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Does $$ \left(\frac{\partial f(x)}{\partial x}\right)^{-1} = \frac{\partial x}{\partial f(x)}?$$

Why?

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  • $\begingroup$ Inversion as an operator or inversion as reciprocal? $\endgroup$ – Randall Apr 8 at 15:05
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    $\begingroup$ What would the partial derivative of $x$ with respect to $f$ even be? $\endgroup$ – Arthur Apr 8 at 15:06
  • $\begingroup$ Inversion as a reciprocal because I do not know what inversion as an operator means. $\endgroup$ – Christina Daniel Apr 8 at 15:21
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Maybe you want to check out the Inverse Function Theorem. It says that if a function $f$ is differentiable at a point $(x,f(x)) = (a,b)$ then there exists "some neighbourhood" around this point where there must exist an invese function $f^{-1}(x)$ which is also differentiable there (around $b$) and for which the following must hold:

$$(f^{-1})'(b) = \frac{1}{f'(a)}$$

This is kind of the same thing you are asking.

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