Question

Using differentials, find the approximate value of $\sqrt{49.5}$

Answer 1

Let $f\left(x\right)=\sqrt{x}$, where x = 49
Let $\delta x=0.5$
Therefore,

$f(x+\delta x)=\sqrt{x+\delta x}=\sqrt{49+0.5}$

Now by definition, approximately we can write,
$f^{\prime }\left(x\right)=\frac{f(x+\delta x)-f\left(x\right)}{\delta x}$--- (1)

Here, $f\left(x\right)=\sqrt{x}=\sqrt{49}=7$
$\delta x=0.5$
$f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{49}}=\frac{1}{14}$

Putting these values in (1), we get
$\frac{1}{14}=\frac{\sqrt{49.5}-7}{0.5}$

$\Rightarrow \sqrt{49.5}=\frac{0.5}{14}+7$

$=\frac{0.5+98}{14}=\frac{98.5}{14}=7.036$

Therefore, the approximate value of $\sqrt{49.5}=7.036$