# Show that F'(x) exists for all x $$\in[a,b]$$

## Homework Statement

Consider
F(x) = x2 sin(1/x2) if 0<x$$\leq1$$
and = 0 if x$$\leq0$$

Show that F'(x) exists for all x $$\in[a,b]$$ but that F':[0,1] $$\rightarrow1$$ is not integrable.

## Homework Equations

So we have to show we do not have F(1)-F(0) = $$\int$$ F'(x)dx
(integral going from 0 to 1)

## The Attempt at a Solution

I'm having trouble proving this statement.
Where should I start?
To show that F'(x) exists, should I just take the derivative or do I have to go under some long theorems of analysis to PROVE?

Most likely you'll have to actually *prove* that this derivative exists - it clearly exists for all x other than 0 because $$x^{2}sin(\frac{1}{x^{2}})$$ is the composition of differentiable functions. The only tricky part is whether F'(0) exists. To show that it does indeed exist you'll need to show that the following limit exists:

$$\lim_{h\to 0}\frac{f(h) - f(0)}{h}$$

But this just equals

$$\lim_{h\to 0}\frac{h^{2}sin(\frac{1}{h^{2}})}{h} = lim_{h\to 0}hsin(\frac{1}{h^{2}}) = 0$$

Then by application of the product rule and the chain rule,

$$F'(x) = 2xsin(\frac{1}{x^{2}}) - \frac{2cos(\frac{1}{x^{2}})}{x^{2}}$$

if x is anything other than 0, and 0 if x=0. So all you need to show is that the above function is unbounded and it will consequently not be integrable.

right,
But I think you made a little mistake with differentiation,
I think it will be:

F'(x)= 2x sin(1/x2) - 2cos(1/x2)/ x

Which still, limits exists.

But How do I prove that F'(x) is unbounded?

HallsofIvy
Take $x= 1/\sqrt{2n\pi}$. Then $1/x^2= 2n\pi$ so $sin(1/x^2)= 0$ and $cos(1/x^2)= 1$. $F'(x)= F'(1/\sqrt{2n\pi})= -2\sqrt{2n\pi}$. x will go to 0 as n goes to infinity. What happens to F'(x)?
F'(x) will be Undefined as x approaches zero $$\rightarrow$$ unbounded $$\rightarrow$$ non integrable ??????