Question

Using differentials, find the sum of digits approximate value of the following up to $3$ places of decimal.

$(26{)}^{\frac{1}{3}}$

Answer 1

Consider $y=x^{\frac{1}{3}}$.

Let $x=27$and $\Delta x=-1.$

Then,

$\Delta y={x+\Delta x}^{\frac{1}{3}}-x^{\frac{1}{3}}=(26{)}^{\frac{1}{3}}-(27{)}^{\frac{1}{3}}=(26{)}^{\frac{1}{3}}-3$

$\Rightarrow (26{)}^{\frac{1}{3}}-)=3+\Delta y$

Now, $dy$ is approximately equal to $\Delta y$ and is given by,

$dy=\left(\frac{dy}{dx}\right)\Delta x=\frac{1}{3(x{)}^{\frac{2}{3}}}\left(\Delta x\right)$ [as $y=x^{\frac{1}{3}}$]

$=\frac{1}{3(27{)}^{2/3}}(-1)=-\frac{1}{27}=-.037$

Hence, the approximate value of $(26{)}^{\frac{1}{3}}$ is $3+(-0.0370)=2.9629\approx 2.963$