Question

Using differentials, find approximate value of $\sqrt{360}$.

Answer 1

Let $y=\sqrt{x}$

and $x=361,△x=-1$

Differentiate $y=\sqrt{x}wrtx$

$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{361}}=\frac{1}{2.19}\frac{1}{38}△y=\left(\frac{dy}{dx}\right)△x=\frac{1}{38}\cdot (-1)=-\frac{1}{38}$

Now, $△y=\sqrt{(x+△x)}-\sqrt{x}(\because △y=f(x+△x)-f(x\left)\right)\Rightarrow -\frac{1}{38}=\sqrt{(361-1)}-\sqrt{361}\Rightarrow \sqrt{360}=\sqrt{361}-\frac{1}{38}=19-\frac{1}{38}\cong 18.97$

So, the appropriate value of $\sqrt{360}=18.97$