Use differentials to approximate (25)^{frac{1 | vedclass

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(B) Let $y = x^{\frac{1}{3}}$. Let $x = 27$ and $\Delta x = -2$.Then $\Delta y = (x + \Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (25)^{\frac{1}{3}} -

Vedclass · Accepted Answer

$2.926$

Vedclass · Answer

(B) Let $y = x^{\frac{1}{3}}$. Let $x = 27$ and $\Delta x = -2$.Then $\Delta y = (x + \Delta x)^{\frac{1}{3}} - x^{\frac{1}{3}} = (25)^{\frac{1}{3}} - (27)^{\frac{1}{3}} = (25)^{\frac{1}{3}} - 3$.So,$(25)^{\frac{1}{3}} = 3 + \Delta y$.Now,$dy$ is approximately equal to $\Delta y$,given by $dy = \left( \frac{dy}{dx} ight) \Delta x$.Since $y = x^{\frac{1}{3}}$,$\frac{dy}{dx} = \frac{1}{3x^{\frac{2}{3}}}$.Substituting $x = 27$ and $\Delta x = -2$:$dy = \frac{1}{3(27)^{\frac{2}{3}}} 	imes (-2) = \frac{-2}{3(3^3)^{\frac{2}{3}}} = \frac{-2}{3(3^2)} = \frac{-2}{27} \approx -0.074$.Thus,the approximate value is $3 + (-0.074) = 2.926$.

Use differentials to approximate $(25)^{\frac{1}{3}}$.

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