# derivative of an integral

• Apr 17th 2008, 08:05 AM
DINOCALC09
derivative of an integral
d/dx of the integral from 0 --> x^2 of sin^2t * dt

can you guys understand that. i don't know how to do that fancy synthax.

and the question i have about this was whether there was some kind of trick to this problem and is there a specific terminology for this type of problem so i can study up further.
• Apr 17th 2008, 08:27 AM
flyingsquirrel
Hello

Let $F(x)=\int_0^x\sin^2t\,\mathrm{d}t$. Then, you know $\frac{\mathrm{d}F}{\mathrm{d}x}$ and you're looking for $\frac{\mathrm{d}F(x^2)}{\mathrm{d}x}$ which simply is the derivative of a composite function.

Quote:

is there a specific terminology for this type of problem
I don't know...
• Apr 17th 2008, 08:32 AM
DINOCALC09

2xcos^2(x^2)

thanks
• Apr 17th 2008, 08:35 AM
flyingsquirrel
No, it's not a cosine.
• Apr 17th 2008, 08:40 AM
Isomorphism
Quote:

Originally Posted by DINOCALC09

2xcos^2(x^2)

thanks

First tell us what is $\frac{dF}{dx}$? I think you are missing this itself!
• Apr 17th 2008, 08:45 AM
DINOCALC09
sin^2(x^2)
• Apr 17th 2008, 08:58 AM
flyingsquirrel
No, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\, \mathrm{d}t\right)=f(x)$ so there shouldn't have any $x^2$ in the expression of $\frac{\mathrm{d}F}{\mathrm{d}x}$
• Apr 17th 2008, 09:02 AM
colby2152
Quote:

Originally Posted by flyingsquirrel
No, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\, \mathrm{d}t\right)=f(x)$ so there shouldn't have any $x^2$ in the expression of $\frac{\mathrm{d}F}{\mathrm{d}x}$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\, \mathrm{d}t\right)$

$\frac{\mathrm{d}}{\mathrm{d}x}(F(x)-F(0))$

$F(0)$ is a constant, so the derivative is zero.

$F'(x) - c'$

$f(x)$
• Apr 17th 2008, 09:20 AM
topsquark
Quote:

Originally Posted by flyingsquirrel
No, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^xf(t)\, \mathrm{d}t\right)=f(x)$ so there shouldn't have any $x^2$ in the expression of $\frac{\mathrm{d}F}{\mathrm{d}x}$

Look at the original: the upper limit is $x^2$.

So the answer will be $f(x^2) \cdot 2x$ by the chain rule.

-Dan
• Apr 17th 2008, 09:21 AM
TheEmptySet
Maybe this will help...

let

$g(u)=\int_{0}^{u}h(t)dt$

by the fundemental theorem of Calculus we get

$\frac{dg}{du}=h(u)$

so for example

$f(x)=\int_{0}^{x}\tan(t)dt$
then
$\frac{df}{dx}=\tan(x)$

let

$f(x)=\int_{0}^{x^2}\sin^2(t)dt$

the problem is your upper limit of integration is not linear so we need to fix it up the the FTC applies

let $g(u)=\int_{0}^{u}\sin^2(t)dt \mbox{ and } u(x)=x^2$

then $f(x)=g(u)=\int_0^{x^2}\sin^2(t)dt$

now using the chain rule and the FTC we get

$\frac{df}{dx}=\frac{dg}{du}\frac{du}{dx}=(\sin^2(u ))(2x)$

now we need to sub the value of u back in and we get

$\frac{df}{dx}=2x\sin^2(x^2)$