Identity for divergence of vector product I'm reviewing this paper and they state the following "vector identity":
$ \nabla \cdot (\textbf{u}\textbf{v}) = \nabla \textbf{u} \cdot \textbf{v} + (\nabla \cdot \textbf{v})\textbf{u}$
I tried to find some material online about this identity but didn't have any luck. I'm pretty sure $\textbf{u}$ and $\textbf{v}$ are both vectors, as they're both bold face in the paper. Can someone help to provide more information on this identity? Is it valid? Can you give the proof?
EDIT
After some discussion, it seems that $\textbf{u}$ should be interpreted as a scalar (remove boldface), not a vector. I will clarify this with the original source and follow up if there is any further amendments. If it is a scalar, the answer below is correct.

For this to make sense $u$ is a scalar function and $\mathbf{v} = (v_1, \ldots, v_n)$ is a vector function.
We can calculate the partial derivatives using the chain rule:
$$\frac{\partial}{\partial x_j}(u\mathbf{v})_i = \frac{\partial}{\partial x_j}(uv_i) = \frac{\partial u}{\partial x_j}v_i + u\frac{\partial v_i}{\partial x_j}$$
Therefore, the Jacobi matrix is given by
$$\nabla(u\mathbf{v}) = \left[\frac{\partial}{\partial x_j}(u\mathbf{v})_i\right]_{1\le i,j\le n} = \left[\frac{\partial u}{\partial x_j}v_i + u\frac{\partial v_i}{\partial x_j}\right]_{1\le i,j\le n} = \mathbf{v}\otimes (\nabla u) + u (\nabla\mathbf{v})$$
The divergence is obtained by taking the trace:
$$\nabla \cdot (u\mathbf{v}) = \operatorname{Tr}\left(\nabla(u\mathbf{v})\right) = \operatorname{Tr}(\mathbf{v}\otimes (\nabla u)) + u \operatorname{Tr}(\nabla\mathbf{v}) = \mathbf{v}\cdot (\nabla u) + u(\nabla\cdot \mathbf{v})$$
