1. Problem 1: Differentiate the function $y = \frac{3}{\sqrt{5x - 2}}$ using the Chain Rule. 2. Rewrite the function in exponent form: $$y = 3(5x - 2)^{-\frac{1}{2}}$$. 3. Apply the Chain Rule: differentiate the outer function and multiply by the derivative of the inner function. 4. The derivative of the outer function is $$-\frac{1}{2} \cdot 3 (5x - 2)^{-\frac{3}{2}} = -\frac{3}{2} (5x - 2)^{-\frac{3}{2}}$$. 5. The derivative of the inner function $5x - 2$ is 5. 6. Multiply to get the derivative: $$\frac{dy}{dx} = -\frac{3}{2} (5x - 2)^{-\frac{3}{2}} \times 5 = -\frac{15}{2} (5x - 2)^{-\frac{3}{2}}$$. --- 7. Problem 2: Solve the integral $$\int \left(2x + \sqrt{x^3(6 - x^2)}\right) dx$$. 8. Simplify the integrand inside the square root: $$\sqrt{x^3(6 - x^2)} = \sqrt{x^3} \cdot \sqrt{6 - x^2} = x^{\frac{3}{2}} (6 - x^2)^{\frac{1}{2}}$$. 9. The integral becomes $$\int 2x \, dx + \int x^{\frac{3}{2}} (6 - x^2)^{\frac{1}{2}} \, dx$$. 10. First integral is straightforward: $$\int 2x \, dx = x^2 + C_1$$. 11. For the second integral $$\int x^{\frac{3}{2}} (6 - x^2)^{\frac{1}{2}} \, dx$$, use substitution: Let $$t = x^2$$ so that $$dt = 2x \, dx$$ or $$x \, dx = \frac{dt}{2}$$. 12. Rewrite powers: $$x^{\frac{3}{2}} = x^{1} \cdot x^{\frac{1}{2}}$$. 13. Substitute into integral: $$\int x^{\frac{3}{2}} (6 - x^2)^{\frac{1}{2}} dx = \int x \cdot x^{\frac{1}{2}} (6 - t)^{\frac{1}{2}} dx$$. 14. Note that $$x \, dx = \frac{dt}{2}$$, so: $$\int x^{\frac{3}{2}} (6 - x^2)^{\frac{1}{2}} dx = \int x^{\frac{1}{2}} (6 - t)^{\frac{1}{2}} \frac{dt}{2}$$. 15. Express $$x^{\frac{1}{2}} = (t)^{\frac{1}{4}}$$ because $$x = \sqrt{t} = t^{\frac{1}{2}}$$ so \($x^{\frac{1}{2}} = t^{\frac{1}{4}}$). 16. Integral becomes: $$\frac{1}{2} \int t^{\frac{1}{4}} (6 - t)^{\frac{1}{2}} dt$$. 17. This integral is non-elementary in simple terms but can be evaluated by substitution or expressed in terms of Beta or hypergeometric functions. 18. For the scope here, final answer will be given as: $$\int \left(2x + \sqrt{x^3(6 - x^2)}\right) dx = x^2 + \frac{1}{2} \int t^{\frac{1}{4}} (6 - t)^{\frac{1}{2}} dt + C$$ where $t = x^2$. --- Final Answers: $$\frac{dy}{dx} = -\frac{15}{2} (5x - 2)^{-\frac{3}{2}}$$ $$\int \left(2x + \sqrt{x^3(6 - x^2)}\right) dx = x^2 + \frac{1}{2} \int t^{\frac{1}{4}} (6 - t)^{\frac{1}{2}} dt + C, \quad t = x^2$$