1. **State the problem:** We are given the implicit equation $$xy^2 - 2xy + x^2y = 12$$ and need to find the slope of the tangent line at the point $$(1,4)$$. 2. **Recall the formula:** The slope of the tangent line to an implicitly defined curve is given by $$\frac{dy}{dx}$$, which we find by implicit differentiation. 3. **Differentiate both sides with respect to $x$: ** - Differentiate $$xy^2$$ using the product rule: $$\frac{d}{dx}(x y^2) = y^2 + x \cdot 2y \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx}$$. - Differentiate $$-2xy$$: $$\frac{d}{dx}(-2xy) = -2y - 2x \frac{dy}{dx}$$. - Differentiate $$x^2 y$$: $$\frac{d}{dx}(x^2 y) = 2x y + x^2 \frac{dy}{dx}$$. - The right side derivative of constant 12 is 0. 4. **Write the differentiated equation:** $$y^2 + 2xy \frac{dy}{dx} - 2y - 2x \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$$ 5. **Group terms with $$\frac{dy}{dx}$$ and without:** $$\left(2xy - 2x + x^2\right) \frac{dy}{dx} + (y^2 - 2y + 2xy) = 0$$ 6. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = - \frac{y^2 - 2y + 2xy}{2xy - 2x + x^2}$$ 7. **Substitute the point $$(x,y) = (1,4)$$:** - Numerator: $$4^2 - 2 \cdot 4 + 2 \cdot 1 \cdot 4 = 16 - 8 + 8 = 16$$ - Denominator: $$2 \cdot 1 \cdot 4 - 2 \cdot 1 + 1^2 = 8 - 2 + 1 = 7$$ 8. **Calculate the slope:** $$\frac{dy}{dx} = - \frac{16}{7} = -\frac{16}{7}$$ **Final answer:** The slope of the tangent line at $$(1,4)$$ is $$-\frac{16}{7}$$.