# Derivatives of Definite Integrals

• Nov 26th 2011, 07:44 PM
cschuster
Derivatives of Definite Integrals
Given: f(x) = the integral from 0 to sin(x) of sqrt.(1 + t^2) dt
and g(y) = the integral from 3 to y of f(x) dx.
Find g''(pi/6)

I'm not sure how to put it all together since there are two different integrals involved.
Thanks!
• Nov 26th 2011, 07:57 PM
Amer
Re: Derivatives of Definite Integrals
in general

$\frac{\delta }{\delta x}\int_{f(x)}^{g(x)} h(t) \cdot dt = h(g(x)) g'(x) - h(f(x)) f'(x)$

where did you stuck ?
can you solve now
can you write down the first derivative of g(y) ?
• Nov 26th 2011, 08:57 PM
cschuster
Re: Derivatives of Definite Integrals
Does g'(y) = f(y) ?
If it does, then what do I do, since I only know what f(x) is?
• Nov 26th 2011, 09:15 PM
uasac
Re: Derivatives of Definite Integrals
By the fundamental theorem of calculus:

$g'(y)=f(y)$

Then, $g''(y)=f'(y)$. You just have to take the derivative of $f(x)$. Use the Leibniz integral rule.
• Nov 26th 2011, 10:10 PM
cschuster
Re: Derivatives of Definite Integrals
Okay, so f'(x) = sqrt (1 + sin^2 (x) ), right?
But I just don't quite understand how you get f'(y) when you only have f'(x). (I did try reading the Leibniz integral page, but I wasn't sure whether I was supposed to use it when differentiating f(x) or what.)
Sorry to ask you to spell out every little step; this doesn't even seem like it should be such a hard problem, I'm just getting hung up with the switching between x and y.
• Nov 26th 2011, 10:23 PM
uasac
Re: Derivatives of Definite Integrals
You shouldn't worry about the switching between x and y, that is just the name of the variable. The real question you have to answer is the derivative of f(x). Using the Leibniz integral rule should look someting like this:

$\begin{matrix}\frac{df}{dx}&=\frac{d}{dx}\int_0^{\ sin x}\sqrt{1+t^2}dt=\frac{d(\sin x)}{dx}.\sqrt{1+\sin^2 x}=\sqrt{1-\sin^4 x}\end{matrix}$
• Nov 27th 2011, 12:46 AM
cschuster
Re: Derivatives of Definite Integrals
I think I've got it. Thanks a lot for all your help!