Derivative of a trace w.r.t to matrix's elements I am checking out the Matrix Cookbook, and I am puzzled by the following:

Assume $F(X)$ to be a differentiable function of each of the elements of
  $X$. It then holds that $$\dfrac{\partial\operatorname{Tr}(F(X))}{\partial X} = f(X)^T,$$ where $f(\cdot)$ is the scalar derivative of $F(\cdot)$.

However, I fail to grasp how does it make any sense. For instance, let's assume that we have the following matrix function:
$$F(x) = \begin{pmatrix} 0 & x \\   0 & 0 \\ \end{pmatrix}$$
Its trace is zero $\forall\,x \in\mathbb{R}$. Furthermore, as $\operatorname{Tr}(F(x))\colon \mathbb{R}^2\times\mathbb{R}^2 \to \mathbb{R}$, I would expect the derivative of $\operatorname{Tr}(F(x))$ to be the same. But according to the equation given above, it is $\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$ .
 A: Part of problem is the confusion of the matrix $X$ with the scalar parameter $x$. If we use $\alpha$ for the scalar parameter, then 
$$X = \left[\begin{array}{ccc}
0 & \alpha \\
0 & 0 \\
\end{array}\right] = N\alpha$$
where $N$ is a $2\times 2\,$ nilpotent matrix.
As concrete examples of the functions, let's use 
$$\eqalign{
F(X) &= X^2 \cr
f(X) &= 2X \cr
}$$
Then what the Cookbook is saying is 
$$\eqalign{
  \frac{\partial\,{\rm tr}(X^2)}{\partial X} &= 2X^T \cr
}$$
Which is different than saying that 
$$\eqalign{
  \frac{\partial\,{\rm tr}(X^2)}{\partial\alpha} &= 2X^T \cr
}$$
since the LHS is a scalar quantity while the RHS is a matrix.
To find an expression for the gradient with respect to $\alpha$, we can use the Cookbook result to write down the differential, and then perform a change of variables 
$$\eqalign{
d\,{\rm tr}(X^2) &= 2X^T:dX \cr &= 2\alpha N^T:N\,d\alpha \cr\cr
\frac{\partial {\rm tr}(X^2)}{\partial\alpha}
 &= 2\alpha N^T:N \cr
 &= 2\alpha\,{\rm tr}(N^2) \cr
 &= 0 \cr
}$$ since $N$ is nilpotent.
Note that colons were used in some of the steps above to denote the trace/Frobenius product, i.e. 
$$A:B = {\rm tr}(A^TB)$$
