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

Question

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.

Answer 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.

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

Answer 2

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

Answer 3

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.

Answer 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.

Answer 5

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.

Answer 6

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".

Answer 7

Let $u=u(z)$ and $v=v(z)\ne0$ be differentiable functions. On page 40 in the book [1] below, the formula \begin{equation}\label{Sitnik-Bourbaki}\tag{1} \frac{\textrm{d}^k}{\textrm{d}z^k}\biggl(\frac{u}{v}\biggr) =\frac{(-1)^k}{v^{k+1}} \begin{vmatrix} u & v & 0 & \dotsm & 0\\ u' & v' & v & \dotsm & 0\\ u'' & v'' & 2v' & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \ddots & \dotsm\\ u^{(k-1)} & v^{(k-1)} & \binom{k-1}1v^{(k-2)} & \dots & v\\ u^{(k)} & v^{(k)} & \binom{k}1v^{(k-1)} & \dots & \binom{k}{k-1}v' \end{vmatrix} \end{equation} for the $k$th derivative of the ratio $\dfrac{u(z)}{v(z)}$ was listed.

The formula \eqref{Sitnik-Bourbaki} can be reformulated as follows.

Let $u(t)$ and $v(t)\ne0$ be differentiable functions, let $U_{(n+1)\times1}(t)$ be an $(n+1)\times1$ matrix whose elements $u_{k,1}(t)=u^{(k-1)}(t)$ for $1\le k\le n+1$, let $V_{(n+1)\times n}(t)$ be an $(n+1)\times n$ matrix whose elements \begin{equation*} v_{i,j}(t)= \begin{cases} \dbinom{i-1}{j-1}v^{(i-j)}(t), & i-j\ge0\\ 0, & i-j<0 \end{cases} \end{equation*} for $1\le i\le n+1$ and $1\le j\le n$, and let $|W_{(n+1)\times(n+1)}(t)|$ denote the determinant of the $(n+1)\times(n+1)$ lower Hessenberg matrix \begin{equation*} W_{(n+1)\times(n+1)}(t)=\begin{bmatrix}U_{(n+1)\times1}(t) & V_{(n+1)\times n}(t)\end{bmatrix}_{(n+1)\times(n+1)}. \end{equation*} Then the $n$th derivative of the ratio $\dfrac{u(t)}{v(t)}$ can be computed by \begin{equation}\label{Sitnik-Bourbaki-reform}\tag{2} \frac{\textrm{d}^n}{\textrm{d}x^n}\biggl[\frac{u(t)}{v(t)}\biggr] =(-1)^n\frac{|W_{(n+1)\times(n+1)}(t)|}{v^{n+1}(t)}. \end{equation}

On 25 September 2014, Professor Sergei M. Sitnik (Voronezh Institute of the Ministry of Internal Affairs of Russia) told me the above derivative formula \eqref{Sitnik-Bourbaki} and the reference [1] via e-mails. Let us appreicate Professor Sergei M. Sitnik sincerey here.

The above derivative formulas \eqref{Sitnik-Bourbaki} and \eqref{Sitnik-Bourbaki-reform} have been applied in the papers from [2] to [26] below, for example.

References

1. N. Bourbaki, Elements of Mathematics: Functions of a Real Variable: Elementary Theory, Translated from the 1976 French original by Philip Spain. Elements of Mathematics (Berlin). Springer-Verlag, Berlin, 2004; available online at http://dx.doi.org/10.1007/978-3-642-59315-4.
2. Feng Qi, Omran Kouba, and Issam Kaddoura, Computation of several Hessenberg determinants, Mathematica Slovaca 70 (2020), no. 6, 1521--1537; available online at https://doi.org/10.1515/ms-2017-0445.
3. Can Kizilates, Wei-Shih Du, and Feng Qi, Several determinantal expressions of generalized Tribonacci polynomials and sequences, Tamkang Journal of Mathematics 53 (2022), no. 3, 275--289; available online at https://doi.org/10.5556/j.tkjm.53.2022.3743.
4. F. Qi, Derivatives of tangent function and tangent numbers, Appl. Math. Comput. 268 (2015), 844--858; available online at http://dx.doi.org/10.1016/j.amc.2015.06.123.
5. F. Qi and R. J. Chapman, Two closed forms for the Bernoulli polynomials, J. Number Theory 159 (2016), 89--100; available online at http://dx.doi.org/10.1016/j.jnt.2015.07.021.
6. Feng Qi, Determinantal expressions and recursive relations of Delannoy polynomials and generalized Fibonacci polynomials, Journal of Nonlinear and Convex Analysis 22 (2021), no. 7, 1225--1239.
7. Bai-Ni Guo, Emrah Polatli, and Feng Qi, Determinantal formulas and recurrent relations for bi-periodic Fibonacci and Lucas polynomials, Chapter 18 in the Springer Proceedings of the International Conference on Advances in Mathematics and Computing (ICAMC-2020) organized by Veer Surendra Sai University of Technology, Odisha, India, during 7-8 February 2020; Susanta Kumar Paikray, Hemen Dutta, and John N. Mordeson (eds.), New Trends in Applied Analysis and Computational Mathematics, pp. 263--276, Springer Book Series Advances in Intelligent Systems and Computing, Volume 1356, Singapore, 2021; available online at https://doi.org/10.1007/978-981-16-1402-6_18.
8. Yan Hong, Bai-Ni Guo, and Feng Qi, Determinantal expressions and recursive relations for the Bessel zeta function and for a sequence originating from a series expansion of the power of modified Bessel function of the first kind, Computer Modeling in Engineering & Sciences 129 (2021), no. 1, 409--423; available online at https://doi.org/10.32604/cmes.2021.016431.
9. Feng Qi, Muhammet Cihat Dagli, and Wei-Shih Du, Determinantal forms and recursive relations of the Delannoy two-functional sequence, Advances in the Theory of Nonlinear Analysis and its Applications 4 (2020), no. 3, 184--193; available online at https://doi.org/10.31197/atnaa.772734.
10. Feng Qi, Determinantal expressions and recurrence relations for Fubini and Eulerian polynomials, Journal of Interdisciplinary Mathematics 22 (2019), no. 3, 317--335; available online at https://doi.org/10.1080/09720502.2019.1624063.
11. Muhammet Cihat Dagli and Feng Qi, Several closed and determinantal forms for convolved Fibonacci numbers, Discrete Mathematics Letters 7 (2021), 14--20; available online at https://doi.org/10.47443/dml.2021.0039.
12. Yan Wang, Muhammet Cihat Dagli, Xi-Min Liu, and Feng Qi, Explicit, determinantal, and recurrent formulas of generalized Eulerian polynomials, Axioms 10 (2021), no. 1, Article 37, 9 pages; available online https://doi.org/10.3390/axioms10010037.
13. Feng Qi, Can Kizilates, and Wei-Shih Du, A closed formula for the Horadam polynomials in terms of a tridiagonal determinant, Symmetry 11 (2019), no. 6, Article 782, 8 pages; available online at https://doi.org/10.3390/sym11060782.
14. Feng Qi, Jing-Lin Wang, and Bai-Ni Guo, A determinantal expression for the Fibonacci polynomials in terms of a tridiagonal determinant, Bulletin of the Iranian Mathematical Society 45 (2019), no. 6, 1821--1829; available online at https://doi.org/10.1007/s41980-019-00232-4.
15. Feng Qi, Jing-Lin Wang, and Bai-Ni Guo, A representation for derangement numbers in terms of a tridiagonal determinant, Kragujevac Journal of Mathematics 42 (2018), no. 1, 7--14; available online at https://doi.org/10.5937/KgJMath1801007F.
16. Feng Qi, Jiao-Lian Zhao, and Bai-Ni Guo, Closed forms for derangement numbers in terms of the Hessenberg determinants, Revista de la Real Academia de Ciencias Exactas Fisicas y Naturales Serie A Matematicas 112 (2018), no. 4, 933--944; available online at https://doi.org/10.1007/s13398-017-0401-z.
17. Feng Qi and Bai-Ni Guo, A determinantal expression and a recurrence relation for the Euler polynomials, Advances and Applications in Mathematical Sciences 16 (2017), no. 9, 297--309.
18. Feng Qi and Bai-Ni Guo, Expressing the generalized Fibonacci polynomials in terms of a tridiagonal determinant, Le Matematiche 72 (2017), no. 1, 167--175; available online at https://doi.org/10.4418/2017.72.1.13.
19. Feng Qi and Bai-Ni Guo, Two nice determinantal expressions and a recurrence relation for the Apostol--Bernoulli polynomials, Journal of the Indonesian Mathematical Society (MIHMI) 23 (2017), no. 1, 81--87; available online at https://doi.org/10.22342/jims.23.1.274.81-87.
20. Feng Qi, A determinantal representation for derangement numbers, Global Journal of Mathematical Analysis 4 (2016), no. 3, 17--17; available online at https://doi.org/10.14419/gjma.v4i3.6574.
21. Feng Qi and Bai-Ni Guo, Some determinantal expressions and recurrence relations of the Bernoulli polynomials, Mathematics 4 (2016), no. 4, Article 65, 11 pages; available online at https://doi.org/10.3390/math4040065.
22. C.-F. Wei and F. Qi, Several closed expressions for the Euler numbers, J. Inequal. Appl. 2015:219, 8 pages; available online at http://dx.doi.org/10.1186/s13660-015-0738-9.
23. Feng Qi and Chuan-Jun Huang, Computing sums in terms of beta, polygamma, and Gauss hypergeometric functions, Revista de la Real Academia de Ciencias Exactas Fisicas y Naturales Serie A Matematicas 114 (2020), Paper No. 191, 9 pages; available online at https://doi.org/10.1007/s13398-020-00927-y.
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25. M. C. Dagli, Closed formulas and determinantal expressions for higher-order Bernoulli and Euler polynomials in terms of Stirling numbers, Rev. R. Acad. Cienc. Exactas Fis. Nat. Ser. A Mat. RACSAM 115 (2021), no. 1, Art. No. 32, 8 pages; available online at https://doi.org/10.1007/s13398-020-00970-9.
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Answer 8

Quotient Rule is actually Product Rule.

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

Answer 9

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

Answer 10

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.