Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 18

(a) $dy = \frac{-2}{(x-1)^2}~dx$ (b) $dy = -0.1$

(a) $y = \frac{x+1}{x-1}$ $\frac{dy}{dx} = \frac{(1)(x-1)-(x+1)(1)}{(x-1)^2}$ $\frac{dy}{dx} = \frac{x-1-x-1}{(x-1)^2}$ $\frac{dy}{dx} = \frac{-2}{(x-1)^2}$ $dy = \frac{-2}{(x-1)^2}~dx$ (b) $dy = \frac{-2}{(x-1)^2}~dx$ $dy = \frac{-2}{(2-1)^2}~(0.05)$ $dy = \frac{-2}{1}~(0.05)$ $dy = -0.1$