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!

Given that $y=\dfrac{e^{ax}}{1+e^{ax}}$ we have 
\begin{eqnarray}y^\prime&=&\frac{ae^{ax}}{1+e^{ax}}\\
&=&a(1-y)y
\end{eqnarray}
from which it follows that
$$ y^{\prime\prime}=a^2(1-2y)(1-y)y $$
etc, for $0< y<1$
So for each $n$ there is a polynomial $P_n(y)$ of degree $n$ such that
$$ y^{(n)}= a^nyP_n(y)\text{ for }y\in(0,1)$$
so the existence of an upper bound is assured.
Addendum: Actually. $(1-y)$ will also be factor of $y^{(n)}$ for $n>0$ so we can make the slightly stronger statement
$$ y^{(n)}= a^ny(1-y)Q_n(y)\text{ for }y\in(0,1) $$
where $Q_n$ is a degree $n-1$ polynomial in $y$.
We have $Q_1(y)=1,\,Q_2(y)=1-2y,\,Q_3(y)=1-6y+6y^2,\cdots$
The sequence of $Q_n(y)$ polynomials are computed recursively by


*

*$Q_1(y)=1$

*$Q_{n+1}(y)=\dfrac{d}{dy}y(1-y)Q_n(y)$
Second Addendum:
\begin{align*}
Q_1(y)&=1\\
Q_2(y)&=1-2y\\
Q_3(y)&=1-6y+6y^2\\
Q_4(y)&=1-14y+36y^2-24y^3\\
Q_5(y)&=1-30y+150y^2-240y^3+120y^4\\
Q_6(y)&=1-62y+540y^2-1560y^3+1800y^4-720y^5\\
Q_7(y)&=1-126y+1806y^2-8400y^3+16800y^4-15120y^5+5040y^6\\
Q_8(y)&=1-254y+5796y^2-40834y^3+126000y^4-191520y^5+141120y^6-40320y^7
\end{align*}
In general
\begin{equation}
Q_n(y)=\sum_{m=0}^{n-1}a(n,m)y^m
\end{equation}
where $\vert a(n,m)\vert$ equals the number of surjections of an $n$-element set onto an $m+1$ element set (according to the Online Encyclopedia of Number Sequences).
Furthermore
\begin{equation}
a(n,m)=\sum_{k=1}^{m+1}(-1)^{k+1}\binom{m+1}{k}k^n
\end{equation}
