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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 $d\mathbf{\Omega^{-1}}=-\mathbf{\Omega^{-1}}d\mathbf{\Omega}\,\,\mathbf{\Omega^{-1}}$?

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You're on the right track, that is the correct product rule.

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 $${\rm Skew}(X) = \tfrac 12\left(X-X^T\right)$$

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