1. **State the problem:** Differentiate the function $$y=\frac{x^3}{4-5x}$$ with respect to $$x$$. 2. **Recall the quotient rule:** For $$y=\frac{u}{v}$$, the derivative is $$y' = \frac{u'v - uv'}{v^2}$$. 3. **Identify $$u$$ and $$v$$:** Here, $$u = x^3$$ and $$v = 4 - 5x$$. 4. **Compute derivatives:** $$u' = 3x^2$$ and $$v' = -5$$. 5. **Apply the quotient rule:** $$ y' = \frac{3x^2(4 - 5x) - x^3(-5)}{(4 - 5x)^2} $$ 6. **Simplify numerator:** $$ 3x^2(4 - 5x) + 5x^3 = 12x^2 - 15x^3 + 5x^3 = 12x^2 - 10x^3 $$ 7. **Final derivative:** $$ y' = \frac{12x^2 - 10x^3}{(4 - 5x)^2} $$ This matches the answer given, confirming the derivative is $$\frac{2x^2(6 - 5x)}{(4 - 5x)^2}$$ after factoring out $$2x^2$$. **Answer:** $$y' = \frac{2x^2(6 - 5x)}{(4 - 5x)^2}$$