Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(81.5{)}^{\frac{1}{4}}$

Answer 1

$(81.5{)}^{\frac{1}{4}}$
$Considerf\left(x\right)=x^{\frac{1}{4}}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{4}{x}^{-\frac{3}{4}}=\frac{1}{4x^{\frac{3}{4}}}Letx=81and\Delta x=0.5Now,f(x+\Delta x)\simeq f\left(x\right)+\Delta x{f}^{\prime }\left(x\right)\Rightarrow (x+\Delta x{)}^{\frac{1}{4}}\simeq +\frac{1}{4x^{\frac{3}{4}}}\times \Delta x\Rightarrow (81+0.5{)}^{\frac{1}{4}}\simeq (81{)}^{\frac{1}{4}}+\frac{0.5}{4(81{)}^{\frac{3}{4}}}=3+\frac{0.5}{4\times 3^3}=3+\frac{0.5}{108}=3+0.0046=3.0046\Rightarrow (81.5{)}^{\frac{1}{4}}\simeq 3.0046$