The speed composition vs the light aberration in SR

The Lorentz's transform:

##x' = k(x - vt), t' = k(t - vx)\ k = \gamma,\ and\ c = 1##



I. The speed composition derivation:



##w' = dx'/dt' = \frac{dx - vdt}{dt - vdx}##

and we divide everything by dt, and:

##w' = \frac{dx/dt - v}{1 - vdx/dt}##



now we assume the dx/dt is some speed u, and the wanted formula is ready:

##w' = \frac{u - v}{1 - vu}##



II. The second part - the relativistic light aberration



##\cos f = \frac{c_x}{c} = \frac{dx}{cdt} = \frac{dx}{dt}## (c = 1)

thus the aberration is:



cosf' = dx'/dt' = ... identical!



we go, and on the stage: cosf' = (dx/dt - v)/(1 - vdx/dt)

and now we don't assume dx/dt is a speed, but it's now just cosf, therefore:



##\cos f' = \frac{\cos f - v}{1 - v\cos f}##

it's a correct result for the relativistic aberration.



Very well, but I have one question: what is it in the SR the quantity dx/dt - a speed or a cosine (of a light ray)?