How to show that $f(x) = x^2 \sin(1/x^2)$ with $f(0)=0$ is differentiable and $f'$ is unbounded on $[-1,1]$? If the function $f$ is defined by 
$$f(x) = \begin{cases}x^2 \sin(\frac{1}{x^2}) &amp; x \ne 0 \\
0 &amp; x = 0
\end{cases}
$$
then $f$ is differentiable and $f'$ is unbounded on $[-1,1]$. 

I know this is true. How should I go about proving this?
The cool point of the problem is that although for any positive value $M$ and positive $\epsilon$ we can find some value of $x$ with $|x|&lt;\epsilon$ such that $|f'(x)| &gt; M$, yet $f$ is differentiable at $x=0$.

In fact, $f'(0) = 0$, and we can see that by applying the definition of derivative:
$$
\left. \frac{df(x)}{dx} \right|_{x=0} = \lim_{h\rightarrow 0} \frac{h^2 \sin\frac{1}{h^2}-f(0)}{h} = \lim_{h\rightarrow 0} {h \sin\frac{1}{h^2}}-0 =0
$$ 
since $|\sin\frac{1}{h^2}|$ is bounded by 1 and that is being multiplied by $h$.

On the other hand, look at the expression for $f'(x)$ when $x \neq 0$.  At points where $\frac{1}{x^2} = (n+\frac{1}{2})\pi$,
$$
|f'(x)| &gt; \frac{2}{x} - 2x
$$
and this is unbounded as $x$ approaches zero.
Using the product and chain rule, one gets that 
$$f'(x) = 2x\sin(1/x^2)-2\frac{\cos(1/x^2)}{x} $$ 
Notice that this function is unbounded on the given interval.