Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis. Suppose a particle P is place in frame S’ and it is moving.

The velocity component of particle P from observer O’ in frame S’ will be:

u’x = dx’/dt’                 (1a)

u’y = dy’/dt’                 (1b)

u’z = dz’/dt’                  (1c)

The velocity component of particle P from observer O in frame S will be:

ux = dx/dt                     (2a)

uy = dy/dt                     (2b)

uz = dz/dt                      (2c)

From Lorentz transformation equations:

x’ = (x – vt)/(√1 – v2/c2)                       (3a)

y’ = y                                       (3b)

z’ = z                                        (3c)

t’ = (t – xv/c2)/(√1 – v2/c2)        (3d)

Differentiate equations (3)

dx’ = (dx – vdt)/(√1 – v2/c2)     (4a)

dy’ =d y                                               (4b)

dz’ = dz                                                (4c)

dt’ = (dt – dxv/c2)/(√1 – v2/c2)  (4d)

Now substitute equations (4a) and (4d) in equation (1a), we get

u’x = (dx – vdt)/ (dt – dxv/c2)

Divide numerator and denominator of R.H.S with dt, we get

u’x = (dx/dt – v)/ (1 – dx/dt(v/c2))

Put equation (2a) in above equation,

u’x = (ux – v)/ (1 – ux(v/c2))       (5)

Similarly by putting equations (4b) and 4d) in equation (1b) and then dividing the numerator and denominator of R.H.S with dt, and then putting equation (2b), we get

u’y = uy(√1 – v2/c2)/ (1 – ux(v/c2))                     (6)

Similarly by putting equations (4c) and 4d) in equation (1c) and then dividing the numerator and denominator of R.H.S with dt, and then putting equation (2c), we get

u’z = uz(√1 – v2/c2)/ (1 – ux(v/c2))                      (7)

Equations (5, 6 and 7) represent the addition of velocity relations as observed by observers O’ from frame S’.

From the observer O in frame S, the relations (5, 6 and 7) will become:

ux = (u’x + v)/ (1 + u’x(v/c2))                             (8)

uy = u’y(√1 – v2/c2)/ (1 + u’x(v/c2))                    (9)

uz = u’z(√1 – v2/c2)/ (1 + u’x(v/c2))                    (10)

Generally equation (8) is written as

u = (u’ + v)/ (1 + u’(v/c2))                                 (11)

Special case: If v <<< c, then v/c2 will get neglected and the equation 11 will become

u = u’ + v