Let $f\left( x \right)=\left( x+\left| x \right| \right) \left| x \right|$. Then for all $x.$

$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} \left( x-x \right) \left( -x \right) =0, & x<0 \end{matrix} \\ \begin{matrix} \left( x+x \right) x=2{ x }^{ 2 }, & x\ge 0 \end{matrix} \end{cases}$

It is evident from the graph of $f(x)$ that is continuous and differentiable for all $x$.

Also $\displaystyle f'\left( x \right) =\begin{cases} \begin{matrix} 0, & x<0 \end{matrix} \\ \begin{matrix} 4x, & x\ge 0 \end{matrix} \end{cases}$

$\therefore f'(x)$ is continuous for all $x$ but it is not differentiable at $x=0$.