Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.

$\sqrt{0.6}$

Answer 1

Consider $y=\sqrt{x}$.

Let $x=1$ and $\Delta x=-\mathrm{0.4.}$

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{0.6}-1$

$\Rightarrow \sqrt{0.6}=1+\Delta y$

Now, dy is approximately equal to $\Delta$y and is given by,

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{2\sqrt{x}}\left(\Delta x\right)$ [as $y=\sqrt{x}$]

$=-\frac{1}{2\sqrt{1}}\left(0.4\right)=\frac{-0.4}{2}=-0.2=\Delta y$

Hence, the approximate value of $\sqrt{0.6}$ is $(1-0.2)=0.8$