$ X:\mathbb{R}\to\mathbb{R}^3$ , $ r:\mathbb{R}\to\mathbb{R}$ ,if $r(t_0) \neq 0$, then $r$ is differentiable at $t_0$ We have $ X:\mathbb{R}\to\mathbb{R}^3$ be a differentiable function, and $ r:\mathbb{R}\to\mathbb{R}$
be the function $r(t) :=\rVert{X(t)}\lVert$, where $\rVert{X(t)}\lVert$ denotes the length of $X(t)$ as measured
in the usual $l^2$ metric. Let $t_0$ be a real number.
Now $r$ will be differentiable at $t_0$ if we have a linear map $L$ such that,
$$\lim_{h\to 0}\frac{\vert\vert{r(t_0+h)-r(t_0)-L(h)}\vert\vert}{\vert\vert{h}\vert\vert}=0$$
But I don't know how to show that.
 A: Let's show that $\lVert \cdot \rVert$ is differentiable at any point except $0$.
Let $x =(a,b,c) \in \mathbb{R}^3 \backslash\{0\}$, then $\lVert x \rVert = \lVert (a,b,c) \rVert = \sqrt{a^2+b^2+c^2} =\sqrt{\langle x|x\rangle}$.
$$
\begin{split}
\lVert x+h \rVert-\lVert x \rVert &=
 \dfrac{\lVert x+h \rVert^2-\lVert x \rVert^2 }{\lVert x+h \rVert+\lVert x \rVert} 
\\
&=\dfrac{2\langle x |h\rangle+\lVert h \rVert^2}{\lVert x+h \rVert+\lVert x \rVert}\\
&=\dfrac{2\left\langle \frac{x}{\lVert x \rVert} \bigg| h\right\rangle+\dfrac{\lVert h \rVert^2}{\lVert x \rVert}}{\dfrac{\lVert x+h \rVert}{\lVert x \rVert}+1}
\end{split}
$$
Therefore by calling $\alpha_h = \dfrac{\lVert x+h \rVert}{\lVert x \rVert}+1$ by continuity of $\sqrt{\cdot} \,$ we have $\alpha_h \underset{h \rightarrow0}{\longrightarrow}2$ and
$$
\begin{split}
\dfrac{\lVert x+h \rVert-\lVert x \rVert - \left\langle \frac{x}{\lVert x \rVert} \bigg| h\right\rangle}{\lVert h \rVert} &= \dfrac{\dfrac{(2-\alpha_h)}{\lVert h \rVert}\left\langle \frac{x}{\lVert x \rVert} \bigg| h\right\rangle}{\alpha_h}+\dfrac{\lVert h \rVert}{\alpha_h \lVert x\rVert}\\
&=\dfrac{(2-\alpha_h)}{\alpha_h}\left\langle \frac{x}{\lVert x \rVert} \bigg| \frac{h}{\lVert h \rVert}\right\rangle+\dfrac{\lVert h\rVert}{\alpha_h \lVert x\rVert}
\end{split}$$
But by the Cauchy-Schwarz inequality, for all $h \neq 0$,
$$\left|\left\langle \frac{x}{\lVert x \rVert} \bigg| \frac{h}{\lVert h \rVert}\right\rangle \right| \leq \left\lVert \frac{x}{\lVert x \rVert}\right\rVert \left\lVert \frac{h}{\lVert h \rVert}\right\rVert =1 $$
Hence by triangular inequality,
$$
\begin{split}
\dfrac{\left|\lVert x+h \rVert-\lVert x \rVert - \left\langle \frac{x}{\lVert x \rVert} \bigg| h\right\rangle \right|}{\lVert h \rVert} &\leq \left|\dfrac{(2-\alpha_h)}{\alpha_h}\left\langle \frac{x}{\lVert x \rVert} \bigg| \frac{h}{\lVert h \rVert}\right\rangle\right|+\left|\dfrac{\lVert h\rVert}{\alpha_h \lVert x\rVert}\right|\\
&\leq \left|\dfrac{2-\alpha_h}{\alpha_h} \right|+\left|\dfrac{\lVert h\rVert}{\alpha_h \lVert x\rVert}\right|
\end{split}$$
But $$\left|\dfrac{2-\alpha_h}{\alpha_h} \right|+\left|\dfrac{\lVert h\rVert}{\alpha_h \lVert x\rVert}\right| \underset{h \rightarrow 0}{\longrightarrow}0$$
Therefore,
$$ \lim_{h\rightarrow 0} \dfrac{\left|\lVert x+h \rVert-\lVert x \rVert - \left\langle \frac{x}{\lVert x \rVert} \bigg| h\right\rangle \right|}{\lVert h \rVert} = 0$$
Therefore as $h \mapsto\left\langle \frac{x}{\lVert x \rVert} \bigg| h\right\rangle  $ is linear, we deduce that $\lVert \cdot \rVert$ is differentiable at $x \neq 0$.
Finally for $t_0$ as $r(t_0) = \lVert X(t_0) \rVert \neq0 $ we have $X(t_0) \neq 0$. So if we call $g : x \in \mathbb{R}^3 \mapsto \lVert x \rVert $ then we have by composition $g \circ X $ differentiable at $t_0$ (because $X$ is differentiable at $t_0$ and $g$ is differentiable at $X(t_0)$).
It means that $r = g \circ X$ is differentiable at $t_0$.
