On the derivative of the inverse of a function

Question

I tried to prove it like this $$y=f(x)$$ $$x=f^{-1}(y)$$ $$dx=(f^{-1})'(y)dy$$ $$dx=(f^{-1})'(y)f'(x)dx$$ $$(f^{-1})'(y)=\frac{1}{f'(x)}$$ Does it lack rigor just because I am using differentials?
Is the last step valid nevertheless?
What about passing from considering $y$ an "independent" variable then reconsidering it as a function of $x$?

Answer 1

Under the right regularity assumptions on $f$ it is in fact a consequence of the chain rule $$(g \circ f)'(x) = f'(g(x)) \cdot g'(x)$$ applied to $g = f^{-1}$: We have that $g \circ f$ is the identity, thus its derivative is constant $1$. On the other hand, by the chain rule, $$(g \circ f)'(x) = g'(y) \cdot f(x),$$ from which we can conclude that $$(f^{-1})'(y) = \frac{1}{f'(x)}.$$

Your answer lacks rigor in that it is just a syntactic manipulation (also the last step). Using the chain rule is kind of the formalization of your last idea.

Answer 2

$$y=f(x)$$ Differentiate with respect to $y$ $$ 1=f'(x)\frac {dx}{dy}$$

$$\frac {dx}{dy}=\frac {1}{f'(x)}$$