Proof of the product rule for the divergence

How can I prove that

$\nabla \cdot (fv) = \nabla f \cdot v + f\nabla \cdot v,$

where $v$ is a vector field and $f$ a scalar valued function?

Many thanks for the help!

Solution 1:

$$ \begin{align} \nabla \cdot (fv) &= \sum \partial_i (f v_i) \\ &= \sum (f \partial_i v_i + v_i \partial_i f) \\ &= f \sum \partial_i v_i + \sum v_i \partial_i f \\ &= f \nabla \cdot v + v \cdot \nabla f \end{align} $$

Solution 2:

Write out both expressions:

$$\nabla\cdot(fv) = \nabla\cdot(fv_1,fv_2,fv_3) = \partial_1(fv_1) + \partial_2(fv_2) + \partial_3(fv_3) = (\partial_1f)v_1 + f(\partial_1 v_1) + \dots$$

and

$$ \nabla f \cdot v = (\partial_1 f, \partial_2 f, \partial_3 f)\cdot(v_1,v_2,v_3) = (\partial_1 f) v_1 + \dots $$ $$ f \nabla \cdot v = f(\partial_1 v_1 + \dots) $$

then compare.