If $f^{i_1...i_p}$ and $g$ are smooth functions on some open set $U\subseteq\mathbb R^n$, with the former symmetrically indexed, then we have the higher product rule $$ \sum_{i_\bullet=1}^n\partial_{i_1}\dots\partial_{i_p}\left(f^{i_1...i_p}g\right)=\sum_{k+l=p}\sum_{j_\bullet,k_\bullet=1}^n{p\choose k}\partial_{j_1}\dots\partial_{j_k}f^{j_1...j_k h_1...h_l}\partial_{h_1}\dots\partial_{h_l}g, $$which can be proven fairly easily by induction.
Let $R=\bigoplus_{p\in\mathbb Z}R^p$ be a graded commutative $k$-ring ($k$ is a field of characteristic 0), and let $d_1,\dots,d_n$ be a set of derivations of degree $-1$, and such that $d_i d_j=-d_j d_i$. Let $a^{i_1...i_p}\in R^p$ be an antisymmetrically indexed family of degree $p$ elements and $b\in R^q$.
I want to find a higher product formula that expresses the $p$-fold derivation of the product as $$ d_{i_1}\dots d_{i_p}\left(a^{i_1...i_p}b\right)=\sum_{k+l=p}c_{(k)(l)} d^{(k)}a\, d^{(l)}b, $$ where the right hand side is of course highly symbolic (for disambiguation, graded commutativity means $ab=(-1)^{pq}ba$).
I would guess induction still works, but the problem is that I need a formula to verify by induction. The combinatorics in this case appear to be much more complicated than in the commutative case, and I am unable to come up with a good formula.