I'm trying to solve this question from the classical Do Carmo's Differential geometry book. Surprisingly for me, I'm stuck on a very simple question on page 25:
MY SOLUTION:
Suppose we have already proved $dt/ds=1/|\alpha'|$ Since he wants the second derivative $d^2t/ds^2$ of $dt/ds$, we have:
$$\frac{d^2t}{ds^2}=(|\alpha'(t)|^{-1})'=-|\alpha'(t)|^{-2}|\alpha'(t)|'=-\frac{|\alpha'(t)|'}{|\alpha'(t)|^2}$$
Since $|\alpha'(t)|'=\frac{\alpha'(t)\alpha''(t)}{|\alpha'(t)|}$, we have
$$d^2t/ds^2=-\frac{\alpha'(t)\alpha''(t)}{|\alpha'(t)|^3}$$
So where is my mistake?
The error is that you differentiate $|\alpha^{\prime}|^{-1}$ with respect to $s$, and you write $|\alpha^{\prime}|^{\prime}$ but the primes mean differentiation with respect to $t$. So your calculation is the derivative of $|\alpha^{\prime}|^{-1}$ with respect to $t$, and not $s$. Correcting this with another chain rule gives the correct answer.