Derivative of $x|x|$ I am trying to find the derivative of $f(x)=x|x|$ using the defition of derivative. For $x > 0$ I found that $f'(x)=2x$ and for $x<0$ the derivative is $f'(x)=-2x$. Everything is fine up to here. Now I want to check what happens when at $x=0$.
By the way, I know that $|x|$ is not differentiable at $x=0$.
So I am checking the left & right limits of $f$ when $x$ approaches $0$.


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*$\lim_{x \to 0^-}\cfrac{x|x|}{x} = \lim_{x \to 0^-}\cfrac{x(-x)}{x}=\lim_{x \to 0^-}\cfrac{(-x)}{1} = -0? = 0. $

*$\lim_{x \to 0^+}\cfrac{x|x|}{x} = \lim_{x \to 0^4}\cfrac{x(x)}{x}=\lim_{x \to 0^+}\cfrac{(x)}{1} = 0. $
I think that $f$ is not differentiable at $x=0$ since $|x|$ is not differentiable at that point. So , what do I do wrong?
Should I write something like $\lim_{x \to 0^-}\cfrac{x|x|}{x} = -0^{-}$ and $\lim_{x \to 0^+}\cfrac{x|x|}{x} =0^{+}$ so that $f'$ does not exist at $x=0$?

You didn't do anything wrong, you in fact shown that $f$ is differentiable at $x = 0$ and $f'(0) = 0$. The fact that $x \mapsto |x|$ is not differentiable at $x = 0$ doesn't mean that if you consider the product of this function with another function then the result won't be differentiable at $x = 0$, as you have just shown. 
However, since you have shown that $f'(x) = 2|x|$ you can see that $f$ is not twice differentiable at $x = 0$.
