Number of Partial Derivatives of Order n

Theorem
Let $u = \map f {x_1, x_2, \ldots, x_m}$ be a function of the $m$ independent variables $x_1, x_2, \ldots, x_m$.

There are $m^n$ partial derivatives of $u$ of order $n$.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * There are $m^n$ partial derivatives of $u$ of order $n$.

$\map P 0$ is the degenerate case where $f$ is not partially differentiated at all:
 * $\map f {x_1, x_2, \ldots, x_m}$

and it is apparent that there is only $1 = m^0$ such.

Thus $\map P 0$ is seen to hold.

Basis for the Induction
$\map P 1$ is the case:
 * There are $m^1 = m$ partial derivatives of $u$ of order $1$.

These can be instanced as:

and it is apparent that there are $m$ such.

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * There are $m^k$ partial derivatives of $u$ of order $k$.

from which it is to be shown that:
 * There are $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$.

Induction Step
This is the induction step:

Let $g$ be one of the partial derivatives of $u$ of order $k$.

Then by the basis for the induction, there are $m$ partial derivatives of $g$.

By the induction hypothesis, there are $m^k$ partial derivatives of $u$ of order $k$.

Thus by the Product Rule for Counting, there are $m \times m^n$ partial derivatives of $u$ of order $k + 1$.

That is, a total of $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}:$ there are $m^n$ partial derivatives of $u$ of order $n$.