General Leibniz rule

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

${\displaystyle (fg)''(x)=\sum \limits _{k=0}^{2}{{\binom {2}{k}}f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).}$

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

${\displaystyle \left(f_{1}f_{2}\cdots f_{m}\right)^{(n)}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}\prod _{1\leq t\leq m}f_{t}^{(k_{t})}\,,}$

where the sum extends over all m-tuples (k₁,...,k_m) of non-negative integers with ${\displaystyle \sum _{t=1}^{m}k_{t}=n,}$ and

${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}}$

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

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

${\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k)}.}$

Then,

${\displaystyle {\begin{aligned}(fg)^{(n+1)}&=\left[\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k)}\right]'\\&=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k+1)}\\&=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n+1}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}\\&={\binom {n}{0}}f^{(n+1)}g+\sum _{k=1}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}+{\binom {n}{n}}fg^{(n+1)}\\&=f^{(n+1)}g+\left(\sum _{k=1}^{n}\left[{\binom {n}{k-1}}+{\binom {n}{k}}\right]f^{(n+1-k)}g^{(k)}\right)+fg^{(n+1)}\\&=f^{(n+1)}g+\sum _{k=1}^{n}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}+fg^{(n+1)}\\&=\sum _{k=0}^{n+1}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}.\end{aligned}}}$

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

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:

${\displaystyle R(x,\xi )=\sum _{\alpha }{1 \over \alpha$ !}\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.

• Binomial theorem
• Derivation (differential algebra)
• Derivative
• Differential algebra
• Pascal's triangle