$\nabla\cdot(\varphi\mathbf v)=\varphi\,\nabla\cdot\mathbf v+\mathbf v\cdot\nabla\varphi$

Initializing...

$\nabla\cdot(\varphi\mathbf v)=\varphi\,\nabla\cdot\mathbf v+\mathbf v\cdot\nabla\varphi$ · Prove2Me