## Calculus: Early Transcendentals 8th Edition

$f'(\theta)=\dfrac{1}{1+\cos\theta}$
$f(\theta)=\dfrac{\sin\theta}{1+\cos\theta}$ Differentiate using the product rule: $f'(\theta)=\dfrac{(1+\cos\theta)(\sin\theta)'-(\sin\theta)(1+\cos\theta)'}{(1+\cos\theta)^{2}}=...$ $...=\dfrac{(1+\cos\theta)(\cos\theta)-(\sin\theta)(-\sin\theta)}{(1+\cos\theta)^{2}}=...$ Simplify: $...=\dfrac{\cos\theta+\cos^{2}\theta+\sin^{2}\theta}{(1+\cos\theta)^{2}}=...$ We know that $\sin^{2}\theta+\cos^{2}\theta=1$, so we get the following expression: $...=\dfrac{1+\cos\theta}{(1+\cos\theta)^{2}}$ So our final answer is: $f'(\theta)=\dfrac{1}{1+\cos\theta}$