Derivative of determinate function w.r.t. matrix with vectors For my research, I need to calculate a derivative of scalar determinate function w.r.t. matrix with vectors
Here is a scalar defined as $c = \sqrt{det(A)}$
$\mathbf{A}=\mathbf{JJ}^T$
where $\mathbf{J}$ is a m-by-n matrix, n $\ge$ m
also $\mathbf{J}(\vec{\theta})$ is a function of vector $\vec{\theta}\in\Bbb{R}^k$
How can I get the derivative $\frac{\partial{c}}{\partial{\vec{\theta}}}$ ?

So far, I had calculated it base on the Matrix cookbook $$\frac{\partial{c}}{\partial{\vec{\theta}}}= 
 \frac{\partial{c}}{\partial{det(\mathbf{A})}} 
 \frac{\partial{det(\mathbf{A})}}{\partial{\mathbf{A}}}
 \frac{\partial{\mathbf{A}}}{\partial{\vec{\theta}}}=
 \frac{1}{\sqrt{det(\mathbf{A})}}
 det(\mathbf{A})(\mathbf{A}^{-1})^T
 \frac{\partial{\mathbf{A}}}{\partial{\vec{\theta}}}$$
The key point I can't get：$\frac{\partial{\mathbf{A}}}{\partial{\vec{\theta}}}$ is a tensor of order 3, right?
How can it multiplicate a scalar and matrix before to become a vector $\frac{\partial{c}}{\partial{\vec{\theta}}}$ ?
I also tried to vectorize $\mathbf{A}$, but there is another question：how to multiplicate a matrix before and the vecterized $\frac{\partial{\mathbf{A}}}{\partial{\vec{\theta}}}$?
Please help me to fiqure out these questions.
Many appreciate any answer and suggestion.
 A: $
\def\a{\alpha}\def\t{\theta}
\def\o{{\tt1}}\def\p{\partial}\def\j{\jmath}
\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\vecc#1{\operatorname{vec}\LR{#1}}
\def\sym#1{\operatorname{sym}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\c#1{\color{red}{#1}}
$First, let's use a convention in which
an uppercase letter denotes a matrix, a lowercase letter a vector, and a Greek letter a scalar. So rename the problem variables to
$\;\{c,\t\} \to \{\a,x\}$
Second, let's introduce some new variables
$$\eqalign{
j &= \vecc{J} \\
B &= \a A^{-1} J &\qiq \;\;\;b\; &= \vecc{B}
   = \a\LR{I_n\otimes A^{-1}} j \\
M &= \grad{j}{x} &\qiq \:dj &= M\,dx \\
A &= JJ^T &\qiq dA &= \LR{dJ\,J^T+J\,dJ^T} \\
}$$
Finally, the Frobenius product is a convenient notation for the trace
$$\eqalign{
A:G &= \sum_{i=1}^m\sum_{k=1}^n A_{ik}G_{ik} \;=\; \trace{A^TG} \\
A:A &= \big\|A\big\|^2_F \\
}$$
The properties of the underlying trace function allow the terms in such
a product to be rearranged in many different but equivalent ways, e.g.
$$\eqalign{
A:G &= G:A \\
A:G &= A^T:G^T \\
H:\LR{AG} &= \LR{HG^T}:A = \LR{A^TH}:G \\
}$$
Square the function and calculate its differential, then perform a change of variables
from $A\to J\to \j\to x$, then recover the desired gradient.
$$\eqalign{
\a &= {\det(A)}^{1/2} \\
\a^2 &= \det(A) \\
2\a\;d\a &= d\det(A) \\
 &= \LR{\a^2A^{-T}}:dA \\
d\a &= \frac 12\LR{\a A^{-\o}}:\LR{dJ\,J^T+J\,dJ^T} \\
 &= B:dJ \\
 &= b:dj \\
 &= b:\LR{M\,dx} \\
 &= \LR{M^Tb}:dx \\
\grad{\a}{x}
 &= {M^Tb} \\
 &= \det(A)^{1/2}
 \,\LR{\grad{\vecc{J}}{x}}^T \LR{I_n\otimes A^{-1}}
 \,\vecc{J} \\
}$$
