1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
 (ix) ( 82) ^{1/4 }

Answers (1)

Let's suppose y = (x)^{\frac{1}{4}} and let x = 81 and \Delta x = 1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({81 + 1})^{\frac{1}{4}} - (81)^{\frac{1}{4}}
\Delta y = ({82})^{\frac{1}{4}} - 3
({82})^{\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}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = 1)\\ dy = \frac{1}{4(81)^{\frac{3}{4}}}.(1)\\ dy = \frac{1}{4\times 27}.(1)\\dy = \frac{1}{108}.(1) \\dy = .009
Now,
(82)^{\frac{1}{4}} = \Delta y +3\\ (82^{\frac{1}{4}} = (0.009) + 3\\ (82)^{\frac{1}{4}} = 3.009
Hence, (82)^{\frac{1}{4}} is approximately equal to 3.009



        

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