Derivative of inverse functions

When we do a composition of a function and its inverse the result is that we do some operation, undo it with the reversed operation, which results in the output and the input being equal to each other. In mathematical notation: [math]\displaystyle{ f(f^{-1}(x)) = x }[/math]. For now we skip the conditions for which a function is invertible. To make the proof easier to read let's write [math]\displaystyle{ f^{-1}(x) = g(x) }[/math]:

The rate of change of [math]\displaystyle{ x }[/math] is trivial, it's 1. If [math]\displaystyle{ \frac{d}{dx}x = 1 }[/math], then [math]\displaystyle{ \frac{d}{dx}f(g(x)) }[/math] is [math]\displaystyle{ g'(x)f'(g(x)) }[/math] by the chain rule.

Now we have this (it's the concept of implicit differentiation):

[math]\displaystyle{ g'(x)f'(g(x)) = 1 }[/math]

Which can be rewritten as

[math]\displaystyle{ g'(x) = \frac{1}{f'(g(x))} }[/math]