# The Nabla Operator

1. Nov 19, 2011

### MCooltA

1. The problem statement, all variables and given/known data
I need to find $\nabla$f(r). I am given r = |R| where R is a vector, R =(x,y,z). I also have the function f(r) which is a differentiable function of r.

2. Relevant equations

So i know $\nabla$(g) = ($\partial$g/$\partial$x, $\partial$g/$\partial$y, $\partial$g/$\partial$z)

3. The attempt at a solution

So I've got;

r=√(x^2 + y^2 + z^2)

$\nabla$(r) = (2x,2y,2z)

Do i need to apply $\nabla$ operator again to the above $\nabla$(r)?

Does $\nabla$($\nabla$(r)) equal $\nabla$f(r)?

2. Nov 19, 2011

### HallsofIvy

Staff Emeritus
No, of course not. For one thing, $\nabla \nabla r$ makes no sense. $\nabla r$ is a vector function and "$\nabla f$ has to be applied to a scalar function. Also, there is NO "f" in that. You would be saying that $nabla f(r)$ is the same for all functions! You need the chain rule:
$$\nabla f(r)= \frac{df}{dr}\nabla r$$
df/dr will be a scalar function aand $\nabla f$ a vector function so this is scalar multiplication.

Or, equivalently, replace r in the function with $(x^2+ y^2+ z^2)^{1/2}$ and find the gradient.

What is f?

3. Nov 19, 2011

### MCooltA

$\nabla$(r) = (1/r) (x,y,z).
And then for df/dr = $\partial$f$_{x}$/$\partial$x + $\partial$f$_{y}$/$\partial$y + $\partial$f$_{z}$/$\partial$z