# Derivative of trace involving hadamard product and product of inverse matrices

I need to find the derivative with respect to $$\mathbf{\Omega}$$ of

$$Tr\left(\left(\left(\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}\right)\circ\mathbf{I}\right)\mathbf{S}\right)$$

In the above, $$\mathbf{\Omega}^{-1}$$ is symmetric, $$\mathbf{C}$$ is symmetric, $$\mathbf{I}$$ is the identity matrix and $$\mathbf{S}$$ is symmetric.

I understand that writing this using $$:$$ notation might help, which if I'm correct allows me to write

\begin{align} \phi&=\left(\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}\right)\circ\mathbf{I}:\mathbf{S}\\ &=\mathbf{\Omega^{-1}}\mathbf{C}\mathbf{\Omega^{-1}}:\mathbf{I}\circ\mathbf{S} \end{align}

However from there I'm unsure - do I need a version of the product rule to deal with the right hand side - in which case how is this written when using $$:$$ notation? Or using the rule that $$d\mathbf{\Omega^{-1}}=-\mathbf{\Omega^{-1}}d\mathbf{\Omega}\,\,\mathbf{\Omega^{-1}}$$?

Here is a useful relationship when $$A,B$$ are symmetric matrices \eqalign{ {\rm Sym}(X) &\doteq \tfrac 12\left(X+X^T\right) \\ d(BAB) &= dB\,AB + BA\,dB \\ &= dB\,AB + (dB\,AB)^T \\ &= 2\;{\rm Sym}(dB\,AB) \\ }
Putting the pieces together yields \eqalign{ \phi &= (\Omega^{-1}C\Omega^{-1}\circ I):S \\ &= (I\circ S):\Omega^{-1}C\Omega^{-1} \\\\ d\phi &= (I\circ S):d\left(\Omega^{-1}C\Omega^{-1}\right) &\big({\rm sym\,rule}\big) \\ &= (I\circ S):2\;{\rm Sym}\left(d\Omega^{-1}C\Omega^{-1}\right) \\ &= 2\;{\rm Sym}(I\circ S):d\Omega^{-1}C\Omega^{-1} \\ &= 2\;(I\circ S):d\Omega^{-1}C\Omega^{-1} \\ &= 2\;(I\circ S)\Omega^{-1}C:d\Omega^{-1} &\big({\rm product\,rule}\big) \\ &= -2\;(I\circ S)\Omega^{-1}C:\Omega^{-1}d\Omega\,\Omega^{-1} \\ &= -2\;\Omega^{-1}(I\circ S)\Omega^{-1}C\Omega^{-1}:d\Omega \\\\ \frac{\partial\phi}{\partial\Omega} &= -2\;\Omega^{-1}(I\circ S)\Omega^{-1}C\Omega^{-1} &\big({\rm gradient\,matrix}\big) \\\\ } Here is another identity which was used above \eqalign{ X:{\rm Sym}(Y) &= X:\tfrac 12(Y+Y^T) \\ &= \tfrac 12X:Y + \tfrac 12X:Y^T \\ &= \tfrac 12X:Y + \tfrac 12X^T:Y \\ &= \tfrac 12(X+X^T):Y \\ &= {\rm Sym}(X):Y \\ } A similar identity exists for the function $${\rm Skew}(X) = \tfrac 12\left(X-X^T\right)$$