Counterexample to the chain-rule I made the following observation
Let $f(t):=\left(\begin{matrix} 0 &e^{it} \\ e^{-it} & 0 \end{matrix}\right),$ then $f(t)^2= \operatorname{id}$.
Thus, we have $\frac{d}{dt}f(t)^2= \frac{d}{dt}\operatorname{id}=0.$
On the other hand yields the chain-rule
$$\frac{d}{dt}f(t)^2= 2 f(t)f'(t)=0.$$
However, $f(t)$ and $f'(t)$ are both matrices with full-rank. So something is very wrong here, no? Can anybody explain to me what just happened?
 A: Trying to use a chain rules leads us into a morass of missing definitions (what does it mean to differentiate a function whose input is a matrix, for example), but we can see something of what happens when the use the product rule:
$$\frac{d}{dt} (f(t)g(t)) = f'(t)g(t) + f(t)g'(t) $$
Here we need to remember that we're talking about matrices, so the order of factors is important. For example, we cannot replace the $f(t)g'(t)$ term with $g'(t)f(t)$ and expect its value to stay the same.
For $f(t)^2$ this gives us
$$ \frac{d}{dt} f(t)^2 = f'(t)f(t) + f(t)f'(t) $$
In contrast to the usual commutative case, the two terms cannot be combined into one. For your particular example we have
$$ f(t) = \begin{pmatrix} 0 & e^{it} \\ e^{-it} & 0 \end{pmatrix} 
\qquad\qquad
f'(t) = \begin{pmatrix} 0 & ie^{it} \\ -ie^{-it} & 0 \end{pmatrix} $$
And we get
$$ f'(t)f(t) = \begin{pmatrix} i&0\\ 0 & -i \end{pmatrix} \qquad\qquad
f(t)f'(t) = \begin{pmatrix} -i&0\\ 0 & i \end{pmatrix} $$
whose sum is clearly the zero matrix.
A: The chain rule obviously applies, and since answers/comments seem to imply that it is ill-defined, I think this answer may be useful.
The problem with your application of the chain rule is that the derivative of $X \mapsto X^2$ at a matrix $X_0$ is not $H \mapsto 2X_0H$, but instead $H \mapsto X_0H+HX_0$.*
This yields the derivative of your $g(t)=f(t)^2$ (identifying the linear map $g':\mathbb{R} \to M_n(2,2)$ with its value at $1$, as is usually done) as
$$f(t)\cdot f'(t)+f'(t) \cdot f(t)$$
by the chain rule, which you can see that makes the computation check out.
*This is a simple computation, given by 
$$(X_0+H)^2=X_0^2+X_0H+HX_0+H^ 2=X_0^2+(X_0H+HX_0)+o(H),$$ and noting that the term in parenthesis is linear on $H$.
