General Leibniz rule

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").

are n-times differentiable functions, then the product

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)},}

is the binomial coefficient and

denotes the jth derivative of f (and in particular

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:

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

where the sum extends over all m-tuples (k1,...,km) of non-negative integers with

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.

-times differentiable functions.

The base case when

which is the usual product rule and is known to be true.

Next, assume that the statement holds for a fixed

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

{\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

, 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

which gives and then dividing both sides by

[2]: 69 With the multi-index notation for partial derivatives of functions of several variables, the Leibniz rule states more generally:

β ≤ α

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

Since R is also a differential operator, the symbol of R is given by:

A direct computation now gives:

It is used to define the composition in the space of symbols, thereby inducing the ring structure.