## University Calculus: Early Transcendentals (3rd Edition)

$$f'(\theta)=\frac{2\sin\theta}{(1+\cos\theta)^2}$$
$$f(\theta)=\Big(\frac{\sin\theta}{1+\cos\theta}\Big)^2$$ According to the Chain Rule, we have $$f'(\theta)=2\Big(\frac{\sin\theta}{1+\cos\theta}\Big)\Big(\frac{\sin\theta}{1+\cos\theta}\Big)'$$ - Calculate $\Big(\frac{\sin\theta}{1+\cos\theta}\Big)'$: $$\Big(\frac{\sin\theta}{1+\cos\theta}\Big)'=\frac{(\sin\theta)'(1+\cos\theta)-\sin\theta(1+\cos\theta)'}{(1+\cos\theta)^2}$$ $$=\frac{\cos\theta(1+\cos\theta)-\sin\theta(-\sin\theta)}{(1+\cos\theta)^2}=\frac{\cos\theta+\cos^2\theta+\sin^2\theta}{(1+\cos\theta)^2}$$ $$=\frac{\cos\theta+1}{(1+\cos\theta)^2}=\frac{1}{1+\cos\theta}$$ Therefore, $$f'(\theta)=2\Big(\frac{\sin\theta}{1+\cos\theta}\Big)\frac{1}{1+\cos\theta}$$ $$f'(\theta)=\frac{2\sin\theta}{(1+\cos\theta)^2}$$