Liouville Transform (ODE change of variable)

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I am reading an article by Y. Chen and V. Rokhlin titled On the inverse scattering problem for the Helmholtz equation in one dimension from 1992. In their Lemma 3.3. they introduce a particular change of variable stated as follows:

Suppose that $q: \mathbb{R} \to \mathbb{R}$ is a $C^2$-function such that $q > -1$ for all $x \in \mathbb{R}$. Suppose further that the functions $n,x,S,\eta,g: \mathbb{R} \to \mathbb{R}$ are defined by the formulae:

$n(x) = \sqrt{1+q(x)}, \quad t(x) = \int_0^x n(\tau) d\tau, \quad S(t) = (1+q(x(t)))^{-1/4}$

$\eta(t) = \frac{S''(t)}{S(t)} - \frac{n'(x)}{2(n(x))^2}, \quad and \quad g(t) = \frac{f(x)}{S(t)}$.

It should be noted that the $f$ appearing is the source of function of the Helmholtz equation and $q$ is the medium. They also state that $\eta(t)$ can be expressed as:

$\eta(t) = \frac{1}{4}(1+q)^{-2}\left( q''(x) - \sqrt{1+q(x)}q'(x)-\frac{5}{4}(1+q)^{-1}q^{'2}(x) \right)$

My question(s) is now... How do they arrive at the above expression? A sub-question is, do they mean $x(t)$ whenever they write just $x$?

No matter how I try, I never arrive at the same result they do. My biggest confusion lies in getting an expression for $x'(t)$ and $x''(t)$ arising when I differentiate $S(t)$ w.r.t. $t$.