Dtjytyk

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:

$A_0=1$

$A_n=nfrac{dleft(g(x)right)}{dx} A_{n-1}-g(x)frac{dleft(A_{n-1}right)}{dx}$

for example let $n=3$:

$frac{d^3}{dx^3} left (frac{f(x)}{g(x)} right ) =frac{1}{g(x)} frac{d^3left(f(x)right)}{dx^3}-frac{3}{g^2(x)}frac{d^2left(f(x)right)}{dx^2}left[frac{dleft(g(x)right)}{d{x}}right] + frac{3}{g^3(x)}frac{dleft(f(x)right)}{d{x}}left[2left(frac{dleft(g(x)right)}{d{x}}right)^2-g(x)frac{d^2left(g(x)right)}{dx^2}right]-frac{f(x)}{g^4(x)}left[6left(frac{dleft(g(x)right)}{d{x}}right)^3-6g(x)frac{dleft(g(x)right)}{d{x}}frac{d^2left(g(x)right)}{dx^2}+g^2(x)frac{d^3left(g(x)right)}{dx^3}right]$

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
$g'=frac{dleft(g(x)right)}{dx}$, $g''=frac{d^2left(g(x)right)}{dx^2}$, 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:

$1+1+1+1 leftrightarrow C_1g'g'g'g'=C_1left(g'right)^4$

$1+1+2+0 leftrightarrow C_2g'g'g''g=C_2gleft(g'right)^2g''$

$1+3+0+0 leftrightarrow C_3g'g'''gg=C_3left(gright)^2g'g'''$

$4+0+0+0 leftrightarrow C_4g''''ggg=C_4left(gright)^3g''''$

$2+2+0+0 leftrightarrow C_5g''g''gg=C_5left(gright)^2left(g''right)^2$

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.

$A_4=24left(g'right)^4-36gleft(g'right)^2g''+8left(gright)^2g'g'''-left(gright)^3g''''+6left(gright)^2left(g''right)^2$

As others have pointed out, the quotient rule is actually a form of a product rule. Just using Leibniz rule for getting higher order derivatives of product of a function-take a look here.

http://en.wikipedia.org/wiki/General_Leibniz_rule

I found a pdf online that had a result for a general formula for $$frac{d^n}{dx^n} left (frac{f(x)}{g(x)} right ). $$

Although I cannot find the resource again (I am looking because it had a proof), one formula is
$$frac{d^n}{dx^n} left (frac{f(x)}{g(x)} right )=frac{1}{g(x)} (f^{(n)}(x))-n! sum_{j=1}^n frac{g^{(n+1-j)}(x)}{(n+1-j!)} frac{ left (frac{f(x)}{g(x)} right)^{{{(j-1)}}}} {(j-1)!}.$$ Now don't attribute this to me, as I referenced from a source I am trying to find again. It, for me, is impractical and apply the product rule for $fcdot g^{-1}$ is a lot easier, but I think the general formula is pretty good to know.

Quotient Rule is actually Product Rule.

$D(u/v) = D(uw)$ where $w = 1/v$.

I'm checking @Mohammad Al Jamal's formula with SymPy, and I can verify it's true (barring a missing $(-1)^k$ term) for up to $n = 16$, at least (it gets really slow after that).

import sympy as sp



k = sp.Symbol('k'); x = sp.Symbol('x'); f = sp.Function('f'); g = sp.Function('g')



n = 0

while True:

    fgn = sp.diff(f(x) / g(x), x, n)

    guess = sp.summation((-1) ** k * sp.binomial(n + 1, k + 1) 

                         * sp.diff(f(x) * (g(x)) ** k, x, n)/(g(x) ** (k + 1)), (k, 0, n))

    print("{} for n = {}".format(sp.expand(guess - fgn) == 0, n))

    n += 1

This is quite surprising to me -- I didn't expect there to be such a simple and straightforward expression for $(f(x)/g(x))^{(n)}$, and haven't seen his formula anywhere before. I tried some inductive proofs, but I haven't succeeded in proving it yet.

i state this without proof, for the proof is tedious and lengthy.
$$left(frac{f(x)}{g(x)}right)^{(n)}=frac{1}{g(x)}sum_{k=0}^{n}(-1)^{k}binom{n+1}{k+1}frac{left(f(x)g^{k}(x)right)^{(n)}}{g^{k}(x)}
$$

The answer by Stopple should be corrected as follows.
$$frac{d^n}{dx^n} left (frac{f(x)}{g(x)} right )=frac{1}{g(x)} left( f^{(n)}(x)-n! sum_{j=1}^n frac{g^{(n+1-j)}(x)}{(n+1-j!)} frac{ left (frac{f(x)}{g(x)} right)^{{{(j-1)}}}} {(j-1)!} right). $$
If we let $h(x)=f(x)/g(x)$, above can be also written as follows suppressing the independent variable $x$
$$ h^{(n)} = frac{1}{g} left( f^{(n)} -sum_{j=1}^{n} binom{n}{j} h^{(n-j)}g^{(j)} right).$$

The proof is straightforward by induction.

The aim is to get a non-recursive expression in dependence of the higher derivatives of the functions in the numerator and denominator. The $n$-th derivative of the reciprocal of an arbitrary function doesn't have a simple expression. It has to be treated as composition of functions and needs Faà di Bruno's formula therefore.

Applying General Leibniz rule and Faà di Bruno's formula, one gets the following Higher Quotient Rule:

$$frac{d^n}{dx^n}frac{f(x)}{g(x)}=sum_{i=0}^{n}{nchoose i}f^{(i)}(x)sum_{k=0}^{n-i}(-1)^{k}k!g(x)^{-k-1}B_{n-i,k}(g(x)).$$

$B_{n,k}(g(x))=B_{n,k}(g^{(1)}(x),g^{(2)}(x),...,g^{(n-k+1)}(x))$ is the partial exponential Bell polynomial of the second kind:

$$B_{n,k}(g(x))=sum_{sum_{t=1}^{n}tk_{t}=natopsum_{t=1}^{n}k_{t}=k}frac{n!}{prod_{i=1}^{n}i!^{k_{i}}k_{i}!}prod_{i=1}^{n}{g^{(i)}(x)}^{k_{i}}.$$

In 2012, I wrote a still unpublished article "On partial Bell polynomials for the higher derivatives of composed functions".