= Solution
Let $\gamma$ be a $C^3$ <regular curve> parametrized by <arc length>, so $|\gamma'(s)|=1$. Define the <Frenet frame>
$$
T=\gamma',
\qquad
\kappa=|T'|,
\qquad
N=\frac{T'}{\kappa},
\qquad
B=T\times N,
$$
and define the torsion by
$$
\tau=-B'\mathbin\cdot N.
$$
Thus curvature requires two derivatives, while $N,B,$ and torsion require $\kappa>0$; the displayed classical torsion requires three derivatives. At a point where $\kappa=0$, $N,B,$ and $\tau$ are not defined by this construction.
Since $(T,N,B)$ is an orthonormal frame, differentiating its inner products shows that its derivative matrix is skew-symmetric. The definition gives $T'=\kappa N$. Next,
$$
N'\mathbin\cdot T=-N\mathbin\cdot T'=-\kappa,
\qquad
N'\mathbin\cdot N=0,
$$
so $N'=-\kappa T+cB$. As $B'=T\times N'$ and $T\times B=-N$, we have $B'=-cN$; hence $c=-B'\cdot N=\tau$. The <Frenet-Serret formulas> are therefore
$$
\boxed{
T'=\kappa N,
\qquad
N'=-\kappa T+\tau B,
\qquad
B'=-\tau N.}
$$
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