1. **State the problem:** Find the derivative of the function $$y=\frac{2x^3 + x}{3x - 1}$$. 2. **Recall the formula:** For a function $$y=\frac{u}{v}$$, the derivative is given by the quotient rule: $$y' = \frac{v u' - u v'}{v^2}$$ where $$u = 2x^3 + x$$ and $$v = 3x - 1$$. 3. **Calculate the derivatives of numerator and denominator:** - $$u' = \frac{d}{dx}(2x^3 + x) = 6x^2 + 1$$ - $$v' = \frac{d}{dx}(3x - 1) = 3$$ 4. **Apply the quotient rule:** $$y' = \frac{(3x - 1)(6x^2 + 1) - (2x^3 + x)(3)}{(3x - 1)^2}$$ 5. **Expand the numerator:** - First term: $$(3x)(6x^2) = 18x^3$$ - $$(3x)(1) = 3x$$ - $$(-1)(6x^2) = -6x^2$$ - $$(-1)(1) = -1$$ So, $$(3x - 1)(6x^2 + 1) = 18x^3 + 3x - 6x^2 - 1$$ - Second term: $$(2x^3 + x)(3) = 6x^3 + 3x$$ 6. **Subtract the second term from the first:** $$18x^3 + 3x - 6x^2 - 1 - (6x^3 + 3x) = (18x^3 - 6x^3) + (3x - 3x) - 6x^2 - 1 = 12x^3 - 6x^2 - 1$$ 7. **Write the final derivative:** $$y' = \frac{12x^3 - 6x^2 - 1}{(3x - 1)^2}$$ **Note:** The derivative you proposed, $$\frac{12x^3 + 6x^2 -1}{(3x -1)^2}$$, has incorrect signs in the numerator. The correct signs are as shown above.