How to compute derivative I'm trying to recall rules of computing a derivative of function like this
$$\dfrac{dx}{d(\log x)}$$
Could you remind me a proper way to compute it and potentially references to read more.
Asking to understand better derivation of formula (11) in this paper: https://arxiv.org/pdf/1911.12487.pdf
 A: Employ chain rule. $\displaystyle \frac{dy}{dx }= \frac 1{\frac{dx}{dy}}$
So if $\displaystyle \frac{d(\log x)}{dx} = \frac 1x$, then $\displaystyle \frac{dx}{d(\log x)} =x$ 
It is unconventional notation, but it seems to cohere with what is in the paper.
A: Doing a "u-substitution" is the most straightforward and standard way to go about this:
$$\frac{dx}{d(\log x)} = \frac{d\left(e^{\log x}\right)}{d(\log x)} = \frac{d\left(e^u\right)}{du} = e^u = x$$
as this the sort of computation that pops up frequently in physics.
A: Observe that, generally $$\frac{d(g(x))}{dx}=g'(x)\implies d(g(x))=g'(x)dx$$
As a corollary, we see that $$\frac{d(f(x))}{d(g(x))}=\frac{d(f(x))}{dx}\cdot\frac{1}{g'(x)}=\frac{f'(x)}{g'(x)}$$
You have $f(x)=x, g(x)=\log(x)$

As an aside, this is the exact trick employed when using u-sub for integration, but it is never taught in a way that makes this clear. We take $$\int_\alpha^\beta g(x)dx=\int_{u(\alpha)}^{u(\beta)} g(u(x))d(u(x))=\int_{u(\alpha)}^{u(\beta)} g(u(x))u'(x)dx$$ of which you should recognise the third as the u-substitution formula.
