Show that $f'^{-1}(x) = \frac{2}{1-x^2}$ for $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{e^x-1}{e^x+1}$

Question

$f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{e^x-1}{e^x+1}$

Given Information: $2\cdot f'(x) = 1 - f(x)^2$

$(a)$ What is the range of $f(x)$

$(b)$ Why is $f(x)$ invertible

$(c)$ Show that $f'^{-1}(x) = \frac{2}{1-x^2}$

---

Now the range is $[-1,1]$ and because it is a monotonically increasing function it is invertible. The question I have is: Can I show $(c)$ without actually finding the inverse function and deriving it? The next exercise would have been to show that $f^{-1}(x) = ln(\frac{1+x}{1-x})$ and because of the order of the questions I assume that it is possible to do it without the inverse function and the derivative.

Answer 1

Let $f(x)=g^{-1}(x)$ Using the formula, $$f(g(x))=x$$ $$f'(g(x))g'(x)=1$$ $$g'(x)=\frac{1}{f'(g(x))}$$

Using the formula,$$f'(x)=\frac{1}{f^{-1'}(f(x))}$$$$1-f(x)^2=\frac{1}{f^{-1'}(f(x))}$$$$f^{-1'}(f(x))=\frac{1}{1-f(x)^2}$$