1. **State the problem:** Find the derivative of the function $$y(x) = \frac{3x^2 - 2x + 2}{-5x^2 + 3x - 1}$$. 2. **Recall the formula:** For a function $$y = \frac{u(x)}{v(x)}$$, the derivative is given by the quotient rule: $$y' = \frac{u'v - uv'}{v^2}$$ where $$u = 3x^2 - 2x + 2$$ and $$v = -5x^2 + 3x - 1$$. 3. **Calculate derivatives of numerator and denominator:** $$u' = \frac{d}{dx}(3x^2 - 2x + 2) = 6x - 2$$ $$v' = \frac{d}{dx}(-5x^2 + 3x - 1) = -10x + 3$$ 4. **Apply the quotient rule:** $$y' = \frac{(6x - 2)(-5x^2 + 3x - 1) - (3x^2 - 2x + 2)(-10x + 3)}{(-5x^2 + 3x - 1)^2}$$ 5. **Expand the terms in the numerator:** First term: $$(6x - 2)(-5x^2 + 3x - 1) = 6x \cdot (-5x^2) + 6x \cdot 3x + 6x \cdot (-1) - 2 \cdot (-5x^2) - 2 \cdot 3x - 2 \cdot (-1)$$ $$= -30x^3 + 18x^2 - 6x + 10x^2 - 6x + 2 = -30x^3 + 28x^2 - 12x + 2$$ Second term: $$(3x^2 - 2x + 2)(-10x + 3) = 3x^2 \cdot (-10x) + 3x^2 \cdot 3 - 2x \cdot (-10x) - 2x \cdot 3 + 2 \cdot (-10x) + 2 \cdot 3$$ $$= -30x^3 + 9x^2 + 20x^2 - 6x - 20x + 6 = -30x^3 + 29x^2 - 26x + 6$$ 6. **Subtract the second term from the first:** $$(-30x^3 + 28x^2 - 12x + 2) - (-30x^3 + 29x^2 - 26x + 6)$$ $$= -30x^3 + 28x^2 - 12x + 2 + 30x^3 - 29x^2 + 26x - 6$$ $$= (\cancel{-30x^3} + \cancel{30x^3}) + (28x^2 - 29x^2) + (-12x + 26x) + (2 - 6)$$ $$= -x^2 + 14x - 4$$ 7. **Write the final derivative:** $$y'(x) = \frac{-x^2 + 14x - 4}{(-5x^2 + 3x - 1)^2}$$ **Answer:** $$y'(x) = \frac{-x^2 + 14x - 4}{(-5x^2 + 3x - 1)^2}$$