## Calculus: Early Transcendentals 8th Edition

$$\frac{dy}{d\theta}=-2\cot(\sin\theta)\csc^2(\sin\theta)\cos\theta$$
$$y=\cot^2(\sin\theta)$$ Then, $$\frac{dy}{d\theta}=\frac{d(\cot^2(\sin\theta))}{d\theta}$$ Let $u=\sin\theta$, then $y=\cot^2 u$. According to the Chain Rule, we have $$\frac{dy}{d\theta}=\frac{dy}{du}\frac{du}{d\theta}$$ $$\frac{dy}{d\theta}=\frac{d(\cot^2u)}{du}\frac{d(\sin\theta)}{d\theta}$$ Now we realize that we don't know any available formula to differentiate $y=\cot^2 u$. But we know how to differentiate $y=\cot u$ and $y=x^2$. That's why we would use the Chain Rule one more time. Let $v=\cot u$, then $y=\cot^2 u=v^2$. According to the Chain Rule, we have $$\frac{dy}{d\theta}=\frac{dy}{dv}\frac{dv}{du}\frac{du}{d\theta}$$ $$\frac{dy}{d\theta}=\frac{d(v^2)}{dv}\frac{d(\cot u)}{du}\frac{d(\sin\theta)}{d\theta}$$ $$\frac{dy}{d\theta}=2v\times(-\csc^2u)\times(\cos\theta)$$ $$\frac{dy}{d\theta}=2(\cot u)\times(-\csc^2(\sin\theta))\times\cos\theta$$ $$\frac{dy}{d\theta}=2\cot(\sin\theta)\times(-\csc^2(\sin\theta))\times\cos\theta$$ $$\frac{dy}{d\theta}=-2\cot(\sin\theta)\csc^2(\sin\theta)\cos\theta$$