Proof that $f(x)=x|x|$ is differentiable on $\mathbb{R}$

Question

I want to show that

$$f(x) = x|x|$$

is differentiable for all reals.

My approach would be:

Since $ \forall x \in \mathbb{R}$ such that $x < 0$ we have $f(x) = -x^2$ which is differentiable. also $\forall x \in \mathbb{R}$ such that $x > 0$ we have $f(x) = x^2$ which is also differentiable.

Last but not least I would have to show that $f$ is differentiable at $x=0$ and from this it would follow that $f$ is differentiable $\forall x \in \mathbb{R}$ and of course also continuous.

Is this correct?

Answer 1

Yes. If you prove the differentiability at $x<0, x>0$ and $x=0$ then $f$ is differentiable.

Answer 2

$f(x)=x|x|= x^2, x>0; -x^2, x<0.$ It being a two piece function only $x=0$ is suspected for a non differentiability. So the right derivative at $x=0$ is $Rf'(0)=0$ and the left derivative at $x=0$ is $Lf'(0)=0$. Since both of them are equal and finite $f(x)$ is differentiable at $x=0$ and everywhere.