Question

Using Differentials, Find The Approximate Value Of The Following Up To $3$ Places Of Decimal.

$\sqrt{0.6}$

Answer 1

Given: $\sqrt{0.6}$

and Consider $y=\sqrt{x}$

Let $x=1$

$△x=-0.4$

Then

$\begin{array}{rcl}∆y& =& \sqrt{(x+∆x)}-\sqrt{x}\\ ∆y& =& \sqrt{0.6}-1\\ \sqrt{0.6}& =& 1+∆y\\ & & \\ & & \end{array}$

Now, $\mathrm{dy}$ is approximately equal to $\mathrm{\Delta y}$ and is given by,

$$\begin{gathered} \mathrm{dy}=\left(\frac{dy}{dx}\right)∆x \\ =\frac{1}{2\sqrt{x}}(∆x) \end{gathered}$$

As $y=\sqrt{x}$

So,

$$\begin{gathered} \mathrm{dy}=\frac{1}{2\times 1}(-0.4) \\ =\frac{-0.4}{2} \\ =-0.2 \end{gathered}$$

Hence, the approximate value of $\sqrt{0.6}=1+(-0.2)=0.8$