Question

Using differential, find the approximate values of the following:
$\left(xxv\right)$. ${25}^{1/3}$

Answer 1

$2.926$

Let $y=x^{1/3}\Rightarrow \frac{dy}{dx}=\frac{1}{3x^{2/3}}$
And $x=27$ and $\Delta x=-2$
then,
$\Delta y=(x+\Delta x{)}^{1/3}-x^{1/3}\Rightarrow \Delta y=(27-3{)}^{1/3}-(27{)}^{1/3}\Rightarrow \Delta y=(25{)}^{1/3}-3\Rightarrow (25{)}^{1/3}=3+\Delta y\dots (1)$
Now, Approximate change in value of $y$ is given by
$\Delta y\approx \left(\frac{dy}{dx}\right)\times \Delta x\Rightarrow \Delta y\approx \frac{1}{3x^{2/3}}\times \Delta x\Rightarrow \Delta y\approx \frac{1}{3(27{)}^{\frac{2}{3}}}\times (-2)$
$\Rightarrow \Delta y\approx -\frac{-2}{3\times 9}\Rightarrow \Delta y\approx -0.074$
From equation $\left(1\right)$
$(25{)}^{1/3}\approx 3-0.074\therefore (25{)}^{1/3}\approx 2.926$