Assume that $\int_{\omega}{\rm div} \ w dx = \int_{\partial \omega}w \hat{n} d\sigma$
Prove that:
$$\int_{\omega}f \ {\rm div} \ v + v \nabla f dx = \int_{\partial \omega} f v \hat{n} d\sigma$$
My try (let $v$ be a vector field and $f$ be a scalar field):
Lets say that $w = f v$, so $${\rm div} \ w = {\rm div} \ (f v) = (\nabla f)v + f(\nabla v) =^{\text{(assumption)}} \int_{\partial \omega} f v \hat{n} d\sigma$$
Is it correct ?