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 $$\frac{\partial\Sigma}{\partial v}$$?

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 }
• I collected all of the terms involving $k$ into the definition of $F$. Now I've made an edit to keep them as separate variables. Hopefully that makes it easier to follow.
• Still reading the derivation. Just at the first few equations, it seems to me that $P$ should be equal to $I+kF$, no?
• Indeed $P=I+kF\;$ Thanks to all the commenters. The answer has been updated.