1. The problem is to differentiate the function $$f(x) = \frac{3x^2 - x}{\sqrt{1-2x}}$$. 2. We use the quotient rule for derivatives: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $$u = 3x^2 - x$$ and $$v = \sqrt{1-2x}$$. 3. Compute the derivatives: - $$u' = \frac{d}{dx}(3x^2 - x) = 6x - 1$$ - $$v = (1-2x)^{1/2}$$ so $$v' = \frac{1}{2}(1-2x)^{-1/2} \cdot (-2) = -\frac{1}{\sqrt{1-2x}}$$ 4. Apply the quotient rule: $$f'(x) = \frac{(6x - 1)\sqrt{1-2x} - (3x^2 - x)\left(-\frac{1}{\sqrt{1-2x}}\right)}{(\sqrt{1-2x})^2}$$ 5. Simplify the denominator: $$(\sqrt{1-2x})^2 = 1-2x$$ 6. Rewrite numerator: $$ (6x - 1)\sqrt{1-2x} + \frac{3x^2 - x}{\sqrt{1-2x}} $$ 7. To combine terms, write both over common denominator $$\sqrt{1-2x}$$: $$ \frac{(6x - 1)(1-2x) + (3x^2 - x)}{\sqrt{1-2x}} $$ 8. Expand numerator: $$ (6x - 1)(1-2x) = 6x - 12x^2 - 1 + 2x = -12x^2 + 8x - 1 $$ 9. Add $$3x^2 - x$$: $$ -12x^2 + 8x - 1 + 3x^2 - x = -9x^2 + 7x - 1 $$ 10. So numerator is: $$ \frac{-9x^2 + 7x - 1}{\sqrt{1-2x}} $$ 11. Therefore, $$ f'(x) = \frac{-9x^2 + 7x - 1}{(1-2x)^{3/2}} $$ This is the derivative of the given function.