Taking the derivative of $A(t)\,x(t).$ Can someone tell me how to interpret $\dfrac{d}{dt}(A(t)\,x(t))$ for some matrix $A$ and a vector $x$? I can't find this sort of derivative anywhere.
 A: $A(t)x(t)$ is a curve $\Bbb R\to \Bbb R^n$, and its derivative is just the tangent vector $A'(t)x(t)+A(t)x'(t)$.
A: Leibniz rule applies here: as already stated by Saucy O'Path, the derivative is
$$\frac{d}{dt}A(t)x(t) = A'(t)x(t) + A(t)x'(t)$$
The exact same formula applies for any bilinear operation. Suppose that $X$, $Y$ and $Z$ are finite-dimensional (for simplicity) vector spaces over $\mathbb R$, and let $\varphi : X \times Y \rightarrow Z$ be any bilinear operator. Then, if $x(t)$ and $y(t)$ are differentiable curves in $X$ and $Y$, respectively, we have
$$\varphi(x(t + \delta),y(t + \delta))\approx\varphi(x(t)+x'(t)\cdot \delta,y(t)+y'(t)\cdot \delta) = \\
= \varphi(x(t),y(t)) + \varphi(x'(t),y(t)) \cdot \delta + \varphi(x(t),y'(t))\cdot \delta + \varphi(x'(t),y'(t))\cdot \delta^2 \approx \\
\approx\varphi(x(t),y(t)) + \left[\varphi(x'(t),y(t)) + \varphi(x(t),y'(t))\right]\cdot \delta$$
(here, we ignore everything of order $\delta^2$ or higher, and make heavy use of the bilinearity)
Substracting and taking the limit, we obtain
$$\frac{d}{dt}\varphi(x(t),y(t))=\varphi(x'(t),y(t)) + \varphi(x(t),y'(t))$$
In your particular case, $X=M_n(\mathbb R)$ is the space of $n\times n$-matrices, and $Y=Z=\mathbb R^n$ is the space of $n$-dimensional column vectors, while $\varphi$ is the operation of multiplication of a vector by a matrix, which is of course bilinear.
