Derivative of the following matrix with respect to a vector $$J=X^T(Q\otimes I) X+U^T(Z\otimes I)U$$
$\frac{\partial J}{\partial U}=? $
where $\otimes$ is the Kronecker product.
$X$ is a vector which depends on vector $U$.
$Q, Z$ is a positive semi-definite matrix,
$I$ is identity matrix.
superscript $T$ is the transpose of the matrices.
My try:
$\frac{\partial U^T(Z\otimes I)U}{\partial U}=(Z\otimes I)U+(Z\otimes I)^TU $
it is right?
$\frac{\partial X^T(Q\otimes I)X}{\partial U}=?$
 A: You've correctly calculated
$$\frac{\partial(U^T(Z\otimes I)U)}{\partial U}=(Z\otimes I)U+(Z\otimes I)^TU\,\,=A$$
Simply replacing $(U,Z)$ by $(X,Q)$ yields
$$\frac{\partial(X^T(Q\otimes I)X)}{\partial X}=(Q\otimes I)X+(Q\otimes I)^TX\,\,=B$$
And although you haven't told us, I assume that you know the relationship between $X$ and $U$ and are able to calculate the matrix $C=\frac{\partial X}{\partial U}$
All that's left is to combine these 3 quantities to obtain the desired gradient. 
$$\eqalign{
dJ
 &= A:dU + B:dX \cr
 &= A:dU + B:(C\,dU) \cr
 &= (A + C^TB):dU \cr
\frac{\partial J}{\partial U} &= A + C^TB \cr
}$$ For ease of typing, I used a colon to denote the inner/scalar product.
A: Derivatives should be understood under the context of linear approximation. Then
\begin{align*}
dJ
&= (dX)^T (Q \otimes I) X + X^T (Q \otimes I) dX \\
&\quad + (dU)^T ( Z \otimes I ) U + U^T ( Z \otimes I) dU
\end{align*}
As for $dX$, we may invoke Jacobian $\operatorname{Jac}(X) = \left( \frac{\partial X^i}{\partial U^j} \right)$ to write $dX = \operatorname{Jac}(X) dU$. Then
$$ dJ = (dU)^T \left[ \operatorname{Jac}(X)^T (Q \otimes I) X + (Z \otimes I) U \right] + \left[ X^T (Q \otimes I) \operatorname{Jac}(X) + U^T (Z \otimes I)\right] dU. $$
If needed, one may read out the partial derivatives $\frac{\partial J}{\partial U^j}$ from the identity above.
