Derivative of covariance w.r.t. inverse covariance when elements are function of a vector I have this equation:
$$\nabla f^T x+ \nabla f^T \Sigma^{-1} (\Sigma \circ Q)x = -\frac{1}{2}\nabla f^T \Sigma^{-1} \nabla \tag{1}f$$
where $\nabla f,x$ are vectors, and
$$\nabla f_i =a_i - E[\nabla r_i]$$
($a_i$ being a scalar)
And $\Sigma$ is the covariance matrix corresponding to the vector $\nabla r$, the "$\circ$" indicates the hadamard product, and the elements of $Q$ are covariance elasticities w.r.t. $x$ defined as:
$$Q_{ij} = \frac{1}{\delta_{ij} + 1}\frac{x_i}{\sigma_{ij}}\frac{\partial \sigma_{ij}}{\partial x_i}$$
(where the factor involving the kronecker delta is there just to get rid of a 2 which appears in the diagonal terms.)
So, it must be kept in mind that each $i,j$ element of $\Sigma$ (and hence of $\Sigma^{-1}$ and $Q$) is a function of $x_i$ and $x_j$.
I want to differentiate equation 1 w.r.t. $\Sigma^{-1}$.
I start with the "trace trick".
$$\nabla f^T x + tr((\nabla f x^T) \Sigma^{-1} (\Sigma \circ E)) = -\frac{1}{2}tr(\nabla f \nabla f^T \Sigma^{-1})$$
Then, I invoke the trace's invariance under cyclic permutations to rewrite the second term on the LHS as follows:
$$\nabla f^T x + tr((\Sigma \circ E) (\nabla f x^T) \Sigma^{-1}) = -\frac{1}{2}tr(\nabla f \nabla f^T \Sigma^{-1})$$
Now everything is set up to apply the basic formula for differentiation of a trace with respect to a matrix (Eq. 100 in the matrix cookbook):
$$(\Sigma \circ E) (\nabla f x^T)=-\frac{1}{2}\nabla f \nabla f^T$$
My main question here is what about $\Sigma \circ Q$? I am basically praying that it does not participate in the differentiation, but that doesn't seem likely since all of these matrices are functions of the same vector $x$.
Secondly, note that, alternatively, you can write equation 1 as:
$$\nabla f^T x+ \nabla f^T \Sigma^{-1} z = -\frac{1}{2}\nabla f^T \Sigma^{-1} \nabla f$$
where $z=(\Sigma \circ Q)x$
Following the same steps as above (and again praying that $\frac{\partial (\Sigma \circ Q)}{\partial \Sigma^{-1}} = 0$) you then arrive at
$$\nabla f z^T=-\frac{1}{2}\nabla f \nabla f^T$$
Which seems to mean that $z=-\frac{1}{2}\nabla f$, and then plugging this back into Eq. 1 gives $\nabla f^Tx =0$?
 A: Let's use a colon to denote the trace/Frobenius product
$$A:B={\rm tr}(A^TB)$$
Define the variables
$$\eqalign{
 g &= \nabla f,\,\,\,\,\,\alpha=g^Tx,\,\,\,\,\,M = M^T = \Sigma^{-1} \cr
\phi &= \alpha + (gx^T):(M(Q\circ M^{-1})) + \tfrac{1}{2}(gg^T):M \,\,\,= 0 \cr
}$$
You wish to find the gradient $\frac{\partial\phi}{\partial M}$, start by finding its differential.
$$\eqalign{
d\phi
 &= (gx^T):(dM(Q\circ M^{-1})) + (gx^T):(M(Q\circ dM^{-1})) + \tfrac{1}{2}(gg^T):dM \cr
 &= (gx^T(Q\circ M^{-1})^T):dM + (Q\circ(Mgx^T)):dM^{-1} + \tfrac{1}{2}(gg^T):dM \cr
 &= \Big(gx^T(Q\circ M^{-1})^T-M^{-1}\big(Q\circ(Mgx^T)\big)M^{-1}+\tfrac{1}{2}gg^T\Big):dM \cr
}$$
Setting the gradient to zero
$$\eqalign{
\frac{\partial\phi}{\partial M}
 &= gx^T(Q\circ M^{-1})^T-M^{-1}\big(Q\circ(Mgx^T)\big)M^{-1}+\tfrac{1}{2}gg^T
  = 0
}$$
leaves us with
$$\eqalign{
-\tfrac{1}{2}gg^T
 &= gx^T(Q\circ M^{-1})^T-M^{-1}\big(Q\circ(Mgx^T)\big)M^{-1} \cr
-\tfrac{1}{2}\nabla f\,\nabla f^T
 &= \nabla fz^T - \Sigma\Big(Q\circ\big(\Sigma^{-1}\nabla f\,x^T\big)\Big)\Sigma \cr
}$$
So you derived the first two terms correctly, but missed the 3rd.
