Question

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

${\left(15\right)}^{\frac{1}{4}}$

Answer 1

Finding The Approximate Value

Given: ${\left(15\right)}^{\frac{1}{4}}$

Consider $y=x^{\frac{1}{4}}$

Let $x=16$

$△x=-1$

Then

$\begin{array}{rcl}∆y& =& {(x+∆x)}^{\frac{1}{4}}-x^{\frac{1}{4}}\\ ∆y& =& {15}^{\frac{1}{4}}-2\\ {15}^{\frac{1}{4}}& =& 2+∆y\\ & & \\ & & \end{array}$

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

$$\begin{gathered} \mathrm{dy}=\left(\frac{dy}{dx}\right)∆x \\ =\frac{1}{4{\left(x\right)}^{{\displaystyle \frac{3}{4}}}}(∆x) \end{gathered}$$

As $y=x^{\frac{1}{4}}$

So,

$$\begin{gathered} \mathrm{dy}=\frac{1}{4\times {\left(2\right)}^3}(-1) \\ =\frac{-1}{32} \\ =-0.03125 \end{gathered}$$

Hence, the approximate value of $$\begin{gathered} {\left(15\right)}^{\frac{1}{4}}=2+(-0.03125) \\ =1.968 \end{gathered}$$