1. Using differentials, find the approximate value of each of the following up to 3 places of decimal.(v)  $( 0.999) ^{1/10 }$

G Gautam harsolia

Lets suppose $y = (x)^{\frac{1}{10}}$ and let x = 1 and $\Delta x = -0.001$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{10}} - (x)^{\frac{1}{10}}$
$\Delta y = ({1 - 0.001})^{\frac{1}{10}} - (1)^{\frac{1}{10}}$
$\Delta y = ({0.999})^{\frac{1}{10}} - 1$
$({0.999})^{\frac{1}{10}} = \Delta y + 1$
Now, we cam say that $\Delta y$  is approximately equals to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{10 (x)^{\frac{9}{10}}}.(-0.001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{10}} \ and \ \Delta x = -0.001)\\ dy = \frac{1}{10(1)^{\frac{9}{10}}}.(-0.001)\\ dy = \frac{1}{10}.(-0.001)\\ dy = -0.0001$
Now,
$(0.999)^{\frac{1}{10}} = \Delta y +1\\ (0.999)^{\frac{1}{10}} = (-0.0001) + 1\\ (0.999)^{\frac{1}{10}} = 0.9999 = 0.999 \ upto \ three\ decimal \ place$
Hence, $(0.999)^{\frac{1}{10}}$ is approximately equal to 0.999  (because we need to answer up to three decimal place)

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