Explanation: $f(x)=x|x|, x\in\mathbb{R}$ is differentiable. $\begin{cases} x^2 & x\ge 0 \\ -x^2 & x<0 \end{cases}$ if $x \neq 0$, then the function is quadratic so is differentiable. The only point to consider is 0 . But since both $x^2$ and $-x^2$ have same derivative at 0 , then it follows that $f$ is differentiable at 0 .