Question

Using differentials the approximate value of $\sqrt{401}$ is

• A. 20.1
• B. 20.025
• C. 20.03
• D. 20.125

Answer 1

Answer: (B)

20.025

$\sqrt{401}=\sqrt{400+1}$
Let $y=f\left(x\right)=\sqrt{x}$
$y+\Delta y=\sqrt{x+\Delta x}$
$\Rightarrow \Delta y=\sqrt{x+\Delta x}-\sqrt{x}$
Also,$\Delta y=f^{\prime }\left(x\right)\Delta x$
$\Rightarrow \sqrt{x+\Delta x}-\sqrt{x}=\frac{1}{2\sqrt{x}}\Delta x$
Put $x=400,\Delta x=1$
$\Rightarrow \sqrt{401}-\sqrt{400}=\frac{1}{40}\left(1\right)$
$\Rightarrow \sqrt{401}=20+0.025=20.025$