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

$\begin{array}{1 1} 0.9999 \\ 0.0999 \\ 0.9000 \\ 0.8999 \end{array}$

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A)
Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
Let $y=x^{\Large\frac{1}{10}}$
Then $y+\Delta y=(x+\Delta x)^{\Large\frac{1}{10}}$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{10}}-x^{\Large\frac{1}{10}}$
$\Delta y=\sqrt{x+\Delta x}-x^{\Large\frac{1}{10}}$
$x+\Delta x=0.999$ and $x=1$
$\therefore \Delta x=-0.001$
Hence $\Delta y=(0.999)^{\Large\frac{1}{10}}-(1)^{\Large\frac{1}{10}}$
$(0.999)^{\Large\frac{1}{10}}$$=1+\Delta y-----(1) Step 2: \Delta y\approx dy=\large\frac{dy}{dx}.$$\Delta x$
$\qquad\;\;\;\;=\large\frac{1}{10}$$x^{\Large\frac{-9}{10}}\Delta x \qquad\;\;\;\;=\large\frac{1}{10}$$1^{\Large\frac{-9}{10}}(-0.001)$
$\qquad\;\;\;\;=-0.0001$
From equ(1) we have
$(0.999)^{\Large\frac{1}{10}}=1-0.0001$
$\qquad\;\;\;\;\;\;\;\;=0.9999$