I need to find the gradient of
$$ X \mapsto \mbox{Tr} \left(\left(AXA\right)^{\frac{1}{2}}\right) $$
where both $X$ and $A$ are symmetric matrices.
I know that the derivative of $\mbox{Tr} \left(AXA\right)$ is $A^{2}$ using equation (101) on page 12 of the Matrix Cookbook. However, I have the trace of the square root of a matrix product, which makes it complicated. Any help is appreciated.
Define two new symmetric matrices $$\eqalign{ M &= AXA \cr Q &= M^{-\frac{1}{2}} \cr }$$ Write the function in terms this new variable, then find its differential and gradient. $$\eqalign{ \phi &= {\rm tr}\Big(M^{\frac{1}{2}}\Big) \cr d\phi &= \tfrac{1}{2}Q:dM = \tfrac{1}{2}Q:A\,dX\,A = \tfrac{1}{2}AQA:dX \cr \frac{\partial\phi}{\partial X} &= \tfrac{1}{2}AQA \cr\cr }$$ In some steps above, a colon was used to denote the trace/Frobenius product, i.e. $$\eqalign{B:C={\rm tr}(B^TC)}$$