Derivative of Frenet-Serret equations I'm somewhat confused with Frenet-Serret equations and its derivatives. The curve $\gamma(s)$ is parametrized by its arc-length and it's such that the following is true:
$$
  \frac{1}{k^2} + \left( \frac{k'}{\tau k^2} \right)^2 = 1.
$$
Then, if $N(s)$ and $B(s)$ represent the Frenet-Serret equations of $\gamma(s)$, I'm asked to prove that $p(s)$ is constant given that 
$$
    p(s)
  =   \gamma(s) + \frac{1}{k(s)} N(s)
    - \left( \frac{k'}{\tau k^2} \right) B.
$$
The first steps aren't hard. I started off with the derivative—if $p(s)$ is constant then its derivative is equal to $0$. By using the Frenet-Serret derivatives I managed to cancel out a lot of terms and ended up with
$$
    p(s)
  = \left(
      \frac{\tau}{k} - \left( \frac{k'}{\tau k^2} \right)'
    \right) B.
$$
I'm currently stuck at this step. I've tried many things like expanding the terms but the derivative of $k'/(\tau k^2)$ is even more confusing than what I just got. Is there a Frenet-Serret relationship I'm ignoring here? 
Also, if you were able to recommend me any books with this kind of problems/views I'd very much appreciate it. I've tried to find one but so far they all simply try to prove the Frenet-Serret formulas instead of working with them.

$$
    p'(s)
  = \left(
      \frac{\tau}{k} - \left( \frac{k'}{\tau k^2} \right)'
    \right) B.
$$
Let's try to make this equation a bit similar to the one given. First, let's focus on $\dfrac{1}{k^2}$. Removing the $\tau$:
$$
    p'(s)
  = \left(
      \frac{1}{k} - \frac{1}{\tau}\left( \frac{k'}{\tau k^2} \right)'
    \right) \tau B.
$$
After a while, I noticed that by doing this, the second part is starting to look a bit similar to the one given. We keep working on that one:
$$
    p'(s)
  = \left(
      \frac{k'}{k^3} - \frac{k'}{\tau k^2}\left( \frac{k'}{\tau k^2} \right)'
    \right) \frac{k^2  \tau}{k'} B.
$$
Integrating the two:
$$
    p'(s)
  = \left(
      \dfrac{-1}{2}(\dfrac{1}{k^2})'  - \dfrac{-1}{2}\left( \left( \frac{k'}{\tau k^2} \right)^2  \right)'
    \right) \frac{k^2  \tau}{k'} B.
$$
$$
    p'(s)
  = \left(
          \left(\dfrac{1}{k^2}\right)'  - \left( \frac{k'}{\tau k^2} \right)^2
    \right)' \frac{k^2  \tau}{-2k'} B.
$$
$$
    p'(s)
  = 0
$$
It took me a while to realize this, so it's clear I need more practice. Book/sites recommendations are very welcome, thank you!
