# 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.

• 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

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.

$\frac{d^n}{dx^n} \left (\frac{f(x)}{g(x)} \right ) = \sum_{k=0}^n {(-1)^k \tbinom{n}{k} \frac{d^{n-k}\left(f(x)\right)}{dx^{n-k}}}\frac{A_k}{g_{(x)}^{k+1}}$

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.

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

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

• Found the pdf, here. – user124862 Feb 28 '14 at 1:52
• Your link doesn't work anymore... – draks ... Apr 11 '17 at 20:17
• Is the little $(j-1)$ in the superscript a power or a derivative? If it's a derivative, this formula is useless. – Abhimanyu Pallavi Sudhir Feb 28 '19 at 15:43
• I mean, it wouldn't be an explicit equation, just a recurrence relation (just missed the comment editing time). – Abhimanyu Pallavi Sudhir Feb 28 '19 at 15:50
• The title of the resource is "A formula for the nth derivative of the quotient of two functions". It currently can be found here. – YellPika Dec 7 '19 at 5:09

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.

• you're correct. there should be a $(-1)^{k}$ factor. i have corrected my answer accordingly. i have the proof somewhere. i will look for it and post it here. – Mohammad Al Jamal Mar 2 '19 at 12:48

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

• What the heck, this seems to be correct (testing upto $n = 2$), barring a missing $(-1)^k$ term. I wasn't expecting there to be such a simple expression. Any hints about how to go about proving this (how'd you find it, anyway)? – Abhimanyu Pallavi Sudhir Mar 1 '19 at 11:26
• you're correct. there should be a $(-1)^{k}$ factor. i have corrected my answer accordingly. i have the proof somewhere. i will look for it and post it here. – Mohammad Al Jamal Mar 2 '19 at 12:48
• Did you find it? – Abhimanyu Pallavi Sudhir Mar 4 '19 at 22:00
• a combination of Faa di bruno's formula, and induction. – Mohammad Al Jamal Mar 7 '19 at 8:29
• Ah, wait -- it seems it can be proven from the expression in giuseppe's answer. – Abhimanyu Pallavi Sudhir Mar 10 '19 at 15:32

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:

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

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

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

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