Let $I = \langle f_1, \ldots f_r \rangle$ be an ideal in $R=k[x_1,\ldots,x_n]$ where $k$ is a field, and put $A = R/I$.
(If $k$ is algebraically closed and $I$ is radical then $A$ is the coordinate ring of an affine variety.)
Let $\mathfrak{X}^p(A) = \operatorname{Der}_k(\wedge^p A, A) $ be skew-symmetric $p$-derivations (derivation in each argument) of $A$.
Question: Is there an algorithm to calculate $\mathfrak{X}^p(A)$ in terms of $f_1,\ldots,f_r$? I am interested in $p=1,2,3$.
What I have tried/observed so far:
The condition that $P \in \mathfrak{X}^p(R)$ descends to (is well-defined on) $\mathfrak{X}^p(A)$ is a system of (ideal membership) equations $$P(x_{i_1}, \ldots, x_{i_{p-1}}, f_j) \in I \quad \text{for each } 1\leq i_1<\cdots<i_{p-1}\leq n \text{ and } 1\leq j\leq r.$$ (Note also every element of $\mathfrak{X}^p(A)$ has a lift to $\mathfrak{X}^p(R)$.)
Letting $P_{i_1,\ldots,i_p} = P(x_{i_1},\ldots,x_{i_p}) \in R$ be the (sought for) coefficients of $P$, the system becomes $$\sum_k P_{i_1,\ldots,i_{p-1},k} \frac{\partial f_j}{\partial x_k} \in I \quad \text{for each } 1\leq i_1<\cdots<i_{p-1}\leq n \text{ and } 1\leq j\leq r.$$
Here one can restrict to the sum over $k \in \{1,\ldots,n\} \setminus \{i_1,\ldots i_{p-1}\}$. Note the equations are not independent, due to skew-symmetry.
The equations above for fixed $j$ suggest to consider the intersection $I \cap \langle \frac{\partial f_j}{\partial x_1}, \ldots \frac{\partial f_j}{\partial x_n} \rangle$ which can be computed using Groebner bases. I am not sure what we can conclude about the coefficients $P_{i_1,\ldots,i_{p-1},k}$. Furthermore, I do not see how one would combine these results for different $(i_1,\ldots,i_{p-1})$'s and $j$'s.
$\mathfrak{X}^p(A)$ is an $A$-module. I think it is finitely generated. (I am interested in the case where it is.)
$\mathfrak{X}^p(A) \cong \operatorname{Hom}_A(\Omega^p(A), A)$ where $\Omega^p(A)$ are Kähler $p$-forms.
$\mathfrak{X}^2(A) \neq \wedge^2 \mathfrak{X}^1(A)$ in general: if $I = \langle yx, yz, y^2 \rangle$ in $R=\mathbb{C}[x,y,z]$ then $0 \neq y \frac{\partial}{\partial y} \wedge \frac{\partial}{\partial z}$ is in $\mathfrak{X}^2(A)$ and not in $\wedge^2 \mathfrak{X}^1(A)$.
I would also be interested in (classes of) examples where one can calculate $\mathfrak{X}^p(A)$ explicitly for $p=1,2,3$.