So say you have a matrix dependent on a variable t:
$A(t)$
How do you compute $e^{A(t)}$ ?
It seems Sylvester's formula, my standard method of computing matrix exponentials can't be applied here given the varying nature of the matrix and furthermore the fact that it may not always have distinct eigenvalues.
One example of where such ordered exponentials arise is in computing the solution to ODEs with time-dependent coefficients: $$ \frac{d\mathbf{x}(t)}{dt} = A(t) \mathbf{x}(t), $$ where $\mathbf{x}(t)=(x_1(t), \ldots, x_n(t))$ and $A(t)$ is an $n\times n $ matrix, the general solution is $$ \begin{equation} \mathbf{x}(t) = \mathcal{T}\{ e^{\int_0^{t} A(t') dt'}\} \mathbf{x}(0), \end{equation} $$ where $\mathcal{T}$ denotes time-ordering,
$\mathcal{T}\{ e^{\int_0^{t} A(t') dt'}\} \equiv \sum\limits_{n=0}^{\infty} \frac{1}{n!}\int_0^t \ldots \int_0^t \mathcal{T}\{A(t_1') \ldots A(t_{n}') \} = \sum\limits_{n=0}^{\infty} \int_0^t \ldots \int_0^{t_{n-1}'} A(t_1') \ldots A(t_{n}')$.
Only when the matrices at different times commute - i.e. $[A(t_1), A(t_2)]=0$, $\forall t_1,t_2$ - does the time-ordered expression simplify to $e^{\int_0^{t} A(t') dt'}$.
Evaluating such exponentials is an arduous task, since even for time-independent matrix exponentials there are articles such as the one mentioned in René's answer. A quick search in the literature revealed a few approaches. Perturbatively, the solution is given by a Magnus series, and there are various papers describing methods to approximate or exactly solve them:
The methods typically involve heavy maths and it would not be appropriate to describe them here with my limited understanding.