= Differentiability does not imply continuity of the derivative
{title2=$f(x)=x^2\sin(1/x),\quad f'(0)=0$}
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.
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