Definition: My motivation for this question stems from the following definition: Define the deterministic finite automata generated by the nonconstant* polynomial $f(x_0 , \dots , x_n) \in \mathbb{Z} [x_0 , x_1 , \dots , x_n]$, denoted $\text{DFA}(f)$, to be the $5$-tuple $$\text{DFA}(f) = (\partial f, \Sigma, f, \delta, \equiv0),$$ where
- $\partial f = \{ \text{all partial derivatives of } f \}$ is the set of states,
- $\text{alphabet } \Sigma = \Bigg \{ \frac{\partial}{\partial x_0}, \frac{\partial}{\partial x_1}, ... , \frac{\partial}{\partial x_n} \Bigg \},$
- $f = f(x_0 , x_1 , \dots , x_n )$ is the unique start state,
- transition (partial, as in not defined for all possible inputs) function $\delta : \Sigma \times \partial f \longrightarrow \partial f, \bigg( \frac{ \partial}{\partial x_i} , p \bigg) \mapsto \frac{\partial p}{\partial x_i}$, where we only keep those elements $\bigg( \frac{\partial}{\partial x_i} , p \bigg)$ that satisfy either $\frac{\partial p}{\partial x_i} \neq 0$ or $\frac{\partial p }{\partial x_i} = 0 \text{ for all } i$ and $p \not\equiv 0$,
- $\text{and } \equiv0$, the zero polynomial, is the unique accept state.
The cardinality of $\partial f$ is the title of the question.
For monomials, this is not to difficult to solve:
In particular, consider $x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}$
Then any sequence $(a_1,a_2,\ldots,a_k) \in \{0,1,...,n_1\}\times\{0,1,...,n_2\}\times \cdots \{0,1,\ldots,n_k\}$
Map it to the monomial $\dfrac{d^{a_1+a_2+\cdots+a_k}}{dx_1^{a_1} dx_2^{a_2}\cdots dx_k^{a_k}}x_1^{n_1}x_2^{n_2}\cdots x_k^{n_k}$
Each monomial is unique and covers every possible monomial, except $0$, is of that form. Hence the total is: $$ (n_1+1)(n_2+1)\cdots(n_k+1)+1$$
In your example, $n_1=n_2=1$ so the total number is $(1+1)(1+1)+1=5$.