nabla f(r), what is the secret?

Hey all!

I'm currently reviewing my multivariable calculus and vector analysis for re-exams in August. This is a problem that has been showing up more than once and back before I thought I got it but now I realize I can not calculate this example simply because I don't know what to do.

The Question:

Solve

(nablaf(r)) * **r**, where r = |**r**| and **r** = xi+yi+zj.

Clarification:

**Bold** is vector

* = dot

Attempt at solution

I know that nabla = (d/dx,d/dy,d/dz) and appying this to f(r) won't work (is my assumption). Thus I assume that nabla = d/dr in this case.

so nablaf(r) = ( d/dr f(r), d/dr f(r) , d/dr f(r) ) = f'(r)(d/dr,d/dr,d/dr)

That is pretty much as far as I get because I don't know what to do. Applying div,grad or curl to any fuction that is simply f(x,y,z) is something i know very well, but this has me baffeld.

I would appreciate any help that I can get on this.

Re: Div of f(r), what is the secret?

Are you sure that it's div? Because divergence is a scalar quantity, and therefore cannot be dotted into the vector **r**. Are you sure it isn't the gradient of the scalar function f?

Re: Div of f(r), what is the secret?

Yes, you are correct. It's supposed to be (nabla f(r)) * **r**

This question in first part of vector differenctial identitiy nabla * (f(r)**r** = (nablaf(r)) * **r** + f(r)(div **r**)

I'll edit my main question so that it is correct. I still need help though :)

Re: Div of f(r), what is the secret?

Right. So I'm assuming it's not the dot product in the expression

that's the problem, but the gradient. Right? Here's the trick: view as a function of the three coordinates thus:

That is, you'll need to use the chain rule on, for example, the x component of the gradient thus:

Can you continue?

Re: Div of f(r), what is the secret?

Indeed I can!

df/dr dr/dx = f'(r) * x/r in the x component and same for the rest. In the end I got **r** * **r** which is r^2 and the answer becomer f'(r)r.

Thanks that really helped me out!