Answer
$f'(\theta)=\dfrac{1}{1+\cos\theta}$
Work Step by Step
$f(\theta)=\dfrac{\sin\theta}{1+\cos\theta}$
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:
$f'(\theta)=\dfrac{1}{1+\cos\theta}$
