1. We are asked to find the first and second derivatives of the function $$f(x) = \frac{1}{2 + \sin x}$$. 2. The first derivative of a function of the form $$\frac{1}{g(x)}$$ is given by the chain rule and quotient rule as: $$f'(x) = -\frac{g'(x)}{(g(x))^2}$$ where $$g(x) = 2 + \sin x$$. 3. Calculate $$g'(x)$$: $$g'(x) = \cos x$$ 4. Substitute into the formula for $$f'(x)$$: $$f'(x) = -\frac{\cos x}{(2 + \sin x)^2}$$ 5. Now, find the second derivative $$f''(x)$$ by differentiating $$f'(x)$$: $$f''(x) = \frac{d}{dx} \left(-\frac{\cos x}{(2 + \sin x)^2}\right)$$ 6. Use the quotient rule or product rule with chain rule. Let: $$u = -\cos x$$ and $$v = (2 + \sin x)^{-2}$$ 7. Differentiate $$u$$ and $$v$$: $$u' = \sin x$$ $$v' = -2(2 + \sin x)^{-3} \cdot \cos x$$ (by chain rule) 8. Apply product rule: $$f''(x) = u'v + uv' = \sin x (2 + \sin x)^{-2} + (-\cos x) \left(-2(2 + \sin x)^{-3} \cos x\right)$$ 9. Simplify: $$f''(x) = \frac{\sin x}{(2 + \sin x)^2} + \frac{2 \cos^2 x}{(2 + \sin x)^3}$$ 10. This is the second derivative of the function. Final answers: $$f'(x) = -\frac{\cos x}{(2 + \sin x)^2}$$ $$f''(x) = \frac{\sin x}{(2 + \sin x)^2} + \frac{2 \cos^2 x}{(2 + \sin x)^3}$$