Leibniz's Rule/Real Valued Functions

Theorem
Let $f,g : \R^n \to \R$ be real valued functions, $k$ times differentiable on some open set $U \subseteq \R^n$.

Let $\alpha = \left(\alpha_1,\ldots,\alpha_n\right)$ be a multiindex indexed by $\left\{1,\ldots,n\right\}$ with $|\alpha| \leq k$.

For $i \in \left\{1,\ldots,n\right\}$ let $\partial_i$ denote the partial derivative $\displaystyle \partial_i = \frac{\partial}{\partial{x_i}}$.

Let $\partial^\alpha$ denote the partial differential operator:
 * $\displaystyle \partial^\alpha = \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n}$

Then as functions on $U$, we have:
 * $\displaystyle \partial^\alpha\left(fg\right) = \sum_{\beta \mathop \le \alpha} \binom \alpha \beta \left(\partial^\beta f\right)\left(\partial^{\alpha - \beta} g\right)$

Proof
First, inserting the definitions; the statement of the theorem reads:
 * $\displaystyle \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n}\left(fg\right) = \sum_{\beta_1 = 0}^{\alpha_1}\cdots\sum_{\beta_n = 0}^{\alpha_n} \binom{\alpha_1}{\beta_1}\cdots\binom{\alpha_n}{\beta_n}\left(\partial_1^{\beta_1}\cdots \partial_n^{\beta_n} f \right) \left(\partial_1^{\alpha_1-\beta_1}\cdots \partial_n^{\alpha_n-\beta_n}g \right)$

We prove this by induction over $n \geq 1$.

Basis for the Induction
If $n = 1$, the result is a simple restatement of Leibniz's Rule in one variable.

This is the basis for the induction.

Induction Step
Suppose now that the result is true for functions of $n-1$ variables.

In particular, we suppose that:
 * $\displaystyle \partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n}\left(fg\right) = \sum_{\beta_2 = 0}^{\alpha_2}\cdots\sum_{\beta_n = 0}^{\alpha_n} \binom{\alpha_2}{\beta_2}\cdots\binom{\alpha_n}{\beta_n}\left(\partial_2^{\beta_2}\cdots \partial_n^{\beta_n} f \right) \left(\partial_2^{\alpha_2-\beta_2}\cdots \partial_n^{\alpha_n-\beta_n}g \right)$

Now let us apply $\partial_1^{\alpha_1}$.

Using the linearity of derivatives:
 * $\displaystyle \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n}\left(fg\right) = \sum_{\beta_2 = 0}^{\alpha_2}\cdots\sum_{\beta_n = 0}^{\alpha_n} \binom{\alpha_2}{\beta_2}\cdots\binom{\alpha_n}{\beta_n} \partial_1^{\alpha_1} \left[ \left(\partial_2^{\beta_2}\cdots \partial_n^{\beta_n} f \right) \left(\partial_2^{\alpha_2-\beta_2}\cdots \partial_n^{\alpha_n-\beta_n}g \right) \right]$

Applying Leibniz's Rule in one variable we have:
 * $\displaystyle \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n}\left(fg\right) = \sum_{\beta_2 = 0}^{\alpha_2}\cdots\sum_{\beta_n = 0}^{\alpha_n} \binom{\alpha_2}{\beta_2}\cdots\binom{\alpha_n}{\beta_n} \sum_{\beta_1 = 0}^{\alpha_1} \binom{\alpha_1}{\beta_1} \left( \partial_1^{\beta_1} \partial_2^{\beta_2}\cdots \partial_n^{\beta_n} f \right) \left(\partial_1^{\alpha_1-\beta_1}\partial_2^{\alpha_2-\beta_2}\cdots \partial_n^{\alpha_n-\beta_n}g \right) $

Finally, since all the sums are finite, we can move the inner sum to the far left Finite Sums Commute, giving:


 * $\displaystyle \partial_1^{\alpha_1}\partial_2^{\alpha_2}\cdots \partial_n^{\alpha_n}\left(fg\right) = \sum_{\beta_1 = 0}^{\alpha_1}\cdots\sum_{\beta_n = 0}^{\alpha_n} \binom{\alpha_1}{\beta_1}\cdots\binom{\alpha_n}{\beta_n}\left(\partial_1^{\beta_1}\cdots \partial_n^{\beta_n} f \right) \left(\partial_1^{\alpha_1-\beta_1}\cdots \partial_n^{\alpha_n-\beta_n}g \right)$

The result now follows by the Principle of Mathematical Induction.