# Faà di Bruno's formula

{{#invoke: Sidebar | collapsible }} Faà di Bruno's formula is an identity in mathematics generalizing the chain rule to higher derivatives, named after Template:Harvs, though he was not the first to state or prove the formula. In 1800, more than 50 years before Faà di Bruno, the French mathematician Louis François Antoine Arbogast stated the formula in a calculus textbook, considered the first published reference on the subject.

Perhaps the most well-known form of Faà di Bruno's formula says that

${d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,1!^{m_{1}}\,m_{2}!\,2!^{m_{2}}\,\cdots \,m_{n}!\,n!^{m_{n}}}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left(g^{(j)}(x)\right)^{m_{j}},$ where the sum is over all n-tuples of nonnegative integers (m1, …, mn) satisfying the constraint

$1\cdot m_{1}+2\cdot m_{2}+3\cdot m_{3}+\cdots +n\cdot m_{n}=n.\,$ Sometimes, to give it a memorable pattern, it is written in a way in which the coefficients that have the combinatorial interpretation discussed below are less explicit:

${d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,m_{2}!\,\cdots \,m_{n}!}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left({\frac {g^{(j)}(x)}{j!}}\right)^{m_{j}}.$ Combining the terms with the same value of m1 + m2 + ... + mn = k and noticing that m j has to be zero for j > n − k + 1 leads to a somewhat simpler formula expressed in terms of Bell polynomials Bn,k(x1,...,xnk+1):

${d^{n} \over dx^{n}}f(g(x))=\sum _{k=1}^{n}f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots ,g^{(n-k+1)}(x)\right).$ ## Combinatorial form

The formula has a "combinatorial" form:

${d^{n} \over dx^{n}}f(g(x))=(f\circ g)^{(n)}(x)=\sum _{\pi \in \Pi }f^{(\left|\pi \right|)}(g(x))\cdot \prod _{B\in \pi }g^{(\left|B\right|)}(x)$ where

• "B ∈ π" means the variable B runs through the list of all of the "blocks" of the partition π, and
• |A| denotes the cardinality of the set A (so that |π| is the number of blocks in the partition π and |B| is the size of the block B).

## Explication via an example

The combinatorial form may initially seem forbidding, so let us examine a concrete case, and see what the pattern is:

{\begin{aligned}(f\circ g)''''(x)&=f''''(g(x))g'(x)^{4}+6f'''(g(x))g''(x)g'(x)^{2}\\[8pt]&{}\quad +\;3f''(g(x))g''(x)^{2}+4f''(g(x))g'''(x)g'(x)\\[8pt]&{}\quad +\;f'(g(x))g''''(x).\end{aligned}} The pattern is

{\begin{aligned}g'(x)^{4}&&\leftrightarrow &&1+1+1+1&&\leftrightarrow &&f''''(g(x))&&\leftrightarrow &&1\\[12pt]g''(x)g'(x)^{2}&&\leftrightarrow &&2+1+1&&\leftrightarrow &&f'''(g(x))&&\leftrightarrow &&6\\[12pt]g''(x)^{2}&&\leftrightarrow &&2+2&&\leftrightarrow &&f''(g(x))&&\leftrightarrow &&3\\[12pt]g'''(x)g'(x)&&\leftrightarrow &&3+1&&\leftrightarrow &&f''(g(x))&&\leftrightarrow &&4\\[12pt]g''''(x)&&\leftrightarrow &&4&&\leftrightarrow &&f'(g(x))&&\leftrightarrow &&1.\end{aligned}} The factor $g''(x)g'(x)^{2}\;$ corresponds to the partition 2 + 1 + 1 of the integer 4, in the obvious way. The factor $f'''(g(x))\;$ that goes with it corresponds to the fact that there are three summands in that partition. The coefficient 6 that goes with those factors corresponds to the fact that there are exactly six partitions of a set of four members that break it into one part of size 2 and two parts of size 1.

Similarly, the factor $g''(x)^{2}\;$ in the third line corresponds to the partition 2 + 2 of the integer 4, (4, because we are finding the fourth derivative), while $f''(g(x))\,\!$ corresponds to the fact that there are two summands (2 + 2) in that partition. The coefficient 3 corresponds to the fact that there are ${\tfrac {1}{2}}{\tbinom {4}{2}}=3$ ways of partitioning 4 objects into groups of 2. The same concept applies to the others.

A memorizable scheme is as follows:

$\displaystyle n=\underbrace {1+\cdots +1} _{m_{1}}\,+\,\underbrace {2+\cdots +2} _{m_{2}}\,+\,\underbrace {3+\cdots +3} _{m_{3}}+\cdots$