Using differentials,find the approximate valu | vedclass

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(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} =

Vedclass · Accepted Answer

$0.677$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = \left(\frac{17}{81}ight)^{\frac{1}{4}} - \left(\frac{16}{81}ight)^{\frac{1}{4}} = \left(\frac{17}{81}ight)^{\frac{1}{4}} - \frac{2}{3}$.Thus,$\left(\frac{17}{81}ight)^{\frac{1}{4}} = \frac{2}{3} + \Delta y$.Using the differential $dy \approx \Delta y$,we have $dy = \frac{dy}{dx} \Delta x = \frac{1}{4} x^{-\frac{3}{4}} \Delta x$.Substituting $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$:$dy = \frac{1}{4 \left(\frac{16}{81}ight)^{\frac{3}{4}}} 	imes \frac{1}{81} = \frac{1}{4 	imes \frac{8}{27}} 	imes \frac{1}{81} = \frac{27}{32} 	imes \frac{1}{81} = \frac{1}{32 	imes 3} = \frac{1}{96} \approx 0.0104$.Therefore,the approximate value is $\frac{2}{3} + 0.0104 = 0.6667 + 0.0104 = 0.6771 \approx 0.677$.

Using differentials,find the approximate value of $\left(\frac{17}{81}\right)^{\frac{1}{4}}$.

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