How to show $\frac{d}{d x}\left(|x|^{1/2}\right)=\frac{x}{2|x|^{3/2}}$? $$\frac{d}{d x}\left(|x|^{\frac{1}{2}}\right)=\frac{x}{2|x|^{\frac{3}{2}}}$$
and also the second derivative $$\frac{d^{2}}{d x^{2}}\left(\sqrt{|x|}\right)=\frac{\delta(x)}{\sqrt{|x|}}-\frac{x^{2}}{4|x|^{\frac{7}{2}}}$$
I know this may seem basic. But I don't get how the Dirac delta (thanks for the comments for correcting me) comes in. Is it just a way someone has decided to write the result for $x >0$, $x<0$ and $x=0$ cleverly in one line. Is it arbitrary as long as the Kronecker term beats the other term?
If anyone has a clean derivation for these I'd be grateful.
EDIT: thanks for a lot of your answers, I've learnt a lot.
Still not entirely sure with the second derivative, the formula is here: https://www.wolframalpha.com/input/?i=second+derivative+of+%7Cx%7C%5E1%2F2, is this wrong? I still think it is probably correct. From taking something like. d/dx($\operatorname{sign}(x) \frac{1}{2|x|^{\frac{1}{2}}} )=1/2 ($$\frac{\delta(x)}{\sqrt{|x|}} -\frac{sign(x)*x}{2|x|^{5 / 2}})$ Don't know why first summand is missin 1/2. I know its distributions, but can't you still differentiate sign(x), you just have to be careful and when evaluating use test functions and integrals

This is not a complete answer, but for the case of the absolute function defined on the real numbers, this might help.

Since $|x|$ is non-differentiable only at $0$, all the derivatives we calculate will have domain $D \subseteq\mathbb{R} \setminus \{0\}$.
If we write $|x|=\sqrt{x^2}$ we see that
$$
\frac{d}{dx} \left(|x| ^\frac{1}{2} \right) = \frac{d}{dx} \left(x^2\right)^\frac{1}{4} = \frac{1}{4}\left(x^2\right)^{-\frac{3}{4}} \left[\frac{d}{dx}x^2 \right]= \frac{1}{4}\left(x^2\right)^{-\frac{3}{4}}2x = \frac{x}{2\left(x^2\right)^\frac{3}{4}}  = \frac{x}{2\left(\sqrt{x^2}\right)^\frac{3}{2}} = \frac{x}{2|x|^\frac{3}{2}}
$$
As for the second part
$$
\frac{d^2}{dx^2} \left(|x| ^\frac{1}{2} \right) = \frac{d}{dx} \left(\frac{x}{2\left(x^2\right)^\frac{3}{4}}\right) = \frac{2\left(x^2\right)^\frac{3}{4} - 2x\left(\frac{3}{4}(x^2)^{-\frac{1}{4}} (2x)\right)}{4\left(x^2\right)^\frac{3}{2}} = \frac{2\left(x^2\right)^\frac{3}{4}}{4\left[\left(x^2\right)^\frac{3}{4}\right]^2} - \left(\frac{3}{4}\right)\frac{4x^2}{4\left(x^2\right)^\frac{3}{2}(x^2)^{\frac{1}{4}}} = \frac{1}{2\left(x^2\right)^\frac{3}{4}}\left(\frac{2x^2}{2x^2}\right) - \frac{3x^2}{4\left(x^2\right)^\frac{7}{4}} =  \frac{2x^2 -3x^2} {4\left(x^2\right)^\frac{7}{4}} = -\frac{x^2}{4 |x|^\frac{7}{2}}
$$
Now, since the Dirac delta turns it's input into $0$ if $x \neq 0$, and the domain on which we're working on doesn't have $0$, our last answer will be equal to
$$
-\frac{x^2}{4 |x|^\frac{7}{2}} = 0 -\frac{x^2}{4 |x|^\frac{7}{2}} = \frac{\delta(x)}{\sqrt{|x|}} -\frac{x^2}{4 |x|^\frac{7}{2}}
$$
since $\frac{\delta(x)}{\sqrt{|x|}}=0$ for all $x \in D$.
I suspect there might be a more general definition of the absolute value you can work with to get the Dirac delta out directly, but I'm not familiar with it. Hope this helps!
