Let $A$ be a matrix of size $n\times n$ and $x$ a $n \times 1$ vector. Consider $y = x'AA'x,$ where $y$ is a scalar. I want to compute the derivatives of $y$ with respect to $A'$.
My attempt : write $y$ as
$$ y = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n x_i a_{ik} a_{jk} x_j$$
and take derivative of it w.r.t $a_{j,i}$, where $a_{j,i}$ is the $i,j$th element of the matrix $A'$. Is it true? If yes, how to get the derivatives?
One possible (tedious) solution?
\begin{align*} \dfrac{\partial x'AA'x}{\partial A'} &= \begin{bmatrix} \dfrac{\partial x'AA'x}{\partial a_{11}} & \dfrac{\partial x'AA'x}{\partial a_{21}} & \cdots & \dfrac{\partial x'AA'x}{\partial a_{n1}}\\ & & \\ \dfrac{\partial x'AA'x}{\partial a_{12}} & \dfrac{\partial x'AA'x}{\partial a_{22}} & \cdots & \dfrac{\partial x'AA'x}{\partial a_{n2}}\\ \vdots & \vdots & \cdots & \vdots\\ \dfrac{\partial x'AA'x}{\partial a_{1n}} & \dfrac{\partial x'AA'x}{\partial a_{2n}} & \cdots & \dfrac{\partial x'AA'x}{\partial a_{nn}} \end{bmatrix}\\ &= \begin{bmatrix} 2x_1\sum\limits_{i=1}^n a_{i1}x_i & 2x_2\sum\limits_{i=1}^n a_{i1}x_i & \cdots & 2x_n\sum\limits_{i=1}^n a_{i1}x_i \\ & & \\ 2x_1\sum\limits_{i=1}^n a_{i2}x_i & 2x_2\sum\limits_{i=1}^n a_{i2}x_i & \cdots & 2x_n\sum\limits_{i=1}^n a_{i2}x_i \\ \vdots & \vdots & \cdots & \vdots\\ 2x_1\sum\limits_{i=1}^n a_{in}x_i & 2x_2\sum\limits_{i=1}^n a_{in}x_i & \cdots & 2x_n\sum\limits_{i=1}^n a_{in}x_i \end{bmatrix}\\ &= 2 \begin{bmatrix} A_{1.}' x x_1 & A_{1.}' x x_2 & \cdots & A_{1.}' x x_n\\ & & \\ A_{2.}' x x_1 & A_{2.}' x x_2 & \cdots & A_{2.}' x x_n\\ \vdots & \vdots & \cdots & \vdots\\ A_{n.}' x x_1 & A_{n.}' x x_2 & \cdots & A_{n.}' x x_n \end{bmatrix}\\ &= 2 A'x [ x_1\; x_2\; \cdots\; x_n]\\ &= 2 A'xx', \end{align*} where $A'_{j.}$ refers to the $j$th row of $A'$.
