# 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

$(fg)^{(n)}=\sum _^f^{(n-k)}g^{(k)},$ where $=$ is the binomial coefficient and $f^{(0)}\equiv f.$ This can be proved 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:

$(fg)''(x)=\sum \limits _^{{\binom }f^{(2-k)}(x)g^{(k)}(x)}=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x).$ ## More than two factors

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

$\left(f_f_\cdots f_\right)^{(n)}=\sum _+k_+\cdots +k_=n},k_,\ldots ,k_}\prod _f_^{(k_)}\,,$ where the sum extends over all m-tuples (k1,...,km) of non-negative integers with $\sum _^k_=n,$ and

$,k_,\ldots ,k_}={\frac !\,k_!\cdots k_!}}$ 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

$(fg)^{(n)}=\sum _^{\binom }f^{(n-k)}g^{(k)}.$ Then,

${\begin(fg)^{(n+1)}&=\left[\sum _^{\binom }f^{(n-k)}g^{(k)}\right]'\\&=\sum _^{\binom }f^{(n+1-k)}g^{(k)}+\sum _^{\binom }f^{(n-k)}g^{(k+1)}\\&=\sum _^{\binom }f^{(n+1-k)}g^{(k)}+\sum _^{\binom }f^{(n+1-k)}g^{(k)}\\&={\binom }f^{(n+1)}g+\sum _^{\binom }f^{(n+1-k)}g^{(k)}+\sum _^{\binom }f^{(n+1-k)}g^{(k)}+{\binom }fg^{(n+1)}\\&=f^{(n+1)}g+\left(\sum _^\left[{\binom }+{\binom }\right]f^{(n+1-k)}g^{(k)}\right)+fg^{(n+1)}\\&=f^{(n+1)}g+\sum _^{\binom }f^{(n+1-k)}g^{(k)}+fg^{(n+1)}\\&=\sum _^{\binom }f^{(n+1-k)}g^{(k)}.\end}$ 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:

$\partial ^{\alpha }(fg)=\sum _{\beta \,:\,\beta \leq \alpha }{\alpha \choose \beta }(\partial ^{\beta }f)(\partial ^{\alpha -\beta }g).$ 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:

$R(x,\xi )=\sum _{\alpha }\left({\partial \over \partial \xi }\right)^{\alpha }P(x,\xi )\left({\partial \over \partial x}\right)^{\alpha }Q(x,\xi ).$ 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.