Question

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

Answer 1

$\left(0.999{)}^{\frac{1}{10}}Considerf\right(x)=x^{\frac{1}{10}},Letx=1and\Delta x=-0.001f^{\prime }(x)=\frac{1}{10}{x}^{\frac{-9}{10}}Now,f(x+\Delta x)\simeq f(x)+\Delta x{f}^{\prime }(x)\Rightarrow (x+\Delta x{)}^{\frac{1}{10}}\simeq x^{\frac{1}{10}}+\frac{1}{10x^{\frac{9}{10}}}\times \Delta x\Rightarrow (1-0.001{)}^{\frac{1}{10}}\simeq (1{)}^{\frac{1}{10}}+\frac{(-0.001)}{10(1{)}^{\frac{9}{10}}}=1-\frac{0.001}{10\times 1}=1-0.0001=0.9999\therefore (0.999{)}^{\frac{1}{10}}=0.9999$