I need to differentiate the following function: $$f(\Sigma)=-\frac{1}{2}\log|\Sigma|+-\sum_i C_i\exp\left(-\frac{1}{2}\frac{\mu_i^2}{1+\Sigma_{ii}}\right)$$
to find, $\frac{\partial f}{\partial \Sigma}$. The first term is a standard result. However its the sum of exponential terms that I'm unsure of. Treat $\mu_i, C_i$ to be constants and $\Sigma_{ii}$ are the diagonal terms in $\Sigma$.
I was thinking rewrite the second term as trace of $EC$ where $E$'s diagonal terms containing the exponential terms and $C$ is a diagonal matrix containing the $C_i$'s. <- just a suggestion, not sure if it will lead to a solution.
The derivative of the function $f_{ii}(\Delta) = [\Delta]_{ii}$ can be computed by noting that $f_{ii}$ is linear and hence is its own derivative, that is $Df_{ii}(\Delta)(H) = f_{ii}(H) = [H]_{ii} = \operatorname{tr} (E^{ii} H)$.
The derivative of $g(t) = e^{-{1 \over 2} { \mu_i^2 \over 1 + t}}$ is given by $g'(t) = {\mu_i^2 \over 2(1+t)^2 }e^{-{1 \over 2} { \mu_i^2 \over 1 + t}} $.
Hence the chain rule gives $D (g \circ f_{ii})(\Sigma)(H) = Dg(f_{ii}(\Sigma))(Df_{ii}(\Sigma)(H)) = {\mu_i^2 \over 2(1+\operatorname{tr} (E^{ii} \Sigma))^2 }e^{-{1 \over 2} { \mu_i^2 \over 1 + \operatorname{tr} (E^{ii} \Sigma)}} \operatorname{tr} (E^{ii} H)$.