The following is paraphrased from Stewart's Calculus text.

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

Proof: We have $\displaystyle g'(a)=\displaystyle{\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}$. Since $\displaystyle f$ and $\displaystyle g$ are inverses we have $\displaystyle g(x)=y$ iff $\displaystyle f(y)=x$ and $\displaystyle g(a)=b$ iff $\displaystyle f(b)=a$.

Since $\displaystyle f$ is differentiable, it is continuous, so $\displaystyle g$ is continuous by earlier theorem. Thus as $\displaystyle x\rightarrow a$ we have $\displaystyle g(x)\rightarrow g(a)$, or $\displaystyle y\rightarrow b$. Therefore:

$\displaystyle g'(a)=\displaystyle{\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}$

**$\displaystyle =\displaystyle{\lim_{y\rightarrow b}\frac{y-b}{f(y)-f(b)}}$

...

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

The starred step is the one I'm having difficulty with understanding fully. I understand the continuity argument that as $\displaystyle x\rightarrow a$ we have $\displaystyle y\rightarrow b$, but it seems like a bit of a leap, at least I can't think of any definitions or theorems to apply, to use that to change $\displaystyle \lim_{x\rightarrow a}$ to $\displaystyle \lim_{y\rightarrow b}$ (although an $\displaystyle \epsilon - \delta$ argument seems to work in the opposite direction--that $\displaystyle \lim_{y\rightarrow b}h(x)=L$ would imply that $\displaystyle \lim_{x\rightarrow a}h(x)=L$). It seems that the claim in Stewart's argument also has to do with the fact that $\displaystyle f$ and $\displaystyle g$ are one-to-one. Any thoughts?