 Differentiability does not imply continuity of the derivative

Extend the displayed function by $f(0)=0$. Its difference quotient at zero is $h\sin (1/h)$, which tends to zero. Away from zero its derivative is $2x\sin (1/x)-\cos (1/x)$. At $x_k=1/(2\pi k)$ this derivative equals $-1$, whereas at $y_k=1/((2k+1)\pi )$ it equals $+1$. Thus differentiability implies continuity of the function but does not imply continuity of its derivative.