How do you solve for the vector $v$?
$v$ = $p \times 1$
$x_i$ = $p \times 1$ constant
$c_i$ = scalar constant
$d$ vector of constants
$0 = \frac{d}{dv} \left[ \sum_{i=1}^n c_i (x_i^{T} v) (v^{T}x_i) \right] + d$
What I have right now, but I am not sure it is correct:
$0 = \sum_{i=1}^n c_i v x_i^{T} x_i + d$
$v = \frac{d}{\sum_{i=1}^n c_i x_i^{T} x_i} $
The objective function is $$\eqalign{ \def\LR#1{\left(#1\right)} \def\m#1{\left[\begin{array}{r|r}#1\end{array}\right]} \def\op#1{\operatorname{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\Si{\sum_{i=1}^n} \phi &= \LR{\Si c_ix_ix_i^T}:vv^T \;=\; XCX^T:vv^T \\ }$$ where $$\eqalign{ &C = \Diag{c_i}, \qquad X &= \m{x_1&x_2&\cdots&x_n}, \quad A:B = \trace{A^TB} \\ }$$ Its gradient is $$\eqalign{ d\phi &= XCX^T:\LR{v\:dv^T+dv\:v^T} \;=\; 2XCX^Tv:dv \\ \grad{\phi}{v} &= 2XCX^Tv \\ }$$ Equating this to the given vector yields $$ v = -\LR{\frac12}\left(XCX^T\right)^{-1}d \qquad\qquad\qquad\qquad $$