Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 11E / ii / Solution

Away from zero, the product and composition of the displayed smooth functions are continuous and differentiable. At zero,so $f$ remains continuous. Hence the continuity set is $\boxed{\mathbb{R}}$.

The difference quotient at zero is $f(h)/h=\sin (1/h)$. For $h_n=(\pi /2+2\pi n{)}^{-1}$ it equals one, and for $k_n=(3\pi /2+2\pi n{)}^{-1}$ it equals minus one. Both sequences tend to zero, so no derivative exists there. For $x\ne 0$, the product rule and chain rule giveThus the differentiability set is $\boxed{\mathbb{R}\setminus \{0\}}$.