Question

using differentials , find the approximate value of $(0.999{)}^{\frac{1}{9}}$.

Answer 1

$Lety=x^{\frac{1}{9}}where,x=1and\Delta x=-0.001now,y=x^{\frac{1}{9}}\Rightarrow \frac{dy}{dx}=\frac{1}{9x^{\frac{8}{9}}}also,\Delta y=\frac{dy}{dx}\Delta x\Rightarrow \Delta y=\frac{1}{9{\left(1\right)}^{\frac{8}{9}}}(-0.001)\Rightarrow \Delta y=-0.0001\left(approx\right)weknowthat\Delta y=f(x+\Delta x)-f\left(x\right)\Rightarrow \Delta y={(x+\Delta x)}^{\frac{1}{9}}-{\left(x\right)}^{\frac{1}{9}}\Rightarrow (-0.0001)={(1-0.001)}^{\frac{1}{9}}-{\left(1\right)}^{\frac{1}{9}}\Rightarrow {\left(0.999\right)}^{\frac{1}{9}}=0.9999\left(approx\right)$