Question

Using differential, find the approximate values of the following:
$\left(xix\right)$. $\sqrt{37}$

Answer 1

Let $y=\sqrt{x}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}$
And $x=36$ and $\Delta x=1$
then,
$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}\Rightarrow \Delta y=\sqrt{36+1}-\sqrt{36}\Rightarrow \Delta y=\sqrt{37}-6\Rightarrow \sqrt{37}=6+\Delta y\dots \left(1\right)$
Now, Approximate change in value of $y$ is given by
$\Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x\Rightarrow \Delta y\approx \frac{1}{2\sqrt{x}}\times \Delta x\Rightarrow \Delta y\approx \frac{1}{2\sqrt{36}}\times 1$
$\Rightarrow \Delta y\approx 0.083$
From equation $\left(1\right)$
$\sqrt{37}\approx 6+0.083\therefore \sqrt{37}\approx 6.083$