find matrix gradient $f: \Omega \rightarrow \mathbb{R}$
defined by
$\Omega=\left\{X \in \mathbb{R}^{m \times n}: X^{\top} A X+B^{\top} X+X^{\top} B+C \succ 0\right\}$
with
$A \in \mathbb{S}^{m}, B \in \mathbb{R}^{m \times n}, C \in \mathbb{S}^{n}$
arbitrary;
$\qquad f(X)=\log \operatorname{det}\left(X^{\top} A X+B^{\top} X+X^{\top} B+C\right)$
Find the gradient.
 A: $\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\B{\big}\def\L{\left}\def\R{\right}\def\o{\operatorname}$Let's
use a colon to denote the trace/Frobenius product, i.e.
$$\eqalign{
A:B &= \o{Tr}\L(A^TB\R) \\
A:A &= \big\|A\big\|_F^2 \\
}$$
And for typing convenience, define the symmetric matrix
$$\eqalign{
M &= X^TAX + B^TX + C \\
dM &= 2\operatorname{Sym}\L(X^TA\,dX\R) + B\,dX \\
}$$
where $\;\operatorname{Sym}\L(X\R)\doteq\tfrac 12\L(X+X^T\R)$
Write the function using the notation above.
Then calculate its differential and gradient.
$$\eqalign{
f &= \log\det(M) \\
df&= d\operatorname{Tr}(\log M) \\
  &= M^{-1}:dM \\
  &= M^{-1}:\big(dX^TAX + X^TA\,dX + B^TdX\big) \\
  &= M^{-1}X^TA:dX^T + AXM^{-1}:dX \,+\, BM^{-1}:dX \\
  &= \B(2AX+B\B)M^{-1}:dX \\
  &= \B(2AX+B\B)\,\B(X^TAX + B^TX + C\B)^{-1}:dX \\
\p{f}{X}
  &= \B(2AX+B\B)\,\B(X^TAX + B^TX + C\B)^{-1} \\\\
}$$

The first step in the derivation makes use of Jacobi's formula.
The next step uses a formula from the Matrix Cookbook
$$\frac{d}{dX}\o{Tr}(f(X)) = f'(X)^{T}$$
And subsequent steps make use of the fact that the terms in a Frobenius product can be rearranged in many equivalent ways, due to the properties of the underlying trace function, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB \\
}$$
