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

1. ## 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.

2. ## 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?

3. ## 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

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

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

$\displaystyle \nabla f(r)\cdot \mathbf{r}$

that's the problem, but the gradient. Right? Here's the trick: view $\displaystyle r=|\mathbf{r}|$ as a function of the three coordinates thus:

$\displaystyle r=\sqrt{x^{2}+y^{2}+z^{2}}.$

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

$\displaystyle \frac{\partial}{\partial x}\,f\!\left(\sqrt{x^{2}+y^{2}+z^{2}}\right).$

Can you continue?

5. ## 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!