Leibniz's Rule/One Variable

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

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

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

Then:
 * $\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$

where $\paren n$ denotes the order of the derivative.

Proof
Proof by induction:

Basis for the Induction
Let $n = 1$.

From Product Rule for Derivatives:
 * $\paren {\map f x \, \map g x}' = \paren {\map f x \, \map {g'} x} + \paren {\map {f'} x \, \map g x}$

Likewise:
 * $\displaystyle \sum_{k \mathop = 0}^1 \binom 1 k \map {f^{\paren k} } x \, \map {g^{\paren {1 - k} } } x = \binom 1 0 \map f x \, \map {g^{\paren {1 - 0} } } x + \binom 1 1 \map {f'} x \, \map {g^{\paren {1 - 1} } } x = \map f x \, \map {g'} x + \map {f'} x \, \map g x$

This is our basis for the induction.

Induction Hypothesis
Let $n \in \N$ be fixed.

We assume the inductive hypothesis:
 * $\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$

We need to show that:
 * $\displaystyle \paren {\map f x \, \map g x}^{\paren {n + 1} } = \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k \map {f^{\paren k} } x \, \map {g^{\paren {n + 1 - k} } } x$

Induction Step
By our inductive hypothesis:

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:

By Pascal's Rule, we finally obtain:

The result follows by the Principle of Mathematical Induction.