1. **Problem statement:** Find the equation(s) of the tangent(s) to the curve $$x^2 y^2 + y^2 + x^3 = 5$$ at points where $$x=1$$ using implicit differentiation. 2. **Formula and rules:** To find the slope of the tangent line, we differentiate both sides of the equation implicitly with respect to $$x$$, treating $$y$$ as a function of $$x$$ (i.e., $$y = y(x)$$). Use the product rule and chain rule: $$\frac{d}{dx}[x^2 y^2] + \frac{d}{dx}[y^2] + \frac{d}{dx}[x^3] = \frac{d}{dx}[5]$$ 3. **Differentiate each term:** - For $$x^2 y^2$$, use product rule: $$\frac{d}{dx}[x^2 y^2] = 2x y^2 + x^2 \cdot 2y \frac{dy}{dx} = 2x y^2 + 2x^2 y \frac{dy}{dx}$$ - For $$y^2$$, use chain rule: $$\frac{d}{dx}[y^2] = 2y \frac{dy}{dx}$$ - For $$x^3$$: $$\frac{d}{dx}[x^3] = 3x^2$$ - Right side derivative is zero: $$\frac{d}{dx}[5] = 0$$ 4. **Combine derivatives:** $$2x y^2 + 2x^2 y \frac{dy}{dx} + 2y \frac{dy}{dx} + 3x^2 = 0$$ 5. **Group terms with $$\frac{dy}{dx}$$:** $$2x^2 y \frac{dy}{dx} + 2y \frac{dy}{dx} = -2x y^2 - 3x^2$$ 6. **Factor out $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} (2x^2 y + 2y) = -2x y^2 - 3x^2$$ 7. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{-2x y^2 - 3x^2}{2y (x^2 + 1)}$$ 8. **Find $$y$$ values at $$x=1$$:** Substitute $$x=1$$ into original equation: $$1^2 y^2 + y^2 + 1^3 = 5 \implies y^2 + y^2 + 1 = 5 \implies 2y^2 = 4 \implies y^2 = 2 \implies y = \pm \sqrt{2}$$ 9. **Calculate slopes at points:** - At $$(1, \sqrt{2})$$: $$\frac{dy}{dx} = \frac{-2(1)(2) - 3(1)^2}{2(\sqrt{2})(1 + 1)} = \frac{-4 - 3}{2 \sqrt{2} \cdot 2} = \frac{-7}{4 \sqrt{2}} = -\frac{7 \sqrt{2}}{8}$$ - At $$(1, -\sqrt{2})$$: $$\frac{dy}{dx} = \frac{-4 - 3}{2(-\sqrt{2})(2)} = \frac{-7}{-4 \sqrt{2}} = \frac{7 \sqrt{2}}{8}$$ 10. **Write tangent line equations:** Use point-slope form $$y - y_1 = m(x - x_1)$$: - For $$(1, \sqrt{2})$$: $$y - \sqrt{2} = -\frac{7 \sqrt{2}}{8} (x - 1)$$ - For $$(1, -\sqrt{2})$$: $$y + \sqrt{2} = \frac{7 \sqrt{2}}{8} (x - 1)$$ **Final answer:** $$\boxed{\begin{cases} y - \sqrt{2} = -\frac{7 \sqrt{2}}{8} (x - 1) \\ y + \sqrt{2} = \frac{7 \sqrt{2}}{8} (x - 1) \end{cases}}$$