Find the approximate value of (0.009)^{frac{1 | vedclass

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(A) Consider the function $y = x^{\frac{1}{3}}$.Let $x = 0.008$ and $\Delta x = 0.001$,so that $x + \Delta x = 0.009$.We know that $(0.008)^{\frac{1}{

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$0.208$

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(A) Consider the function $y = x^{\frac{1}{3}}$.Let $x = 0.008$ and $\Delta x = 0.001$,so that $x + \Delta x = 0.009$.We know that $(0.008)^{\frac{1}{3}} = 0.2$.Using the differential formula,$\Delta y \approx dy = \left( \frac{dy}{dx} ight) \Delta x$.Since $y = x^{\frac{1}{3}}$,we have $\frac{dy}{dx} = \frac{1}{3} x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$.At $x = 0.008$,$\frac{dy}{dx} = \frac{1}{3(0.008)^{\frac{2}{3}}} = \frac{1}{3(0.2)^2} = \frac{1}{3(0.04)} = \frac{1}{0.12} = \frac{100}{12} = \frac{25}{3} \approx 8.333$.Now,$dy = \left( \frac{25}{3} ight) 	imes 0.001 = \frac{0.025}{3} \approx 0.00833$.The approximate value is $y + dy = 0.2 + 0.00833 = 0.20833$.

Find the approximate value of $(0.009)^{\frac{1}{3}}$.

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