Use of chain rule in deriving Lorentz velocity transformation When deriving Lorentz velocity transformation in a boosted frame of reference, we are obviously going to use the Lorentz coordinate transformation and then differentiate it with respect to $t'$. My question is: Since $x'$ is a function of both $x$ and $t$ and they are both functions of $x'$ and $t'$, I'm wondering why we use the chain rule for normal derivatives (single variable) e.g. $\text{d}x'/\text{d}t' = (\text{d}x'/\text{d}t)\times(\text{d}t/\text{d}t')$? The reason this confuses me is because since its a multivariable function, wouldn't we have to take partial derivatives? Hopefully my question makes sense. Thanks in advance!
 A: It is important to remember what the symbols $x$ and $t$ actually stand for, as a great number of sources tend to overload them which becomes confusing.
In the original Lorentz transformations
$$t'=\gamma\left(t-\frac{vx}{c^2}\right) \\ x'=\gamma\left(x-vt\right)$$
the symbols $x$, $t$, $x'$ and $t'$ are the coordinates of a single event in spacetime. That is, one point in space and one instant in time.
However, when deriving the velocity addition formula, you are giving the symbols $x$ and $x'$ a slightly different meaning: the position of an object as a function of time. Suppose the object is traveling with velocity $u$ in $S$ and $u'$ in $S'$, with $S'$ moving with $v$ relative to $S$.
Therefore, we have
$$x=x(t)=ut\\ x'=x'(t')=u't'$$
This is why $u=\text{d}x/\text{d}t$ and $u'=\text{d}x'/\text{d}t'$ are ordinary derivatives, because they represent the position of an object instead of an event in spacetime. The transformations then become
$$t'=\gamma\left(1-\frac{uv}{c^2}\right)t \\ x'=\gamma\left(u-v\right)t$$ from which we can see that $\text{d}t'/\text{d}t$ and $\text{d}x'/\text{d}t$ are ordinary derivatives too. The desired formula $u'=(u-v)/(1-uv/c^2)$ is also apparent.
In summary, once we use $x$ and $x'$ to represent the positions of the moving object, $x$ and $t$ are no longer independent; neither are $x'$ and $t'$. The Lorentz transformations become functions of a single variable.
A: $\require{\cancel}$
I don't think that the chain rule is necessary. Simply express the Lorentz transformation in differential form
\begin{align}
\mathrm d x' & = \gamma\left(\mathrm dx-v\mathrm dt\right)
\tag{01a}\\
\mathrm d t' & = \gamma\left(\mathrm dt-\frac{v}{c^2}\mathrm dx\right)
\tag{01b}
\end{align}
divide side by side
\begin{equation}
    \dfrac{\mathrm d x'}{\mathrm d t'}=\dfrac{\gamma\left(\mathrm dx-v\mathrm dt\right)}{\gamma\left(\mathrm dt-\dfrac{v}{c^2}\mathrm dx\right)}=\dfrac{\left(\dfrac{\mathrm dx}{\mathrm dt}-v\right)\cancel{\gamma\mathrm dt}}{\left(1-\dfrac{v}{c^2}\dfrac{\mathrm dx}{\mathrm dt}\right)\cancel{\gamma\mathrm dt}}
\tag{02}\label{02}
\end{equation}
to have
\begin{equation}
    u'=\dfrac{\left(u-v\right)}{\left(1-\dfrac{uv}{c^2}\right)}
\tag{03}\label{03}
\end{equation}
