i have trouble to calculate the derivation of the the norm.
I have given a curve $\gamma :[0,L] \rightarrow \mathbb{R^3}$ with $\gamma(0)=\gamma(L)$, $\Vert \gamma^´(s) \Vert=1$ and a second function $\beta(s)=\Vert \gamma(s)-\gamma(0)\Vert$.
I´d like to calculate $\beta'(s)$
My approach: I set $f(\cdot):= \Vert \cdot\Vert$ and $g(s):=\gamma(s)-\gamma(0)$, then i get by chain rule $\frac{d}{ds}f(g(s)) =f'(g(s))g'(s)=\frac{\gamma(s)-\gamma(0)}{\Vert \gamma(s)-\gamma(0)\Vert}\gamma'(s) $ where i use that $f'(\vec{v})=\frac{\vec{v}}{\Vert \vec{v} \Vert}$
Is it correct or did i miss something ?
$$\beta^2(s)=||\gamma(s)-\gamma(0)||^2=\langle\gamma(s)-\gamma(0),\gamma(s)-\gamma(0)\rangle\\=\langle\gamma(s),\gamma(s)\rangle-2\langle\gamma(s),\gamma(0)\rangle+\langle\gamma(0),\gamma(0)\rangle$$
Therefore $$2\beta(s)\beta'(s)=2\langle\gamma'(s),\gamma(s)\rangle-2\langle\gamma'(s),\gamma(0)\rangle=2\langle\gamma'(s),\gamma(s)-\gamma(0)\rangle$$ Hence $$\beta'(s)=\frac{\langle\gamma'(s),\gamma(s)-\gamma(0)\rangle}{\beta(s)}=\frac{\langle\gamma'(s),\gamma(s)-\gamma(0)\rangle}{||\gamma(s)-\gamma(0)||}$$