Derivative with respect to diagonal of diagonal matrix Suppose I have a diagonal matrix $\pmb{D}$ and a symmetric matrix $\pmb{X}$ that is not a function of $\pmb{D}$, and I wish to find the following derivative:
$$
\frac{\partial}{\partial \mathrm{diag}(\pmb{D})} \mathrm{vec}\left(\pmb{D}\pmb{X}\pmb{D}\right),
$$
in which $\mathrm{diag}(\pmb{D})$ represents the diagonal of $\pmb{D}$. I know the following derivative:
$$
\frac{\partial}{\partial \mathrm{vec}(\pmb{D})} \mathrm{vec}\left(\pmb{D}\pmb{X}\pmb{D}\right) = \left(\pmb{D}\pmb{X} \otimes \pmb{I} \right) + \left(\pmb{I} \otimes \pmb{X}\pmb{D}\right)
$$
So I guess I can find the answer by multyplying this with $\frac{\partial \mathrm{vec}\left( \pmb{D} \right)}{\partial \mathrm{diag}(\pmb{D})}$, which should be some straightforward matrix with zeroes and ones. So my question is, (a) does this matrix have a name and can it easily be determined? and (b) isn't there some simpler way to do this?
 A: Although "D" is a great mnemonic for "diagonal" it is easily confused with "derivative" operations, which also use a mnemonic "D". 
Instead let's use a convention where upper/lower case letters are related by a diagonal operation:  uppercase is the matrix, lowercase is the vector.    
Given an arbitrary matrix $X$ and a diagonal matrix 
$\,A={\rm Diag}(a)$
the product in question can be expanded as
$$\eqalign{
P
 &= AXA \cr
 &= X\odot aa^T \cr
{\rm vec}(P) &= {\rm vec}(X)\odot{\rm vec}(aa^T) \cr
v_p
 &= v_x\odot(a\otimes a) \cr
 &= V_x\,(a\otimes a) \cr
}$$
where $\odot$ denotes the elementwise/Hadamard product
and $\otimes$ denotes the Kronecker product.
The differential and gradient of the vectorized product are 
$$\eqalign{
dv_p
 &= V_x\,(a\otimes da+da\otimes a) \cr
 &= V_x\,(a\otimes E+E\otimes a)\,da \cr
\frac{\partial v_p}{\partial a}
 &= V_x\,(a\otimes E+E\otimes a) \cr
 &= {\rm Diag}\Big(v_x\Big)\,\big(a\otimes E+E\otimes a\big) \cr
 &= \Big(v_xe^T\Big)\odot\big(a\otimes E+E\otimes a\big) \cr
 &= \Big({\rm vec}(X)\,e^T\Big)\odot\big(a\otimes E+E\otimes a\big) \cr
}$$
where $E$ is the identity matrix and $e$ is the vector of all ones, i.e. $\,E={\rm Diag}(e)$
A: I found a solution, but will not accept this answer yet in case someone has an easier solution. My solution to find $\frac{\partial \mathrm{vec}\left( \pmb{D} \right)}{\partial \mathrm{diag}(\pmb{D})}$ was to reconize that for $n \times n$ diagonal matrix $\pmb{D}$:
$$
\pmb{D} = \sum_{i=1}^{n} \pmb{E}_i \delta_{ii},
$$
with $\pmb{E}_i$ being an $n \times n$ matrix with a 1 on the $i$th diagonal and zeroes otherwise:
$$
e_{jk} = \begin{cases}
1 & \text{if } j = k = i \\
0 & \text{otherwise}
\end{cases}
$$
The derivative can then be derived as:
$$
\frac{\partial \mathrm{vec}\left( \pmb{D} \right)}{\partial \mathrm{diag}(\pmb{D})} = \begin{bmatrix}
\mathrm{vec}(\pmb{E}_1) & \mathrm{vec}(\pmb{E}_2) & \ldots & \mathrm{vec}(\pmb{E}_n)
\end{bmatrix}
$$
Denoting this result $\pmb{E}^*$ I obtain:
$$
\frac{\partial}{\partial \mathrm{diag}(\pmb{D})} \mathrm{vec}\left(\pmb{D}\pmb{X}\pmb{D}\right) = \left( \left(\pmb{D}\pmb{X} \otimes \pmb{I} \right) + \left(\pmb{I} \otimes \pmb{X}\pmb{D}\right) \right) \pmb{E}^*
$$
