Answer
$$\frac{dr}{ds}=\tan2s(2r-1)$$
Work Step by Step
$$r\cos2s+\sin^2s=\pi$$
To find $dr/ds$, we would use the methods of implicit differentiation.
Differentiate both sides of the equation with respect to $s$: $$\Big(\frac{dr}{ds}\cos2s+r\frac{d(\cos2s)}{ds}\Big)+\frac{d(\sin^2s)}{ds}=\frac{d\pi}{ds}$$
$$\cos2s\frac{dr}{ds}+r(-\sin2s)\frac{d(2s)}{ds}+2\sin s\frac{d(\sin s)}{ds}=0$$
$$\cos2s\frac{dr}{ds}-2r\sin2s+2\sin s\cos s=0$$
$$\cos2s\frac{dr}{ds}-2r\sin2s+\sin2s=0$$
$$\cos2s\frac{dr}{ds}+\sin2s(1-2r)=0$$
Next, separate the elements with $dr/ds$ and those without it into 2 sides of the equation: $$\cos2s\frac{dr}{ds}=-\sin2s(1-2r)$$
$$\cos2s\frac{dr}{ds}=\sin2s(2r-1)$$
Then calculate for $dr/ds$: $$\frac{dr}{ds}=\frac{\sin2s(2r-1)}{\cos2s}$$
$$\frac{dr}{ds}=\tan2s(2r-1)$$
