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

I hardly understand English.
I wrote a question using translation.
I want to generalize.

$a$ 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}} $
$=frac{a}{1+e^{-at}}left(1-frac{1}{1+e^{-at}}right)$
$=ax(t)(1-x(t))$

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

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

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

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

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

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

$x'=ap$

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

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

C are constant
$n = 0,1,2,3....$

$x^{(m)}$ $(m=1,3,5...)$

$(p(C+Cp+cdots+Cp^{n}))'$
$=p'(C+Cp+cdots+Cp^{n})+p(C+Cp+cdots+Cp^{n})'$
$=apq(C+Cp+cdots+Cp^{n})+p(Capq+cdots+nCp^{n-1}apq)$
$=qpq(C+Cp+cdots+Cp^{n})+apq(Cp+cdots+nCp^{n})$
$=apq(C+2Cp+cdots+(n+1)Cp^{n})$
$=apq(C+Cp+cdots+Cp^{n})$ $----(1)$

$x^{(m)}$ $(m=2,4,6...)$

$(pq(C+Cp+cdots+Cp^{n}))'$
$=(pq)'(C+Cp+cdots+Cp^{n})+pq(C+Cp+cdots+Cp^{n})'$
$=(pq)'(C+Cp+cdots+Cp^{n})+pq(C+Cp+cdots+Cp^{n})'$
$=ap(1-6p)(C+Cp+cdots+Cp^{n})+pq(Capq+cdots+nCp^{n-1}apq)$
$=ap(1-6p)(C+Cp+cdots+Cp^{n})+apq^{2}(Cp+cdots+nCp^{n})$
$=apleft{(1-6p)(C+Cp+cdots+Cp^{n})+(1-4p)(Cp+cdots+nCp^{n})right}$
$=apleft{(1-6p)C+(2-10p)Cp+cdots+((n+1)-2(2n+3)p)Cp^{n}right}$
$=apleft{C+(-4)Cp+cdots+(-(3n+1))Cp^{n}+(-2(2n+3))Cp^{n+1}right}$
$=ap(C+Cp+cdots+Cp^{n+1})$ $----(2)$

$(1)->(2)->(1)->(2)->.........$

$x^{(m)}=?$

Postscript

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

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

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

$$x^{(m)}=a^{m}bBiggl((b')^{m-1} + left{sum_{k=2}^{m-1} 2^{k-2}(m-k) right}b(b')^{m-3}b'' + left{sum_{k=4}^{m-1} 3^{k-4}S_{left[1right]{m-k+2}}(m-k) right}b^2(b')^{m-5}(b'')^2 + left{sum_{k=6}^{m-1} 4^{k-6}S_{left[2right]{m-k+4}}(m-k)right}b^3(b')^{m-7}(b'')^3 + ...... Biggr)$$

$$S_{left[1right]{m}} = left{sum_{k=2}^{m-1} 2^{k-2}(m-k) right}$$
$$S_{left[2right]{m}} = left{sum_{k=4}^{m-1} 3^{k-4}S_{left[1right]{m-k+2}}(m-k) right}$$

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.

$$x(t)=1/(1+e^{-at})$$
$$x(t)=sum_{k=0}^infty (-1)^k e^{-akt}$$
$$x^{(m)}(t)=sum_{k=0}^infty (-1)^{k} (-ak)^m e^{-akt}$$
$$x^{(m)}(t)=(-a)^msum_{k=0}^infty (-1)^{k} k^m e^{-akt}$$
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 :
$$x^{(m)}(t)=(-a)^m Phileft(-e^{-at}:,:-m:,:0 right)$$
$$x^{(m)}(t)=(-a)^m text{Li}_{-m}left(-e^{-at} right)$$
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