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Using differentials, find the approximate value of each of the following up to 3 places of decimal 32 15 raised to 1 by 5

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(xv) ( 32.15 ) ^{1/5}

Answers (1)
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Lets suppose y = (x)^{\frac{1}{5}} and let x = 32 and \Delta x = 0.15
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{5}} - (x)^{\frac{1}{5}}
\Delta y = ({32 + 0.15})^{\frac{1}{5}} - (32)^{\frac{1}{5}}
\Delta y = ({32.15})^{\frac{1}{5}} - 2
({32.15})^{\frac{1}{5}} = \Delta y + 2
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{5 (x)^{\frac{4}{5}}}.(0.15) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{5}} \ and \ \Delta x = 0.15)\\ dy = \frac{1 }{5 (32)^{\frac{4}{5}}}.(0.15)\\ dy = \frac{1}{5\times16}.(0.15)\\\\dy = \frac{0.15}{80}\\ dy = 0.001
Now,
(32.15)^{\frac{1}{5}} = \Delta y +2\\ (32.15)^{\frac{1}{5}} = (0.001) + 2\\ (32.15)^{\frac{1}{5}} = 2.001
Hence, (32.15)^{\frac{1}{5}} is approximately equal to 2.001

 

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