Tangent Line 55C3F3

1. **State the problem:** Find the equation of the tangent line to the curve given by $$\frac{1}{3} y^3 + 2xy - 100 = 0$$ at the point $(-2, 2)$. 2. **Implicit differentiation:** Differentiate both sides with respect to $x$. Treat $y$ as a function of $x$ (i.e., $y = y(x)$). $$\frac{d}{dx}\left(\frac{1}{3} y^3\right) + \frac{d}{dx}(2xy) - \frac{d}{dx}(100) = 0$$ Using the chain and product rules: $$\frac{1}{3} \cdot 3y^2 \frac{dy}{dx} + 2 \left( y + x \frac{dy}{dx} \right) = 0$$ Simplify: $$y^2 \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} = 0$$ 3. **Group terms with $\frac{dy}{dx}$:** $$\left(y^2 + 2x\right) \frac{dy}{dx} = -2y$$ 4. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{-2y}{y^2 + 2x}$$ 5. **Evaluate the derivative at the point $(-2, 2)$:** $$\frac{dy}{dx}\bigg|_{(-2,2)} = \frac{-2 \cdot 2}{2^2 + 2 \cdot (-2)} = \frac{-4}{4 - 4} = \frac{-4}{0}$$ Since the denominator is zero, the slope is undefined, indicating a vertical tangent line at $x = -2$. 6. **Equation of the tangent line:** Since the tangent is vertical at $x = -2$, the equation is $$x = -2$$ **Final answer:** The tangent line at $(-2, 2)$ is $$x = -2$$.