What's the generalisation of the quotient rule for higher derivatives? I know that the product rule is generalised by Leibniz's general rule and the chain rule by Faà di Bruno's formula, but what about the quotient rule? Is there a generalisation for it analogous to these? Wikipedia mentions both  Leibniz's general rule and Faà di Bruno's formula for the product and the chain rule, but rather nothing for the quotient rule. 

The answer is:
$\frac{d^n}{dx^n}  \left (\frac{f(x)}{g(x)} \right ) = \sum_{k=0}^n {(-1)^k \tbinom{n}{k} \frac{d^{n-k}\left(f(x)\right)}{dx^{n-k}}}\frac{A_k}{g_{(x)}^{k+1}} $
where:
$A_0=1$
$A_n=n\frac{d\left(g(x)\right)}{dx}\ A_{n-1}-g(x)\frac{d\left(A_{n-1}\right)}{dx}$
for example let $n=3$:
$\frac{d^3}{dx^3}  \left (\frac{f(x)}{g(x)} \right ) =\frac{1}{g(x)} \frac{d^3\left(f(x)\right)}{dx^3}-\frac{3}{g^2(x)}\frac{d^2\left(f(x)\right)}{dx^2}\left[\frac{d\left(g(x)\right)}{d{x}}\right] + \frac{3}{g^3(x)}\frac{d\left(f(x)\right)}{d{x}}\left[2\left(\frac{d\left(g(x)\right)}{d{x}}\right)^2-g(x)\frac{d^2\left(g(x)\right)}{dx^2}\right]-\frac{f(x)}{g^4(x)}\left[6\left(\frac{d\left(g(x)\right)}{d{x}}\right)^3-6g(x)\frac{d\left(g(x)\right)}{d{x}}\frac{d^2\left(g(x)\right)}{dx^2}+g^2(x)\frac{d^3\left(g(x)\right)}{dx^3}\right]$
Relation with Faa' di Bruno coefficents:
The $A_n$ have also a combinatorial form, similar to the Faa' di Bruno coefficents (ref http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno).
An explication via an example (with for shortness
$g'=\frac{d\left(g(x)\right)}{dx}$, $g''=\frac{d^2\left(g(x)\right)}{dx^2}$, etc.):
Let we want to find $A_4$.
The partitions of 4 are: $1+1+1+1, 1+1+2, 1+3, 4, 2+2$.
Now for each partition we can use the following pattern:
$1+1+1+1 \leftrightarrow C_1g'g'g'g'=C_1\left(g'\right)^4$
$1+1+2+0 \leftrightarrow C_2g'g'g''g=C_2g\left(g'\right)^2g''$
$1+3+0+0 \leftrightarrow C_3g'g'''gg=C_3\left(g\right)^2g'g'''$
$4+0+0+0 \leftrightarrow C_4g''''ggg=C_4\left(g\right)^3g''''$
$2+2+0+0 \leftrightarrow C_5g''g''gg=C_5\left(g\right)^2\left(g''\right)^2$
with $C_i=(-1)^{(4-t)}\frac{4!t!}{m_1!\,m_2!\,m_3!\,\cdots 1!^{m_1}\,2!^{m_2}\,3!^{m_3}\,\cdots}$ (ref. closed-form of the Faà di Bruno coefficents)
where $t$ is the numers of partition items different of $0$, and $m_i$ is the numer of i.
We have $C_1=24$ (with $m_1=4, t=4$), $C_2=-36$ (with $m_1=2, m_2=1, t=3$), $C_3=8$ (with $m_1=1, m_3=1, t=2$), $C_4=-1$ (with $m_4=2, t=1$), $C_5=6$ (with $m_2=2,t=2$).
Finally $A_4$ is the sum of the formula found for each partition, i.e.
$A_4=24\left(g'\right)^4-36g\left(g'\right)^2g''+8\left(g\right)^2g'g'''-\left(g\right)^3g''''+6\left(g\right)^2\left(g''\right)^2$
