Derivative of $|x|^\alpha$ I want to calculate the derivative of $|x|^\alpha$ with respect to $x$, where $1<\alpha<2$.
$\frac{d|x|^\alpha}{dx}=\alpha|x|^{\alpha-1}\mathrm{sign}(x)$
Is it correct?
And is it true that
$\lim_{x \to 0} \alpha|x|^\alpha\mathrm{sign}(x)=0$?
What with the following derivative $|x|^\alpha\rm{sign}(x)$. I would be grateful for any hints. 
 A: Assume $x_0 > 0$. We can assume that $|h| < x_0$ so that $x_0 + h > 0$. We get
$$\lim_{h \to 0}\frac{|x_0+h|^\alpha-|x_0|^\alpha}{h} = \lim_{h \to 0}\frac{(x_0+h)^\alpha-x_0^\alpha}{h} = \alpha x_0^{\alpha-1} = \alpha |x_0|^{\alpha-1}$$
since the latter limit is simply the derivative of $x \mapsto x^\alpha$ at $x_0$.
Similarly, if $x_0 < 0$ and $|h| < -x_0$, we have $x_0 + h < 0$ so
$$\lim_{h \to 0}\frac{|x_0+h|^\alpha-|x_0|^\alpha}{h} = \lim_{h \to 0}\frac{(-x_0-h)^\alpha-(-x_0)^\alpha}{h} = -\alpha (-x_0)^{\alpha-1} =-\alpha |x_0|^{\alpha-1}$$
since the latter limit is simply the derivative of $x \mapsto (-x)^\alpha$ at $x_0$.
If $x_0 = 0$, we have
$$\lim_{h \to 0}\left|\frac{|h|^\alpha}{h}\right| = \lim_{h \to 0}\frac{|h|^\alpha}{|h|} = \lim_{h\to 0} |h|^{\alpha-1} = 0$$
and hence $\lim_{h \to 0}\frac{|h|^\alpha}{h}=0$.
Putting everything together, we get that the derivative of $x\mapsto |x|^\alpha$ is $x \mapsto \alpha|x|^{\alpha-1}\operatorname{sign}(x)$.

Yes, we have $\lim_{x\to 0}\alpha|x|^\alpha\operatorname{sign}(x) = 0$ because
$$\lim_{x\to 0}\big|\alpha|x|^\alpha\operatorname{sign}(x)\big| = \lim_{x\to 0}\alpha|x|^\alpha = 0$$

For the function $x \mapsto |x|^\alpha\operatorname{sign}(x)$ we obtain the derivative $x \mapsto \alpha|x|^{\alpha-1}$ similarly as above.
A: Hint:
$$\frac{d}{dx}\mathrm{sign} (x)=2\delta (x)$$
Where $\delta (x)$ is the Dirac delta function. You can reach the required result by using the product rule.
