Schubert polynomials are polynomials in the ring $\mathbb{Z}[x]$ where $x$ is an infinite set of variables indexed by the positive integers and they can be expressed in terms of "standard elementary monomials" which are products of elementary symmetric polynomials $e_i^j$ in varying initial sets of variables. Quantum Schubert polynomials lie in the ring $\mathbb{Z}[x,q]$ where $q$ is another infinite set of variables and can be obtained from the Schubert polynomials by taking the expression in terms of SEMs and whenever $e_i^j$ appears in such a monomial replace it with the "quantized" $E_i^j$ which involves both sets of variables $x$ and $q$. This miraculously results in polynomials that represent the small quantum cohomology ring of the complete flag variety.
Fulton defines universal Schubert polynomials in terms of "divided difference operators" that simply re-express standard elementary monomials as the standard elementary monomials you'd get when you apply a divided difference in the usual polynomial ring $\mathbb{Z}[x]$, then when you deform them to quantum SEM's you get the quantum Schubert polynomials.
We can define Fulton's divided difference operators, say $D^i$ for $i\in\mathbb{N}$, in $\mathbb{Z}[x,q]$ and they act on the quantum Schubert polynomials, at least linearly, like the usual divided difference operators act on Schubert polynomials. For completeness, I will add the formula: say we are defining $D^k$. Abbreviate $E_a^{k-1}E_b^{k}$ as $[a,b]$, where $E_b^k$, etc. is the quantum elementary polynomial of degree $b$ in $k$ $x$ variables. Then if $a\geq b-1$, then $$D^k([a,b])=\sum_{i\geq 0}[a+i,b-1-i]-\sum_{i\geq 1}[b-1-i,a+i]$$ and if $a\leq b-2$ then $$D^k([a,b])=\sum_{i\geq 0} [b-1-i,a-i]-\sum_{i\geq 1}[a-i,b-1-i]$$
These are clearly not the same operators as the standard divided difference operators, which would have the same formula but replacing $E_a^k$ with $e_a^k$. This is fine, but my question is: is there a Leibniz formula for these operators?
My entire dissertation was about the Leibniz formula for divided difference operators, but prior to that the following Leibniz formula was known and is easy to see: the divided difference operator $\partial^i$ is defined as $$\partial^i(p)=\frac1{x_{i}-x_{i+1}}(1-s_i)$$ and the Leibniz formula is $$\partial^i(fg)=\partial^i(f)g+s_i(f)\partial^i(g)$$
A Leibniz formula could look something like this: there is an operator $S_i$ such that $$D^i(fg)=D^i(f)g+S_i(f)D^i(g)$$ or it could have $\mathbb{Z}[x,q]$ coefficients like, just for a random example that is not correct because it has the wrong degree, $$D^i(fg)=D^i(f)g+fD^i(g)+(x_{i+1}-x_i+q_i)D^i(f)D^i(g)$$
Does anyone know of any such Leibniz formula or anything like it?
