Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(32.15{)}^{\frac{1}{5}}$

Answer 1

$\left(32.15{)}^{\frac{1}{5}}Considerf\right(x)=x^{\frac{1}{5}}\Rightarrow f^{\prime }(x)=\frac{1}{5}{x}^{\frac{-4}{5}}=\frac{1}{5x^{\frac{4}{5}}}Letx=32and\Delta x=0.15Now,f(x+\Delta x)\simeq f(x)+\Delta x{f}^{\prime }(x)\Rightarrow (x+\Delta x{)}^{\frac{1}{5}}\simeq x^{\frac{1}{5}}+\frac{1}{5x^{\frac{1}{5}}}\simeq x^{\frac{1}{5}}+\frac{1}{5x^{\frac{4}{5}}}\times \Delta x\Rightarrow (32.15{)}^{\frac{1}{5}}\simeq (32{)}^{\frac{1}{5}}+\frac{0.15}{5(32{)}^{\frac{4}{5}}}=2+\frac{0.15}{5\times 2^4}=2+\frac{0.15}{80}=2+0.0019=2.0019\Rightarrow (32.15{)}^{\frac{1}{5}}\simeq 2.0019$