Question

Using differential, find the approximate values of the following:
$\left(xx\right)$. $\sqrt{0.48}$

Answer 1

$0.69286$

Let $y=\sqrt{x}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}$
And $x=0.49$ and $\Delta x=-0.01$
then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\Rightarrow \Delta y=\sqrt{0.49-0.01}-\sqrt{0.49}\Rightarrow \Delta y=\sqrt{0.48}-0.7\Rightarrow \sqrt{0.48}=0.7+\Delta y\dots \left(1\right)$
Now, Approximate change in value of $y$ is given by
$\Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x\Rightarrow \Delta y\approx \frac{1}{2\sqrt{x}}\times \Delta x\Rightarrow \Delta y\approx \frac{1}{2\sqrt{0.49}}\times (-0.01)$
$\Rightarrow \Delta y\approx -0.00714$
From equation $\left(1\right)$
$\sqrt{0.48}\approx 0.7+0.00714\therefore \sqrt{0.48}\approx 0.69286$