Product Rule for Time Varying Matrices $M(t) B(t)$ How would one evaluate $\frac{d}{dt} \left( M(t) B(t) \right)$ where $M$ and $B$ are square $n \times n$ matrices defined as follows.
$$M = \begin{bmatrix} m_{11} (t) & \dots & m_{1n}(t) \\ m_{21}(t) & \dots & m_{2n}(t) \\ \vdots & \ddots & \vdots \\ m_{n1} (t)& \dots & m_{nn} (t) \end{bmatrix}$$
Each $m_{ij}(t): \mathbb{R} \to \mathbb{C}$, and $B$ is defined in an equivalent fashion. As multiplication is not commutative, I do not believe a formula $M'(t) B(t) + M(t) B'(t)$ would suffice as the choice of direction in multiplication would completely change the result.
Would the proper method of differentiation rely on a commutator or Lie bracket?
 A: If you're thinking of differentiating entrywise, we can think of it as follows:
Given $n\times n$ matrices $M(t),B(t)$ as you've defined them, then $M(t)B(t)$ is defined to be the matrix whose $ij$ entry is given by $$\sum_{k=1}^{n}m_{ik}(t)b_{kj}(t).$$  So, the matrix $\frac{d}{dt}(M(t)B(t))$ would then be the matrix whose $ij$ entry is given by $$\sum_{k=1}^{n}\frac{d}{dt}(m_{ik}(t)b_{kj}(t)) = \sum_{k=1}^{n}\bigl(m_{ik}'(t)b_{kj}(t)+m_{ik}(t)b'_{kj}(t)\bigr).$$
Note that we can split this up into $$\sum_{k=1}^{n}m_{ik}'(t)b_{kj}(t) + \sum_{k=1}^{n}m_{ik}(t)b'_{kj}(t).$$  We can recognize the first term as being the $ij$ entry in the matrix $M'(t)B(t)$, where the derivative is taken entrywise, and likewise the second term is $M(t)B'(t)$.  So in this case, yes, we do recover the product rule $$\frac{d}{dt}(M(t)B(t)) = M'(t)B(t) + M(t)B'(t).$$
A: Does not have to be commutative, has to be bilinear (or close to it). The usual proof of product rule goes through if you are careful about multiplication order:
$$(MB)'(t) = \lim_{h\to 0} \frac{M(t+h)B(t+h)-M(t)B(t)}{h}.$$
Now:
$$M(t+h)B(t+h)-M(t)B(t)=M(t+h)[B(t+h)-B(t)]+[M(t+h)-M(t)]B(t)$$
Divide by $h$ and take limit $h\to 0$ in the above, and (using $\lim M(t+h)=M(t)$) voila:
$$(MB)'(t)=M(t)B'(t)+M'(t)B(t).$$
Discussion:
Same proof for dot product, cross product, matrix product, you name it (where by "you name it" I mean any binary operation that is continuous and distributive over addition in both arguments).  The reason that commutativity is irrelevant is that you never commute things; the influence of the time change on the product is due to it influencing each of the multiplicands (in differentiation everything is linearized, so interactions of these influences -- which are of higher order -- do not appear).
If that seems too tautological, think this way: derivative is a response to a small perturbation - to first order. When you perturb a product, the changes are due to perturbing each of the multiplicands. Formally, this is just chain rule: if $f(M, B)=MB$ then $\frac{d}{dt} f(M, B)=\frac{Df}{DM}(\frac{d}{dt}M)+\frac{Df}{DB}(\frac{d}{dt}B)$. Now, what are these derivatives that appear? $\frac{Df}{DM}$ says how $f(M, B)$ changes (to first order) when we add $\delta M$ to $M$. Of course, it changes by $(\delta M) B$. So $\frac{Df}{DM}$ is the operator that multiplies by $B$ on the right. Similarly, $\frac{Df}{DB}$ is the operator that multiplies by $M$ on the left. Overall, $\frac{d}{dt} f(M, B)=(\frac{d}{dt}M)B+M(\frac{d}{dt}B)$, as before.
