Answer
$\sqrt{4.02} \approx 2.005$
Work Step by Step
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 $\sqrt{4+0.02}$:
$\sqrt{4.02} \approx y+dy$
$\sqrt{4.02} \approx \sqrt{4}+0.005$
$\sqrt{4.02} \approx 2.005$
