I'm reading the proof for the following theorem.

If f is a one-to-one differentiable function with inverse function $\displaystyle g=f^{-1}$ and $\displaystyle f'(g(a)) \neq 0$, then the inverse function is differentiable at

*a* and

$\displaystyle g'(a)=\frac{1}{f'(g(a))}$

The Calculus textbook that I'm reading gives two proofs, the first one uses Newton's difference quotient, which I can understand.

The 2nd goes like this:

Replacing

*a* by the general number

*x* in the theorem, we get

$\displaystyle g'(x)=\frac{1}{f'(g(x))}$

If we write $\displaystyle y=g(x)$, then $\displaystyle f(y)=x$, so the equation can be expressed in Leibniz notation.

$\displaystyle \frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$

IF IT IS KNOWN IN ADVANCE THAT $\displaystyle f^{-1}$ IS DIFFERENTIABLE, then its derivative can be computed more easily by using implicit differentiation.

$\displaystyle f(y)=x$

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

$\displaystyle \frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{\frac{dx}{d y}}$

My question is this, I don't understand which step of the proof requires the fact that $\displaystyle f^{-1}$ is differentiable

Thanks a lot if you can help