Q:

the derivative for F-norm of a matrix

From the matrix cookbook, I got:
$\frac{\partial}{\partial X} ||X||_F^2 = \frac{\partial}{\partial X} Tr(XX^H) = 2X$ 
Now I want to compute $\frac{\partial}{\partial X} ||X-Y||_F^2 = \frac{\partial}{\partial X} Tr\left((X-Y)(X-Y)^H\right)$, but I am stuck here, any idea? 

A:

For convenience, define a new variable $$\eqalign{
 W &= X-Y \cr
}$$
Write the function in terms of this new variable and the Frobenius (:) Inner Product and find its differential
$$\eqalign{
 f &= W:W \cr
df &= 2W:dW \cr
   &= 2W:dX\cr
}$$Since $df=\big(\frac{\partial f}{\partial X}:dX\big),\,$ we can identify
$$\eqalign{
 \frac{\partial f}{\partial X} &= 2\,W \cr
  &= 2\,(X-Y) \cr
}$$
Note that setting $Y=0,\,$ recovers the Cookbook result.

