Question

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

$(401{)}^{\frac{1}{2}}$

Answer 1

Consider $y=x^{\frac{1}{2}}$.

Let $x=400$ and $\Delta x=1$.

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{401}-\sqrt{400}=\sqrt{401}-20$

$\Rightarrow \sqrt{401}=20+\Delta y$

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 $y=x^{\frac{1}{2}}$]

$=\frac{1}{2\sqrt{400}}\left(1\right)=\frac{1}{40}=.025$

Hence, the approximate value of $\sqrt{401}$ is $20+.025=20.025$