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

where ${n \choose k}={n! \over k!(n-k)!}$ is the binomial coefficient and $f^{(0)}(x)=f(x)$ .

This can be proved by using the product rule and mathematical induction (see proof below).

Second derivative

In case $n=2$ :

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

The binomial coefficients can be deduced thanks to the Pascal's triangle.

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

Proof

Show that the equality holds for any functions $f$ and $g$ that are $n$ -times differentiable functions.

Basis At rank $m=1$ we get: $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$

and $\sum _{k=0}^{1}{1 \choose k}f^{(1-k)}(x)g^{(k)}(x)=f'(x)g(x)+f(x)g'(x)$ by the product rule.

Hence, the equality holds at the initial rank.

Inductive step We assume that the equality

$(fg)^{(m)}(x)=\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k)}(x)$

holds for $m\in \mathbb {N} ^{*}$ .

Therefore, at rank $(m+1)$ we get:

${\begin{aligned}(fg)^{(m+1)}(x)&={\frac {\text{d}}{{\text{d}}x}}\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k)}(x)\\&=\sum _{k=0}^{m}{{m} \choose {k}}f^{(m-k)}(x)g^{(k+1)}(x)+\sum _{k=0}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)\\&=\sum _{k=1}^{m+1}{{m} \choose {k-1}}f^{(m+1-k)}(x)g^{(k)}(x)+\sum _{k=0}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)\\&={{m} \choose {m}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}{{m} \choose {k-1}}f^{(m+1-k)}(x)g^{(k)}(x)+\sum _{k=1}^{m}{{m} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)+{{m} \choose {0}}f^{(m+1)}(x)g(x)\\&={{m} \choose {m}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}\left[{{m} \choose {k-1}}+{{m} \choose {k}}\right]f^{(m+1-k)}(x)g^{(k)}(x)+{{m} \choose {0}}f^{(m+1)}(x)g(x)\\&={{m+1} \choose {m+1}}f(x)g^{(m+1)}(x)+\sum _{k=1}^{m}{{m+1} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x)+{{m+1} \choose {0}}f^{(m+1)}(x)g(x)\\&=\sum _{k=0}^{m+1}{{m+1} \choose {k}}f^{(m+1-k)}(x)g^{(k)}(x).\end{aligned}}$

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, Applications of Lie groups to differential equations, page 318

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[1] Olver, Applications of Lie groups to differential equations, page 318