The chain rule and velocity transformation in relativity From elementary calculus, we have that the chain rule occurs when we differentiate a function like $f(y(x)) \equiv f(x)$:
$$\frac{\mathrm{d}}{\mathrm{dx}}[f] = \frac{\mathrm{d}}{\mathrm{dx}}[f(y(x))] = \frac{df}{dy}\frac{dy}{dx} \tag{1}$$
But, consider now a well-known [1] expression about velocity transformation in elemetary Special Relativity context:
$$\frac{\mathrm{d}}{\mathrm{dt'}}[x'] = \frac{dx'}{dt}\frac{dt}{dt'} \tag{2}$$
So, I'm struggling to understand this expression because suppose the situation:
I'm a observer in reference frame $S'$, so I would construct quantities like velocity "with my proper primed coordinates", ($x'$, $t'$). Hence, the spatial coordinates will be parametrized by "my proper primed coordinate $t'$" 
$$U' =: \frac{\mathrm{d}}{\mathrm{dt'}}[x'(t')] \tag{3}$$ 
Of course that if I want to know the velocity in a reference $S$, we have to transform $(1)$ accordingly Lorentz Transformations (LT). My problem isn't acctualy to derive the velocity transformations between $S'$ and $S$ (vice-versa), but to deal properly with the chain rule.
Note that if $(1)$ is the chain rule, and $(2)$ is a valid expression, then the function $x'$ must have the form (I guess) of:
$$x' \equiv x'[t(t')] \tag{4}$$
And my question arise here in $(4)$: What suppose to mean, physically, $t(t')$? . Because it's occur naturally that if we perform a LT like $S \to S'$ time will have the dependency like $t'(t)$ because , $t'(t) \equiv t' = \gamma(t-xv/c^2)$.
 A: I'm not sure exactly what your question is, but perhaps we can start with the following, and work from there....
From the Lorentz transform we know x' = F(x, t). Specifically:
$ x' = \gamma \left(x - vt \right)$.
We also know
$t' = \gamma \left(t - \frac{v x}{c^2}\right)$
So,
$v' = \frac{\partial x'}{\partial t'} = \frac{\partial x'}{\partial t}\frac{\partial t}{\partial t'} + \frac{\partial x'}{\partial x}\frac{\partial x}{\partial t'}$
So we arrive at
$v' = \frac{\partial x'}{\partial t'} = F(x', t') = G(x, t)$
Specifically, $x'$ is a function of both $t$ and $x$, which are independent variables. In the same way $t'$ is also a function of $x$ and $t$, again, two independent variables. Does this help at all?
A: $t$ and $t'$ are just coordinates and you should treat them the same way as spatial coordinates. In Euclidean geometry you can write any curve parametrically as $x(s), y(s)$, but in special cases you may be able to write $y(x)$ or $x(y)$. Likewise you can write a curve in Minkowski space as $x(s), t(s)$ or in special cases as $x(t)$. In point of fact you can almost always write $x(t)$ since the curves are usually causal worldlines, but just because you can doesn't mean you should. It's usually less hassle to use a non-coordinate parameter (which may be proper time).
Given a curve $x(τ), t(τ)$, its Lorentz-boosted form is $$X(τ) = γ\,x(τ)+γβ\,t(τ),\;T(τ) = γ\,t(τ)+γβ\,x(τ),$$ where I'm using upper case instead of primes to avoid confusion with the derivative. If you need $T$ as a function of $t$, it's $T\circ t^{-1}$, where the $T$ and $t$ in that formula are treated as functions of $τ$. If you want the derivative of $T$ with respect to $t$, that's $T'(t) = T'(τ)/t'(τ)$, or $\displaystyle \frac{dT}{dt}=\frac{dT/dτ}{dt/dτ}$ if you prefer that notation. All of this is exactly the same as the Euclidean case, and the fact that it's a time coordinate is not relevant.
In most cases you shouldn't need any of those things; you should just uniformly treat all of the coordinates as functions of proper time (or some other non-coordinate parameter).
A: For your equation 2, the chain rule should be:
$\frac{dx'}{dt'} = \frac{\partial x'}{\partial t}\frac{\partial t}{\partial t'} + \frac{\partial x'}{\partial x}\frac{\partial x}{\partial t'}$
