I'm looking for Leibniz rule for directional derivatives of higher order.
Higher order directional derivatives are connected to symmetric tensors. There is an algebra homomorphism $D$ from the space of symmetric tensors $Sym(V)$ over vector space $V$ to the space $L(C^\infty(V))$, space of linear transformations of smooth functions on $V$.
$$ D: Sym(V) \rightarrow L(C^\infty(V)) $$ defined such that for any $u\in V$ and $A,B \in Sym(V)$ \begin{align} D(u) &= \partial_u \\ D(A\odot B) &= D(A)D(B) \\ D(A+B) &= D(A)+D(B) \end{align} where $\partial_u$ is derivative in the direction $u$.
This algebra homomorphism is well defined because $V$ generates $Sym(V)$ and partial derivatives commute.
If I have two smooth function $f,g \in C^\infty(V)$ and symmetric tensor $A\in Sym(V)$, is there something like Leibniz rule for $$ D(A)(fg) = ? $$
For example for $A=u\odot v$. I have $$ D(u\odot v)(fg) = D(u\odot v)(f)g + D(u)(f) D(v)(g) + D(v)(f)D(u)(g) + fD(u\odot v)(g) $$ The above formula depends on an explicit decomposition of $A$, do we need an explicit decomposition for such a formula?