This question is from baby Rudin, Ch 5, exer 2:
Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=\frac1{f'(x)}$.
The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'
Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]\subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])\to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.
So, how to show that $g$ indeed continuous on $(a,b)$?
It is sufficient to show that $f$ is an open map. So for any $(c,d)\subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.