Question

Use differential to approximate $\sqrt{36.6}$.

Answer 1

Let $y=\sqrt{x}\cdots \left(i\right)$
Where $x=36\&\Delta x=0.6$
Now, we know
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}$
$\Rightarrow \Delta y=\sqrt{36.6}-\sqrt{36}$
$\Rightarrow \Delta y+6=\sqrt{36.6}\cdots \left(ii\right)$
We know that $\Delta y=\left(\frac{dy}{dx}\right)\Delta x$
and
$\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt{x}\right)=\frac{1}{2\sqrt{x}}$
Then
$\Delta y=\left(\frac{1}{2\sqrt{x}}\right)\Delta x$
$=\frac{0.6}{2\times \sqrt{36}}$
$=\frac{0.6}{2\times 6}=0.05$
From equation (2), we get
$\Delta y+6=\sqrt{36.6}$
$\therefore \sqrt{36.6}=6.05$
Hence, the approximate value of $\sqrt{36.6}$ is $6.05$.