Does there exist a $\nabla$-notation variant of the product rule applied to $\nabla[\mathbf{f}(\mathbf{x})\otimes\mathbf{g}(\mathbf{x})]$? This is a vector-calculus notation question; as a disclaimer, I am working in rectilinear space! 
For vector functions $\mathbf{f},\mathbf{g}:\mathbb{R}^n\rightarrow\mathbb{R}^n$, the chain rule for $\mathbf{f}(\mathbf{g}(\mathbf{x}))$ in $\nabla$-notation reads
$$\nabla\left[\mathbf{f}(\mathbf{g}(\mathbf{x}))\right]=\nabla\mathbf{f}(\mathbf{g}(\mathbf{x}))\cdot\nabla\mathbf{g}(\mathbf{x})$$
where the usual convention that $\nabla$ appends an additional index to an array has been used.
My question is, is there a similar version for the product rule? Ie, 
$$\nabla[\mathbf{f}(\mathbf{x})\otimes\mathbf{g}(\mathbf{x})]=\mbox{}?$$
It is reasonably simple in indicial comma-derivative notation, whereupon we have $[f_i(\mathbf{x})g_j(\mathbf{x})]_{,k}=f_i(\mathbf{x})g_{j,k}(\mathbf{x})+f_{i,k}(\mathbf{x})g_{j}(\mathbf{x})$ which is reminiscent of the product rule in single-variable calculus. When I try rewriting this in $\nabla$-notation, the best I can come up with is 
$$\nabla[\mathbf{f}(\mathbf{x})\otimes\mathbf{g}(\mathbf{x})]=\mathbf{f}(\mathbf{x})\otimes\nabla\mathbf{g}(\mathbf{x})+\mbox{Transpose}[\nabla\mathbf{f}(\mathbf{x})\otimes\mathbf{g}(\mathbf{x}),\{1,3,2\}]$$
where the Transpose operator permutes the indices of an array in the same manner as implemented in Mathematica. 
I find this a bit unsightly, although still usable for handwritten work. Is there a less hideous way of expressing this identity using $\nabla$ notation, or is indicial notation the only way to express this identity without resorting to expressions involving array index permutations?
 A: $$\nabla[\mathbf{f}(\mathbf{x})\otimes\mathbf{g}(\mathbf{x})]$$ is not well defined as $\mathbf{f}:\mathbb{R}^m\rightarrow\mathbb{R}^n$, $\mathbf{g}:\mathbb{R}^p\rightarrow\mathbb{R}^m$ and $x\in \mathbb{R}^p$ (in other words, $\mathbf{f}(\mathbf{x})$ is not well defined if $p\neq m$). 
On the other hand, $\nabla\left[\mathbf{f}(\mathbf{g}(\mathbf{x}))\right]$ is understood as the application of the $\nabla$-operator to the composition $\mathbf{f}\circ\mathbf{g}$ at the point $\mathbf{x}$.
In a more general context, if you consider two complexes $(A,d_A)$ and $(B,d_B)$ with differentials $d_A$, and $d_B$, then the tensor product $A\otimes B$ is again a complex with differential $d_{A\otimes B}=d_A\otimes 1_B+1_A\otimes d_B.$
These formulae can be useful when dealing with vector-valued differential forms, for example. I do not know if this is relevant for your applications, though.


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*EDIT
After a notational edit in the OP, I add this section. Let $f,g\in A$, with $A:= C^\infty(\mathbb R^n;\mathbb R^n)$. This is an $\mathbb R$-bimodule, i.e. $\lambda f\in A$, for all $f\in A$ and $\lambda \in\mathbb R$ (with a straightforward compatibility condition). Then $\nabla: A\otimes_{\mathbb R} A\rightarrow A\otimes_\mathbb R A$ is the operator defined as follows
$$\nabla (f\otimes g)(x):= \nabla f(x)\otimes g(x)+ f(x)\otimes \nabla g(x), $$
for all $f,g\in A$ and $x\in\mathbb R^n$. Note that $\nabla (f\otimes 1)= \nabla f\otimes 1$ and $\nabla (1\otimes g)= 1\otimes \nabla g$.
