Is the function $x|x|$ differentiable at $0$? Is the function $x|x|$ differentiable at $0$? I have calculated the left and right derivatives and both are $0$, so it is differentiable. am I right?  
 A: In general, if the function $g(x)$ is continuous at a point $x_0$, then the function $f(x)=(x-x_0)g(x)$ is differentiable at $x_0$ because $f'(x_0)=\lim\limits _{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim\limits _{x\to x_0}g(x)=g(x_0).$
In particular take $g(x)=|x|, x_0=0$
A: The function $f(x)=x\left|x\right|$
  is defined by $\begin{cases}
x^{2} & x\geq0\\
-x^{2} & x<0
\end{cases}$.
 If $x\neq0$
 , then the function is a quadratic, so it is differentiable. The only point you need to worry about is $0$. But since both $x^{2}$
  and $-x^{2}$
  have the same derivative at $0$
 , then it follows that $f$
  is differentiable at $0$
 .
A: Yes if the left derivate and right derivates at a point coincides then the function is derivable at that point with that derivate. As you've (correctly) identified these as both being $0$ then it follows the function is derivable at $0$ with the derivate $0$.
The left derivate is defined as
$$ f'(a) = \lim_{x\to a^-} {f(x)-f(a)\over x-a}$$
and the right derivate is defined similarily while the derivate is defined as
$$ f'(a) = \lim_{x\to a} {f(x)-f(a)\over x-a}$$
So if the LD and RD is the same then the normal derivate is defined and the same too. This relies on that if L-limit and R-limit exists and is the same then the limit is defined and the same too (which in turn follows directly from the definition of limits).
A: More generally: for any $f$ with $|f(x)|\le Ax^2$:
$$f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{f(x)}x = 0$$
by squeeze theorem:
$$x\ne 0\implies -A|x|\le\frac{f(x)}x\le A|x|.$$
A: Yes, you are right. 
You can define function as $$f(x)=x^2 ;x>0 ~~~-x^2; x<0$$
It's LHD and RHD are both $=0$
