Let $\displaystyle f(x)=x^2 \sin(\frac{1}{x})$ for $\displaystyle x \not = 0$ and $\displaystyle f(0)=0.$

a) Prove that $\displaystyle f(x)$ is differentiable on $\displaystyle \mathbb{R}$.

b) Prove that $\displaystyle f'(x)$ is not continuous on $\displaystyle \mathbb{R}$.

I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

Attempt:a) Clearly $\displaystyle f(x)$ is differentiable at $\displaystyle x \not = 0$. [How do I show this?] In fact, $\displaystyle f'(x)=2x \sin(x) - \cos(\frac{1}{x})$ [valid for $\displaystyle x \not =0$]. At $\displaystyle x=0$, $\displaystyle f'(0)=\lim_{x \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim_{h \rightarrow 0} h \sin(\frac{1}{h})=0$. Hence, $\displaystyle f(x)$ is differentiable on $\displaystyle \mathbb{R}$.

b) $\displaystyle f'(x)=2x \sin(x) - \cos(\frac{1}{x})$. Let $\displaystyle p(x)=2x \sin(x)$. Then $\displaystyle p(0)=0$ because $\displaystyle p(0)=2 \cdot \lim_{x \rightarrow 0} x \sin(\frac{1}{x})=0$. So $\displaystyle p(0)=0$. So by FTC $\displaystyle \exists P(x)$ s.t. $\displaystyle P'(x)=p(x)$. So $\displaystyle f'(x)=P'(x)-cos(\frac{1}{x})$. But then $\displaystyle \cos(\frac{1}{x})$ is not a continuous function. But by FTC, $\displaystyle P(x)$ is differentiable. Then $\displaystyle P(x)$, $\displaystyle f(x)$ are differentiable functions whose derivatives are not continuous.

Thanks.