1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xiii)  $( 81.5 ) ^{1/4 }$

Answers (1)

Lets suppose $y = (x)^{\frac{1}{4}}$ and let x = 81 and 0.5
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}$
$\Delta y = ({81 + 0.5})^{\frac{1}{4}} - (81)^{\frac{1}{4}}$
$\Delta y = ({81.5})^{\frac{1}{4}} - 3$
$({81.5})^{\frac{1}{4}} = \Delta y + 3$
Now, we can say that $\Delta y$  is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 0.5)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(0.5)\\ dy = \frac{1}{4\times 27}.(0.5)\\dy = \frac{1}{108}.(0.5) \\dy = .004$
Now,
$(81.5)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.004) + 3\\ (82)^{\frac{1}{4}} = 3.004$
Hence, $(81.5)^{\frac{1}{4}}$ is approximately equal to 3.004

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