I was looking at the logit function, $\operatorname{logit}(p)=\log\dfrac{p}{1-p}$, and I just realized that the sigmoid function, $S(x) = \dfrac{1}{1+e^{-x}}$, was its inverse function.
I could not come up with a way to get to the inverse from the original function, and it got me thinking if there was a way to derive a function inverse from the function itself?
$$y=\log\frac{x}{1-x}=\log\frac1{1-\dfrac1x}$$
$$e^y=\frac1{1-\dfrac1x}$$
$$e^{-y}=1-\dfrac1x$$
$$1-e^{-y}=\dfrac1x$$
$$\frac1{1-e^{-y}}=x$$
The trick I used is to isolate a single instance of the variable and see the function as a composition of elementary functions with known inverses.