1. **Problem Statement:** Find the second derivative of the function $$f(\theta) = \frac{1}{3 + 2\cos\theta}$$ with respect to $$\theta$$. 2. **Recall the formula:** To find the second derivative, we first find the first derivative $$f'(\theta)$$ and then differentiate it again to get $$f''(\theta)$$. 3. **First derivative:** Use the quotient rule or rewrite $$f(\theta)$$ as $$f(\theta) = (3 + 2\cos\theta)^{-1}$$ and apply the chain rule. $$f'(\theta) = -1 \times (3 + 2\cos\theta)^{-2} \times (-2\sin\theta) = \frac{2\sin\theta}{(3 + 2\cos\theta)^2}$$ 4. **Second derivative:** Differentiate $$f'(\theta)$$ using the quotient and product rules. Let $$u = 2\sin\theta$$ and $$v = (3 + 2\cos\theta)^2$$. Then, $$u' = 2\cos\theta$$ $$v' = 2 \times (3 + 2\cos\theta) \times (-2\sin\theta) = -4\sin\theta (3 + 2\cos\theta)$$ Using the quotient rule: $$f''(\theta) = \frac{u'v - uv'}{v^2} = \frac{2\cos\theta (3 + 2\cos\theta)^2 - 2\sin\theta (-4\sin\theta (3 + 2\cos\theta))}{(3 + 2\cos\theta)^4}$$ Simplify numerator: $$2\cos\theta (3 + 2\cos\theta)^2 + 8\sin^2\theta (3 + 2\cos\theta)$$ Factor out $$2(3 + 2\cos\theta)$$: $$2(3 + 2\cos\theta) \left[ \cos\theta (3 + 2\cos\theta) + 4\sin^2\theta \right]$$ 5. **Final expression:** $$f''(\theta) = \frac{2(3 + 2\cos\theta) \left[ \cos\theta (3 + 2\cos\theta) + 4\sin^2\theta \right]}{(3 + 2\cos\theta)^4} = \frac{2 \left[ \cos\theta (3 + 2\cos\theta) + 4\sin^2\theta \right]}{(3 + 2\cos\theta)^3}$$ This is the second derivative of $$f(\theta)$$. **Summary:** $$f''(\theta) = \frac{2 \left[ \cos\theta (3 + 2\cos\theta) + 4\sin^2\theta \right]}{(3 + 2\cos\theta)^3}$$