General Leibniz rule

More than two factors

The formula can be generalized to the product of m differentiable functions f₁,...,f_m. $${\displaystyle {\left(f_1{f}_2\cdots f_m\right)}^{(n)}=\sum _{k_1+k_2+\cdots +k_m=n}(\genfrac{}{}{0ex}{}{n}{k_1,k_2,\dots ,k_m})\prod _{1\le t\le m}{f}_t^{(k_t)},}$$ where the sum extends over all m-tuples (k₁,...,k_m) of non-negative integers with $\sum _{t=1}^m{k}_t=n,$ and $${\displaystyle (\genfrac{}{}{0ex}{}{n}{k_1,k_2,\dots ,k_m})=\frac{n!}{k_1!k_2!\cdots k_m!}}$$ 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 ${\displaystyle f}$ and ${\displaystyle g}$ be ${\displaystyle n}$-times differentiable functions. The base case when ${\displaystyle n=1}$ claims that: $${\displaystyle (fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime },}$$ which is the usual product rule and is known to be true. Next, assume that the statement holds for a fixed ${\displaystyle n\ge 1,}$ that is, that $${\displaystyle (fg{)}^{(n)}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(n-k)}{g}^{(k)}.}$$

Then, $${\displaystyle \begin{array}{cc}(fg{)}^{(n+1)}& ={\left[\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(n-k)}{g}^{(k)}\right]}^{\prime }\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(n+1-k)}{g}^{(k)}+\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(n-k)}{g}^{(k+1)}\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(n+1-k)}{g}^{(k)}+\sum _{k=1}^{n+1}(\genfrac{}{}{0ex}{}{n}{k-1})f^{(n+1-k)}{g}^{(k)}\\ & =(\genfrac{}{}{0ex}{}{n}{0})f^{(n+1)}{g}^{(0)}+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(n+1-k)}{g}^{(k)}+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n}{k-1})f^{(n+1-k)}{g}^{(k)}+(\genfrac{}{}{0ex}{}{n}{n})f^{(0)}{g}^{(n+1)}\\ & =(\genfrac{}{}{0ex}{}{n+1}{0})f^{(n+1)}{g}^{(0)}+\left(\sum _{k=1}^n\left[(\genfrac{}{}{0ex}{}{n}{k-1})+(\genfrac{}{}{0ex}{}{n}{k})\right]f^{(n+1-k)}{g}^{(k)}\right)+(\genfrac{}{}{0ex}{}{n+1}{n+1})f^{(0)}{g}^{(n+1)}\\ & =(\genfrac{}{}{0ex}{}{n+1}{0})f^{(n+1)}{g}^{(0)}+\sum _{k=1}^n(\genfrac{}{}{0ex}{}{n+1}{k})f^{(n+1-k)}{g}^{(k)}+(\genfrac{}{}{0ex}{}{n+1}{n+1})f^{(0)}{g}^{(n+1)}\\ & =\sum _{k=0}^{n+1}(\genfrac{}{}{0ex}{}{n+1}{k})f^{(n+1-k)}{g}^{(k)}.\end{array}}$$ And so the statement holds for ${\displaystyle n+1}$, and the proof is complete.

Relationship to the binomial theorem

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 $${\displaystyle (a+b{)}^n{e}^{(a+b)x}=e^{(a+b)x}\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^{n-k}{b}^k,}$$ and then dividing both sides by ${\displaystyle e^{(a+b)x}.}$

Multivariable calculus

With the multi-index notation for partial derivatives of functions of several variables, the Leibniz rule states more generally: $${\displaystyle {\mathrm{\partial }}^{\alpha }(fg)=\sum _{\beta :\beta \le \alpha }(\genfrac{}{}{0ex}{}{\alpha }{\beta })({\mathrm{\partial }}^{\beta }f)({\mathrm{\partial }}^{\alpha -\beta }g).}$$

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: $${\displaystyle R(x,\xi )=e^{-⟨x,\xi ⟩}R(e^{⟨x,\xi ⟩}).}$$

A direct computation now gives: $${\displaystyle R(x,\xi )=\sum _{\alpha }\frac{1}{\alpha !}{\left(\frac{\mathrm{\partial }}{\mathrm{\partial }\xi }\right)}^{\alpha }P(x,\xi ){\left(\frac{\mathrm{\partial }}{\mathrm{\partial }x}\right)}^{\alpha }Q(x,\xi ).}$$

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.