关于迹的矩阵函数求导

秩的交换性，转置不影响
egin{align*}
frac{partialmathrm{tr} AX}{partial X}&=A^T,&frac{partialmathrm{tr} left(AXB ight)}{partial X}&=A^TB^T,\
frac{partialmathrm{tr} left(X^TAX ight)}{partial X}&=left(A^T+A ight)X,&frac{partialmathrm{tr} left(X^TAXB ight)}{partial X}&=AXB+A^TXB^T,
end{align*}