Question

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

${\left(401\right)}^{\frac{1}{2}}$

Answer 1

Given: ${\left(401\right)}^{\frac{1}{2}}$

Consider $y=x^{\frac{1}{2}}$

Let $x=400$

$△x=1$

Then

$\begin{array}{rcl}∆y& =& {(x+∆x)}^{\frac{1}{2}}-x^{\frac{1}{2}}\\ ∆y& =& {401}^{\frac{1}{2}}-20\\ {401}^{\frac{1}{2}}& =& 20+∆y\\ & & \\ & & \end{array}$

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

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

As $y=x^{\frac{1}{2}}$

So,

$$\begin{gathered} \mathrm{dy}=\frac{1}{2\times \left(20\right)}\left(1\right) \\ =\frac{1}{40} \\ =0.025 \end{gathered}$$

Hence, the approximate value of $$\begin{gathered} {\left(401\right)}^{\frac{1}{2}}=20+0.025 \\ =20.025 \end{gathered}$$