Consider a trajectory $x(t)$ and a time-scaling $dt = \gamma(x(t)) ds$ . The first new-time derivative simply gives: $\frac{dx}{ds}=\frac{dx}{dt}\cdot\frac{dt}{ds}=\gamma(x) \frac{dx}{dt}$ . What about the second derivative?
My attempt:
$\frac{d^2x}{ds^2}=\frac{d}{ds}(\frac{dx}{ds})=\frac{d}{ds}(\frac{dx}{dt}\frac{dt}{ds})=\frac{d}{ds}(\frac{dx}{dt})\cdot\frac{dt}{ds}+\frac{dx}{dt}\cdot\frac{d}{ds}(\frac{dt}{ds})==(\frac{d}{dt}\cdot\frac{dt}{ds})(\frac{dx}{dt})\cdot\frac{dt}{ds}+\frac{dx}{dt}\cdot(\frac{d}{dt}\cdot\frac{dt}{ds})(\frac{dt}{ds})=\gamma^2(x)\cdot\frac{d^2x}{dt}+\frac{dx}{dt}\frac{d\gamma}{dt}=\gamma^2(x)\cdot\frac{d^2x}{dt}+\frac{\partial \gamma}{\partial x}\frac{dx}{dt}\frac{dx}{dt}$,
but I believe I made some shenanigans, in particular after $==$. Can you help me please? All Best!
Some, yes. Mostly you dropped a gamma somewhere along the way.
$$\def\d{\mathrm d}\begin{align}\dfrac{\d^2x}{\d s~^2}&=\dfrac{\d~~}{\d s}\left(\dfrac{\d x}{\d s}\right)\\&=\dfrac{\d~~}{\d s}\left(\dfrac{\d x}{\d t}\cdotp\dfrac{\d t}{\d s}\right)\\&=\dfrac{\d~~}{\d s}\left(\dfrac{\d x}{\d t}\right)\cdot\dfrac{\d t}{\d s}+\dfrac{\d x}{\d t}\cdotp\dfrac{\d~~}{\d s}\left(\dfrac{\d t}{\d s}\right)\\&=\left[\frac{\d t}{\d s}\cdot\dfrac{\d~~}{\d t}\right]\!\left(\dfrac{\d x}{\d t}\right)\cdot\dfrac{\d t}{\d s}+\dfrac{\d x}{\d t}\cdot\dfrac{\d ~~}{\d s}\left(\dfrac{\d t}{\d s}\right)\\&=\left(\gamma(x)\cdot\dfrac{\d^2 x}{\d t~^2}\right)\cdot\gamma(x)+\dfrac{\d x}{\d t}\cdot\dfrac{\d \gamma(x)}{\d s}\\&=\gamma^2(x)\cdot\dfrac{\d^2 x}{\d t~^2}~+~\dfrac{\d x}{\d t}\cdot\dfrac{\d \gamma(x)}{\d x}\cdot\dfrac{\d x}{\d t}\cdot\dfrac{\d t}{\d s}\\&=\gamma^2(x)\cdot\dfrac{\d^2 x}{\d t~^2}+\gamma(x)\cdot\dfrac{\d\gamma(x)}{\d x}\cdot\left(\dfrac{\d x}{\d t}\right)^2\end{align}$$
Note: $\d t= \gamma(x)~\d s$ means $\dfrac{\d~~}{\d s}= \gamma(x)~\dfrac{\d~~}{\d t}$ so quickly:
$$\begin{align}\dfrac{\d^2 x}{\d s~^2}&=\left[\gamma(x)~\dfrac{\d~~}{\d t}\right]\left(\gamma(x)~\dfrac{\d x}{\d t}\right)\\&=\gamma(x)~\left(\gamma(x)~\dfrac{\d^2 x}{\d t~^2}+\dfrac{\d\gamma(x)}{\d t}\dfrac{\d x}{\d t}\right)\\&=\gamma(x)~\left(\gamma(x)~\dfrac{\d^2 x}{\d t~^2}+\dfrac{\d\gamma(x)}{\d x}\left(\dfrac{\d x}{\d t}\right)^2\right)\\&= \gamma^2(x)~\dfrac{\d^2 x}{\d t~^2}+\gamma(x)~\dfrac{\d\gamma(x)}{\d x}\left(\dfrac{\d x}{\d t}\right)^2\end{align}$$