# Proof with the nabla operator

• November 15th 2009, 06:48 AM
chella182
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 $\nabla(fg)=f\nabla g +g\nabla f$.
• November 15th 2009, 08:44 AM
Scott H
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?
• November 16th 2009, 03:08 AM
chella182
Ahh I see it now, cheers.