In a very natural sense, you can! If $\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = L$ is some real number, then it makes sense to define $f(\infty) = L$, where we identify $\infty$ and $-\infty$ in something called the one-point compactification of the real numbers (making it look like a circle).
In that case, $f'(\infty)$ can be defined as
$$f'(\infty) = \lim_{x \to \infty} x \big(f(x) - f(\infty)\big).$$
When you learn something about analytic functions and Taylor series, it will be helpful to notice that this is the same as differentiating $f(1/x)$ at zero.
Notice that this is actually not the same as $\lim_{x \to \infty} f'(x)$.
These ideas actually show up quite a bit in analytic capacity, so this is a rather nice idea to have.
I wanted to expand this answer a bit to give some explanation about why this is the "correct" generalization of differentiation at infinity. and hopefully address some points raised in the comments.
Although $\lim_{x \to \infty} f'(x)$ might feel like the natural object to study, it is quite badly behaved. There are functions which decay very quickly to zero and have horizontal asymptotes, but where $f'$ is unbounded as we tend to infinity; consider something like $\sin(x^a) / x^b$ for various $a, b$. Furthermore, $\lim_{x \to \infty} f'(x) = 0$ is not sufficient to guarantee a horizontal asymptote, as $\sqrt{x}$ shows.
So why should we consider the definition I proposed above? Consider the natural change of variables interchanging zero and infinity*, swapping $x$ and $1/x$. Then if $g(x) := f(1/x)$ we have the relationship
$$\lim_{x \to 0} \frac{g(x) - g(0)}{x} = \lim_{x \to \infty} x \big(f(x) - f(\infty)\big).$$
That is to say, $g'(0) = f'(\infty)$. Now via this change of variables, neighborhoods of zero for $g$ correspond to neighborhoods of $\infty$ for $f$. So if we think of the derivative as a measure of local variation, we now have something that actually plays the correct role.
Finally, we can see from this that this definition of $f'(\infty)$ gives the coefficient $a_1$ in the Laurent series $\sum_{i \ge 0} a_i x^{-i}$ of $f$. Again, this corresponds to our idea of what the derivative really is.
* This is one of the reasons why I used the one-point compactification above. Otherwise, everything that follows must be a one-sided limit or a one-sided derivative.
You are not missing anything.
$$2x(t)\cdot x'(t)=2s(t)\cdot s'(t)$$
$$ \iff x(t)\cdot x'(t)=s(t)\cdot s'(t)$$
$$\iff x\frac{dx}{dt} = s\frac{ds}{dt}$$
They divided by $2$ and used the $ \frac{dx}{dt}$ for $x'(t)$.
Best Answer
Yes!
If $f(\mathbf{x}) = \mathbf{x}\bullet \mathbf{y}$, then $f^\prime(\mathbf{x}) = \mathbf{y}$.
To see this, first use the definition of derivative: $$f^\prime(\mathbf{x}) = \left[\frac{\partial}{\partial x_k} f(\mathbf{x})\right]_k$$
(The total derivative can be defined as a vector of partial derivatives.)
Then look at a specific component $k$ of the derivative and expand the definition of dot product: $$\frac{\partial}{\partial x_k} f(\mathbf{x}) = \frac{\partial}{\partial x_k} \mathbf{x}\bullet \mathbf{y} = \frac{\partial}{\partial x_k} \sum_i x_iy_i $$
Now, only one of those terms in the sum depends on $x_k$. It is the term $x_ky_k$ (which occurs when $i=k$), and its derivative with respect to $x_k$ is just $y_k$.
Hence the total derivative $f^\prime(\mathbf{x})$ is the vector $[y_k]_k$, or in other words $[y_1, y_2, \ldots] = \mathbf{y}$, as required.
And in general, the rules of calculus (such as the product rule) tend to generalize to matrix equations.