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I have encountered here that sometimes people write $dx/df$. For example, here, the author of the answer writes down the chain rule as $$\frac{dg}{df}=\frac{dg}{dx}\frac{dx}{df},$$ which seems to be an abuse of notation. And $dx/df$ is found to be just $\frac{1}{df/dx}$.

Here, I learned that when someone writes "differentiate $f$ with respect to $g$", what they mean is to differentiate some hidden composition, which seems to rely upon an inverse of what is said to be a quantity with respect to which we want to differentiate (there, they have $\phi(t)=2t$, whereas they differentiate $f(x)=x^2$ with respect to $x/2$, and the composition is $f\circ\phi$), but I may be wrong here.

So, if the goal is to "differentiate $x$ with respect to $f$", do we actually have $$(x\circ h)'(u)=x'(h(u))h'(u),$$ where $h(u):=f^{-1}(u)$ and $x(u):=u$? This seems to agree with what $dx/df$ is to be found: $$(x\circ h)'(u)=\frac{1}{f'(f^{-1}(u))}=\frac{dx}{df}.$$

noballpointpen
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    Correct, $x$ needs to be interpreted as the identity map ($x(u)=u$) and $h=f^{-1}$, so $x\circ h=f^{-1}$, and this is the function you differentiate. But really though, I would strongly urge you to not think in terms of ‘differentiating one function with respect to another’. – peek-a-boo Feb 23 '24 at 14:20
  • @peek-a-boo, yeah, it made a lot of headache for me revolving around strict concepts of differentiation to even understand what that answering user even did. – noballpointpen Feb 23 '24 at 19:16
  • haha that “answering user” is me :) – peek-a-boo Feb 23 '24 at 19:29
  • @peek-a-boo, no, actually your answer there is very thorough and nice. I meant Deepak, https://math.stackexchange.com/a/954085/1174491, the first link in my question. – noballpointpen Feb 23 '24 at 19:31
  • @peek-a-boo, also, I think you may amend your answer there with explicitly saying that $\phi$ is an inverse in general, at least for reals. It was the only one point which made some doubt. – noballpointpen Feb 23 '24 at 19:33
  • Right, having an inverse is a special case, but not the general one which is why I intentionally avoided saying that (and instead gave a paragraph about the three sets $X,Y,Z$ and functions between them). Oh and another reason why “differentiating a function with respect to another” (atleast a naive interpretation) is bad is because it doesn’t generalize to higher dimensions. See this answer of mine for some further details. – peek-a-boo Feb 23 '24 at 19:39

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Your confusion seems to originate in the notation. Let's have an independent variable $x$ and a dependent one, namely $y = f(x)$, hence the "standard" derivative $\frac{\mathrm{d}y}{\mathrm{d}x} = f'(x)$. Then, their roles can be reversed, so that $x = f^{-1}(y)$. In consequence, one has $\frac{\mathrm{d}x}{\mathrm{d}y} = (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} = \frac{1}{f'(x)}$, which is what you found. However, this development may become messy when the same symbol is used for both $y$ and $f$ (which itself isn't $y$ but a map describing the relation between $x$ and $y$).

Abezhiko
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