Higher-order derivative of $1/(1+e^x)$ Let 
$$
f(x) = \frac{1}{1+e^x}
$$
1) Is there a closed formula for the $k$-derivative $f^{(k)}(x)$?
2) Is there a simple argument which shows for $k \geq 1$ that
$$
\lim_{x \to \pm \infty} f^{(k)}(x) = 0 ?
$$
Thanks!
 A: Write $e^x=:u$ for short. Then there is a sequence $(p_n)_{n\geq0}$ of polynomials such that
$$f^{(n)}(x)={p_n(u)\over(1+u)^{n+1}}\qquad(n\geq0)\ .$$
It is easy to show that the $p_n$ satisfy the recursion
$$p_0=1,\qquad p_{n+1}=(u+u^2)\, p_n'-(n+1) u\, p_n\qquad(n\geq0)\ .$$
E.g., one obtains
$$p_6(u)=u (-1+57 u-302 u^2+302 u^3-57 u^4+u^5)\ .$$
A: Is not full answer only for 1):
Note that:
$$\sum _{j=0}^{\infty } (-1)^j e^{j x}=\frac{1}{1+e^x}$$
so,
$$\color{red}{\frac{\partial ^k}{\partial x^k}\frac{1}{1+e^x}}=\sum _{j=0}^{\infty } \frac{\partial
   ^k\left((-1)^j e^{j x}\right)}{\partial x^k}=\sum _{j=0}^{\infty } (-1)^j e^{j x}
   j^k=\color{red}{\Phi \left(-e^x,-k,0\right)}$$
where $\Phi \left(-e^x,-k,0\right)$ is: Hurwitz-Lerch transcendent function.
A: Probably cheating, but (2) is clear from a little complex analysis: It's easy to see that $$\lim_{x\to\infty}f(x+iy)=0$$uniformly for $-1\le y\le 1$; hence Cauchy's Estimates show that $\lim_{x\to\infty}f^{(k)}(x)=0$.
The nicest calculus-only proof of (2) is implicit in one of the comments: Since $f'=f^2-f$ it follows by induction that $f^{(k)}$ is a linear combination of $f,f^2,\dots$.
A: 1) The derivatives of $\frac{1}{e^t+1}$ can be computed from Euler polynomials and reindexing;
2) $$\frac{1}{e^t+1}=\frac{1-\tanh(t/2)}{2}=\frac{1}{2}-\sum_{n\geq 0}\frac{2t}{\pi^2(2n+1)^2+t^2}\\=\frac{1}{2}-\sum_{n\geq 0}\left(\frac{1}{t+\pi i(2n+1)}+\frac{1}{t-\pi i(2n+1)}\right)$$
by considering the logarithmic derivative of the Weierstrass product for the $\cosh$ function. By applying termwise differentiation to the last series, $\lim_{t\to \pm\infty} f^{(n)}(t) = 0$ simply follows from the dominated convergence theorem. $\frac{1}{e^t+1}$ only has simple poles away from the real axis, equally spaced on the imaginary axis. As a shortcut, $\frac{1}{\cosh^2(t)}$ clearly is an element of the Schwartz space $\mathcal{S}(\mathbb{R})$.
