I wanted to validate whether my differentiation is correct here
Let $\mathcal{R}(\mathbf{x}) = \sqrt{\mathbf{x}^T \mathbf{\Sigma} \mathbf{x}}$. Note that $\mathbf{x}$ is $(n \times 1)$ vector and $\mathbf{\Sigma} = \mathbf{\Sigma}^T$, then:
$$\frac{\partial \mathcal{R} (\mathbf{x})}{\partial \mathbf{x}^T} = \frac{1}{2} \left( \mathbf{x}^T \mathbf{\Sigma}\mathbf{x} \right)^{-1/2} \frac{\partial}{\partial \mathbf{x}^T} \mathbf{x}^T \mathbf{\Sigma} \mathbf{x}^ = \frac{ \mathbf{\Sigma} \mathbf{x}}{\sqrt{\mathbf{x}^T \mathbf{\Sigma}\mathbf{x}} }$$
This boils down to knowing what is: $$\frac{\partial}{\partial \mathbf{x}^T} \mathbf{x}^T \mathbf{\Sigma} \mathbf{x}$$
noone knows? :)
EDIT:
it was actually:
$\mathcal{R}(\mathbf{x}) = \sqrt{\mathbf{x}^T \mathbf{\Sigma} \mathbf{x}}$
And I cannot use the tool from comments, because when I set $y=x^T$, then the tool is saying that it cannot multiply column vector by a matrix, since it deems $y$ to be column vector by default...
I know for sure that that answer above is incorrect :)
$\Sigma$ is a symmetric matrix.
EDIT EDIT:
Since I know that the answer to $\frac{\partial \mathcal{R} (\mathbf{x})}{\partial \mathbf{x}}$ is $\frac{ \mathbf{\Sigma} \mathbf{x}}{\sqrt{\mathbf{x}^T \mathbf{\Sigma}\mathbf{x}} }$, and since differentiating wrt to transpose converts you from column vector answer to row vector answer, then this must generalize to my case too, hence, I believe the answer is:
$$\frac{\mathbf{x}^T \mathbf{\Sigma}}{\sqrt{\mathbf{x}^T \mathbf{\Sigma} \mathbf{x}}}$$
EDIT EDIT EDIT:
Nope, cannot be correct, because then my dimensions aren't right. The answer must be $n \times 1$, like in the original solution.