# Show that $\nabla(fg)=f\nabla g+g\nabla f$ [closed]

This looks very much like the product rule to me. However, is this technically a valid answer to the question? How is it best answered?

• If everything else fails, write each side out in coordinates, and use the plain old product rule for each component separately. Commented Feb 3, 2015 at 13:58
• You can find a proof everywhere, and I'm sure that this had been asked here before
– user207710
Commented Feb 3, 2015 at 14:00

It's definitely the product rule here. Since $\nabla = \left<\partial/\partial x,\ \ \partial/\partial y,\ \ \partial/\partial z\right>$,
\begin{align} \nabla(fg) &= \left<\frac{\partial}{\partial x} fg,\ \ \frac{\partial}{\partial y}fg,\ \ \frac{\partial}{\partial z}fg\right>&\mathrm{(Definition)}\\ &=\left<g\frac{\partial}{\partial x} f + f\frac{\partial}{\partial x} g,\ \ g\frac{\partial}{\partial y} f + f\frac{\partial}{\partial y} g,\ \ g\frac{\partial}{\partial z} f + f\frac{\partial}{\partial z} g\right> & \mathrm{(Product\ Rule)}\\ &= g \left<\frac{\partial}{\partial x} f,\ \ \frac{\partial}{\partial y}f,\ \ \frac{\partial}{\partial z}f\right> + f \left<\frac{\partial}{\partial x} g,\ \ \frac{\partial}{\partial y}g,\ \ \frac{\partial}{\partial z}g\right>&\mathrm{(Grouping)}\\ &=g\nabla{}f + f\nabla{}g &\mathrm{(Definition)} \end{align}
So to answer your question, this is a direct result of the product rule. As for "product rule" being a valid answer to the question, it depends on the context. While that answer would be correct, it definitely doesn't hurt to mathematically show why it's correct. And the best way to do that is break the $\nabla$ operator into it's components.