Let $f(x) = x^2 \sin (1/x^2),\,x\ne 0,$ and $ f(0)=0.$ Prove $f$ is differentiable on $\mathbb R$
So $f'(x)=2x\sin 1/x^2 -\frac{2\cos 1/x^2}{x}$ when $x\ne 0$, but what about at $0$? My prof in the assignment solutions just says "$f'(0) = 0$." (though his solutions tend to be quite abbreviated). Am I missing something obvious? Is he skipping steps or can you go right from $f(0) = 0$ to $f'(0)=0$?