∇ ⋅ (a b) = a ⋅ ∇b + b(∇ ⋅ a)
Please note.....on the left side, that is NOT a dot product between vectors a and b. I'm assuming the problem wants them multiplied together to make a vector "a1b1, a2b2, a3b3".
I think you're supposed to use the product rule in this proof, but I'm not positive. If anybody can provide a step by step of how the heck you would prove this, that would be amazing!
On
I'm guessing that $a $ is a vector field and $b$ is a scalar field. $\nabla \cdot (ab)$ stands for the divergence of the vector field $ab$. $$ a = (a_1, \cdots, a_n), \quad ab = (b a_1, \cdots, b a_n) $$
\begin{align*} \nabla \cdot (ab)= & \frac{\partial}{\partial x_1}(b a_1) + \cdots +\frac{\partial}{\partial x_n}(b a_n) = \frac{\partial b }{\partial x_1} a_1 + b \frac{\partial a_1}{\partial x_1} + \cdots +\frac{\partial b }{\partial x_n} a_n + b \frac{\partial a_n}{\partial x_n} \\ =& (\frac{\partial b }{\partial x_1} a_1 + \cdots + \frac{\partial b }{\partial x_n} a_n)+( b \frac{\partial a_1}{\partial x_1} + \cdots b \frac{\partial a_n}{\partial x_n})\\ =& a \cdot \nabla b + b (\nabla \cdot a) \end{align*}
You want to prove $\partial_i (a_i b_j) = a_i\partial_i b_j +b_j \partial_i a_i$ (summation over the repeated index $i$ is implicit). So yes, you just need the product rule.