Gradient of ${\bf x}^\top {\bf A}^{1/2} {\bf x}$ with respect to $\bf A$ How to calculate the gradient $\nabla_{\bf A} \left( {\bf x}^\top {\bf A}^{1/2} {\bf x} \right)$, where $\bf x$ is $N \times 1$ column vector and $\bf A$ is $N \times N$ symmetric positive matrix?
The difficulty is that there is ${\bf x}^\top {\bf A}^{1/2} {\bf x}$ rather than ${\bf x}^\top {\bf A} {\bf x}$.

Motivation
I want to calculate the gradient of the vector Gaussian distribution $\mathcal{N} \left( \mathbf{y} \mid \mathbf{A}^{\frac{1}{2}}\mathbf{x}, \mathbf{I} \right)$ w.r.t. $\mathbf{A}$, where $\mathbf{I}$ is an identity matrix, and
$$ \mathcal{N} \left( \mathbf{y} \mid \mathbf{A}^{\frac{1}{2}},\mathbf{I} \right) = (2\pi)^{-\frac{N}{2}} \exp \left( -\left\|\mathbf{y}-\mathbf{A}^{\frac{1}{2}}\mathbf{x} \right\|_2^2 \right) $$
where $\|\cdot\|_2$ denotes the $\ell_2$ norm. Its difficulty is to calculate the term
$$\nabla_{\mathbf{A}} \left( \mathbf{y}^T\mathbf{A}^{\frac{1}{2}}\mathbf{x} \right)$$
This term can be similarly solved by $\nabla_{\mathbf{A}}\left( \mathbf{x}^T\mathbf{A}^{\frac{1}{2}}\mathbf{x} \right)$. And I believe this term exists.

$
\def\o{{\tt1}}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big(#1\Big)}
\def\bR#1{\big(#1\big)}
\def\vc#1{\operatorname{vec}\LR{#1}}
\def\dvc#1{\operatorname{unvec}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\gradLR#1#2{\LR{\grad{#1}{#2}}}
$For typing convenience, define the variables
$$\eqalign{
Y &= xx^T &\qiq y = \vc{Y} \,\doteq\, \LR{x\otimes I}x \\
B &= A^{1/2} &\qiq b = \vc{B} \\
}$$
and the Frobenius product, which is a concise notation for the trace
$$\eqalign{
Y:Z &= \sum_{i=1}^n\sum_{j=1}^n Y_{ij}Z_{ij} \;=\; \trace{Y^TZ} \\
Y:Y &= \|Y\|^2_F \\
}$$
The properties of the underlying trace function allow the terms in a
Frobenius product to be rearranged in many different ways, e.g.
$$\eqalign{
X:Y &= Y:X \\
X:Y &= X^T:Y^T \\
Z:\LR{XY} &= \LR{ZY^T}:X \\&= \LR{X^TZ}:Y \\
}$$
Exploring the relationship between the symmetric matrices $A\,{\rm and}\,B$
$$\eqalign{
A &= B^2 \\
dA &= B\,dB + dB\,B \\
   &= B\,dB\,I + I\,dB\,B \\
da &= \LR{I\otimes B+B\otimes I} db \\
   &= \LR{B\oplus B} db \qquad \big\{{\rm Kronecker\:Sum}\big\} \\
db &= \LR{B\oplus B}^{-1}da \\
}$$
Use the above notation to write the objective function. Then calculate its differential and gradient.
$$\eqalign{
\phi &= Y:B \\
d\phi &= Y:dB \\
  &= y:db \\
  &= y:\LR{B\oplus B}^{-1}da \\
  &= \LR{B\oplus B}^{-1}y:da \\
\grad{\phi}{a}  &= \LR{B\oplus B}^{-1}y \\
}$$
This is the vectorized form of the gradient,
but there's a clever formula involving the identity matrix and Kronecker products which recovers the matrix form
$$\eqalign{
\grad{\phi}{A}
 &= \LR{\vc{I}^T\otimes I}\LR{I\otimes\gradLR{\phi}{a}} \\
}$$
which can be written entirely in terms of $(I,A,x)$ as
$$\eqalign{
\grad{\phi}{A}
 &= \LR{\vc{I}^T\otimes I}\LR{I\otimes\LR{\LR{I\otimes A^{1/2}+A^{1/2}\otimes I}^{-1}\LR{x\otimes I}x}} \\
}$$
