Derivative of a matrix function with respect to a matrix I am trying to differentiate the following function, with respect to a matrix $X$:
$$
\operatorname{tr}(AX(X^TX)^{-1}B)
$$
where tr corresponds to the trace. Is there an easy way to see what the derivative will be? I've come across rules for $\operatorname{tr}(AXB)$ but not the form above with inverses etc . 
$A$ and $B$ are known (constant) matrices. 

Define $W \equiv (X'X)^{-1}$, then $f = BAX : W$ and its derivative is
$$
\frac {\partial f} {\partial X} = A'B'W -  XW(BAX+X'A'B')W
$$
The algebra to arive at this result is tedious but straight-forward. The only tricky part is knowing that
$$ \eqalign {
  dW &= - W\,\,dY\,\,W \cr
} $$
where $Y \!\equiv W^{-1}\!= X'X$
It's worth noting that both $Y$ and $W$ are symmetric. 
Then you just expand the differential of $f$
$$ \eqalign{
 df &= BA\,dX : W    +  BAX : dW  \cr
    &= dX : (BA)'W -  BAX : W\big[dX'X + X'dX\big]W \cr
    &= dX : A'B'W  -  WBAXW : \big[dX'X + X'dX\big] \cr
    &= A'B'W : dX  -  WBAXW : \big[dX'X + X'dX\big] \cr
    &= A'B'W : dX  -  WBAXW : dX'X -  WBAXW : X'dX \cr
    &= A'B'W : dX  -  WBAXWX' : dX' -  XWBAXW : dX \cr
    &= A'B'W : dX  -  XWX'A'B'W : dX - XWBAXW : dX \cr
    &=[A'B'W       -  XWX'A'B'W      - XWBAXW]: dX \cr
} $$
So the expression in brackets must be the derivative.
If you dislike the Frobenius product, you can carry out the above steps using the trace 
$$
\text{tr}(A'B) = A:B
$$
