1. **Stating the problem:** Given the implicit equation $$xy + 2x + 3x^2 = -8$$ we need to: (a) Find $y'$ (the derivative of $y$ with respect to $x$) implicitly. (b) Solve for $y$ explicitly and differentiate to find $y'$. 2. **Part (a) - Implicit differentiation:** Differentiate both sides with respect to $x$: $$\frac{d}{dx}(xy) + \frac{d}{dx}(2x) + \frac{d}{dx}(3x^2) = \frac{d}{dx}(-8)$$ Using product rule on $xy$: $$\frac{d}{dx}(xy) = y + x \frac{dy}{dx} = y + x y'$$ Also, $$\frac{d}{dx}(2x) = 2, \quad \frac{d}{dx}(3x^2) = 6x, \quad \frac{d}{dx}(-8) = 0$$ So the differentiated equation is: $$y + xy' + 2 + 6x = 0$$ Isolate $y'$: $$xy' = -y - 2 - 6x$$ $$y' = \frac{-y - 2 - 6x}{x}$$ This is $y'$ implicitly in terms of $x$ and $y$. 3. **Part (b) - Solve explicitly and differentiate:** Rewrite original equation: $$xy + 2x + 3x^2 = -8$$ $$xy = -8 - 2x -3x^2$$ Solve for $y$: $$y = \frac{-8 - 2x - 3x^2}{x} = \frac{-8}{x} - 2 - 3x$$ Differentiate term-by-term: $$y' = \frac{d}{dx}\left(-8x^{-1} - 2 - 3x\right)$$ $$y' = 8x^{-2} - 0 - 3 = \frac{8}{x^2} - 3$$ 4. **Consistency of results:** From (a): $$y' = \frac{-y - 2 - 6x}{x}$$ Substitute $y = \frac{-8}{x} - 2 - 3x$: $$y' = \frac{-\left(\frac{-8}{x} - 2 - 3x\right) - 2 - 6x}{x} = \frac{\frac{8}{x} + 2 + 3x - 2 - 6x}{x} = \frac{\frac{8}{x} - 3x}{x} = \frac{8}{x^2} - 3$$ This matches the explicit derivative exactly. **Final answers:** (a) $$y' = \frac{-y - 2 - 6x}{x}$$ (b) $$y' = \frac{8}{x^2} - 3$$