Derivative at the origin For which $a\in {\bf R}$ is $|x|^a$ differentiable at $x=0$ and what is the derivative there?
 A: Hint: Let $a\gt 1$. Calculate $\displaystyle\lim_{x\to 0} \frac{|x|^a-0}{x-0}$. It is useful to work separately with the limit from the right and the limit from the left.
If $x\gt 0$, then $\dfrac{|x|^a-0}{x-0}=\dfrac{x^a}{x}=x^{a-1}$.
If $x\lt 0$, then $\dfrac{|x|^a-0}{x-0}=\dfrac{|x|^a}{-|x|}=-|x|^{a-1}$.
Now we need to show that the limit does not exist when $a\le 1$. For $a=1$, we are working with the familiar absolute value function, and the limit of the differential quotient from the right is $1$, while the limit from the left is $-1$.
If $a\lt 1$, the differential quotient blows up as $x$ approaches $0$ from the right.
A: Hint: take the definition of differential and separate it to left and right limits.
A: By definition, $f$ is called differentiable at the point $x_0$, if its increment $Δf=f(x_0+h)−f(x_0)$ may be represented as $$f(x_0+h)−f(x_0)=f'(x_0)h+α(x_0,h),$$
where $\alpha$ satisfies 
$|\alpha(x_0,h)|=o(|h|),\quad h\rightarrow 0$. Thus is sufficient to write increment for $f(x)=|x|^\alpha$ at point $x_0=0$ for $h>0$ and  $h<0$
