I have find some reason why $\frac{\partial y_{mn}}{\partial \mathbf X}= \mathbf B \mathbf I_{mn}^T\mathbf A$ when $\mathbf Y=\mathbf A \mathbf X^T \mathbf B$ in some books,but i can't still understand why,can anyone teach me how to calculate it in detail?
$\mathbf A$ is $m*k$ matrix,$\mathbf B$ is $i*n$ matrix,$\mathbf X$ is $i*k$ matrix
Starting with $$\eqalign{ Y &= AX^TB \cr }$$ Take the Frobenius product with the single-entry matrix to isolate the $(m,n)$ element. $$\eqalign{ y_{mn} &= E_{mn}:Y \cr &= E_{mn}:AX^TB \cr &= E_{mn}^T:B^TXA^T \cr &= BE_{mn}^TA:X \cr }$$ Now find the differential and gradient of this element wrt $X$ $$\eqalign{ dy_{mn} &= BE_{mn}^TA:dX \cr \frac{\partial y_{mn}}{\partial X} &= BE^T_{mn}A \cr }$$ There is no standard notation for the single-entry matrix. I use the letter ${\mathbf E}$ others use ${\mathbf J}$, but ${\mathbf I}$ is a confusing choice since it usually denotes the identity matrix.
NB: The Frobenius product is just a convenient way to write the trace function $$A:B = {\rm Tr}(A^TB)$$