## Thomas' Calculus 13th Edition

$\dfrac{2\sin\theta}{(1+\cos\theta)^2}$
Apply the Chain Rule, we get $f'(\theta)=2(\dfrac{\sin\theta}{1+\cos\theta})(\dfrac{\sin\theta}{1+\cos\theta})'$ ...(1) and $(\dfrac{\sin\theta}{1+\cos\theta})'=\dfrac{(\sin\theta)'(1+\cos\theta)-\sin\theta(1+\cos\theta)'}{(1+\cos\theta)^2}$ or, $\dfrac{\cos\theta(1+\cos\theta)-\sin\theta(-\sin\theta)}{(1+\cos\theta)^2}=\dfrac{1}{1+\cos\theta}$ Now, equation (1) becomes: $f'(\theta)=2(\dfrac{\sin\theta}{1+\cos\theta})[\dfrac{1}{1+\cos\theta}]=\dfrac{2\sin\theta}{(1+\cos\theta)^2}$