What's the generalisation of the quotient rule for higher derivatives?

Solution 1:

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.

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

Solution 2:

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.

Solution 3:

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

Solution 4:

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.