1. **State the problem**: Given the parametric equations $$ x = \sqrt{3t - 2} $$ and $$ y = \sqrt{4t + 1} $$, find the derivative $$ \frac{dy}{dx} $$ at $$ t = 2 $$. 2. **Differentiate each function with respect to $$ t $$**: For $$ x $$, use the chain rule: $$ x = (3t - 2)^{1/2} $$ $$ \frac{dx}{dt} = \frac{1}{2} (3t - 2)^{-1/2} \times 3 = \frac{3}{2\sqrt{3t - 2}} $$ For $$ y $$: $$ y = (4t + 1)^{1/2} $$ $$ \frac{dy}{dt} = \frac{1}{2} (4t + 1)^{-1/2} \times 4 = \frac{2}{\sqrt{4t + 1}} $$ 3. **Apply the formula for $$ \frac{dy}{dx} $$:** $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2}{\sqrt{4t + 1}}}{\frac{3}{2\sqrt{3t - 2}}} = \frac{2}{\sqrt{4t + 1}} \times \frac{2\sqrt{3t - 2}}{3} = \frac{4\sqrt{3t - 2}}{3\sqrt{4t + 1}} $$ 4. **Evaluate at $$ t = 2 $$:** Calculate radicands: $$ 3(2) - 2 = 6 - 2 = 4, \quad 4(2) + 1 = 8 + 1 = 9 $$ Substitute: $$ \frac{dy}{dx}\bigg|_{t=2} = \frac{4 \sqrt{4}}{3 \sqrt{9}} = \frac{4 \times 2}{3 \times 3} = \frac{8}{9} $$ 5. **Final answer:** $$ \boxed{\frac{8}{9}} $$, which corresponds to option D.