Why does the Sigmoid function use e?

Question

From what I can guess, because the Sigmoid function uses $S(x)=\frac{1}{1+e^{-x}}$, I would think that there would be some connection with the fact that the derivative of $e^x$ is itself. What that connection is (assuming I'm not just on the wrong track all together), I am uncertain of.

Answer 1

If you use a different constant that is the same as rescaling $x$

$$ b^{-x} = (e^{\ln b})^{-x} = e^{ - x\ln b} $$

So rename $x\ln b \to x$. Might as well use $e$ because that is the one that makes the differential equation look nicer.