# Thread: showing function differentiable

1. ## showing function differentiable

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

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

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

2. 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) = $\displaystyle\int^{x^2}_0 f \,$

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

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

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

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

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

I used G instead of F for the left side so you wouldn't confuse the two!

3. am i missing something here or did you assume that $F(0) = 0$

4. Originally Posted by bobak
am i missing something here or did you assume that $F(0) = 0$
Yes, I added that assumption in.

5. Originally Posted by colby2152
Yes, I added that assumption in.
surely $F(0)$ is constant so its differentiates to give 0

6. Originally Posted by bobak
surely $F(0)$ is constant so its differentiates to give 0
Good point, because $F(0)=\int_0^0 fdx \Rightarrow 0$

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

isnt that $G(0)$ ?

8. Originally Posted by bobak
isnt that $G(0)$ ?
Yes, in my example, that is $G(0)$

Cite back to my original reply.

9. 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) = $\displaystyle\int^{x^2}_0 f \,$

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

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

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