Question

Using differentials, find the approximate value of the following:
$\sqrt{0.082}$

Answer 1

Approx value of $\sqrt{0.082}$

Let function $y=f\left(x\right)=\sqrt{x}$

Let $x=0.081x+\Delta x=0.082$

$\Delta x=0.001$

for $x=0.081y=0.009$

Let $dx=\Delta x=0.001$

$y=\sqrt{x}=\frac{1}{2\sqrt{x}}(\frac{dy}{dx}{)}_{x=0.081}=\frac{1}{2\times 0.009}=\frac{1}{0.081}=55.5$

$dy=\frac{dy}{dx}dxdy=55.5\times 0.001$

$\Delta y=0.0555$

Hence $\sqrt{0.082}=0.009+0.0555=0.065$