Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 30

$\sqrt{4.02} \approx 2.005$

Let $y = \sqrt{x}$ We can find $dy$: $\frac{dy}{dx} = \frac{1}{2}x^{-1/2}$ $\frac{dy}{dx} = \frac{1}{2~\sqrt{x}}$ $dy = \frac{1}{2~\sqrt{x}}~dx$ When $x=4$ and $dx = 0.02$: $y = \sqrt{4} = 2$ $dy = \frac{1}{2~\sqrt{4}}~(0.02) = 0.005$ We can approximate $\sqrt{4+0.02}$: $\sqrt{4.02} \approx y+dy$ $\sqrt{4.02} \approx \sqrt{4}+0.005$ $\sqrt{4.02} \approx 2.005$