Derivative as a fraction in deriving the Lorentz transformation for velocity Consider a frame $S$ and $S'$ which is coincides at $t=0$ and then $S'$ starts moving with velocity $v$ in $+x$ direction.
By Lorentz transformation equation,
\begin{align}
x'&=\gamma(x-vt)  \\   
t'&=\gamma\left(t-\frac{vx}{c^2}\right)   \\  
x&=\gamma(x'+vt')    \\
t&=\gamma\left(t'+\frac{vx'}{c^2}\right)\end{align}
where $\gamma=\sqrt{1-\frac{v^2}{c^2}}$ .
Consider a body $A$ having coordinates $(x,y,z,t)$ in $S$ frame and $(x',y',z',t')$ in $S'$ frame.
In time $dt$, coordinates become $(x+dx,y+dy,z+dz,t+dt)$ in $S$ frame.
In $S'$ frame, in interval $dt'$, coordinate become $(x'+dx',y'+dy',z'+dz',t'+dt')$
In books the derivation for velocity transformation equations are given as

$u_x=\frac{dx}{dt}$
$u'_x=\frac{dx'}{dt'}$
$dx'=\gamma(dx-vdt)$
$dt'=\gamma(dt-\frac{vdx}{c^2})$
$u'_x=\frac{\gamma(dx-vdt)}{\gamma(dt-\frac{vdx}{c^2})}$
Dividing the numerator and denominator by $dt$, we get
$\displaystyle u'_x=\frac{\frac{dx}{dt}-v}{1-\frac{v(dx/dt)}{c^2}}$
$\displaystyle u'_x=\frac{u_x-v}{1-\frac{vu_x}{c^2}}$

I have tried the derivation using chain rule of differentiation,
As $x'=x'(x,t)$
$\frac{dx'}{dt'}=\frac{\partial x'}{\partial x}\Big|_t\frac{dx}{dt'}+\frac{\partial x'}{\partial t}\Big|_x\frac{dt}{dt'}\tag{1}$
As the transformation equations are invertible so,
$x=x(x',t')$
$t=t(x',t')$
So, $\frac{dx}{dt'}=\frac{\partial x}{\partial x'}\Big |_{t'}\frac{dx'}{dt'}+\frac{\partial x}{\partial t'}\Big |_{x'}\frac{dt'}{dt'}\tag{2}$
$\frac{dt}{dt'}=\frac{\partial t}{\partial x'}\Big|_{t'}\frac{dx'}{dt'}+\frac{\partial t}{\partial t'}\Big |_{x'}\frac{dt'}{dt'}\tag{3}$
Plugging $(2)$ and $(3)$ in $(1)$,
$\frac{dx'}{dt'}=\frac{\partial x'}{\partial x}\Big|_t\Big(\frac{\partial x}{\partial x'}\Big |_{t'}\frac{dx'}{dt'}+\frac{\partial x}{\partial t'}\Big |_{x'}\Big)+\frac{\partial x'}{\partial t}\Big|_{x}\Big(\frac{\partial t}{\partial x'}\Big|_{t'}\frac{dx'}{dt'}+\frac{\partial t}{\partial t'}\Big |_{x'}\Big)$
I have calculated all the above partial derivatives from Lorentz transformation. And I get
$\Big(\gamma^2(1-\frac{v^2}{c^2})-1\Big)u'_x=0$
$\implies 1-1=0$
$\implies 0=0$
I have two doubts,
i) If I use chain rule for differentiation then why I am not able to get transformation equation for velocity. Why I get $0=0$ instead of getting an expression for $u'_x$? Have I made some mistake or there is some other way to derive it using chain rule?
ii) The way used in books is that they treated derivative as fraction. They write expression for $dx'$ and $dt'$ and divide them. But isn't derivative a sort of operation. In elementary calculus courses, we have been told that $\frac{dx}{dt}$ is actually $\frac{d}{dt}(x)$. It is not like dividing $dx$ and $dt$. I get very confused how derivative as a fraction is justified.
Please help!
 A: The particle is moving in the 1+1-dimensional Minkowski space-time on a (world-line) curve which could be represented as function of the parameter $\,\tau$, the proper time. More precisely $\,s=c\,\tau\,$ is the relativistic $''$arc length$''$, a scalar invariant under Lorentz transformations. It corresponds to the arc length (natural) parameter  $\,s\,$ of Euclidean curves, a scalar invariant under space transformations.
So, for the parametric representation of the curve we have
\begin{align}
\texttt{in system  } \mathrm S & : \mathbf X\left(\tau\right)\boldsymbol{=}\left[x\left(\tau\right),t\left(\tau\right)\right]
\tag{01a}\label{01a}\\
\texttt{in system  } \mathrm S'\! & : \mathbf X'\!\left(\tau\right)\boldsymbol{=}\left[x'\!\left(\tau\right),t'\!\left(\tau\right)\right]
\tag{01b}\label{01b}
\end{align}
The space-time coordinates are related by a Lorentz boost transformation with velocity $\,\upsilon \in \left(-c,c\right)\,$ along the common $\,x,x'-$axis in differential form
\begin{align}
\mathrm dx' & \boldsymbol{=} \gamma_v\left(\mathrm dx\boldsymbol{-}\upsilon \mathrm dt\right)
\tag{02a}\label{02a}\\
\mathrm dt' & \boldsymbol{=}\gamma_v\left(\mathrm dt\boldsymbol{-}\dfrac{\upsilon}{c^2} \mathrm dx\right)
\tag{02b}\label{02b}\\
\gamma_v & \boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}\frac12}
\tag{02c}\label{02c}
\end{align}
Now by the chain rule and differentiation with respect to $\,\tau\,$  we have
\begin{align}
u'_x& \boldsymbol{=}\dfrac{\mathrm dx'}{\mathrm dt'}  \boldsymbol{=}\dfrac{\dfrac{\mathrm dx'}{\mathrm d\tau}}{\dfrac{\mathrm dt'}{\mathrm d\tau}} \boldsymbol{=}\dfrac{\dfrac{\partial x'}{\partial x}\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\dfrac{\partial x'}{\partial t}\dfrac{\mathrm dt}{\mathrm d\tau}}{\dfrac{\partial t'}{\partial x}\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\dfrac{\partial t'}{\partial t}\dfrac{\mathrm dt}{\mathrm d\tau}}\boldsymbol{=}\dfrac{\gamma_v\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\left(\boldsymbol{-}\gamma_v\upsilon\right)\dfrac{\mathrm dt}{\mathrm d\tau}}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\dfrac{\mathrm dx}{\mathrm d\tau}\boldsymbol{+}\gamma_v\dfrac{\mathrm dt}{\mathrm d\tau}}
\nonumber\\
& \boldsymbol{=}\dfrac{\gamma_v\left(\dfrac{\mathrm dx}{\mathrm d\tau}/\dfrac{\mathrm dt}{\mathrm d\tau}\right)\boldsymbol{-}\gamma_v\upsilon}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\left(\dfrac{\mathrm dx}{\mathrm d\tau}/\dfrac{\mathrm dt}{\mathrm d\tau}\right)\boldsymbol{+}\gamma_v} \boldsymbol{=}\dfrac{\gamma_v\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{-}\gamma_v\upsilon}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\gamma_v}
\nonumber\\
& \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{\boldsymbol{-}\dfrac{\upsilon}{c^2} u_x\boldsymbol{+}1}
\tag{03}\label{03}
\end{align}
that is
\begin{equation}
u'_x \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{1\boldsymbol{-}\dfrac{\upsilon u_x}{c^2} }
\tag{04}\label{04}   
\end{equation}
$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
ADDENDUM
An other way :
\begin{align}
\require{cancel}
u'_x & \boldsymbol{=}\dfrac{\mathrm dx'}{\mathrm dt'}  \boldsymbol{=}\dfrac{\dfrac{\partial x'}{\partial x}\mathrm dx\boldsymbol{+}\dfrac{\partial x'}{\partial t}\mathrm dt}{\dfrac{\partial t'}{\partial x}\mathrm dx\boldsymbol{+}\dfrac{\partial t'}{\partial t}\mathrm dt}\boldsymbol{=}\dfrac{\left(\dfrac{\partial x'}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\dfrac{\partial x'}{\partial t}\right)\cancel{\mathrm dt}}{\left(\dfrac{\partial t'}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\dfrac{\partial t'}{\partial t}\right)\cancel{\mathrm dt}}
\nonumber\\
& \boldsymbol{=}\dfrac{\gamma_v\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{-}\gamma_v\upsilon}{\left(\boldsymbol{-}\gamma_v\dfrac{\upsilon}{c^2} \right)\dfrac{\mathrm dx}{\mathrm dt}\boldsymbol{+}\gamma_v}
\boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{\boldsymbol{-}\dfrac{\upsilon}{c^2} u_x\boldsymbol{+}1}
\tag{A-01}\label{A-01}
\end{align}
that is
\begin{equation}
u'_x \boldsymbol{=}\dfrac{u_x\boldsymbol{-}\upsilon}{1\boldsymbol{-}\dfrac{\upsilon u_x}{c^2} }
\tag{A-02}\label{A-02}  
\end{equation}
Any way we use is equivalent to the division of the Lorentz equations \eqref{02a},\eqref{02b} side by side so I don't think that there exists any sense of the chain rule use. It doesn't give to us a better way or something new.
A: Let the velocity of $S′$ in relation to $S$ in x-direction be $v_0$, let $\gamma_0=\gamma(v_0)=\frac{1}{\sqrt{1-v_0^2/c^2}}$ and let a position and velocity in spacetime be $$x^\mu=(ct,x,y,z)^T,\,\,\,\,\,u^\mu=(c,\vec v)^T=(c,v_x,v_y,v_z)^T.$$
If you want to solve it using differentials (which is formally and mathematically not the best way) you can first calculate the lorentztransformation of the spacetime coordinates $x$: $$\begin{pmatrix}ct'\\x'\\y'\\z'\end{pmatrix}\equiv x'^\mu=\Lambda^\mu\,_\nu x^\nu=\begin{pmatrix} c\gamma_0(t-\frac{v_0}{c^2}x)\\ \gamma_0(x-v_0t)\\ y\\ z\end{pmatrix}.\,\,\,\,\,\,\,\,\,(1)$$ Thus we get for the velocity $$\gamma'\begin{pmatrix}c\\ v'_x\\v'_y\\v'_z\end{pmatrix}\equiv u'^\mu=\frac{dx'^\mu}{d\tau}=\gamma'\frac{dx'^\mu}{dt'}\stackrel{(1)}{=}\gamma'\frac{d}{d(\gamma_0(t-\frac{v_0}{c^2}x))}\begin{pmatrix} c\gamma_0(t-\frac{v_0}{c^2}x)\\ \gamma_0(x-v_0t)\\ y\\ z\end{pmatrix}=\gamma'\begin{pmatrix}c\\\frac{\gamma_0(dx-v_0dt)}{\gamma_0(dt-\frac{v_0}{c^2}dx)}\\ \frac{dy}{\gamma_0(dt-\frac{v_0}{c^2}dx)}\\ \frac{dz}{{\gamma_0(dt-\frac{v_0}{c^2}dx)}}\end{pmatrix},$$
where $\tau$ is the proper time. Dividing the numerators and denominators by $dt$ and simplifying, the space components become
$$\gamma'\begin{pmatrix}v'_x\\v'_y\\v'_z\end{pmatrix}=\gamma'\begin{pmatrix}\frac{v_x-v_0}{1-\frac{v_0}{c^2}v_x}\\\frac{v_y}{\gamma_0(1-\frac{v_0}{c^2}v_x)}\\\frac{v_z}{\gamma_0(1-\frac{v_0}{c^2}v_x)}\end{pmatrix}$$ yielding the same result as in my other formally correcter answer:
$$\vec v'(\vec v)=\frac{1}{1-\frac{v_0}{c^2}v_x}\begin{pmatrix}v_x-v_0\\ v_y/\gamma_0\\ v_z/\gamma_0\end{pmatrix}.$$
A: Let the velocity of $S'$ in relation to $S$ in x-direction be $v_0$, let $\gamma_0=\gamma(v_0)=\frac{1}{\sqrt{1-v_0^2/c^2}}$ and let the velocity in a System be $$(u^\mu)=\gamma(c,v_x,v_y,v_z)^T.$$
The velocity $u$ transforms as follows:$$u'^\mu=\Lambda^\mu\,_\nu u^\nu=\begin{pmatrix}
\gamma_0 & -\gamma_0v_0/c & 0&0\\
-\gamma_0v_0/c & \gamma_0 & 0&0\\
0&0&1&0\\
0&0&0&1
\end{pmatrix}(u^\mu)$$ $$\Leftrightarrow \gamma'\begin{pmatrix}c\\ v_x'\\ v_y'\\ v_z' \end{pmatrix}=\gamma\begin{pmatrix} \gamma_0(c-v_0v_x/c)\\ \gamma_0(-v_0+v_x)\\ v_y\\ v_z\end{pmatrix}$$
From the first component we get $$\gamma'=\gamma\gamma_0(1-\frac{v_0v_x}{c^2}).\,\,\,\,\,\,(1)$$ For the second component we get $$v_x'=\frac{\gamma}{\gamma'}\gamma_0(v_x-v_0)\stackrel{(1)}{=}\frac{v_x-v_0}{1-\frac{v_0v_x}{c^2}}.$$
For the third component we get $$v_y'=\frac{\gamma}{\gamma'}v_y\stackrel{(1)}{=}\frac{v_y/\gamma_0}{1-\frac{v_0v_x}{c^2}}.$$ For the fourth component we get $$v_z'=\frac{\gamma}{\gamma'}v_z\stackrel{(1)}{=}\frac{v_z/\gamma_0}{1-\frac{v_0v_x}{c^2}}.$$
