chain rule for outer products involving the gradient operator?

Hi, I'm gonna calculate the Jacobian matrix of the following function:

$\displaystyle f(r) = \frac{(r\cdot v)r}{r^2}$

where $\displaystyle r$ is the position. Since all variables in this expression are vectors I haven't bothered making them bold. I realized that the the Jacobian can be expressed as

$\displaystyle J_f(r) = (\nabla\otimes f(r))^{\text{T}},$

where $\displaystyle \otimes$ is the outer product, so we have

$\displaystyle (J_f(r))^{\text{T}} = \nabla\otimes f(r) = \nabla\otimes \frac{(r\cdot v)r}{r^2}$

I don't know how to simplify this without making really ugly expansions of all the elements in the vectors, but I've kind of already figured out (I think) that this is equals to

$\displaystyle \frac{r\cdot v}{r^2}\,I + \frac{r\otimes v}{r^2} - \frac{2(r\cdot v)r}{r^4}\otimes r$

where $\displaystyle I$ is the identity matrix. As I said, I think this is correct, but I don't know how to show it! Can anyone please help me to show that this is really the case? Which product rules should I use here?

Re: chain rule for outer products involving the gradient operator?

So to be clear, the question is: *How do I calculate the Jacobian matrix of $\displaystyle f(r)$*?

**Edit:** To start off simpler, can someone please tell me how to calculate the Jacobian martix of $\displaystyle (r\cdot v)r$? The Jacobian of r is the unity matrix, $\displaystyle I$, and the Jacobian of v is the zero matrix.

**Edit 2:** The final expression in my previous post is supposed to equal the Jacobian matrix, not the transpose of the Jacobian matrix.

Re: chain rule for outer products involving the gradient operator?

What I finally did was that I calculated an expression for element_{i,j}. I eventually ended up with an expression containing only one matrix from which element_{i,j} was extracted; this matrix could then be identified as the Jacobian matrix since the equation I had held for every element_{i,j}. Kind of like the process you normally follow when you're using Einstein notation (I guess).