There is a scaler-by-matrix derivative identity:
$$\frac{\partial}{\partial X}trace\left(AXBX'C\right)=B'X'A'C'+BX'CA$$
How does this change if instead I am trying to find
$$\frac{\partial}{\partial x}trace\left(Adiag(x)Bdiag(x)'C\right)$$
where $x$ is a vector rather than a matrix.
My thinking is that all I have to do is multiply the original identity by a vector of ones as that would be the derivative of $diag(x)$. However, I'm not sure how the chain rule interacts with traces.
I ask as I am trying to calculate. $$\frac{\partial}{\partial w}trace\left(Ddiag(w)\Omega diag(w)D'\right)$$
where $w \mathbb{\in R^{N}}$, $D\mathbb{\in R^{M\times N}}$, and $\Omega\mathbb{\in R^{N\times N}}$. Also $\Omega$ can be assumed to be positive definite.
This implies the result would be
$$\left(2\Omega diag(w)D'D\right)e$$
where $e \mathbb{\in R^{N}}$ is a vector of ones.