Let $\alpha:I\rightarrow \mathbb{R}$ be a twice differentiable curve such that $\alpha '(t)$ and $\alpha ''(t)$ are linearly independent for every $t$. Let
$$s:=\int_{t_0}^{t}|a'(u)|du$$
and $\beta = \alpha \circ s^{-1}$, that is, $\beta$ is an arc length parametrization of $\alpha$.
One defines the Frenet Trihedron of the curve $\beta$ as
$$T_{\beta}(t):=\beta'(t)$$ $$N_{\beta}(t):=\frac{T'_{\beta}(t)}{|T'_{\beta}(t)|}=\frac{\alpha''(t)}{|\alpha''(t)|}$$ $$B_{\beta}(t):=T_{\beta}(t)\times N_{\beta}(t)$$
My goal is to derive a Frenet Trihedron for the curve $\alpha$, so as to not need $\beta$. Deriving the tangent vector $T_{\alpha}$ is easy enough, since we would like $T_{\alpha}(t)=T_{\beta}(s(t))$, thus
$$T_{\alpha}(t):=T_{\beta}(s(t))=\beta '(s(t))=\frac{\alpha '(t)}{|\alpha '(t)|}$$
yet I have trouble deriving the normal vector $N_{\alpha}$. I begin with
$$N_{\alpha}(t):=N_{\beta}(s(t))=\frac{T'_{\beta}(s(t))}{|T'_{\beta}(s(t)|)}=\frac{\left(\frac{d}{dt}\frac{\alpha'(t)}{|\alpha'(t)|}\right)}{\Big|\frac{d}{dt}\frac{\alpha'(t)}{|\alpha'(t)|}\Big|}$$
A couple of applications of the chain rule give me
$$\frac{d}{dt}\frac{\alpha'(t)}{|\alpha'(t)|}=\frac{\alpha'(t)\cdot \alpha''(t)}{|\alpha'(t)|^3}$$
so that
$$N_{\alpha}(t)=\frac{\alpha'(t)\cdot \alpha''(t)}{|\alpha'(t)\cdot \alpha''(t)|}$$
yet this seems wrong, since it is possible for $\alpha$ to be an arc-length parametrization, in which case $|\alpha'(t)|=1$ and $\alpha'(t)\cdot \alpha''(t)=0$ so that $N_{\alpha}(t)$ would not be well defined.
$1.$ What did I do wrong?
$2.$ How could a formula for the normal vector $N_{\alpha}$ be derived?