The question I am trying to answer states that: Given a curve $\alpha (s)$ of constant width $R$, then $\beta (s) = \alpha (s) + R \mathbf{n}(s)$ is its opposite. Show that, $$\frac{1}{\kappa_\alpha}+\frac{1}{\kappa_\beta}$$ is a constant.
I have shown in a previous part of the question that for a parallel curve $\beta (s) = \alpha (s) -r \mathbf{n}(s)$, we have $\kappa_\beta=\frac{\kappa_\alpha}{1+r\kappa_\alpha}$.
So I took $r=-R$ in this formula, then attempted to show that $$(\frac{1}{\kappa_\alpha}+\frac{1}{\kappa_\beta})'=-\frac{\kappa'_\alpha}{\kappa_\alpha^2}-\frac{\kappa'_\beta}{\kappa_\beta^2}=0$$ which would imply that $\frac{1}{\kappa_\alpha}+\frac{1}{\kappa_\beta}$ is indeed constant.
However, I have found that $\kappa'_\beta=\frac{\kappa'_\alpha}{(1-R\kappa_\alpha)^2}$ and so $\frac{\kappa'_\beta}{\kappa_\beta^2}=\frac{\kappa'_\alpha}{\kappa_\alpha^2}$ which clearly doesn't give the desired result.
Any help to point me in the right direction would be much appreciated.