Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 31

$\frac{1}{9.98} \approx 0.1002$

Let $y = \frac{1}{x}$ We can find $dy$: $\frac{dy}{dx} = -\frac{1}{x^2}$ $dy = -\frac{1}{x^2}~dx$ When $x=10$ and $dx = -0.02$: $y = \frac{1}{10} = 0.1$ $dy =-\frac{1}{10^2}~(-0.02) = 0.0002$ We can approximate $\frac{1}{9.98}$: $\frac{1}{9.98} \approx y+dy$ $\frac{1}{9.98} \approx 0.1+0.0002$ $\frac{1}{9.98} \approx 0.1002$