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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.

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that's because you can apply Faa di Bruno's formula to g(x)^{-1} and then the product rule to f(x) and g(x)^{-1}. – Qiaochu Yuan Sep 24 '10 at 5:00
up vote 1 down vote accepted

As others have already said, you just apply the product rule to $f.g^{-1}.$ However, the is an American Mathematical Monthly article on how NOT to do it, which you may find instructive.

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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 $f\cdot g^{-1}$ is a lot easier, but I think the general formula is pretty good to know.

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Found the pdf, here. – M. Knight Feb 28 '14 at 1:52

Quotient Rule is actually Product Rule.

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

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Please help me with the notation. – Pratik Deoghare Sep 24 '10 at 5:02

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.

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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}} $



$A_n=n\frac{d\left(g(x)\right)}{dx}\ A_{n-1}-g(x)\frac{d\left(A_{n-1}\right)}{dx}$

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

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.


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