Compute the $m$th derivative of the sigmoid function $x(t)=1/(1+e^{-at})$

Question

Compute the $m$th derivative of the sigmoid function $x(t)=1/(1+e^{-at})$.

Can you express the $m$th derivative of a sigmoid function like a recursive function?

$a$ is a constant

$$x(t) = \frac{1}{1+e^{-at}}$$

$x'=\left(\frac{1}{1+e^{-at}}\right)'$
$=((1+e^{-at})^{-1})'$ $=(-(1+e^{-at})^{-2})(-ae^{-at})$
$=\frac{ae^{-at}}{(1+e^{-at})^{2}}$ $=\frac{a}{1+e^{-at}}\frac{e^{-at}+1-1}{1+e^{-at}} $
$=\frac{a}{1+e^{-at}}\left(1-\frac{1}{1+e^{-at}}\right)$ $=ax(t)(1-x(t))$

$x''=(ax(1-x))'$
$=a(x'(1-x)+x(1-x)')$
$=a\left\{ax(1-x)(1-x)+x(-ax(1-x))\right\}$
$=a(ax(1-x)^{2}-ax^{2}(1-x))$
$=a^{2}x(1-x)(1-x-x)$
$=a^{2}x(1-x)(1-2x)$

$p=x(1-x),q=(1-2x)$

$p'=ax(1-x)(1-2x)=apq$

$q'=(1-2x)'=-2ax(1-x)=-2ap$

$q^{2} $ $=(1-2x)^{2}$ $=(1-4x+4x^{2})$ $=(1-4x(1-x))$ $=(1-4p)$

$x^{(3)}=(a^{2}pq)'$
$=a^{2}(p'q+pq')$
$=a^{2}(apqq+p(-2ap))$
$=a^{2}(apq^{2}-2ap^{2})$
$=a^{2}(ap(1-4p)+p(-2ap))$
$=a^{3}p((1-4p)+(-2p))$
$=a^{3}p(1-6p)$

$x'=ap$

$x''=a^{2}pq$

$x^{(3)}=a^{3}p(1-6p)$
$x^{(4)}=a^{4}pq(1-12p)$
$x^{(5)}=a^{5}p(1-30p+120p^{2})$
$x^{(6)}=a^{6}pq(1-60p+360p^{2})$
$x^{(7)}=a^{7}p(1-126p+1680p^{2}-5040p^{3})$
$x^{(8)}=a^{8}pq(1-252p+5040p^{2}-20160p^{3})$

Postscript

redefinition

$\frac{d^m x}{dt^m}=a^m P_m(x)$

$P_0=x,\,P_{m+1}=x(1-x)P_m'$

$b=x(1-x)$
$b'=(1-2x)$
$b''=-2$
$b'''=0$

$P_1=x(1-x)P_0'=x(1-x)x'=b$
$P_2=x(1-x)P_1'=x(1-x){x(1-x)}'=x(1-x)(1-2x)=bb'$
$P_3=x(1-x)P_2'=x(1-x)(x(1-x)(1-2x))'=b(bb')'=b(b'b'+bb'')$
$P_4=x(1-x)P_3'=b(b((b')^2+bb''))'=b((b')^3+bb'b''+2bb'+bb'b'')=b((b')^3+4bb'b'')$

$\frac{d^1 x}{dt^1}=a^1 P_1(x)=ax(1-x)=ab$
$\frac{d^2 x}{dt^2}=a^2 P_2(x)=a^{2}bb'$
$\frac{d^3 x}{dt^3}=a^3 P_3(x)=a^{3}b((b')^2+bb'')$
$\frac{d^4 x}{dt^4}=a^4 P_4(x)=a^{4}b((b')^3+4b(b')^1b'')$

$x'=ab$
$x''=a^{2}b(b')$
$x^{(3)}=a^{3}b((b')^2+b(b')^0b'')$
$x^{(4)}=a^{4}b((b')^3+4b(b')^1b'')$
$x^{(5)}=a^{5}b((b')^4+11b(b')^2b''+4b^2(b')^0(b'')^2)$
$x^{(6)}=a^{6}b((b')^5+26b(b')^3b''+34b^2(b')^1(b'')^2)$
$x^{(7)}=a^{7}b((b')^6+57b(b')^4b''+180b^2(b')^2(b'')^2+34b^3(b')^0(b'')^3)$

$x^{(1)}=abC_{0,0}S_{0,0,0}$
$x^{(2)}=a^{2}bC_{1,0}S_{0,1,0}$
$x^{(3)}=a^{3}b(C_{2,0}S_{0,2,0} + C_{2,1}S_{1,0,1})$
$x^{(4)}=a^{4}b(C_{3,0}S_{0,3,0} + C_{3,1}S_{1,1,1})$
$x^{(5)}=a^{5}b(C_{4,0}S_{0,4,0} + C_{4,1}S_{1,2,1} + C_{4,2}S_{2,0,2})$
$x^{(6)}=a^{6}b(C_{5,0}S_{0,5,0} + C_{5,1}S_{1,3,1} + C_{5,2}S_{2,1,2})$
$x^{(7)}=a^{7}b(C_{6,0}S_{0,6,0} + C_{6,1}S_{1,4,1} + C_{6,2}S_{2,2,2} + C_{6,3}S_{3,0,3})$

$(m\geqq0),(n\geqq0)$

$ C_{m-1,n}= \left\{ \begin{array}{ll} 0 & (n=0) \\ 1 & (m=2n) \\ (m-2n-2)C_{m-2,n-1}+(n+1)C_{m-2,n} & (else) \\ \end{array} \right. $

$S_{p,q,r} = (b)^{p}(b')^{q}(b'')^{r}$

$K_m = \frac{2m+(-1)^{m+1}-3}{4}$

$$x^{(m)} = a^{m}b\left\{\sum_{n=0}^{K_m}C_{m-1,n}S_{n,m-2n-1,n} \right\}$$

Answer 1

Applying the higher chain rule (Faà di Bruno's formula) in its form with Partial Exponential Bell Polynomials ($B_{n,k}$), we get:

$$x(t)=\frac{1}{1+e^{-at}}$$

$$x(t)=\frac{1}{f(t)};\ f(t)=1+e^{-at}$$

$$\frac{d^n}{dt^n}x(t)=\frac{d^n}{dt^n}\frac{1}{f(t)}=\sum_{k=0}^n(-1)^kk!f(t)^{-(k+1)}B_{n,k}(f(t))$$

$$\frac{d^k}{dt^k}f(t)=(1-k)_k+(-a)^ke^{-at}$$

$$B_{n,k}(f(t))=(-1)^nS_{n,k}a^ne^{-kat}$$

$$\frac{d^n}{dt^n}x(t)=\sum_{k=0}^n(-1)^{n+k}k!S_{n,k}a^n(1+e^{-at})^{-(k+1)}e^{-kat}$$

$$\frac{d^n}{dt^n}x(t)=\sum_{k=0}^n(-1)^{n+k}k!S_{n,k}a^nx(t)(1-x(t))^k$$

$$x^{(m)}=\sum_{k=0}^m(-1)^{m+k}k!S_{m,k}a^mx(1-x)^k$$

$(r)_k$ are the ascending factorials (Pochhammer function), $S_{n,k}$ are the Stirling numbers of the second kind.

Answer 2

$$x(t)=1/(1+e^{-at})$$ $$x(t)=\sum_{k=0}^\infty (-1)^k e^{-akt}$$ $$x^{(m)}(t)=\sum_{k=0}^\infty (-1)^{k} (-ak)^m e^{-akt}$$ $$x^{(m)}(t)=(-a)^m\sum_{k=0}^\infty (-1)^{k} k^m e^{-akt}$$ This cannot be expressed in terms of a finite number of elementary functions. A closed form requiers a special function, that is the Lerch function or a Polylogarithm function : $$x^{(m)}(t)=(-a)^m \Phi\left(-e^{-at}\:,\:-m\:,\:0 \right)$$ $$x^{(m)}(t)=(-a)^m \text{Li}_{-m}\left(-e^{-at} \right)$$ This comes directly from the series definition of the Lerch function : http://mathworld.wolfram.com/LerchTranscendent.html and the series definition of the polylogarithm functions : http://mathworld.wolfram.com/Polylogarithm.html