Leibniz's Rule

Theorem
Let $f, g$ be continuous, differentiable functions. Let $k$ and $n$ be integers such that $ 0 \leq k \leq n $.

For all $n \geq 1$:
 * $\displaystyle [f(x)g(x)]^{(n)} = \sum_{k=0}^n \binom n k f^{(k)}(x)g^{(n-k)}(x)$

where $(n)$ is the order of the derivative.

Proof
Proof by induction:

Base Case
Let $n = 1$.

By the Product Rule for Derivatives, we know that $[f(x)g(x)]' = f(x)g'(x)+f'(x)g(x)$.

Likewise:
 * $\displaystyle \sum_{k=0}^1 \binom 1 k f^{(k)}(x)g^{(1-k)}(x) = \binom 1 0 f(x)g^{(1-0)}(x) + \binom 1 1 f'(x)g^{(1-1)}(x) = f(x)g'(x)+ f'(x)g(x)$.

This is our base case.

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

for $ 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:

Thus, by the Principle of Mathematical Induction, we have proven the Leibniz Rule for all $n \geq 1$.