Computing $(\frac{\alpha''(s)}{|\alpha''(s)|})'$ Let $\alpha:I\rightarrow\mathbb{R}^3$ be a curve parameterized by arc length $s$ with curvature $k(s)\ne0$ for all $s\in I$.
I am attempting to compute $(\frac{\alpha''(s)}{|\alpha''(s)|})'$. This ought to equal $\frac{\alpha'''(s)}{|\alpha''(s)|^2}$. 
Attempt: 
$(\frac{\alpha''(s)}{|\alpha''(s)|})'=(\frac{t'}{k})'=\frac{kt''-t'k'}{k^2}=\frac{k(kn)'-(kn)k'}{k^2}=\frac{k(k'n+kn')-knk'}{k^2}=\frac{k(k'n+kn'-nk')}{k^2}=\frac{k(kn')}{k^2}=\frac{k^2n'}{k^2}$.
I tried to see if one of these intermediary expressions could be transformed into the desired result, but I'm simply not seeing it. I've spent longer than I care to admit following dead ends.
Any help is appreciated.
 A: Not to be the bearer of evil tidings, but for a unit-speed curve $\alpha(s)$ with non-vanishing curvature $k(s)$, the formula
$\left (\dfrac{\alpha''(s)}{\vert \alpha''(s) \vert} \right )' = \dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert^2} \tag 1$
does not, in general, bind.  We may see that this is so by direct calcuation in terms of the Frenet-Serret formulas; indeed, since $\alpha(s)$ is a unit-speed curve, we have 
$\alpha'(s) = T(s), \tag 2$
where $T(s)$ is the unit tangent vector field to $\alpha(s)$; then
$\alpha''(s) = T'(s) = k(s)N(s), \tag 3$
$N(s)$ being the unit normal vector field to $\alpha(s)$ and $k(s)$ being the curvature; this implies
$\vert \alpha''(s) \vert = \vert k(s) N(s) \vert = k(s) \vert N(s) \vert = k(s), \tag 4$
since
$\vert N(s) \vert = 1; \tag 5$
combining (3) and (4) we see that
$\dfrac{\alpha''(s)}{\vert \alpha''(s) \vert} = \dfrac{k(s)N(s)}{k(s)} = N(s), \tag 6$
from which
$\left ( \dfrac{\alpha''(s)}{\vert \alpha''(s) \vert} \right )' = N'(s) = -k(s)T(s) + \tau(s) B(s), \tag 7$
where the right-hand equality is in fact a well-known Frenet-Serret equation with
$B(s) = T(s) \times N(s), \tag 8$
and $\tau(s)$ the torsion of the curve $\alpha(s)$.  Now, differentiating (3) we find
$\alpha'''(s) = (k(s)N(s))' = k'(s)N(s) + k(s)N'(s)$
$= k'(s)N(s) + k(s)(-k(s)T(s) + \tau(s) B(s)) = k'(s)N(s) -k^2(s)T(s) + k(s)\tau(s) B(s); \tag 9$
via (4),
$\vert \alpha''(s) \vert^2 = k^2(s), \tag{10}$
whence
$\dfrac{\alpha'''(s)}{\vert \alpha''(s) \vert^2} = \dfrac{1}{\vert \alpha''(s) \vert^2}( k'(s)N(s) -k^2(s)T(s) + k(s)\tau(s) B(s))$
$= \dfrac{k'(s)}{k^2(s)}N(s) - T + \dfrac{\tau(s)}{k(s)}B(s). \tag{11}$
Comparing (7) and (11) we see they cannot be the same if $k'(s) \ne 0$, since (11) contains the term
$\dfrac{k'(s)}{k^2(s)}N(s) \ne 0, \tag{12}$
whereas (7) has no component proportional to $N(s)$.  Indeed, if we assume (1), then we obtain through (7) and (11)
$-k(s)T(s) + \tau(s) B(s) = \dfrac{k'(s)}{k^2(s)}N(s) - T + \dfrac{\tau(s)}{k(s)}B(s); \tag{13}$
comparing the coefficients of the orthonormal vectors $T(s)$, $N(s)$, and $B(s)$, we write
$\dfrac{k'(s)}{k^2(s)} = 0 \Longrightarrow k'(s) = 0, \tag{14}$
$k(s) = 1, \tag{15}$
and $\tau(s)$ is apparently arbitrary.
In the above it has been shown that any unit-speed curve $\alpha(s)$ with $k(s) \ne 0$ that satisfies (1) must have constant curvature $1$ though its torsion is unconstrained; likewise, a curve $\alpha(s)$ with $k(s) = 1$ satisfies (1) no matter what its torsion may be.  So the curves $\alpha(s)$ such that (1) binds are precisely the curves on unit curvature.
