second order derivative of log det of matrix I need to find the second order derivative of a matrix.
$f(\pmb{\Delta})=\log \det(\pmb{I}+k\pmb{V^T \Delta V})$
Where $ \pmb{\Delta}$ is a triangular matrix.
I did the first order derivative but not sure if is correct.
$$\frac{\partial f(\pmb{\Delta})}{\partial  [\pmb{\Delta}]_{ii}} =  
\left[ k\pmb{V} \left( \pmb{I}+k\pmb{V^H}\pmb{\Delta}\pmb{V} \right)^{-1}\pmb{V^H} \right]_{ii}  $$
This is correct? How about the second order derivative?
 A: Define the matrix variables
$$\eqalign{
 X &= \Delta \cr
 A &= I + kV^TXV \implies dA &= kV^T\,dX\,V \cr
 Y &= kVA^{-T}V^T \cr
}$$
Write the function in terms of these variables and calculate its differential and gradient.
$$\eqalign{
 f &= \log\det A \cr
df &= A^{-T}:dA \cr
   &= A^{-T}:kV^T\,dX\,V \cr
   &= kVA^{-T}V^T:dX \cr
G = \frac{\partial f}{\partial X} &= kVA^{-T}V^T \cr
}$$
Repeat the process for the gradient.
$$\eqalign{
dG &= kV\,dA^{-T}\,V^T \cr
   &= -kVA^{-T}\,dA^T\,A^{-T}V^T \cr
   &= -k^2VA^{-T}V^T\,dX^T\,VA^{-T}V^T \cr
   &= -Y\,dX^T\,Y \cr
   &= -Y{\mathcal E}Y:{\mathcal L}:dX \cr
{\mathcal H} = \frac{\partial G}{\partial X}
   &= -Y{\mathcal E}Y:{\mathcal L} \cr
}$$
where $({\mathcal E},{\mathcal L})$ are 4th order tensors whose components can be written in terms of Kronecker deltas
$$\eqalign{
  {\mathcal E}_{ijkl} &= {\delta}_{ik} \, {\delta}_{jl} \cr
  {\mathcal L}_{ijkl} &= {\delta}_{il} \, {\delta}_{jk} \cr
}$$
Note that the second derivative (aka Hessian) is itself a 4th order tensor.
Double and single contraction products were used in some of the steps above, which can be expressed in component form as 
$$\eqalign{
{\mathcal B}_{ikln} &= \big(-Y{\mathcal E}Y\big)_{ikln}
   &= \sum_j\sum_m -Y_{ij}{\mathcal E}_{jklm}Y_{mn} \cr
{\mathcal H}_{ijmn} &= \big({\mathcal B}:{\mathcal L}\big)_{ijmn}
   &= \sum_k\sum_l {\mathcal B}_{ijkl}{\mathcal L}_{klmn} \cr
A:X &= \sum_i\sum_j A_{ij}X_{ij}
   &= {\,\rm Tr}\big(A^TX\big) \cr
}$$
