Given $A$ and $X$ matrices, is this true: $dAX/dX = dXA/dX = A$? I know that $dAX/dX = A$, but what about $dXA/dX$?
 A: To calculate $d/dX(AX)$ or $d/dX( XA)$ I will assume you mean for us to calculate the usual Frechet derivative. In particular, let $F(X)=AX$ and consider:
$$ F(X+H) = A(X+H) = AX+AH $$
thus $F(X+H)-F(X) = AH$. It follows that 
$$\frac{F(X+H)-F(X)}{||H||} = \frac{AH}{||H||} \rightarrow 0 $$
as $H \rightarrow 0$ and thus $F'(X) = A$ as $H \mapsto AH$ defines the differential of the map $F$ at $X$. Notice, I have defined $F'(X)$ to be the matrix which appears in the formula for the differential $d_XF(H) = F'(X)H$. I can't have it both ways though...
For $G(X) = XA$ by nearly the same argument we'll find $d_XG(H) = HA \neq AH$ generally. So, modulo some additional conventions, we will not get a nice formula for both. If you allowed left and right derivatives then you could have your cake and eat it to, or have no cake on either side. But, the pesky reality is that the non-commutativity of matrix multiplication spoils it.
A: *

*derivative of $AX$ is the linear map $H\mapsto AH$

*derivative of $XA$ is the linear map $H\mapsto HA$

