I am doing a lab in seismic ray tracing and I am trying to understand how the following equation is derived: $$ \dfrac{\text{d}}{\text{d}\theta}\left( \dfrac{1}{r}\dfrac{ \text{d} r}{ \text{d} \theta } \right) = \left( \dfrac{ r }{ pv } \right)^2 \left( 1- \dfrac{ r }{ v } \dfrac{ \text{d} v }{ \text{d} r } \right) $$
Apparently, this is found by taking the derivative with respect to $\theta$ of this equation: $$ \left( \dfrac{ r }{ pv } \right)^2 = 1+ \left( \dfrac{ 1 }{ r } \dfrac{ \text{d} r }{ \text{d} \theta } \right)^2 $$
where the ray parameter $p = \dfrac{r_k\sin(i)_k}{v_k} = \text{constant} \hspace{7px}$ for any $k$.
Here is a diagram that shows the problem:
The $v_k$'s are the velocities of the propagating ray/wave through the material.
You are right. Let us differentiate the second equation with respect to $\theta$ :
$$\frac{d}{d\theta} (\frac{r}{pv})^2 = \frac{d}{d\theta} (1 + (\frac{1}{r} \frac{dr}{d\theta})^2)$$ $$\implies \frac{r}{pv}\frac{d}{d\theta}(\frac{r}{pv}) = \frac{1}{r} \frac{dr}{d\theta} \frac{d}{d\theta}(\frac{1}{r}\frac{dr}{d\theta})$$ $$\implies \frac{1}{r}\frac{dr}{d\theta} \frac{d}{d\theta}(\frac{1}{r}\frac{dr}{d\theta}) = \frac{r}{pv}\frac{1}{p} \frac{1}{v^2}(v \frac{dr}{d\theta} - r \frac{dv}{d\theta})$$ $$\implies \frac{d}{d\theta}(\frac{1}{r} \frac{dr}{d\theta}) = (\frac{r}{pv})^2 (1 - \frac{r}{v}\frac{dv}{dr})$$