General Leibniz rule

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In calculus, the general Leibniz rule,^[1] named after Gottfried Wilhelm Leibniz, generalizes the product rule for the derivative of the product of two (which is also known as "Leibniz's rule"). It states that if ${\displaystyle f}$ and ${\displaystyle g}$ are n-times differentiable functions, then the product ${\displaystyle fg}$ is also n-times differentiable and its n-th derivative is given by $${\displaystyle (fg)^{(n)}=\sum _{k=0}^{n}{n \choose k}f^{(n-k)}g^{(k)},}$$ where ${\displaystyle {n \choose k}={n! \over k!(n-k)!}}$ is the binomial coefficient and ${\displaystyle f^{(j)}}$ denotes the jth derivative of f (and in particular ${\displaystyle f^{(0)}=f}$).

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

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 ${\textstyle \sum _{t=1}^{m}k_{t}=n,}$ 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^{[2]: 68–69} 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, $${\displaystyle {\begin{aligned}(fg)^{(n+1)}&=\left[\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k)}\right]'\\&=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=0}^{n}{\binom {n}{k}}f^{(n-k)}g^{(k+1)}\\&=\sum _{k=0}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n+1}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}\\&={\binom {n}{0}}f^{(n+1)}g^{(0)}+\sum _{k=1}^{n}{\binom {n}{k}}f^{(n+1-k)}g^{(k)}+\sum _{k=1}^{n}{\binom {n}{k-1}}f^{(n+1-k)}g^{(k)}+{\binom {n}{n}}f^{(0)}g^{(n+1)}\\&={\binom {n+1}{0}}f^{(n+1)}g^{(0)}+\left(\sum _{k=1}^{n}\left[{\binom {n}{k-1}}+{\binom {n}{k}}\right]f^{(n+1-k)}g^{(k)}\right)+{\binom {n+1}{n+1}}f^{(0)}g^{(n+1)}\\&={\binom {n+1}{0}}f^{(n+1)}g^{(0)}+\sum _{k=1}^{n}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}+{\binom {n+1}{n+1}}f^{(0)}g^{(n+1)}\\&=\sum _{k=0}^{n+1}{\binom {n+1}{k}}f^{(n+1-k)}g^{(k)}.\end{aligned}}}$$ And so the statement holds for ${\displaystyle n+1}$, and the proof is complete.

The Leibniz rule bears a strong resemblance to the binomial theorem, and in fact the binomial theorem can be proven directly from the Leibniz rule by taking ${\displaystyle f(x)=e^{ax}}$ and ${\displaystyle g(x)=e^{bx},}$ which gives

${\displaystyle (a+b)^{n}e^{(a+b)x}=e^{(a+b)x}\sum _{k=0}^{n}{\binom {n}{k}}a^{n-k}b^{k},}$

and then dividing both sides by ${\displaystyle e^{(a+b)x}.}$^[2]: 69

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.

• Derivation (differential algebra)
• Umbral calculus