Question

Using differential, find the approximate value of the following:
${25}^{1/3}$

Answer 1

Let $y=f\left(x\right)=x^{1/3}$

Let $x=27$ $\Delta x=-2$

$\therefore y=(27{)}^{1/3}=3$

$\therefore \frac{dy}{dx}=\frac{1}{3}{x}^{2/3}\Rightarrow dy=\frac{1}{3}{x}^{2/3}dx$

$\Rightarrow dy=\frac{1}{3(27{)}^{2/3}}(-2)=\frac{1}{27}(-2)$

$\Rightarrow dy=\frac{-2}{27}=-0.0741$

$\Rightarrow y+\Delta y=(x+\Delta x{)}^{1/3}$

$\Rightarrow 3-0.0741=(27-2{)}^{1/3}$

$\Rightarrow (25{)}^{1/3}=2.9259$