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## Main Question or Discussion Point

I'm reading a text on special relativity (Core Principles of Special and General Relativity), in which we start with the equation for composition of velocities in non-standard configuration. Frame ##S'## velocity w.r.t. ##S## is ##\vec v##, and the velocity of some particle in ##S'## is ##\vec u##. Then the particle's velocity in ##S'## is:

$$\vec u'=\frac{\vec u-\vec v}{1-\vec u.\vec v\ /\ c^2}+\frac{\gamma}{c^2(1+\gamma)}\frac{\vec v\times(\vec v\times\vec u)}{(1-\vec u.\vec v\ /\ c^2)}$$

where ##\gamma=\frac{1}{\sqrt{1-v^2/c^2}}## and ##\vec v## and ##c## are constants. Then the text states that "differentiating" the above equation gives us

$$d\vec u'=\frac{1}{\gamma(1-\vec u.\vec v\ /\ c^2)^2}\bigg[d\vec u-\frac{\gamma}{c^2(1+\gamma)}(\vec v.d\vec u)\ \vec v+\frac{1}{c^2}\vec v\times\vec u\times d\vec u\bigg]$$

I'm struggling with proving this. Just to reduce some of the notational headache, if we denote ##f(\vec u)=\frac{1}{1-\vec u.\vec v\ /\ c^2}##, then

$$df(\vec u)=\frac{f(\vec u)^2\vec v.d\vec u}{c^2}$$

Also let ##K\equiv \frac{\gamma}{c^2(1+\gamma)}##. Then the original equation is:

$$\vec u'=f(\vec u)(\vec u-\vec v)+Kf(\vec u)\ (\vec v\times (\vec v\times\vec u))$$

Differentiating (writing ##f## without its argument for convenience),

$$d\vec u'=(\vec u-\vec v)\ df+fd\vec u + K\ df\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$

$$=\frac{f^2(\vec u-\vec v)\vec v.d\vec u}{c^2}+fd\vec u + K\ \frac{f^2\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$

$$=f^2\bigg[\frac{(\vec u-\vec v)\vec v.d\vec u}{c^2}+\frac{d\vec u}{f} + K\ \frac{\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + \frac{K}{f}\vec v\ (\vec v.d\vec u)-\frac{K}{f}v^2d\vec u\bigg]$$

Beyond this, I'm really not able to get to the final result despite trying a bunch of times. Not sure if I'm overcomplicating things or missing some magical identity that simplifies everything. Would appreciate any help. Thanks!

$$\vec u'=\frac{\vec u-\vec v}{1-\vec u.\vec v\ /\ c^2}+\frac{\gamma}{c^2(1+\gamma)}\frac{\vec v\times(\vec v\times\vec u)}{(1-\vec u.\vec v\ /\ c^2)}$$

where ##\gamma=\frac{1}{\sqrt{1-v^2/c^2}}## and ##\vec v## and ##c## are constants. Then the text states that "differentiating" the above equation gives us

$$d\vec u'=\frac{1}{\gamma(1-\vec u.\vec v\ /\ c^2)^2}\bigg[d\vec u-\frac{\gamma}{c^2(1+\gamma)}(\vec v.d\vec u)\ \vec v+\frac{1}{c^2}\vec v\times\vec u\times d\vec u\bigg]$$

I'm struggling with proving this. Just to reduce some of the notational headache, if we denote ##f(\vec u)=\frac{1}{1-\vec u.\vec v\ /\ c^2}##, then

$$df(\vec u)=\frac{f(\vec u)^2\vec v.d\vec u}{c^2}$$

Also let ##K\equiv \frac{\gamma}{c^2(1+\gamma)}##. Then the original equation is:

$$\vec u'=f(\vec u)(\vec u-\vec v)+Kf(\vec u)\ (\vec v\times (\vec v\times\vec u))$$

Differentiating (writing ##f## without its argument for convenience),

$$d\vec u'=(\vec u-\vec v)\ df+fd\vec u + K\ df\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$

$$=\frac{f^2(\vec u-\vec v)\vec v.d\vec u}{c^2}+fd\vec u + K\ \frac{f^2\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + Kf\vec v\ (\vec v.d\vec u)-Kfv^2d\vec u$$

$$=f^2\bigg[\frac{(\vec u-\vec v)\vec v.d\vec u}{c^2}+\frac{d\vec u}{f} + K\ \frac{\vec v.d\vec u}{c^2}\ (\vec v\times (\vec v\times\vec u)) + \frac{K}{f}\vec v\ (\vec v.d\vec u)-\frac{K}{f}v^2d\vec u\bigg]$$

Beyond this, I'm really not able to get to the final result despite trying a bunch of times. Not sure if I'm overcomplicating things or missing some magical identity that simplifies everything. Would appreciate any help. Thanks!