Faà di Bruno's formula

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Faà di Bruno's formula is an identity in mathematics generalizing the chain rule to higher derivatives. It is named after Francesco Faà di Bruno (1855, 1857), although 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 had stated the formula in a calculus textbook, which is considered to be the first published reference on the subject.

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

$${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,1!^{m_{1}}\,m_{2}!\,2!^{m_{2}}\,\cdots \,m_{n}!\,n!^{m_{n}}}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left(g^{(j)}(x)\right)^{m_{j}},}$$

where the sum is over all ${\displaystyle n}$-tuples of nonnegative integers ${\displaystyle (m_{1},\ldots ,m_{n})}$ satisfying the constraint

$${\displaystyle 1\cdot m_{1}+2\cdot m_{2}+3\cdot m_{3}+\cdots +n\cdot m_{n}=n.}$$

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:

${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum {\frac {n!}{m_{1}!\,m_{2}!\,\cdots \,m_{n}!}}\cdot f^{(m_{1}+\cdots +m_{n})}(g(x))\cdot \prod _{j=1}^{n}\left({\frac {g^{(j)}(x)}{j!}}\right)^{m_{j}}.}$

Combining the terms with the same value of ${\displaystyle m_{1}+m_{2}+\cdots +m_{n}=k}$ and noticing that ${\displaystyle m_{j}}$ has to be zero for ${\displaystyle j>n-k+1}$ leads to a somewhat simpler formula expressed in terms of partial (or incomplete) exponential Bell polynomials ${\displaystyle B_{n,k}(x_{1},\ldots ,x_{n-k+1})}$:

${\displaystyle {d^{n} \over dx^{n}}f(g(x))=\sum _{k=0}^{n}f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots ,g^{(n-k+1)}(x)\right).}$

This formula works for all ${\displaystyle n\geq 0}$, however for ${\displaystyle n>0}$ the polynomials ${\displaystyle B_{n,0}}$ are zero and thus summation in the formula can start with ${\displaystyle k=1}$.