Question

Use differential to approximate $(25{)}^{\frac{1}{3}}$

Answer 1

Let $y=x^{\frac{1}{3}}\cdots \left(i\right)$
Where $x=27$ and $\Delta x=-2$
$\frac{dy}{dx}=\frac{d\left(x^{\frac{1}{3}}\right)}{dx}$
$=\frac{1}{3}{x}^{\frac{-2}{3}}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{3x^{\frac{2}{3}}}\cdots \left(ii\right)$
Now, we know
$\Delta y=\frac{dy}{dx}\Delta x$
$\Rightarrow (x+\Delta x{)}^{\frac{1}{3}}-(x{)}^{\frac{1}{3}}=\left(\frac{1}{3x^{\frac{2}{3}}}\right)\Delta x$
$\Rightarrow (27-2{)}^{\frac{1}{3}}-(27{)}^{\frac{1}{3}}=\left(\frac{1}{3(27{)}^{\frac{2}{3}}}\right)(-2)$
$\Rightarrow (25{)}^{\frac{1}{3}}-3=\left(\frac{1}{3(27{)}^{\frac{2}{3}}}\right)(-2)$
$\Rightarrow (25{)}^{\frac{1}{3}}=3-\frac{2}{27}$
$\Rightarrow (25{)}^{\frac{1}{3}}=\frac{79}{27}=2.926$
Hence, approximate value of $(25{)}^{\frac{1}{3}}$ is $\mathrm{2.926.}$