# showing function differentiable

• Feb 1st 2008, 04:03 AM
hunkydory19
showing function differentiable
Let [0,1] --->R be continuous. (little f above arrow)
Define [0,1] --->R (capital F above arrow) by F(x) = $\displaystyle \displaystyle\int^{x^2}_0 f \,$

Show that F is differentiable with $\displaystyle F'(x) = 2x f(x^2).$

I seriously have worked on this question for hours and have got nowhere! Can anyone give me any help at all?!

• Feb 1st 2008, 05:01 AM
colby2152
Quote:

Originally Posted by hunkydory19
Let [0,1] --->R be continuous. (little f above arrow)
Define [0,1] --->R (capital F above arrow) by F(x) = $\displaystyle \displaystyle\int^{x^2}_0 f \,$

Show that F is differentiable with $\displaystyle F'(x) = 2x f(x^2).$

I seriously have worked on this question for hours and have got nowhere! Can anyone give me any help at all?!

Let $\displaystyle G(x) = \displaystyle\int^{x^2}_0 f \,$

$\displaystyle G(x)=F(x^2)-F(0)$

$\displaystyle G'(x)=f(x^2)2x-f(0)$, assume $\displaystyle f(0)=0$ to get: $\displaystyle G'(x)=f(x^2)2x$

I used G instead of F for the left side so you wouldn't confuse the two!
• Feb 1st 2008, 06:20 AM
bobak
am i missing something here or did you assume that $\displaystyle F(0) = 0$
• Feb 1st 2008, 06:22 AM
colby2152
Quote:

Originally Posted by bobak
am i missing something here or did you assume that $\displaystyle F(0) = 0$

Yes, I added that assumption in.
• Feb 1st 2008, 06:25 AM
bobak
Quote:

Originally Posted by colby2152
Yes, I added that assumption in.

surely $\displaystyle F(0)$ is constant so its differentiates to give 0
• Feb 1st 2008, 06:27 AM
colby2152
Quote:

Originally Posted by bobak
surely $\displaystyle F(0)$ is constant so its differentiates to give 0

Good point, because $\displaystyle F(0)=\int_0^0 fdx \Rightarrow 0$
• Feb 1st 2008, 06:39 AM
bobak
Quote:

Originally Posted by colby2152
Good point, because $\displaystyle F(0)=\int_0^0 fdx \Rightarrow 0$

isnt that $\displaystyle G(0)$ ?
• Feb 1st 2008, 06:44 AM
colby2152
Quote:

Originally Posted by bobak
isnt that $\displaystyle G(0)$ ?

Yes, in my example, that is $\displaystyle G(0)$ :rolleyes:

Cite back to my original reply.
• Feb 1st 2008, 07:49 AM
ThePerfectHacker
Quote:

Originally Posted by hunkydory19
Let [0,1] --->R be continuous. (little f above arrow)
Define [0,1] --->R (capital F above arrow) by F(x) = $\displaystyle \displaystyle\int^{x^2}_0 f \,$

Show that F is differentiable with $\displaystyle F'(x) = 2x f(x^2).$

I seriously have worked on this question for hours and have got nowhere! Can anyone give me any help at all?!

Let $\displaystyle \chi (x) = \int_0^x f(x) dx$ for $\displaystyle x\in [0,1]$.
Let $\displaystyle \mu (x) = \int_0^{x^2} f(x) dx$ for $\displaystyle x\in [0,1]$.
We wish to show $\displaystyle \mu$ is differenciable and find its derivative.
Note, $\displaystyle \mu (x) = \chi (x^2)$. Now $\displaystyle \chi$ and $\displaystyle x^2$ are differenciable on $\displaystyle (0,1)$ by the fundamental theorem. This means, $\displaystyle \mu ' (x) = 2x \chi ' (x^2) = 2x f(x^2)$.