Bounding the derivative of a function Let $f(x) = \frac{\exp(ax)}{1+\exp(ax)}$ for some $a > 0$ be a logistic function.
I am looking for an upper bound on the $k$-th derivative of $f$, i.e. $|f^{(k)}(x)|$ on $\mathbb{R}$.
Using Cauchy's integral formula I only obtain a bound of order $k! / (\pi-\epsilon)^{k+1}$ for $a=1$. Is it possible to get below $k!$ and obtain a growth rate that is just exponential?
Thanks!

Using a CAS to get derivatives and OEIS, I conjecture that the $k$th derivative of $f(x)$ is:
$$f^{(k)}(x) = \frac{a^ke^{ax}}{(e^{ax}+1)^{k+1}}\sum_{m=0}^{k-1} (-1)^{m}A(k, m)e^{axm}$$
Where $A(k, m)$ are the Eulerian numbers. Does this help?
