Let me provide the context for my question with the geometric statement that I'm trying to prove, although my question is not geometrical:
Let $\beta(s)$ and $\gamma(\widetilde{s})$ be two unit-speed reparametrizations of a regular curve $\alpha:(a,b) \to \mathbb{R}^n$. Prove that $\widetilde{s}=s+C$ for some constant $C$ and that $\frac{d^2\beta}{ds^2}=\frac{d^2\gamma}{d\widetilde{s}^2}$.
The assumptions I'm making here are that $s:(a,b)\to (c_{1},d_{1})$ and $\widetilde{s}:(a,b)\to (c_{2},d_{2})$ are diffeomorphisms, and that $\beta:(c_{1},d_{1}) \to \mathbb{R}^n$ and $\gamma:(c_{2},d_{2}) \to \mathbb{R}^n$ are functions with the listed domains and ranges.
I was easily able to prove the first part, $(\forall t \in (a,b)) \widetilde{s}=s+C$ by taking derivatives of $\alpha$ in two different ways, as $\alpha = \beta \circ s$ and $\alpha = \gamma \circ \widetilde{s}$, but the second part of the statement has some confusing notation for me: what exactly is $\frac{d^2\beta}{ds^2}$? $\beta$ and $s$ are both functions, with different domains; what exactly does it mean to take the derivative of $\beta$ with respect to $s$? I took a look at a recommended question that came out, Derivative of a function with respect to another function., but in that example it was implied that the functions had the same domains, so in my case I don't think I can do $\frac{d\beta}{ds}=\frac{\frac{d\beta}{dt}}{\frac{ds}{dt}}$.
I'd appreciate any clarification on what exactly $\frac{d\beta}{ds}$ and $\frac{d^2\beta}{ds^2}$ mean, as well as a proof of the second part of the statement.
