How do you prove that $x|x|$ is differentiable at all points?

Question

In general how do you prove that if two functions are not differentiable at a point, then it is not necessary that their product is not differentiable at that point. (Which in $x|x|$ is $0$ )

Answer 1

In this particular case, of course, we can find the derivative.

\begin{eqnarray} \frac{d}{dx}x|x|&=&|x|\frac{dx}{dx}+x\frac{d|x|}{dx}\\ &=&|x|+x\cdot\frac{|x|}{x}\\ &=&|x|\left(1+\frac{x}{x}\right) \end{eqnarray} which is undefined at $x=0$ but which has limit $0$ at $x=0$. Since the function is continuous at $x=0$ and has a horizontal tangent at $x=0$ it is the case that $$ \frac{d}{dx}x|x|=2|x| $$ for all $x$.

ADDENDUM: Here is a complete proof that $\frac{d}{dx}(x|x|)=2|x|$ using the definition $$ f^\prime(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} $$

\begin{eqnarray} f^\prime(x)&=&\lim_{h\to0}\frac{(x+h)|x+h|-x|x|}{h}\cdot{\color{blue}{\frac{(x+h)|x+h|+x|x|}{(x+h)|x+h|+x|x|}}}\\ &=&\text{a few omitted algebra steps}\\ &=&\lim_{h\to0}\frac{4(x+h)^3+6(x+h)^2h+4(x+h)h+2+h^3}{(x+h)|x+h|+x|x|}\\ &=&\frac{4x^3}{2x|x|}\\ &=&2|x| \end{eqnarray}

Answer 2

You only need to check at $x = 0$ in which case $$f'(0) = \lim_{h\to 0}{h|h|\over h} = \lim_{h\to 0} |h| = 0.$$ The rest is clear.