Show that $\nabla(fg)=f\nabla g+g\nabla f$

Question

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?

Answer 1

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.

Answer 2

It is definitely a product rule. And you write in coordinates it is as many product rules as coordinates