1. We are given the function $$f(x) = \left(\frac{\sin x}{1 + \cos x}\right)^2$$ and need to find its derivative. 2. Start by using the chain rule: $$f'(x) = 2 \cdot \frac{\sin x}{1 + \cos x} \cdot \frac{d}{dx} \left( \frac{\sin x}{1 + \cos x} \right)$$ 3. Use the quotient rule to differentiate $$\frac{\sin x}{1 + \cos x}$$: $$\frac{d}{dx} \left( \frac{\sin x}{1 + \cos x} \right) = \frac{(\cos x)(1 + \cos x) - (\sin x)(-\sin x)}{(1 + \cos x)^2}$$ 4. Simplify the numerator: $$\cos x (1 + \cos x) + \sin^2 x = \cos x + \cos^2 x + \sin^2 x = \cos x + 1$$ because $$\sin^2 x + \cos^2 x = 1$$ 5. Thus, we have: $$\frac{d}{dx} \left( \frac{\sin x}{1 + \cos x} \right) = \frac{\cos x + 1}{(1 + \cos x)^2} = \frac{1 + \cos x}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$$ 6. Substitute back into the derivative expression: $$f'(x) = 2 \cdot \frac{\sin x}{1 + \cos x} \cdot \frac{1}{1 + \cos x} = 2 \cdot \frac{\sin x}{(1 + \cos x)^2}$$ Final answer: $$f'(x) = \frac{2 \sin x}{(1 + \cos x)^2}$$