I have a vector $V=[v_0,v_1...v_n]'$ which is a function of another Vector $W=[w_0,w_1,...,w_n]'$. Here $'$ denotes the transpose.
The scalar field is $f=V'(W) V(W)$ where $V$ and $W$ are of the same dimension $(n+1), n=0,1,...,n$.
I need the gradient. So I think this is the chain rule and I need the Jacobian matrix - is this right?
$$del(f) = 2JV$$
where $J$ is the Jacobion representing the differentiation of the vector $V$ wrt the vector $W$.
Also I am confused about the form of the Jacobian. Standard form has each row as follows eg. row $0$ would be
$$[dv_0/dw_0, \, dv_0/dv_1, \, \cdots, \, dv_0/dv_n]$$
Now to prove this I set up a summation $$f=\sum_{i=0}^n v(w_0,w_1,...,w_n)(i)^2 $$
$$ \frac{df}{v(i)} = 2\sum_{i=0}^n v\cdot \frac{dv(j)}{di}, \quad j=0,1,...,n$$
Does this look right? - sorry the text may put my equation off skew