Does $$ \left(\frac{\partial f(x)}{\partial x}\right)^{-1} = \frac{\partial x}{\partial f(x)}?$$
Why?
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Inversion as an operator or inversion as reciprocal? – Randall Apr 08 '19 at 15:05
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3What would the partial derivative of $x$ with respect to $f$ even be? – Arthur Apr 08 '19 at 15:06
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Inversion as a reciprocal because I do not know what inversion as an operator means. – Christina Daniel Apr 08 '19 at 15:21
1 Answers
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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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