For a non-square matrix $X$ of size $n \times p$ ($n>p$) and another non-square matrix $A$ of size $p \times n$, what is the derivative of $\det(X^TA)$ w.r.t. $X$? i.e.,
$\frac{\partial\det(X^TA)}{\partial X}$
It'd be a matrix of size $n \times p$ while I got stuck in some intermediate steps of using chain rules.
Update: I think I got the solution. Thanks to AlexR and Bob.
Reference: The Matrix Cookbook; Section 2.1.2; Equation (45)
$\frac{\partial\det(AXB)}{\partial X} = \det(AXB){(X^{-1})}^T = \det(AXB){(X^T)}^{-1}$
This is, actually, a special case when A and B are non-singular matrices.
The more general form, according to matrix reference manual, is
(URL: http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/calculus.html#deriv_det)
$\frac{\partial\det(A^TXB)}{\partial X} = \frac{\partial\det(B^TX^TA)}{\partial X} = A\operatorname{adj}(A^TXB)^TB^T = A(\det(A^TXB)(A^TXB)^{-1})^TB^T = \det(A^TXB)A(B^TX^TA)^{-1}B^T$
When $B = B^T= I$
$\frac{\partial\det(X^TA)}{\partial X} = A\operatorname{adj}(A^TX)^T = A(\det(A^TX)(A^TX)^{-1})^T = \det(X^TA)A(X^TA)^{-1}$
We know that $\frac{\partial\det(XA)}{\partial X}=\det(XA)(X^T)^{-1} $ (see The Matrix Cookbook, for example). Thus,
$\frac{\partial\det(X^TA)}{\partial X}=\det(XA)(X)^{-1}$.