Consider the ordinary velocity of a moving object, defined by \begin{equation} u=\frac{d}{dt}\,x \end{equation} This transforms in a complicated way, since \begin{equation} \cm \frac{dx'}{dt'} = \frac{\cm\frac{dx}{dt}-\vc}{1-\vcsq\frac{dx}{dt}} \label{dadd} \end{equation} The reason for this is that both the numerator and the denominator need to be transformed. Note that ($\ref{dadd}$) is just the Einstein addition formula for velocities, which we have therefore independently derived using Lorentz transformations! But that's not quite what we are looking for here.

The invariance of proper time suggests that we should instead differentiate with respect to proper time, since of course \begin{equation} \frac{d}{d\tau}\,x'=\frac{dx'}{d\tau} \end{equation} In other words, the operator $\frac{d}{d\tau}$ pulls through the Lorentz transformation; only the numerator is transformed when changing reference frames.

Furthermore, the same argument can be applied to $t$, which suggests that there are (in two dimensions) two components to the velocity. We therefore consider the “2-velocity” \begin{equation} \bu = \frac{d}{d\tau}\pmatrix{\cc t\cr x\cr} = \pmatrix{\cc\frac{dt}{d\tau}\cr \frac{dx}{d\tau}} \end{equation}

But since \begin{equation} dt=\cosh\alpha\,d\tau \end{equation} and \begin{equation} dx^2 - \csq dt^2 = - \csq d\tau^2 \end{equation} we also have \begin{equation} dx = \cc\,\sinh\alpha\,d\tau \end{equation} so that \begin{equation} \bu = \cc \pmatrix{\cosh\alpha\cr \sinh\alpha\cr} \end{equation}

Note that $\cm\,\bu$ is a unit vector, that is \begin{equation} \cmsq \> \bu\cdot\bu = -1 \end{equation} and further that \begin{equation} \uc=\frac{dx}{\cc\,dt}=\tanh\alpha \end{equation} as expected.