Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 9C / iv / Solution

The function $f(x)=x^2\sin(1/x^2)$ for $x\ne0$, extended by $f(0)=0$, is <differentiable> everywhere. At zero its <derivative> is zero by the difference quotient, while away from zero $f'(x)=2x\sin(1/x^2)-2x^{-1}\cos(1/x^2)$ is unbounded near zero. Differentiability of a function is therefore weaker than continuity of its derivative.

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 Incoming links: Differentiability does not imply continuous derivatives