1. **Problem statement:** Find the derivative $\frac{dy}{dx}$ of the function $$y = \frac{x + 2}{\sqrt{3x + 4}}$$. 2. **Formula and rules:** We use the quotient rule for derivatives: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where $u = x + 2$ and $v = \sqrt{3x + 4} = (3x + 4)^{1/2}$. 3. **Calculate derivatives of numerator and denominator:** $$\frac{du}{dx} = 1$$ $$\frac{dv}{dx} = \frac{1}{2} (3x + 4)^{-1/2} \cdot 3 = \frac{3}{2 \sqrt{3x + 4}}$$ 4. **Apply quotient rule:** $$\frac{dy}{dx} = \frac{\sqrt{3x + 4} \cdot 1 - (x + 2) \cdot \frac{3}{2 \sqrt{3x + 4}}}{(\sqrt{3x + 4})^2}$$ 5. **Simplify denominator:** $$(\sqrt{3x + 4})^2 = 3x + 4$$ 6. **Rewrite numerator with common denominator:** $$\frac{2(3x + 4)}{2 \sqrt{3x + 4}} - \frac{3(x + 2)}{2 \sqrt{3x + 4}} = \frac{2(3x + 4) - 3(x + 2)}{2 \sqrt{3x + 4}}$$ 7. **Simplify numerator expression:** $$2(3x + 4) - 3(x + 2) = 6x + 8 - 3x - 6 = 3x + 2$$ 8. **So numerator is:** $$\frac{3x + 2}{2 \sqrt{3x + 4}}$$ 9. **Putting it all together:** $$\frac{dy}{dx} = \frac{\frac{3x + 2}{2 \sqrt{3x + 4}}}{3x + 4} = \frac{3x + 2}{2 \sqrt{3x + 4} (3x + 4)}$$ 10. **Final derivative:** $$\boxed{\frac{dy}{dx} = \frac{3x + 2}{2 (3x + 4)^{3/2}}}$$ --- Since the second question is not solved here due to instructions, the count of distinct questions is 2.