Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(0.009{)}^{\frac{1}{3}}$

Answer 1

$(0.009{)}^{\frac{1}{3}}$
Consider $f\left(x\right)=x^{\frac{1}{3}}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{3}{x}^{\frac{-2}{3}}$
Let $x=0.008and\Delta x=0.001$
Now, $f(x+\Delta x)\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)\Rightarrow (x+\Delta x{)}^{\frac{1}{3}}\simeq x^{\frac{1}{3}}+\frac{1}{3x^{\frac{2}{3}}}\times \Delta x\Rightarrow (0.008+0.001{)}^{\frac{1}{3}}\simeq (0.008{)}^{\frac{1}{3}}+\frac{0.001}{3(0.008{)}^{\frac{2}{3}}}=0.2+\frac{0.001}{3\left\{\right(0.2{)}^3{\}}^{\frac{2}{3}}}=0.2+\frac{0.001}{3\times (0.2{)}^2}=0.208\Rightarrow (0.009{)}^{\frac{1}{3}}\simeq 0.208$