Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.
$(0.999{)}^{\frac{1}{10}}$

Answer 1

Consider $y=x^{\frac{1}{10}}$. Let $x=1$ and $\Delta x=\mathrm{0.001.}$

Then,

$\Delta y=(x+\Delta x{)}^{\frac{1}{10}}-x^{\frac{1}{10}}=(0.009{)}^{\frac{1}{10}}-)-1$

$\Rightarrow (0.009{)}^{\frac{1}{10}}=1+\Delta y$

Now, $dy$ is approximately equal to $\Delta$y and is given by,

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{3(x{)}^{\frac{9}{10}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{10}}$]

$=\frac{1}{10}(-0.001)=-0.0001$

Hence, the approximate value of $(0.999{)}^{\frac{1}{10}}$ is $1+(-0.0001)=\mathrm{0.9999.}$