Proving ∇·F = r̂·dF/dr using the chain rule

[yy205001]

Homework Statement

$\widetilde{F}$(r)=F₁(r)i+F₂(r)j+F₃(r)k
$\hat{r}$=r/r
r(x,y,z)=xi+yj+zk, r=abs(r)=sqrt(x²+y²+z²)
(Hint: The chain rule will be helpful for this question.)

Show that:
$\nabla$$\cdot$F = $\hat{r}$$\cdot$dF/dr.

Homework Equations

The Attempt at a Solution

My attempt:
First, $\nabla$$\cdot$F=(dF₁/dr,dF₂/dr,dF₃/dr)

Then, Start on the RHS.
$\hat{r}$$\cdot$dF/dr
=$\hat{r}$$\cdot$(dF₁/dr,dF₂/dr,dF₃/dr)
=((x,y,z)/r)$\cdot$(dF₁/dr,dF₂/dr,dF₃/dr)
Now, i use the chain rule here.
=((x,y,z)/r)$\cdot$(dF₁/dx*dx/dr, dF₂/dy*dy/dr, dF₃/dz*dz/dr)

And i can't do further more here, can anyone help me on this?
Thanks in advanced!

 

[CompuChip]

The Attempt at a Solution

My attempt:
First, $\nabla$$\cdot$F=(dF₁/dr,dF₂/dr,dF₃/dr)

I'm pretty sure that's not how the gradient works.
First of all, the "definition" is
$$\vec\nabla = \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right)$$

Secondly, there is a dot in between, which means that you should get a scalar and not a vector like you have written.

 

[yy205001]

ops! So my definition is wrong! That's why i can't do it further more!
I can prove it now!
thanks CompuChip