Answer
$$\frac{dR}{d\theta}=ks\left(\frac{r_2^{4}-r_1^{4}\cos(\theta)}{r_1^{4}r_2^{4}\sin^{2}(\theta)}\right)$$
Work Step by Step
The equation of $R$ from the previous part:
$$R=k\left(\frac{L_{0}-s\cot(\theta)}{r_1^{4}}+\frac{s}{\sin(\theta)r_2^{4}}\right)$$
$$\frac{dR}{d\theta}=k\left(\frac{d}{d\theta}\left(\frac{L_{0}-s\cot(\theta)}{r_1^{4}}\right)+\frac{d}{d\theta}\left(\frac{s}{\sin(\theta)r_2^{4}}\right)\right)$$
$$\frac{dR}{d\theta}=k\left(\frac{-s}{r_1^{4}}\frac{d}{d\theta}\left(\cot(\theta)\right)+\frac{s}{r_2^{4}}\frac{d}{d\theta}\left(\frac{1}{\sin(\theta)}\right)\right)$$
$$\frac{dR}{d\theta}=k\left(\frac{-s}{r_1^{4}}\frac{d}{d\theta}\left(\cot(\theta)\right)+\frac{s}{r_2^{4}}\frac{d}{d\theta}\left(\csc(\theta)\right)\right)$$
$$\frac{dR}{d\theta}=k\left(\frac{s}{r_1^{4}}\csc^{2}(\theta)-\frac{s}{r_2^{4}}\csc(\theta)\cot(\theta)\right)$$
$$\frac{dR}{d\theta}=k\left(\frac{s}{r_1^{4}}\frac{1}{\sin^{2}(\theta)}-\frac{s}{r_2^{4}}\frac{\cos(\theta)}{\sin^{2}(\theta)}\right)$$
$$\frac{dR}{d\theta}=k\left(\frac{sr_2^{4}-sr_1^{4}\cos(\theta)}{r_1^{4}r_2^{4}\sin^{2}(\theta)}\right)$$
$$\frac{dR}{d\theta}=ks\left(\frac{r_2^{4}-r_1^{4}\cos(\theta)}{r_1^{4}r_2^{4}\sin^{2}(\theta)}\right)$$
