Question

Using differentials find the approximate value of $(25{)}^{\frac{1}{3}}$.

Answer 1

Let $y=x^{\frac{1}{3}}$

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

Since, $y=x^{\frac{1}{3}}$

$⟹\frac{dy}{dx}=\frac{1}{3}{x}^{\frac{1}{3}-1}$

$⟹\frac{dy}{dx}=\frac{1}{3x^{\frac{2}{3}}}$

$\Delta y=\frac{dy}{dx}\Delta x$

$⟹\Delta y=\frac{1}{3x^{\frac{2}{3}}}\Delta x$

$⟹\Delta y=\frac{1}{3(27{)}^{\frac{2}{3}}}(-2)$

$⟹\Delta y=\frac{-2}{3(3^3{)}^{\frac{2}{3}}}$

$⟹\Delta y=\frac{-2}{3\cdot 3^2}$

$⟹\Delta y=\frac{-2}{27}=-0.074$