Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.
$(82{)}^{\frac{1}{4}}$

Answer 1

Consider $y=x^{\frac{1}{4}}$. Let $x=81$ and $\Delta x=1$.

Then,

$\Delta y={(x+\Delta x)}^{\frac{1}{4}}-x^{\frac{1}{4}}=(82{)}^{\frac{1}{4}}-(81{)}^{\frac{1}{4}}=(82{)}^{\frac{1}{4}}-3$

$\Rightarrow (82{)}^{\frac{1}{4}}=\Delta y+3$

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

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{4(x{)}^{\frac{3}{4}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{4}}$]

$=\frac{1}{4(81{)}^{3/4}}\left(1\right)=\frac{1}{4\left(27\right)}\left(1\right)=\frac{1}{108}=0.009$

Hence, the approximate value of $(82{)}^{\frac{1}{4}}$ is $3+\left(0.009\right)=\mathrm{3.009.}$