Question

Use differentiation to approximate $(26{)}^{\frac{1}{3}}$.

Answer 1

A] Let $y=(x{)}^{1/3}$

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 value s,

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

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

$\Delta y=-0.037037$

we know,

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

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

putting values,

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

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

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

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

$2.9629={26}^{1/3}$

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