Second Derivative Implicit 6645Cc

1. **State the problem:** We need to find the second derivative $$\frac{d^2y}{dx^2}$$ given the implicit equation $$\sqrt{y} + xy = 3$$. 2. **Rewrite the equation:** $$\sqrt{y} + xy = 3$$ can be written as $$y^{1/2} + xy = 3$$. 3. **Differentiate implicitly once:** Differentiate both sides with respect to $$x$$: $$\frac{d}{dx}(y^{1/2}) + \frac{d}{dx}(xy) = \frac{d}{dx}(3)$$ Using the chain and product rules: $$\frac{1}{2}y^{-1/2} \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$$ 4. **Group $$\frac{dy}{dx}$$ terms:** $$\left(\frac{1}{2} y^{-1/2} + x\right) \frac{dy}{dx} = -y$$ 5. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{-y}{\frac{1}{2} y^{-1/2} + x} = \frac{-y}{\frac{1}{2\sqrt{y}} + x}$$ 6. **Simplify denominator:** Multiply numerator and denominator by $$2\sqrt{y}$$: $$\frac{dy}{dx} = \frac{-y \cdot 2\sqrt{y}}{1 + 2x\sqrt{y}} = \frac{-2 y^{3/2}}{1 + 2x\sqrt{y}}$$ 7. **Differentiate $$\frac{dy}{dx}$$ to find $$\frac{d^2y}{dx^2}$$:** Use quotient rule: $$\frac{d^2y}{dx^2} = \frac{(1 + 2x\sqrt{y}) \frac{d}{dx}(-2 y^{3/2}) - (-2 y^{3/2}) \frac{d}{dx}(1 + 2x\sqrt{y})}{(1 + 2x\sqrt{y})^2}$$ 8. **Calculate derivatives inside numerator:** - $$\frac{d}{dx}(-2 y^{3/2}) = -2 \cdot \frac{3}{2} y^{1/2} \frac{dy}{dx} = -3 y^{1/2} \frac{dy}{dx}$$ - $$\frac{d}{dx}(1 + 2x\sqrt{y}) = 2 \sqrt{y} + 2x \cdot \frac{1}{2} y^{-1/2} \frac{dy}{dx} = 2 \sqrt{y} + x y^{-1/2} \frac{dy}{dx}$$ 9. **Substitute $$\frac{dy}{dx}$$ from step 6:** $$\frac{dy}{dx} = \frac{-2 y^{3/2}}{1 + 2x\sqrt{y}}$$ 10. **Plug in and simplify numerator:** $$N = (1 + 2x\sqrt{y})(-3 y^{1/2} \frac{dy}{dx}) + 2 y^{3/2} (2 \sqrt{y} + x y^{-1/2} \frac{dy}{dx})$$ Replace $$\frac{dy}{dx}$$: $$N = (1 + 2x\sqrt{y})(-3 y^{1/2} \cdot \frac{-2 y^{3/2}}{1 + 2x\sqrt{y}}) + 2 y^{3/2} \left(2 \sqrt{y} + x y^{-1/2} \cdot \frac{-2 y^{3/2}}{1 + 2x\sqrt{y}}\right)$$ Simplify first term: $$= (1 + 2x\sqrt{y}) \cdot \frac{6 y^2}{1 + 2x\sqrt{y}} = 6 y^2$$ Simplify second term inside parentheses: $$2 \sqrt{y} + x y^{-1/2} \cdot \frac{-2 y^{3/2}}{1 + 2x\sqrt{y}} = 2 \sqrt{y} - \frac{2 x y}{1 + 2x\sqrt{y}}$$ Multiply by $$2 y^{3/2}$$: $$2 y^{3/2} \cdot 2 \sqrt{y} = 4 y^2$$ and $$2 y^{3/2} \cdot \left(- \frac{2 x y}{1 + 2x\sqrt{y}}\right) = - \frac{4 x y^{5/2}}{1 + 2x\sqrt{y}}$$ So second term total: $$4 y^2 - \frac{4 x y^{5/2}}{1 + 2x\sqrt{y}}$$ 11. **Combine numerator terms:** $$N = 6 y^2 + 4 y^2 - \frac{4 x y^{5/2}}{1 + 2x\sqrt{y}} = 10 y^2 - \frac{4 x y^{5/2}}{1 + 2x\sqrt{y}}$$ Rewrite $$y^{5/2} = y^2 \sqrt{y}$$: $$N = 10 y^2 - \frac{4 x y^2 \sqrt{y}}{1 + 2x\sqrt{y}}$$ Get common denominator: $$N = \frac{10 y^2 (1 + 2x\sqrt{y}) - 4 x y^2 \sqrt{y}}{1 + 2x\sqrt{y}} = \frac{10 y^2 + 20 x y^2 \sqrt{y} - 4 x y^2 \sqrt{y}}{1 + 2x\sqrt{y}} = \frac{10 y^2 + 16 x y^2 \sqrt{y}}{1 + 2x\sqrt{y}}$$ 12. **Final expression for $$\frac{d^2y}{dx^2}$$:** $$\frac{d^2y}{dx^2} = \frac{N}{(1 + 2x\sqrt{y})^2} = \frac{10 y^2 + 16 x y^2 \sqrt{y}}{(1 + 2x\sqrt{y})^3} = \frac{2 y^2 (5 + 8 x \sqrt{y})}{(1 + 2x\sqrt{y})^3}$$ **Answer:** $$\boxed{\frac{d^2y}{dx^2} = \frac{2 y^2 (5 + 8 x \sqrt{y})}{(1 + 2x\sqrt{y})^3}}$$