According to "Nonlinear fiber optics" from Agrawal the mode propagation constant $\beta = n\cdot\frac{\omega}{c}$ can be taylored into $$\beta = \beta_0+\beta_1(\omega-\omega_0)+\frac{1}{2}\beta_2(\omega-\omega_0)^2+\ldots$$ with $$\beta_n = \left.\frac{d^n\beta}{d\omega^n}\right\vert_{\omega=\omega_0}$$ He also gives two examples for the first two derivations: $$\beta_1 = \frac{1}{c}\left(n+\omega\frac{dn}{d\omega}\right)$$ $$\beta_2 = \frac{1}{c}\left(2\frac{dn}{d\omega}+\omega\frac{d^2n}{d\omega^2}\right)\\$$
Now I wanted to do the calculations myself, and started: $$\begin{align} \beta_2&=\frac{d\beta_1}{d\omega}\\ \frac{d\beta_1}{d\omega} &= \frac{d}{d\omega}\left(\frac{\omega}{c}\frac{dn}{d\omega}\right)\\ &= \frac{1}{c}\left(\frac{dn}{d\omega}+\omega\frac{dn^2}{d\omega^2}\right)\\ &\neq\frac{1}{c}\left(2\frac{dn}{d\omega}+\omega\frac{d^2n}{d\omega^2}\right) \end{align}$$ which is not the same as above. Where is my error? And how can I calculate it correctly? Possible Error: I can not remove the $n$ from $\beta_1$, but rather have to include it as $\frac{dn}{d\omega}$ in my equation (resulting in $2\frac{dn}{d\omega}$ instead of $\frac{dn}{d\omega}$)?
