Meaning of derivative of function by absolute value: $\frac{d}{d|x|} f(|x|)$ What does the following Leibniz's notation of derivative mean?
$$\frac{d}{d|x|} f(|x|)$$
$f(|x|)$ is a function of absolute value of variable x.
I am OK with this notation and I know how to treat it:
$$\frac{d}{dx} f(|x|)$$
What is the difference between those two?
 A: You'd need to define its meaning. Given that the idea of the derivative is based on making small changes $dx$ to a function and analyzing the resulting difference quotient
$$\frac{f(x+dx)-f(x)}{dx}$$
a reasonable definition could be
$$\frac{d}{d|x|}f(x)=\lim_{|h|\to 0}\frac{f(x+|h|)-f(x)}{|h|}$$
provided that the "limit" in question exists. By substituting $|x_0|$ for $x$, your “derivative” emerges as a special case.
Limit operations like $\lim_{|h|\to 0}$ are not addressed by the usual $(\varepsilon,\delta)$ definition, so we would also need to clarify what we mean by something like $\lim_{|h|\to 0}$. Using the standard $(\varepsilon,\delta)$ definition as a model, one possibility is to declare the following:

For real functions $f$ and $g$ with "suitable" domains, we say that $\lim_{g(x)\to a}f(x)=L$ if and only if for every $\varepsilon>0$, there is a $\delta>0$ such that for every $x\in\text{dom}[f]\cup\text{dom}[g]$, if $0<|g(x)-a|<\delta$, then $|f(x)-L|<\varepsilon$.

Expanding $\frac{d}{d|x|}f(|x|)=\lim_{|h|\to 0}\frac{f(|x|+|h|)-f(|x|)}{|h|}$ with this definition, $\frac{d}{d|x|}f(|x|)$ would be the real number with the property that for every $\varepsilon>0$, there is a $\delta>0$ such that for every $h\in\mathbb{R}$ (the domain of the absolute value is $\mathbb{R}$),
$$0<||h|-0|<\delta\implies\left|\frac{f(|x|+|h|)-f(|x|)}{|h|}-\frac{d}{d|x|}f(|x|)\right|<\varepsilon$$
Of course, there's nothing special about the absolute value here. For a general function $g$ and some $x_0\in\text{dom}[f]$, you could define
$$\frac{df}{dg}(x_0):=\lim_{g(h)\to 0}\frac{f(x_0+g(h))-f(x_0)}{g(h)}$$
