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Using differentials, find the approximate value of each of the following up to 3 places of decimal 81 5 ^1 / 4

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

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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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