General Leibniz rule

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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 $f$ and $g$ are $n$ -times differentiable functions, then the product $fg$ is also $n$ -times differentiable and its $n$ -th derivative is given by

(fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{(k)},

where

{n \choose k}={n! \over k!(n-k)!}

is the binomial coefficient and

f^{(j)}

denotes the jth derivative of f (and in particular

f^{(0)}=f

The rule can be proven by using the product rule and mathematical induction.

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Transcription

Second derivative

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

(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).

More than two factors

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

\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

{\textstyle \sum _{t=1}^{m}k_{t}=n,}

and

{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.

Proof

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

(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

n\geq 1,

that is, that

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

Then,

{\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^{(0)}+\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}}f^{(0)}g^{(n+1)}\\&={\binom {n+1}{0}}f^{(n+1)}g^{(0)}+\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}}f^{(0)}g^{(n+1)}\\&={\binom {n+1}{0}}f^{(n+1)}g^{(0)}+\sum _{k=1}^{n}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}+{\binom {n+1}{n+1}}f^{(0)}g^{(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

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:

\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 $R=P\circ Q.$ Since R is also a differential operator, the symbol of R is given by:

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

A direct computation now gives:

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.

References

^ Olver, Peter J. (2000). Applications of Lie Groups to Differential Equations. Springer. pp. 318–319. ISBN 9780387950006.

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