# General Leibniz rule

In calculus, the general Leibniz rule,[1] named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). It states that if ${\displaystyle f}$ and ${\displaystyle g}$ are ${\displaystyle n}$-times differentiable functions, then the product ${\displaystyle fg}$ is also ${\displaystyle n}$-times differentiable and its ${\displaystyle n}$th derivative is given by

${\displaystyle (fg)^{(n)}=\sum _^f^{(n-k)}g^{(k)},}$

where ${\displaystyle =}$ is the binomial coefficient and ${\displaystyle f^{(0)}\equiv f.}$

This can be proved by using the product rule and mathematical induction.

## Second derivative

If, for example, n = 2, the rule gives an expression for the second derivative of a product of two functions:

${\displaystyle (fg)''(x)=\sum \limits _^{{\binom }f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).}$

## More than two factors

The formula can be generalized to the product of m differentiable functions f1,...,fm.

${\displaystyle \left(f_f_\cdots f_\right)^{(n)}=\sum _+k_+\cdots +k_=n},k_,\ldots ,k_}\prod _f_^{(k_)}\,,}$

where the sum extends over all m-tuples (k1,...,km) of non-negative integers with ${\displaystyle \sum _^k_=n,}$ and

${\displaystyle ,k_,\ldots ,k_}={\frac !\,k_!\cdots k_!}}}$

are the multinomial coefficients. This is akin to the multinomial formula from algebra.

## Proof

The proof of the general Leibniz rule proceeds by induction. Let ${\displaystyle f}$ and ${\displaystyle g}$ be ${\displaystyle n}$-times differentiable functions. The base case when ${\displaystyle n=1}$ claims that:

${\displaystyle (fg)'=f'g+fg',}$

which is the usual product rule and is known to be true. Next, assume that the statement holds for a fixed ${\displaystyle n\geq 1,}$ that is, that

${\displaystyle (fg)^{(n)}=\sum _^{\binom }f^{(n-k)}g^{(k)}.}$

Then,

${\displaystyle {\begin(fg)^{(n+1)}&=\left[\sum _^{\binom }f^{(n-k)}g^{(k)}\right]'\\&=\sum _^{\binom }f^{(n+1-k)}g^{(k)}+\sum _^{\binom }f^{(n-k)}g^{(k+1)}\\&=\sum _^{\binom }f^{(n+1-k)}g^{(k)}+\sum _^{\binom }f^{(n+1-k)}g^{(k)}\\&={\binom }f^{(n+1)}g+\sum _^{\binom }f^{(n+1-k)}g^{(k)}+\sum _^{\binom }f^{(n+1-k)}g^{(k)}+{\binom }fg^{(n+1)}\\&=f^{(n+1)}g+\left(\sum _^\left[{\binom }+{\binom }\right]f^{(n+1-k)}g^{(k)}\right)+fg^{(n+1)}\\&=f^{(n+1)}g+\sum _^{\binom }f^{(n+1-k)}g^{(k)}+fg^{(n+1)}\\&=\sum _^{\binom }f^{(n+1-k)}g^{(k)}.\end}}$

And so the statement holds for ${\displaystyle n+1,}$ and the proof is complete.

## Multivariable calculus

With the multi-index notation for partial derivatives of functions of several variables, the Leibniz rule states more generally:

${\displaystyle \partial ^{\alpha }(fg)=\sum _{\beta \,:\,\beta \leq \alpha }{\alpha \choose \beta }(\partial ^{\beta }f)(\partial ^{\alpha -\beta }g).}$

This formula can be used to derive a formula that computes the symbol of the composition of differential operators. In fact, let P and Q be differential operators (with coefficients that are differentiable sufficiently many times) and ${\displaystyle R=P\circ Q.}$ Since R is also a differential operator, the symbol of R is given by:

${\displaystyle R(x,\xi )=e^{-{\langle x,\xi \rangle }}R(e^{\langle x,\xi \rangle }).}$

A direct computation now gives:

${\displaystyle R(x,\xi )=\sum _{\alpha }\left({\partial \over \partial \xi }\right)^{\alpha }P(x,\xi )\left({\partial \over \partial x}\right)^{\alpha }Q(x,\xi ).}$

This formula is usually known as the Leibniz formula. It is used to define the composition in the space of symbols, thereby inducing the ring structure.