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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) $( 32.15 ) ^{1/5}$

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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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