Question

Using differentials , find the approximate value of each of the following up to 3 places of decimal .
$\sqrt{0.6}$

Answer 1

Let $y=\sqrt{x}$ where $x=64$ and hence
$y+\Delta y=\sqrt{x+\Delta x}\Delta x=.04$
$7+\Delta y=\sqrt{.6}...\left(1\right)$$
$\Delta y\frac{dy}{dx}\times \Delta x=\frac{1}{2\sqrt{x}}\times \Delta x=\frac{1}{2\times .8}\times -.04$
$=\frac{-.04}{1.6}$
from (1) $\sqrt{6}=8.025=775$