How to invert the derivative of the logistic function?

Question

I am trying to find the inverse of the logistic function:

we have $\sigma(x) = \frac{1}{1 + e^x}$

$f(x) = \sigma'(x) = \sigma(x)(1 - \sigma(x)) = \frac{e^x}{(1 + e^x)^2}$

$f^{-1}(x) = ?$

The result is easily visible from wolfram alpha, but it just provides the result without proof. I am wondering how it is possible to isolate for x given some value of f(x). Taking logarithm of both sides leads me to trouble with the $(1+e^x)^2$ term being hard to isolate for x.

Answer 1

Following the advice from Bernard Masse's comment I am able to successfully invert the function by substituting $w = e^x$ and solving the equation using the quadratic formula since you can divide both sides by $\sigma'(x)$, as $\sigma'(x) > 0$ for all x

Answer 2

Since $\sigma(x)=\tfrac12(1+\tanh\tfrac{x}{2})$, $f(x)=\tfrac14\operatorname{sech}^2\tfrac{x}{2}$, so $x=2\operatorname{arcosh}\tfrac{1}{2\sqrt{f}}=2\ln\tfrac{1+\sqrt{1-4f}}{2\sqrt{f}}$.