1. ## Grad of a scalar field problem

Hi,
I am having trouble understanding the following question:
There is a scalar field f(r) depending on r = (x,y,z) through r = $\sqrt[2]{x^2+y^2+z^2}$.
Show that $\nabla f = \frac{f'}{r}$r

I know that grad f is normally each partial derivative i.e. (df/dx, df/dy, df/dz) but how do I calculate it when it only has r as the parameter?
I'm also unsure as to how you would calculate f' in this situation, as I can't see an equation to differentiate.

Anything that helps me understand this better is greatly appreciated.
Thanks.

2. This is a confusing problem because of bad notation.

It is given that $f$ is a function whose domain is $\mathbb{R}^3$, but whose value only depends on $r$. So what is meant, really, is that there is a function $g:\mathbb{R}\to \mathbb{R}$ such that $f(x,y,z)=g(r(x,y,z))=g(\sqrt{x^2+y^2+z^2})$.

Now by $f'(x,y,z)$, what is meant is $g'(r(x,y,z))$, where $g'$ is the usual derivative

Essentially, you just have to take the gradient on both sides of the equation relating $f$ to $g$, using the chain rule.

3. That makes it so much clearer - thanks a lot for your help