General Leibniz rule

In calculus, the general Leibniz rule, 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.

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,...,fm. $$\left(f_1 f_2 \cdots f_m\right)^{(n)}=\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}f_{t}^{(k_{t})}\,,$$ where the sum extends over all m-tuples (k1,...,km) 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
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, $$\begin{align} (fg)^{(n+1)} &= \left[ \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k)} \right]' \\ &= \sum_{k=0}^n \binom{n}{k} f^{(n+1-k)} g^{(k)} + \sum_{k=0}^n \binom{n}{k} f^{(n-k)} g^{(k+1)} \\ &= \sum_{k=0}^n \binom{n}{k} f^{(n+1-k)} g^{(k)} + \sum_{k=1}^{n+1} \binom{n}{k-1} f^{(n+1-k)} g^{(k)} \\ &= \binom{n}{0} f^{(n+1)} g^{(0)} + \sum_{k=1}^{n} \binom{n}{k} f^{(n+1-k)} g^{(k)} + \sum_{k=1}^n \binom{n}{k-1} f^{(n+1-k)} g^{(k)} + \binom{n}{n} f^{(0)} g^{(n+1)} \\ &= \binom{n+1}{0} f^{(n+1)} g^{(0)} + \left( \sum_{k=1}^n \left[\binom{n}{k-1} + \binom{n}{k} \right]f^{(n+1-k)} g^{(k)} \right) + \binom{n+1}{n+1} f^{(0)} g^{(n+1)} \\ &= \binom{n+1}{0} f^{(n+1)} g^{(0)} + \sum_{k=1}^n \binom{n+1}{k}  f^{(n+1-k)} g^{(k)} + \binom{n+1}{n+1}f^{(0)} g^{(n+1)} \\ &= \sum_{k=0}^{n+1} \binom{n+1}{k} f^{(n+1-k)} g^{(k)}. \end{align}$$ 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: $$\partial^\alpha (fg) = \sum_{ \beta\,:\,\beta \le \alpha } {\alpha \choose \beta} (\partial^{\beta} f) (\partial^{\alpha - \beta}  g).$$

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.