# Math Help - Proof with the nabla operator

1. ## Proof with the nabla operator

I've tried fiddling with this but I'm getting no where.

If $f$ and $g$ are two differentiable functions of $x$, $y$ and $z$, show that $\nabla(fg)=f\nabla g +g\nabla f$.

2. Hint:

$\nabla (fg)=\frac{\partial}{\partial x}(fg)\mathbf{i}+\frac{\partial}{\partial y}(fg)\mathbf{j}+\frac{\partial}{\partial z}(fg)\mathbf{k}.$

What happens when you apply the Product Rule on each of the three components?

3. Ahh I see it now, cheers.