Differential geometry and frenet formula Given a curve C by its arclenght (vector $r(s)$), prove that $\frac{dT(s)}{ds} \times \frac{d^2 T(s)}{ds^2} = k^2 \omega$ where k is the curvature and $\omega$ is the darboux vector.

I tried using the formula that
  $\frac {dT}{ds} = kN =\frac {d^2r}{ds^2}$
  And also by chain rule 
  $\frac {dr}{ds} = \frac {dr}{dt} \frac {dt} {ds}$ So if I am not incorrect $\frac{d^2r}{ds^2} = \frac {d^2r}{dt^2} + \frac {dr}{dt} \frac {d^2t}{ds^2} $
  And $\frac {d^3r}{ds^3}$ = $\frac {\frac{d^3r}{dt^3} + \frac {d^2r}{dt^2} \frac {d^2t}{ds^2}}{\frac {ds}{dt}} + \frac {dr} {dt} \frac {d^3t}{ds^3}$ but it doesn’t seem to go anywhere to do this.
  Is any practical way to evaluate $\frac{d^2 T(s)}{ds^2}$?

 A: Herein, "dot" notation is used for derivatives with respect to the arc-length $s$, thusly:
$\dfrac{dT}{ds} = \dot T, \text{etc}. \tag 0$
We start by differentiating the first of the Frenet-Serret equations,
$\dot T = \kappa N, \tag 1$
and obtain
$\ddot T = \dot \kappa N + \kappa \dot N; \tag 2$
we may then substitute in $\dot N$ from the second Frenet-Serret equation,
$\dot N = -\kappa T + \tau B, \tag 3$
yielding
$\ddot T = \dot \kappa N -\kappa^2 T + \kappa \tau B; \tag 4$
we take the $\times$ product of (1) and (4), recalling that $N \times N = 0$:
$\dot T \times \ddot T = -\kappa^3 N \times T + \kappa^2 \tau N \times B; \tag 5$ 
the definition of $B$ is
$B = T \times N = -N \times T,\tag 6$
equivalent under cyclic permutation to
$T = N \times B; \tag 7$
via (6) and (7), (5) becomes
$\dot T \times \ddot T = \kappa^3 B + \kappa^2 \tau T = \kappa^2(\kappa B + \tau T) = \kappa^2 \omega, \tag 8$
where $\omega = \kappa B + \tau T$ is the Darboux vector.
A: Don't worry about the chain rule and $t$. Just use the Frenet equations to find $kN\times \dfrac d{ds}(kN)$.
