1. **State the problem:** Differentiate the curve given by $$y = \frac{(3x^2 - 5)^{\frac{1}{3}}}{x+4}$$ and express the derivative \(\frac{dy}{dx}\) in the form $$\frac{Ax^2 + Bx + C}{(3x^2 - 5)^{\frac{2}{3}} (x+4)^2}$$ where \(A\), \(B\), and \(C\) are integers. 2. **Rewrite y:** $$y = (3x^2 - 5)^{\frac{1}{3}} (x+4)^{-1}$$ so that the product and chain rule can be applied easily. 3. **Differentiate y using the product rule:** Let $$u = (3x^2 - 5)^{\frac{1}{3}}, \quad v = (x+4)^{-1}$$ Then, $$\frac{dy}{dx} = u'v + uv'$$ 4. **Compute \(u'\):** By the chain rule, $$u' = \frac{1}{3}(3x^2 - 5)^{-\frac{2}{3}} \cdot (6x) = 2x (3x^2 - 5)^{-\frac{2}{3}}$$ 5. **Compute \(v'\):** $$v' = -1 (x+4)^{-2} = - (x+4)^{-2}$$ 6. **Substitute back:** $$\frac{dy}{dx} = 2x (3x^2 - 5)^{-\frac{2}{3}} (x+4)^{-1} - (3x^2 - 5)^{\frac{1}{3}} (x+4)^{-2}$$ 7. **Find common denominator:** The derivative can be written as $$\frac{dy}{dx} = \frac{2x}{(3x^2 - 5)^{\frac{2}{3}} (x+4)} - \frac{(3x^2 - 5)^{\frac{1}{3}}}{(x+4)^2}$$ Multiplying top and bottom to get common denominator $$(3x^2 - 5)^{\frac{2}{3}} (x+4)^2$$ 8. **Rewrite second term with the denominator:** Note that $$(3x^2 - 5)^{\frac{1}{3}} = (3x^2 - 5)^{\frac{2}{3}} (3x^2 - 5)^{-\frac{1}{3}}$$ but it is simpler to multiply numerator and denominator appropriately: Multiply the first term by \((x+4)\) and the second term by \((3x^2 - 5)^{\frac{1}{3}}\) to write: $$\frac{dy}{dx} = \frac{2x (x+4) - (3x^2 - 5)^{1/3} (3x^2 - 5)^{2/3}}{(3x^2 - 5)^{2/3} (x+4)^2}$$ Simplify numerator: $$2x (x+4) - (3x^2 - 5) = 2x^2 + 8x - 3x^2 + 5 = -x^2 + 8x + 5$$ 9. **Final expression:** $$\frac{dy}{dx} = \frac{-x^2 + 8x + 5}{(3x^2 - 5)^{\frac{2}{3}} (x+4)^2}$$ Hence, $$A = -1, \quad B = 8, \quad C = 5$$