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 $x &gt;0$, $x&lt;0$ 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 ($$\frac{\delta(x)}{\sqrt{|x|}} -\frac{sign(x)*x}{2|x|^{5 / 2}})$ Don't know why first summand is missin 1/2. I know its distributions, but can't you still differentiate sign(x), you just have to be careful and when evaluating use test functions and integrals
Let $f(x) = |x|^{1/2}$. Of course, away from $x=0$, the answer of Robert Lee (without any Dirac delta) is correct, and you will just have
$$
\begin{align*}
f'(x) &amp;= \frac{x}{2\,|x|^{3/2}}
\\
f''(x) &amp;= \frac{-1}{4\,|x|^{3/2}}
\end{align*}
$$
for any $x\in\mathbb{R}\setminus\{0\}$ (since $x^2/|x|^{7/2} = 1/|x|^{3/2}$).

So, to know if there is a Dirac delta, you have to take the derivative in the sense of distributions. To get the first derivative, since $f'(x) = \frac{x}{2\,|x|^{3/2}}$
everywhere except at $x=0$, we see that $f'(x) = \frac{x}{2\,|x|^{3/2}}$ almost everywhere. Moreover, since $|f'| = \frac{1}{2\,|x|^{1/2}}$ is locally integrable, we obtain that the identity $f'(x) = \frac{x}{2\,|x|^{3/2}}$ is also valid locally in $L^1$ and so also in the sense of distributions.
This means that to compute $f''$ in the sense of distributions, we just need to compute the derivative of $f'$ in the sense of distributions. Let $\varphi$ be a smooth compactly supported (let say $\varphi = 0$ in $[-a,a]$ for some $a&gt;0$) test function. Then by definition and the above computation
$$
\langle f'',\varphi\rangle = \langle(f')',\varphi\rangle = -\langle f',\varphi'\rangle = -\langle \tfrac{x}{2\,|x|^{3/2}},\varphi'\rangle
$$
and we can write this as an integral since both functions are locally integrable. Therefore
$$
\begin{align*}
\langle f'',\varphi\rangle &amp;= -\int_{\mathbb{R}} \tfrac{x}{2\,|x|^{3/2}}\,\varphi'(x)\,\mathrm{d}x = -\int_{-a}^a \tfrac{x}{2\,|x|^{3/2}}\,\varphi'(x)\,\mathrm{d}x
\\
&amp;= -\int_{-a}^a \tfrac{x}{2\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))'\,\mathrm{d}x
\end{align*}
$$
Since $\varphi(x)-\varphi(0)$ is vanishing at $x=0$ and even better, $|\varphi(x)-\varphi(0)| ≤ C\,|x|$ (by the fundamental theorem of calculus, with $C= \sup |\varphi'|$ for instance), we can integrate by parts and use the value of $f''$ that was already computed for $x≠0$ to get
$$
\begin{align*}
\langle f'',\varphi\rangle &amp;= \left[-\tfrac{x}{2\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\right]_{-a}^a - \int_{-a}^a \tfrac{1}{4\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x
\\
&amp;= - \int_{-a}^a \tfrac{1}{4\,|x|^{3/2}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x
\end{align*}
$$
since $\varphi(a) = \varphi(-a) = 0$. The distribution $\mathrm{pf}(\frac{1}{|x|^a})$ for $a\in(1,2)$ is known as the finite part of $\frac{1}{|x|^a}$ and is defined exactly by
$$
\begin{align*}
\langle \mathrm{pf}(\tfrac{1}{|x|^a}),\varphi\rangle &amp;= \int_{\mathbb{R}} \tfrac{1}{|x|^{a}}\,(\varphi(x)-\varphi(0))\,\mathrm{d}x
\end{align*}
$$
so in the sense of distributions
$$
f'' = \frac{-1}{4} \mathrm{pf}\left(\frac{1}{|x|^{3/2}}\right)
$$

Remarks: I do not know where you found the formula with a Dirac delta but it is obviously wrong, since there is no meaning to $\frac{\delta(x)}{\sqrt{x}}$.
Even if it is true that $\mathrm{sign}' = \delta_0$ in the sense of distributions, the following computation
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x}{|x|^{3/2}}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{sign}(x)\frac{1}{|x|^{1/2}}\right) = 2\delta_0(x)\left(\frac{1}{|x|^{1/2}}\right) + \mathrm{sign}(x)\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{|x|^{1/2}}\right)
$$
is not valid since one cannot multiply singular distributions in general.
Okay.... second go.....
If $1 =\int_{-\infty}^{\infty} \delta(x) dx$ then if we have a function $f(x)$ where $f(x)$ &quot;equals $\frac 10$&quot; when $x=0$  I guess can say:
$f(x) = 1*f(x) = [\int_{-\infty}^{\infty} \delta(x) dx]f(x)$ so
$f'(x)= [\int_{-\infty}^{\infty} \delta(x) dx]'f(x) + [\int_{-\infty}^{\infty} \delta(x) dx]f(x)= \delta(x)f(x) + f'(x)$.
When $x\ne 0$ then this is $f'(x)$.  If $x=0$ then... well I'm waving my hands wildly... but as $f(x)$ is assymptotic at $x=0$ then the derivative is infinite.
So.... $|x|^{\frac 12}$ is defined everywhere but has a &quot;divot&quot; at $x=0$.
the first derivative if $sign(x) \frac 1{2|x|^{\frac 12}} = \frac x{2|x|^{\frac 32}}$.
Not the first derivative &quot;equals $\frac 10$ at $x = 0$ so if we take that the first derivative is $[\int_{-\infty}^{\infty} \delta(x) dx]\cdot\frac x{2|x|^{\frac 32}}$ then the second derivative will be $\delta(x)sign(x) \frac 1{2|x|^{\frac 12}} + [sign(x) \frac 1{2|x|^{\frac 12}}]'=$
$\pm \delta(x)sign(x) \frac 1{2|x|^{\frac 12}} -\frac{x^{2}}{4|x|^{\frac{7}{2}}}$
.....
Okay, somehow in my not really knowing what I was doing and handwaving I handwaved $\frac {sign(x)}2$ in whereas the author doing whatever s/he was doing and knowing what she was doing probably had a reason they &quot;canceled&quot; out.
Actually, I think I can almost see why.  $sign(x) \frac 1{2|x|^{\frac 12}} = \frac x{2|x|^{\frac 32}}$ is negative of $x &lt; 0$ and positive otherwise so so we don't really have the first derivative of $[\int_{-\infty}^{\infty} \delta(x) dx]\cdot\frac x{2|x|^{\frac 32}}$ but the $sign(x)[\int_{-\infty}^{\infty} \delta(x) dx]\cdot\frac x{2|x|^{\frac 32}}$ and so the signs cancel out. And as we are doing it for negative and positive values of $x$ we ... are doing it twice????  Okay.... I'm handwaving again.
....
I don't know what the author was doing but I can see that as the first derivative is infinite at $x=0$ and as $\delta(x) =0$ whe $x \ne 0$ so the result is consistant and that weird chain rule for differentiating $f(x)_{x\ne 0}= 1 = \int \delta(x)dx$ I can squint and almost see what the author had in mind.