I'm going through various vector calculus identities and I was curious that they listed no way of expressing $\nabla\cdot(\vec{F}\circ\vec{G})$, where $\vec{F} : \mathbb{R}^m\to\mathbb{R}^n$, and $\vec{G} : \mathbb{R}^n\to\mathbb{R}^m$.
Using the total derivative formula, we can derive the sum formula $$ \nabla\cdot (\vec{F}\circ\vec{G}) = \sum_{i=1}^n\sum_{j=1}^m \frac{\partial G_j}{\partial x_i}\left(\frac{\partial F_i}{\partial x_j}\circ G\right) $$ but I can't find a way to express this in a simpler, non-sum fashion. I had the impression it was some expression involving the Jacobian, seeing as all the partial derivatives of all the components of $\vec{F}$ and $\vec{G}$ are involved, but I can't find a way of combining $\text{J}_{\vec{G}}$ and $\text{J}_{\vec{F}}\circ G$ that results in a scalar.
Any help would be appreciated!
What about $\text{tr}(J_GJ_F)$ (where each of the jacobians are evaluated at the point corresponding to the chain rule)?
This describes exactly that sum, but using only the Jacobian matrices of each vector field.