1. **State the problem:** Given the function $$y = \frac{x+1}{\sqrt{x-1}}$$, show that $$2\sqrt{x-1} \left(\sqrt{x-1} \frac{dy}{dx} - 1\right) = -y = -\frac{x+1}{\sqrt{x-1}}$$. 2. **Recall the formula for the derivative of a quotient:** If $$y = \frac{u}{v}$$, then $$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$. 3. **Identify $$u$$ and $$v$$:** Here, $$u = x+1$$ and $$v = \sqrt{x-1} = (x-1)^{1/2}$$. 4. **Compute derivatives:** - $$\frac{du}{dx} = 1$$ - $$\frac{dv}{dx} = \frac{1}{2}(x-1)^{-1/2}$$ 5. **Apply the quotient rule:** $$\frac{dy}{dx} = \frac{(x-1)^{1/2} \cdot 1 - (x+1) \cdot \frac{1}{2} (x-1)^{-1/2}}{(x-1)}$$ 6. **Rewrite numerator:** $$= \frac{(x-1)^{1/2} - \frac{1}{2} (x+1)(x-1)^{-1/2}}{(x-1)}$$ 7. **Multiply numerator and denominator by $$ (x-1)^{1/2} $$ to simplify:** $$\frac{dy}{dx} = \frac{(x-1)^{1/2} \cdot (x-1)^{1/2} - \frac{1}{2} (x+1)(x-1)^{-1/2} \cdot (x-1)^{1/2}}{(x-1) \cdot (x-1)^{1/2}}$$ 8. **Simplify powers:** $$= \frac{(x-1) - \frac{1}{2} (x+1)}{(x-1)^{3/2}}$$ 9. **Simplify numerator:** $$= \frac{(x-1) - \frac{1}{2} (x+1)}{(x-1)^{3/2}} = \frac{\frac{2(x-1) - (x+1)}{2}}{(x-1)^{3/2}} = \frac{\frac{2x - 2 - x - 1}{2}}{(x-1)^{3/2}} = \frac{\frac{x - 3}{2}}{(x-1)^{3/2}} = \frac{x - 3}{2 (x-1)^{3/2}}$$ 10. **Calculate $$\sqrt{x-1} \frac{dy}{dx} - 1$$:** $$\sqrt{x-1} \frac{dy}{dx} - 1 = (x-1)^{1/2} \cdot \frac{x - 3}{2 (x-1)^{3/2}} - 1 = \frac{x - 3}{2 (x-1)} - 1 = \frac{x - 3 - 2(x-1)}{2 (x-1)} = \frac{x - 3 - 2x + 2}{2 (x-1)} = \frac{-x - 1}{2 (x-1)}$$ 11. **Multiply by $$2 \sqrt{x-1}$$:** $$2 \sqrt{x-1} \left(\sqrt{x-1} \frac{dy}{dx} - 1\right) = 2 (x-1)^{1/2} \cdot \frac{-x - 1}{2 (x-1)} = (x-1)^{1/2} \cdot \frac{-x - 1}{(x-1)} = \frac{-(x+1)}{\sqrt{x-1}} = -y$$ **Final answer:** $$2\sqrt{x-1} \left(\sqrt{x-1} \frac{dy}{dx} - 1\right) = -y = -\frac{x+1}{\sqrt{x-1}}$$