1. **Stating the problem:** We are given the implicit equation $$y^2 + 3xy = 2x + 3$$ and asked to find the derivative $$\frac{dy}{dx}$$ and the equation of the tangent line at a specific point. 2. **Implicit differentiation:** Differentiate both sides with respect to $$x$$: $$\frac{d}{dx}(y^2 + 3xy) = \frac{d}{dx}(2x + 3)$$ Using the product rule and chain rule: $$2y y' + 3(y + x y') = 2$$ 3. **Group terms with $$y'$$:** $$2y y' + 3y + 3x y' = 2$$ $$y'(2y + 3x) = 2 - 3y$$ 4. **Solve for $$y'$$:** $$y' = \frac{2 - 3y}{2y + 3x}$$ 5. **Find slope at point $$(1,2)$$:** Substitute $$x=1$$ and $$y=2$$: $$m = \frac{2 - 3(2)}{2(2) + 3(1)} = \frac{2 - 6}{4 + 3} = \frac{-4}{7} = -\frac{4}{7}$$ 6. **Equation of tangent line:** Using point-slope form: $$y = mx + b$$ Substitute $$x=1$$, $$y=2$$, and $$m = -\frac{4}{7}$$: $$2 = -\frac{4}{7} \times 1 + b$$ $$b = 2 + \frac{4}{7} = \frac{14}{7} + \frac{4}{7} = \frac{18}{7}$$ So the tangent line is: $$y = -\frac{4}{7} x + \frac{18}{7}$$ 7. **Second problem: Analyze derivative $$y' = 3x^2 - 6x$$** 8. **Factor derivative:** $$y' = 3x(x - 2)$$ 9. **Find critical points:** Set $$y' = 0$$: $$3x(x - 2) = 0 \Rightarrow x=0 \text{ or } x=2$$ 10. **Sign analysis of $$y'$$:** - For $$x < 0$$, $$3x < 0$$ and $$x-2 < 0$$, so $$y' = (+)$$ (since negative times negative is positive, but 3 is positive, so negative * negative * positive = positive) - For $$0 < x < 2$$, $$x > 0$$ but $$x-2 < 0$$, so $$y' < 0$$ - For $$x > 2$$, both $$x > 0$$ and $$x-2 > 0$$, so $$y' > 0$$ 11. **Summary table:** $$\begin{array}{c|cccc} x & -\infty & 0 & 2 & +\infty \\ y' & + & 0 & 0 & + \\ \end{array}$$ 12. **Interpretation:** - The function is increasing on $$(-\infty, 0)$$ and $$(2, +\infty)$$. - The function is decreasing on $$(0, 2)$$. --- **Final answers:** - Derivative: $$y' = \frac{2 - 3y}{2y + 3x}$$ - Tangent line at $$(1,2)$$: $$y = -\frac{4}{7} x + \frac{18}{7}$$ - Derivative for second problem: $$y' = 3x(x - 2)$$ with critical points at $$x=0$$ and $$x=2$$ and sign analysis as above.