Differential pair output voltage formula derivation
HW_SW_engineer - I think, the simplest way is to start with the currents:

In general, we have: Ic=Ico[(exp(Vbe/Vt)-1]

Ic1/Ic2=exp[Vbe1-Vbe2)/Vt]=exp[Vd/Vt]

with Iee=Ic1+Ic2 we arrive at

Ic1=Iee/[1+exp(-Vd/Vt)] and Ic2=Iee/[1+exp(+Vd/Vt)]

From mathematics: 2/[1+exp(-x)]=....[1+th(x/2)]

Ic1=0.5*Iee[1+th(Vd/2Vt)]  and  Ic2=0.5*Iee[1-th(Vd/2Vt)]

Ic1-Ic2=Iee*th(Vd/2Vt)

Vo1=Vdc - RcIc1  and Vo2=Vdc - RcIc2 (Vdc: DC quiescent voltage)

Vd=Vo1-Vo2=-IeeRc*th(Vd/2Vt)

Comment: I have assumed Iee=Ic1+Ic2 (for each transistor: coll=emitter current); hence the factor "alpha" does not appear in the equations)