Leibniz's Rule

Theorem
Let $$f, g$$ be real functions which are $n$ times differentiable at a point $$\xi$$.

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

at the point $$\xi$$, where $$(n)$$ is the order of the derivative.

Proof
To prove the Leibniz Rule, we use the Principle of Mathematical Induction.

Basis Step: The result holds when $$n = 1$$ by an application of the Product Rule for Derivatives. Thus, we may hold the inductive hypothesis that $$[f(x)g(x)]^{(n)} = \sum_{k=0}^n \binom n k f^{(k)}(x)g^{(n-k)}(x)$$ for all natural numbers.

Inductive Step: By our inductive hypothesis,$$ [f(x)g(x)]^{(n+1)} = \left[ \sum_{k=0}^n \binom n k f^{(k)}(x)g^{(n-k)}(x) \right]' = \sum_{k=0}^n \binom n k f^{(k+1)}(x)g^{(n-k)}(x) + f^{(k)}(x)g^{(n+1-k)}(x)$$ by the Product Rule for Derivatives. Expanding this further, we obtain $$ \sum_{k=0}^n \binom n k f^{(k+1)}(x)g^{(n-k)}(x) + \sum_{k=0}^n \binom n k f^{(k)}(x)g^{(n+1-k)}(x)$$ $$= \sum_{k=1}^n \binom n k  f^{(k)}(x)g^{(n+1-k)}(x) + \sum_{k=1}^n \binom n {k-1} f^{(k)}(x)g^{(n+1-k)}(x) + \binom n 0 f(x)g^{(n+1)}(x) + \binom n n f^{(n+1)}(x)g(x)$$.

By Pascal's Rule, we obtain $$ \sum_{k=1}^n \binom {n+1} k f^{(k)}(x)g^{(n+1-k)}(x) + \binom n 0 f(x)g^{(n+1)}(x) + \binom n n f^{(n+1)}(x)g(x)$$.

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