Question

Using differentiation find approximate value of ${\left(26\right)}^{\frac{1}{3}}$

Answer 1

Let $y=(x{)}^{2}3$

Here $x=26$

$\Delta x=-1$

Differentiating w.r.t x

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

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

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

Putting values,

$\Delta y=\frac{1}{3x^{2/3}}\times -1$

$=-\frac{1}{3\times 3^2}=-\frac{1}{3\times 9}=-\frac{1}{27}$

$\Delta y=-0.037037$

We know,

$\Delta y=6(x+\Delta x)-6\left(x\right)$

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

Putting values,

$-0.037037=(27+(-1){)}^{1/3}-(27{)}^{1/3}$

$={26}^{1/3}-(27{)}^{1/3}$

$={26}^{1/3}-3$

$-0.037037+3={26}^{1/3}$

$2.9629={26}^{1/2}$

$\therefore$ Approximate value of ${26}^{1/3}$ is $2.9629$.