Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.

$(255{)}^{\frac{1}{4}}$

Answer 1

Consider $y=x^{\frac{1}{4}}$. Let $x=256$ and $\Delta x=-1$.

Then,

$\Delta y={(x+\Delta x)}^{\frac{1}{4}}-x^{\frac{1}{4}}=(255{)}^{\frac{1}{4}}-(256{)}^{\frac{1}{4}}=(255{)}^{\frac{1}{4}}-4$

$\Rightarrow (255{)}^{\frac{1}{4}}=4+\Delta y$

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

$=\frac{1}{4(256{)}^{3/4}}(-1)=\frac{1}{4\left(64\right)}(-1)=-\frac{1}{256}=-.0039$

Hence, the approximate value of $(255{)}^{\frac{1}{4}}$ is $3+(-0.0039)=\mathrm{3.996.}$