Differential form of the velocity equation in non-standard configuration

Question

I'm reading a text on special relativity ($^{\prime\prime}$Core Principles of Special and General Relativity$^{\prime\prime}$, by James H. Luscombe, Edition 2019), in which we start with the equation for composition of velocities in non-standard configuration. Frame $S^{\prime}$ is moving w.r.t. $S$ with constant velocity $\boldsymbol{\upsilon}$ and the velocity of a particle in $S$ is $\boldsymbol{u}$. Then the velocity of the particle in $S^{\prime}$ is \begin{equation} \boldsymbol{u^{\prime}=}\dfrac{\boldsymbol{u-\upsilon}}{1\boldsymbol{-\upsilon\cdot u}/c^2}\boldsymbol{+}\dfrac{\gamma}{c^2\left(1\boldsymbol{+}\gamma\right)}\dfrac{\boldsymbol{\upsilon\times}\left(\boldsymbol{\upsilon\times u}\right)}{\left(1\boldsymbol{-\upsilon\cdot u}/c^2\right)} \tag{3.26}\label{3.26} \end{equation} where \begin{equation} \gamma\boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-\frac12}} \nonumber \end{equation} Then the text states that "differentiating" the above equation \eqref{3.26} gives us \begin{equation} \mathrm{d}\boldsymbol{u^{\prime}=}\dfrac{1}{\gamma\left(1\boldsymbol{-\upsilon\cdot u}/c^2\right)^2}\left[\mathrm{d}\boldsymbol{u-}\dfrac{\gamma}{c^2\left(1\boldsymbol{+}\gamma\right)}\left(\boldsymbol{\upsilon\cdot \mathrm{d}u}\right)\boldsymbol{\upsilon}\boldsymbol{+}\dfrac{1}{c^2}\boldsymbol{\upsilon\times}\left(\boldsymbol{u\times} \mathrm{d}\boldsymbol{u}\right) \right] \tag{3.32}\label{3.32} \end{equation} I'm struggling with proving this. Just to reduce some of the notational headache, if we denote \begin{equation} f\left(\boldsymbol{u}\right)\boldsymbol{=}\dfrac{1}{1\boldsymbol{-\upsilon\cdot u}/c^2} \tag{01}\label{01} \end{equation} then \begin{equation} \mathrm d f\left(\boldsymbol{u}\right)\boldsymbol{=}\dfrac{ f^2\left(\boldsymbol{u}\right)\left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)}{c^2} \tag{02}\label{02} \end{equation} Also let \begin{equation} K\boldsymbol{\equiv}\dfrac{\gamma}{c^2\left(1\boldsymbol{+}\gamma\right)} \tag{03}\label{03} \end{equation} Then the original equation \eqref{3.26} is: \begin{equation} \boldsymbol{u^{\prime}=}f\left(\boldsymbol{u}\right)\left(\boldsymbol{u-\upsilon}\right)\boldsymbol{+}K f\left(\boldsymbol{u}\right)\left[\boldsymbol{\upsilon\times}\left(\boldsymbol{\upsilon\times u}\right)\right] \tag{04}\label{04} \end{equation} Differentiating (writing $\,f\,$ without its argument for convenience), \begin{align} \mathrm{d}\boldsymbol{u^{\prime}}& \boldsymbol{=}\left(\boldsymbol{u-\upsilon}\right)\mathrm{d}f\boldsymbol{+}f\mathrm{d}\boldsymbol{u}\boldsymbol{+}K \mathrm{d}f\left[\boldsymbol{\upsilon\times}\left(\boldsymbol{\upsilon\times u}\right)\right]\boldsymbol{+}K f \left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)\boldsymbol{\upsilon}\boldsymbol{-}K f\upsilon^2 \mathrm{d}\boldsymbol{u} \nonumber\\ &\boldsymbol{=}\dfrac{ f^2\left(\boldsymbol{u-\upsilon}\right)\left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)}{c^2}\boldsymbol{+}f\mathrm{d}\boldsymbol{u}\boldsymbol{+}K \dfrac{ f^2\left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)}{c^2}\left[\boldsymbol{\upsilon\times}\left(\boldsymbol{\upsilon\times u}\right)\right]\boldsymbol{+}K f \left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)\boldsymbol{\upsilon}\boldsymbol{-}K f\upsilon^2 \mathrm{d}\boldsymbol{u} \nonumber\\ &\boldsymbol{=} f^2\Biggl[\dfrac{ \left(\boldsymbol{u-\upsilon}\right)\left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)}{c^2}\boldsymbol{+}\dfrac{\mathrm{d}\boldsymbol{u}}{f}\boldsymbol{+}K \dfrac{\left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)}{c^2}\left[\boldsymbol{\upsilon\times}\left(\boldsymbol{\upsilon\times u}\right)\right]\boldsymbol{+}\dfrac{K}{f} \left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)\boldsymbol{\upsilon}\boldsymbol{-}\dfrac{K}{f}\upsilon^2 \mathrm{d}\boldsymbol{u}\Biggr] \nonumber \end{align} 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.

Answer 1

Hints :

1. In the brackets of the last line of your equation replace all \begin{equation} \dfrac{1}{f} \quad \boldsymbol{\longrightarrow} \quad \left(1\boldsymbol{-}\dfrac{\boldsymbol{\upsilon\cdot u} }{c^2}\right) \tag{a-01}\label{a-01} \end{equation}

2. In the brackets of the last line of your equation expand \begin{equation} \boldsymbol{\upsilon\times}\left(\boldsymbol{\upsilon\times u}\right) \quad \boldsymbol{\longrightarrow} \quad \left[\left(\boldsymbol{\upsilon\cdot u} \right)\boldsymbol{\upsilon}\boldsymbol{-}\upsilon^2\boldsymbol{u}\right] \tag{a-02}\label{a-02} \end{equation}

3. Expand the last item in the rhs of equation \eqref{3.32} \begin{equation} \boldsymbol{\upsilon\times}\left(\boldsymbol{u\times} \mathrm{d}\boldsymbol{u}\right)\boldsymbol{=}\left(\boldsymbol{\upsilon\cdot} \mathrm{d}{\boldsymbol{u}}\right)\boldsymbol{u}\boldsymbol{-}\left(\boldsymbol{\upsilon\cdot u} \right)\mathrm{d}{\boldsymbol{u}} \tag{a-03}\label{a-03} \end{equation}

4. Keep $\,K\,$ as it is until the end and don't replace it by its expression \eqref{03} in order to avoid lengthy equations

5. In the next steps you must realize that \begin{equation} \left(1\boldsymbol{-}K\upsilon^2\right)\boldsymbol{=}\dfrac{1}{\gamma} \quad \text{and} \quad \left(K\boldsymbol{-}\dfrac{1}{c^2}\right)\boldsymbol{=-}\dfrac{1}{c^2\left(1\boldsymbol{+}\gamma\right)}\boldsymbol{=-}\dfrac{K}{\gamma} \tag{a-04}\label{a-04} \end{equation}