How can I differentiate the following using Matrix Calculus? The mathematical expression is like this: 
$f(\mu_{q(\beta)}, \Sigma_{q(\beta)}) = \mathbf{1}_{n}^{T} \exp \left \{ X \mu_{q(\beta)} + \frac{1}{2} \operatorname{diagonal}(X \Sigma_{q(\beta)} X^{T}) \right\}$ where $\exp$ indicates element-wise exponentiation and $\operatorname{diagonal}$ is a vector with the diagonal entries of the matrix as its components. Additionally, $\mathbf{1}_{n}^{T}$ is a vector in $\mathbb{R}^{n}$ with only ones as its entries and for notational ease, I used lower-case for vectors whereas matrices are in upper-case. $\Sigma_{q(\beta)}$ is symmetric whereas $X$ is not.
And I want to differentiate this by $\mu_{q(\beta)}$ and $\Sigma_{q(\beta)}$ each. So $\frac{\partial f}{\partial \mu_{q(\beta)}}$ and $\frac{\partial f}{\partial \Sigma_{q(\beta)}}$ are what I need.
 A: Your elaborate subscripts are too hard to type, so I'm going to use simpler variables and write the problem as
$$\eqalign{
  w &= \mu_{q(\beta)} \cr
  S &= \Sigma_{q(\beta)} \cr
  y &= Xw+\frac{1}{2}\,{\rm diag}(XSX^T) \cr
  e &= \exp(y) \cr
  f &= 1^Te \cr
}$$
The differentials of these quantities can be expressed using the Hadamard ($\circ$) product as 
$$\eqalign{
 dy &= X\,dw+\frac{1}{2}\,{\rm diag}(X\,dS\,X^T) \cr
 de &= e\circ dy \cr
 df &= 1^Tde \cr
    &= 1^T(e\circ dy) \cr
    &= e^Tdy \cr
    &= e^T(X\,dw+\frac{1}{2}\,{\rm diag}(X\,dS\,X^T)) \cr
}$$
Now set $dS=0$ and find the gradient with respect to $w$
$$\eqalign{
 \frac{\partial f}{\partial w} &= e^TX \cr
}$$
Working out the gradient with respect to $S$ is a bit harder
$$\eqalign{
 df &= \frac{1}{2}\,e^T{\rm diag}(X\,dS\,X^T)) \cr
  &= \frac{1}{2}\,{\rm diag}(e^T):X\,dS\,X^T \cr
  &= \frac{1}{2}\,E:X\,dS\,X^T \cr
  &= \frac{1}{2}\,X^TEX:dS \cr\cr
 \frac{\partial f}{\partial S} &= \frac{1}{2}\,X^TEX \cr
   &= \frac{1}{2}\,X^T\,{\rm diag}(\exp(y))\,X \cr
}$$
