Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 10F / c / Solution

For $x\ne 0$, $f(x)=|x{|}^x=e^{x\log |x|}$, so $f^{\prime }(x)=|x{|}^x(\log |x|+1)$. Although $x\log |x|\to 0$ makes $f$ continuous at zero, $(f(x)-1)/x\sim \log |x|\to -\infty$, so it is not differentiable there.

Since the cosine function is even, $g(x)=\cos |x|=\cos x$, so it is differentiable everywhere with $g^{\prime }(x)=-\sin x$. Finally, $h(x)=x|x|$ is differentiable everywhere, including at zero, and $h^{\prime }(x)=2|x|$. Therefore