1. **State the problem:** Given the implicit equation $$x + xy - 2x^3 = 2$$, we need to find \(\frac{dy}{dx}\) by implicit differentiation (part a), solve for \(y\) as a function of \(x\) and find \(\frac{dy}{dx}\) explicitly (part b), and confirm the consistency of both derivatives by expressing the implicit derivative as a function of \(x\) alone (part c). 2. **Part (a): Differentiate implicitly** - Differentiate both sides with respect to \(x\): $$\frac{d}{dx}(x) + \frac{d}{dx}(xy) - \frac{d}{dx}(2x^3) = \frac{d}{dx}(2)$$ - Using the product rule on \(xy\): $$1 + \left(x \frac{dy}{dx} + y \right) - 6x^2 = 0$$ - Rearranged: $$1 + x \frac{dy}{dx} + y - 6x^2 = 0$$ - Solve for \(\frac{dy}{dx}\): $$x \frac{dy}{dx} = 6x^2 - y - 1$$ $$\frac{dy}{dx} = \frac{6x^2 - y - 1}{x}$$ 3. **Part (b): Solve for \(y\) explicitly and find \(\frac{dy}{dx}\)** - From the original equation: $$x + xy - 2x^3 = 2$$ - Isolate \(y\): $$xy = 2 + 2x^3 - x$$ $$y = \frac{2 + 2x^3 - x}{x} = \frac{2}{x} + 2x^2 - 1$$ - Differentiate \(y\) with respect to \(x\): $$\frac{dy}{dx} = -\frac{2}{x^2} + 4x$$ 4. **Part (c): Confirm consistency by expressing implicit derivative as function of \(x\) only** - Substitute \(y = \frac{2}{x} + 2x^2 - 1\) into the implicit derivative: $$\frac{dy}{dx} = \frac{6x^2 - \left(\frac{2}{x} + 2x^2 - 1\right) - 1}{x}$$ - Simplify numerator: $$6x^2 - \frac{2}{x} - 2x^2 + 1 - 1 = 4x^2 - \frac{2}{x}$$ - So: $$\frac{dy}{dx} = \frac{4x^2 - \frac{2}{x}}{x} = \frac{4x^2}{x} - \frac{2}{x \cdot x} = 4x - \frac{2}{x^2}$$ - This matches the explicit derivative found in part (b). **Final answer:** $$\frac{dy}{dx} = 4x - \frac{2}{x^2}$$