How to prove this theorem about partial derivative and how to write this theorem with Newton's notation? How to prove this theorem about partial derivative and how to re-write this theorem with Newton's notation? i can't understand totally without newton's notation! i accept if you just generalize this from product rule.
From wikipedia section 6.3 of http://en.wikipedia.org/wiki/Product_rule:
For partial derivatives, we have
$${\partial^n \over \partial x_1\,\cdots\,\partial x_n} (uv)
= \sum_S {\partial^{|S|} u \over \prod_{i\in S} \partial x_i} \cdot {\partial^{n-|S|} v \over \prod_{i\not\in S} \partial x_i}$$
where the index ''S'' runs through the whole list of 2''n'' subsets of {1, ..., ''n''}.  For example, when ''n'' = 3, then
$\begin{align} &{}\quad {\partial^3 \over \partial x_1\,\partial x_2\,\partial x_3} (uv)  \\  \\
&{}= u \cdot{\partial^3 v \over \partial x_1\,\partial x_2\,\partial x_3} + {\partial u \over \partial x_1}\cdot{\partial^2 v \over \partial x_2\,\partial x_3} +  {\partial u \over \partial x_2}\cdot{\partial^2 v \over \partial x_1\,\partial x_3} + {\partial u \over \partial x_3}\cdot{\partial^2 v \over \partial x_1\,\partial x_2} \\  \\
&{}\qquad + {\partial^2 u \over \partial x_1\,\partial x_2}\cdot{\partial v \over \partial x_3}
+ {\partial^2 u \over \partial x_1\,\partial x_3}\cdot{\partial v \over \partial x_2}
+ {\partial^2 u \over \partial x_2\,\partial x_3}\cdot{\partial v \over \partial x_1}
+ {\partial^3 u \over \partial x_1\,\partial x_2\,\partial x_3}\cdot v. \end{align}$
 A: We can do this through induction.
For the first partial, it is certainly true through the product rule that
$$\frac{\partial}{\partial x_{1}}(uv) = \frac{\partial u}{\partial x_1}\cdot v + \frac{\partial v}{\partial x_1}\cdot u$$
where the sum moves through $S\in\left\{\emptyset, 1 \right\}$ which is the powerset of $\{1\}$. Suppose then that this holds for $n-1$. Then we can apply the product rule to each term of the sum
$$\frac{\partial^{n}}{\partial x_{1} \cdots \partial x_{n-1} \partial x_n}(uv)=\sum_{S}\frac{\partial}{\partial x_n}\bigg(\frac{\partial^{\mid S\mid}u}{\prod_{i\in S}\partial x_i}\cdot \frac{\partial^{n-1-\mid S\mid}v}{\prod_{i\notin S}\partial x_i}\bigg)$$ 
$$= \sum_{S}\frac{\partial^{\mid S\mid + 1}u}{(\prod_{i\in S}\partial x_i)\partial x_n}\cdot  \frac{\partial^{n-1-\mid S\mid}v}{\prod_{i\notin S}\partial x_i} + \sum_S \frac{\partial^{n-\mid S\mid}v}{(\prod_{i\notin S}\partial x_i)\partial x_n}\cdot \frac{\partial^{\mid S\mid}u}{\prod_{i\in S}\partial x_i}$$
Letting $S'$ be the elements of the powerset of $\{1, 2, \cdots, n\}$, we now want to sum over $S'$. 
Note that what the product rule has done is effectively seperated the sum into two. The first sum contains the sets which contain $n$, where the product rule has placed $n$ into $S'$. In the second sum, $S = S'$ remains unchanged and the product rule places $n$ into the "unwanted pile". 
In the first sum, the cardinality of the set increases by 1 through the inclusion of $n$. In the second sum, the cardinality does not change. So in the first, $\mid S'\mid = \mid S\mid + 1$ and in the second $\mid S' \mid = \mid S\mid$. We can then write
$$= \sum_{S'}\frac{\partial^{\mid S'\mid}u}{\prod_{i\in S'}\partial x_i}\cdot  \frac{\partial^{n-\mid S'\mid}v}{\prod_{i\notin S'}\partial x_i} + \sum_{S'} \frac{\partial^{n-\mid S'\mid}v}{\prod_{i\notin S'}\partial x_i}\cdot \frac{\partial^{\mid S'\mid}u}{\prod_{i\in S'}\partial x_i}$$
it is easy to see that the first sum constitute all the sets which contain contain $n$ while the second sum constitutes all the sets which do not. Together, they give all elements of the powerset, so we can simply write
$$\frac{\partial^{n}}{\partial x_{1} \cdots \partial x_{n-1} \partial x_n}(uv)=\sum_{S'}\frac{\partial^{\mid S'\mid}u}{\prod_{i\in S'}\partial x_i}\cdot \frac{\partial^{n-\mid S'\mid}v}{\prod_{i\notin S'}\partial x_i}$$
where the sum is taken through all elements $S'$ of the powerset of $\{1, 2, \cdots, n\}$ as required.
