# Can we replace the sufficient conditions by the new ones in the following theorem?

Theorem: Let $$I\subset \mathbb R$$ be an interval and a function $$f:I \to \mathbb R$$ be strictly monotone and continuous on $$I$$.Let $$g:=f^{-1}:f(I)\to \mathbb R$$.If $$f$$ is differentiable at $$c\in I$$ and $$f'(c)\neq0$$ then $$g$$ is differentiable at $$f(c)$$ and $$g'(f(c))=\frac{1}{f'(c)}$$ .

In this theorem can we replace 'strictly monotone and continuous' by the assumptions that $$f$$ is 'invertible and continuous' or by 'injective and continuous' or by 'injective and having IVP on $$I$$'.

I think any of these conditions will act the same i.e. they will all provide me with the facts I need to prove the original version of the theorem. Another question is how to generalize this theorem for arbitrary domain $$X$$.

Assume $$x are points in $$I$$ and, w.l.o.g, let's assume $$f(x) < f(y)$$. ($$f(x)=f(y)$$ would contradict the fact that $$f$$ is invertible).
Now assume $$x < r < y$$. Claim: $$f(x) < f(r) < f(y)$$. Assume that this is not true. The $$f(x)\ge f(r)$$ of $$f(r)\ge f(y)$$. Since both cases are similar, I'll only look at the first one. $$f(x)=f(r)$$ is impossible, so $$f(x) > f(r)$$. But then be the mean value theorem there is $$s \in (r,y)$$ sucht that $$f(s) = f(x)$$, contradicting the invertibility of $$f$$.
• And the fact $f'(c)\neq 0$ is essential because we want to avoid the situation of infinite derivative of inverse(although it is a bit informal to say,because we do not consider the differential quotient's infinite limit as a derivative).e.g.$f(x)=x^3$ with $f^{-1}(x)=x^{\frac{1}{3}}$. – Kishalay Sarkar Dec 23 '19 at 9:07
• How to generalize this theorem for an arbitrary domain $X$ in place of interval domain $I$. – Kishalay Sarkar Dec 23 '19 at 9:48