Question
Using differentials, find the approximate values of the following:
$(255)^{\frac{1}{4}}$

Answer

Let $\text{x}=256,\text {x}+\triangle\text{x}=255$ $\triangle\text{x}=255 -256$
$\triangle\text{x}=1$
Let $\text{y}=\text{x}^ {\frac{1}{4}}$
$\frac{\text{dy}} {\text{dx}}=\frac{1}{4\text{x}^{\frac{3} {4}}}$
$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}\Rightarrow256}=\frac{1} {4(256)^{\frac{3}{4}}}$
$=\frac{1}{256}$
$=0.00391$
Now,
$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}-256}\times\triangle\text{x}$
$=(0.00391)(-1)$
$\triangle\text{y}=- 0.00391$
$(255)^{\frac{1}{4}}= \text{y}+\triangle\text{y}$
$=(\text{x})^{\frac{1} {4}}+(-0.00391)$
$=4-0.00391$
$(255)^{\frac{1}{4}} =3.99609$

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