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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

Deep Learner
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    You mean d(log(x))/dx ? – Jean Marie Apr 03 '20 at 05:06
  • The derivative of $\ln(x)$ (I switch to this notation) is $1/x$. You can obtain it directly by consideration of $\lim_{h \to 0} (\ln(x+h)-\ln(x))/h$, or with less calculations as the derivative of the inverse of exponential function $\exp$. – Jean Marie Apr 03 '20 at 05:32
  • No, I mean d(x)/d(log(x)), which is a part of formula (11) in this paper https://arxiv.org/pdf/1911.12487.pdf – Deep Learner Apr 03 '20 at 06:01
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    I have had a look at formula (11). IMHO, it is a very dangerous notation, prone to subsequent errors, and certainly not correct from the point of view of mathematics. They should have done firstly a change of variable $u=\ln(....)$ and then take the derivative with respect to $u$. – Jean Marie Apr 03 '20 at 06:10
  • @DeepLearner : You should add your source and background to your Question. You already have a nonresponsive answer because you did not. – Eric Towers Apr 03 '20 at 06:12
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    Jean Marie: it is a perfectly good notation, many people use it. I myself also do not like it that much, but sometimes it is convenient. To say that "it is a very dangerous notation" and "certainly not correct" is off the chart, though. $d \log(x) = \frac{1}{x} dx$, so it is perfectly fine to say that $dx/d \log(x) = dx/(\frac{1}{x} dx) = x$. I am even more baffled by the 3 votes to close. This is a decent post, the question is well-phrased. – A. Pongrácz Apr 03 '20 at 09:07

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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.

Deepak
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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.

Ninad Munshi
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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.

Rhys Hughes
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