If $f=f(y)$ and $x=x(y)$. Sometimes when computing a chain rule in physics I see this kind of calculations:
$$\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}=\frac{df}{dy}\left(\frac{dx}{dy}\right)^{-1},$$
when you can't easily express $y(x)$. So when is $\frac{dy}{dx}=\left(\frac{dx}{dy}\right)^{-1}$ true?
First of all, when is this true from a mathematically rigorous point of view? do you need f(y) and x(y) to have some strong properties? or is it an approximation?
Finally, as a physicist do I need to worry about this? or are functions that do not respect this equivalence rare?
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$\begingroup$You actually are talking about the inverse function theorem in one-dimension. The theorem is correctly stated as:
Let $f : \mathbb{R}\to \mathbb{R}$ be of class $C^1$ with nonzero derivative at the point $a$, that is, $f'(a)\neq 0$. Then $f$ is invertible in a neighborhood of $a$, its inverse $f^{-1}$ is $C^1$ and satisfies
$$(f^{-1})'(f(a))=\dfrac{1}{f'(a)}.$$
So these are the complete hypothesis and the correct notation.
Most of time in Physics functions are $C^\infty$, so physicists don't worry too much with this. Still, it won't work when the function is not $C^1$ and when $f'(a)=0$.
By the way, notice that a very important aspect of the rigorous approach is that the inverse exists and the formula holds in the neighborhood of a point. It isn't in the whole real line, as the physicist's notation might suggest.
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