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
We assume the inductive hypothesis:
 * $\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n}\right)} = \sum_{k=0}^n \binom n k f^{(k)}(x)g^{(n-k)}(x)$

for all $n \ge 1$.

We need to show that:
 * $\displaystyle \left({f \left({x}\right) g \left({x}\right)}\right)^{\left({n + 1}\right)} = \sum_{k=0}^{n+1} \binom {n+1} k f^{(k)}(x)g^{(n+1-k)}(x)$

for all $n \ge 1$.

Induction Step
By our inductive hypothesis:

Expanding this further, we obtain:

We then 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 $1$.

By Pascal's Rule, we obtain:

The result follows by the Principle of Mathematical Induction.