Let $I_1,I_2$ be intervals. Let $f:I_1\rightarrow I_2$ be a bijective function, and $g:I_2\rightarrow I_1$ be the inverse. Suppose that both $f$ is differentiable at $c\in I_1$ and $f^\prime (c) \not = 0$ and $g$ is differentiable at $f(c)$. Use the chain rule to find a formula for $g^\prime (f(c))$ (in terms of $f^\prime (c)).$
Attempt: $g \circ f = g(f(x))=h(x)$. Then, $h(x)$ is differentiable at $c$. $h^\prime (c)= g^\prime (f(c))f^\prime (c) \rightarrow g^\prime (f(c))= \frac {h^\prime (c)}{f^\prime (c)}$. I have to express $h^\prime (c)$ in terms of $f^\prime (c)$, but don't know how to do it. Could you give some hint for this?
Thank you in advance!
Note that $h$ is the identify, $h(x)=x$. Thus $h'(c)=1$, which is all you need.