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

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

Why?

• Inversion as an operator or inversion as reciprocal? – Randall Apr 8 at 15:05
• What would the partial derivative of $x$ with respect to $f$ even be? – Arthur Apr 8 at 15:06
• Inversion as a reciprocal because I do not know what inversion as an operator means. – Christina Daniel Apr 8 at 15:21

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