Differential form of the velocity equation in non-standard configuration 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.
 A: Hints :

*

*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}


*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}


*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}


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


*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}
