Leibniz's Rule

Theorem

Let $f$ and $g$ be real functions defined on the open interval $I$.

Let $n \in \Z: n > 0$ be a positive integer.

Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are $n$ times differentiable.

Then:

$\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n}\right)} = \sum_{k=0}^n \binom n k f^{\left({k}\right)} \left({x}\right) g^{\left({n - k}\right)} \left({x}\right)$

where $\left({n}\right)$ denotes the order of the derivative.

Proof

Proof by induction:

Base Case

Let $n = 1$.

By the Product Rule for Derivatives, we know that:

$\left({f \left({x}\right) g \left({x}\right)}\right)' = \left({f \left({x}\right) g' \left({x}\right)}\right) + \left({f' \left({x}\right) g \left({x}\right)}\right)$

Likewise:

$\displaystyle \sum_{k=0}^1 \binom 1 k f^{(k)} \left({x}\right) g^{(1-k)} \left({x}\right) = \binom 1 0 f \left({x}\right) g^{(1-0)} \left({x}\right) + \binom 1 1 f' \left({x}\right) g^{(1-1)} \left({x}\right) = f \left({x}\right) g' \left({x}\right) + f' \left({x}\right) g \left({x}\right)$

This is our base case.

Induction Hypothesis

Let $n \in \N$ be fixed.

We assume the inductive hypothesis:

$\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n}\right)} = \sum_{k=0}^n \binom n k f^{(k)} \left({x}\right) g^{(n-k)} \left({x}\right)$

We need to show that:

$\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n + 1}\right)} = \sum_{k=0}^{n+1} \binom {n+1} k f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right)$

Induction Step

By our inductive hypothesis:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n + 1}\right)}\)	\(=\)	\(\displaystyle \left({\sum_{k=0}^n \binom n k f^{(k)} \left({x}\right) g^{(n-k)} \left({x}\right) }\right)'\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^n \binom n k \left({\left({f^{(k)} \left({x}\right) g^{(n-k)} \left({x}\right) }\right)'}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Applications of Linear Combination of Derivatives
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^n \binom n k \left({f^{(k+1)} \left({x}\right) g^{(n-k)} \left({x}\right) + f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right) }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From the base case
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^n \binom n k f^{(k+1)} \left({x}\right) g^{(n-k)} \left({x}\right) + \sum_{k=0}^n \binom n k f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Splitting the summation

Subsequently, we separate the $k=0$ case from the second summation.

For the first summation, we separate the case $k=n$ and then shift the indices up by $1$.

These manipulations give us the following:

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n + 1}\right)}\)

\(=\)

\(\displaystyle \sum_{k=1}^n \binom n k f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right) + \sum_{k=1}^n \binom n {k-1} f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right) + \binom n 0 f \left({x}\right) g^{(n+1)} \left({x}\right) + \binom n n f^{(n+1)} \left({x}\right) g \left({x}\right)\)

\(\displaystyle \)

\(\displaystyle \)

\(\displaystyle \)

By Pascal's Rule, we finally obtain:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n + 1}\right)}\)	\(=\)	\(\displaystyle \sum_{k=1}^n \binom {n+1} k f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right) + \binom n 0 f \left({x}\right) g^{(n+1)} \left({x}\right) + \binom n n f^{(n+1)} \left({x}\right) g \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=1}^n \binom {n+1} k f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right) + \binom {n+1} 0 f \left({x}\right) g^{(n+1)} \left({x}\right) + \binom {n+1} {n+1} f^{(n+1)} \left({x}\right) g \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^{n+1} \binom {n+1} k f^{(k)} \left({x}\right) g^{(n+1-k)} \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

The result follows by the Principle of Mathematical Induction.

$\blacksquare$

Source of Name

This entry was named for Gottfried Wilhelm von Leibniz.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $13.46$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 10.11 \ (6)$