General Leibniz rule

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Short description: Generalization of the product rule in calculus

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 [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math] are [math]\displaystyle{ n }[/math]-times differentiable functions, then the product [math]\displaystyle{ fg }[/math] is also [math]\displaystyle{ n }[/math]-times differentiable and its [math]\displaystyle{ n }[/math]th derivative is given by

[math]\displaystyle{ (fg)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(n-k)} g^{(k)}, }[/math]

where [math]\displaystyle{ {n \choose k}={n!\over k! (n-k)!} }[/math] is the binomial coefficient and [math]\displaystyle{ f^{(j)} }[/math] denotes the jth derivative of f (and in particular [math]\displaystyle{ f^{(0)}= f }[/math]).

The rule can be proven 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:

[math]\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). }[/math]

More than two factors

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

[math]\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\le t\le m}f_{t}^{(k_{t})}\,, }[/math]

where the sum extends over all m-tuples (k1,...,km) of non-negative integers with [math]\displaystyle{ \sum_{t=1}^m k_t=n, }[/math] and

[math]\displaystyle{ {n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!} }[/math]

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


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

[math]\displaystyle{ (fg)'=f'g+fg', }[/math]

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

[math]\displaystyle{ (fg)^{(n)}=\sum_{k=0}^n\binom{n}{k} f^{(n-k)}g^{(k)}. }[/math]


[math]\displaystyle{ \begin{align} (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)} \\ &= \binom{n+1}{0} 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) + \binom{n+1}{n+1} fg^{(n+1)} \\ &= \binom{n+1}{0} f^{(n+1)} g + \sum_{k=1}^n \binom{n+1}{k} f^{(n+1-k)} g^{(k)} + \binom{n+1}{n+1}fg^{(n+1)} \\ &= \sum_{k=0}^{n+1} \binom{n+1}{k} f^{(n+1-k)} g^{(k)} . \end{align} }[/math]

And so the statement holds for [math]\displaystyle{ n+1, }[/math] 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:

[math]\displaystyle{ \partial^\alpha (fg) = \sum_{ \beta\,:\,\beta \le \alpha } {\alpha \choose \beta} (\partial^{\beta} f) (\partial^{\alpha - \beta} g). }[/math]

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 [math]\displaystyle{ R = P \circ Q. }[/math] Since R is also a differential operator, the symbol of R is given by:

[math]\displaystyle{ R(x, \xi) = e^{-{\langle x, \xi \rangle}} R (e^{\langle x, \xi \rangle}). }[/math]

A direct computation now gives:

[math]\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). }[/math]

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.

See also