x^2, so du/dx = 2x

Actually the question is same as

` f'(x)=d/dx int_(1)^(x^2)1 dt`

Now we are going to use chain rule and get the following:

` f'(x)=d/dx int_(1)^(x^2)1 dt =d/(du) int_(1)^(u)1 dt ((du)/dx)`

According to the fundamental theorem of calculus, we get f'(x) = 1 * du/dx

And at the beginning we get du/dx = 2x, we can put this back it and get:

f'(x) = 1 * du/dx = 1 * 2x =2x

**Subject :**Math**Topic :**Calculus-
**Posted By :**Jason

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