The following is paraphrased from Stewart's Calculus text.
Theorem: If is a one-to-one differentiable function with inverse function and , then the inverse function is differentiable at and
Proof: We have . Since and are inverses we have iff and iff .
Since is differentiable, it is continuous, so is continuous by earlier theorem. Thus as we have , or . Therefore:
The starred step is the one I'm having difficulty with understanding fully. I understand the continuity argument that as we have , 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 to (although an argument seems to work in the opposite direction--that would imply that ). It seems that the claim in Stewart's argument also has to do with the fact that and are one-to-one. Any thoughts?
Snowtea, that makes sense, and helps to alleviate the problem I was having.
What I was thinking was that, for example, though implies ,
whereas doesn't make sense since is not a function of on . However, since is a continuous function of here, the switch is justified.
So I suppose the more rigorous argument I was searching for would go as follows:
Given one to one continuous function such that exists and is equal to , we claim that . Given , choose such that
implies that by continuity and injectivity of ,
where is small enough so that
by the assumed limit.