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

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

Proof
To prove the Leibniz Rule, we use the Principle of Mathematical 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:
 * $$ \sum_{k=0}^1 \binom 1 k f^{(k)}(x)g^{(1-k)}(x) = \binom 1 0 f(x)g^{(1-0)}(x) + \binom 1 1 f'(x)g^{(1-1)}(x) = f(x)g'(x)+ f'(x)g(x)$$.

This is our base case.

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

for all $$ n \ge 1 $$.

Induction Step
By our inductive hypothesis:

$$ $$ $$

Expanding this further, we obtain:

$$ $$

By Pascal's Rule, we obtain:

$$ $$ $$

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