I have been attempting to prove the following identity:
$\frac{\partial [W^{\frac{1}{2}} K W^{\frac{1}{2}}]}{\partial \hat{f_i}} = K \frac{\partial W}{\partial \hat{f_i}}$
where $W^{\frac{1}{2}}$ is a diagonal matrix with all-positive elements (the matrix square-root of $W$) which is a function of $\hat{f_i}$, and $K$ is a symmetric positive-definite matrix that is not a function of $\hat{f_i}$.
If it does hold, I would appreciate a proof or a link to a reference. Thanks!
It does not hold. Here is a counter-example: $W = \begin{bmatrix} f_1 & 0 \\ 0 & f_2\end{bmatrix}, K = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$. Then, $$W^{1/2} K W^{1/2} = \begin{bmatrix} f_1 & -\sqrt{f_1f_2} \\ -\sqrt{f_1f_2} & f_2\end{bmatrix}$$ and hence $$\frac{\partial}{\partial f_1} W^{1/2} K W^{1/2} = \begin{bmatrix} 1 & -\frac 1 2 \sqrt{\frac{f_2}{f_1}} \\ -\frac 1 2 \sqrt{\frac{f_2}{f_1}} & 0\end{bmatrix}.$$ On the other hand, $$K \frac{\partial W}{\partial f_1} = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}.$$
It does work if $K$ is diagonal (or $W$ is a scaled identity) since then $W^{1/2} K W^{1/2} = K W$.
Let $$\eqalign{ W &= A^2 \cr }$$ where both $(W,A)$ are diagonal matrices.
Taking derivatives wrt $f_i$ $$\eqalign{ d_iW &= 2A\,\,d_iA \implies d_iA = \tfrac{1}{2}A^{-1}d_iW \cr }$$ Write the function in terms of $A$ and calculate its derivative $$\eqalign{ Y &= AKA \cr d_iY &= AK\,d_iA + d_iA\,KA \cr &= \tfrac{1}{2}AKA^{-1}\,d_iW + \tfrac{1}{2}d_iW\,A^{-1}KA \cr }$$ If we define the matrix $$\eqalign{ M &= AKA^{-1}\,d_iW \cr }$$ then the symmetry of $K$ and commutivity of the diagonal matrices yields $$\eqalign{ d_iY &= \tfrac{1}{2}(M+M^T)\cr }$$
