Question

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

Answer 1

$\sqrt{0.6}$
Consider $f\left(x\right)=\sqrt{x}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$
Let $x=0.64and\Delta x=-0.04$
$Now,f(x+\Delta x)=f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)\Rightarrow \sqrt{x+\Delta x}=\sqrt{x}+\frac{1}{2\sqrt{x}}\times \Delta x\Rightarrow \sqrt{0.64-0.04}=\sqrt{0.64}+\frac{(-0.04)}{2\sqrt{0.64}}=0.8-\frac{0.04}{2\times 0.8}=0.8-0.025=0.775\Rightarrow \sqrt{0.6}\simeq 0.775$