Derivative of the log determinant of the covariance matrix I have a covariance matrix defined as a rank-one matrix plus a diagonal matrix, with free parameters including a scalar $k$ and a column vector $\vec{v}$. This covariance matrix can be written as $\Sigma=kvv^T+(1-k)I\circ{vv^T}$. I am interested in the derivative of the log-determinant of this covariance matrix $\Sigma$ with respect to each of the element in $\vec{v}$ and with respect to $k$.
Checking on some online materials, I found the derivation and formula that $\frac{\partial\ln|A|}{\partial{x}}=Tr(A^{-1}\frac{\partial{A}}{\partial{x}})$. However, I still get stuck at deriving $\frac{\partial\Sigma}{\partial v}$ and $\frac{\partial\Sigma}{\partial{k}}$.
A related question, if I am interested in the derivative of $\Sigma$ with respect to each of the element in vector $\vec{v}$, should I still think of it as a question in which one is interested in the derivative of a matrix with respect to a vector $\frac{\partial\Sigma}{\partial v}$?
 A: Define the variables
$$\eqalign{
 F &= {\large\tt 1} - I \cr
 P &= I+kF \cr
 S &= \Sigma = P\circ vv^T \cr
}$$
Write the function of interest in terms of these variables and find its differential.
$$\eqalign{
\phi &= \log(\det(S)) \cr
d\phi
 &= S^{-1}:dS \cr
}$$
Expand $dS$ in terms of $dv$ and find the gradient wrt $v$.
$$\eqalign{
d\phi &= S^{-1}:P\circ(dv\,v^T+v\,dv^T) \cr
 &= 2P\circ S^{-1}:dv\,v^T \cr
 &= 2(P\circ S^{-1})v:dv \cr
\frac{\partial \phi}{\partial v} &= 2(P\circ S^{-1})v \cr
}$$
Expand $dS$ in terms of $dk$ and find the gradient wrt $k$.
$$\eqalign{
d\phi &= S^{-1}:dP\circ vv^T \cr
 &= S^{-1}:(dk\,F)\circ vv^T \cr
 &= S^{-1}:({\large\tt 1}-I)\circ vv^T\,dk \cr
 &= S^{-1}:(vv^T-I\circ vv^T)\,dk \cr
\frac{\partial\phi}{\partial k}
 &= S^{-1}:(vv^T-I\circ vv^T) \cr\cr
}$$
In the steps above, a colon denotes the trace/Frobenius product, i.e.
$$A:B = {\rm Tr}(A^TB)$$
which commutes with the elementwise/Hadamard product
$$A:B\circ C = A\circ B:C$$
and, of course, the Hadamard and Frobenius products are themselves commutative
$$\eqalign{
A:B &= B:A \cr
A\circ B &= B\circ A \cr
}$$
