Using differentials,find the approximate valu | vedclass

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(B) Let $y = f(x) = x^{-\frac{1}{5}}$.Let $x = 32$ and $\Delta x = 1$,so that $x + \Delta x = 33$.We know that $f(x) = x^{-\frac{1}{5}} = (32)^{-\frac

Vedclass · Accepted Answer

$0.497$

Vedclass · Answer

(B) Let $y = f(x) = x^{-\frac{1}{5}}$.Let $x = 32$ and $\Delta x = 1$,so that $x + \Delta x = 33$.We know that $f(x) = x^{-\frac{1}{5}} = (32)^{-\frac{1}{5}} = (2^5)^{-\frac{1}{5}} = 2^{-1} = \frac{1}{2} = 0.5$.Using the differential formula,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.First,find the derivative: $f'(x) = -\frac{1}{5} x^{-\frac{6}{5}}$.At $x = 32$,$f'(32) = -\frac{1}{5} (32)^{-\frac{6}{5}} = -\frac{1}{5} (2^5)^{-\frac{6}{5}} = -\frac{1}{5} (2^{-6}) = -\frac{1}{5 	imes 64} = -\frac{1}{320}$.Now,calculate the approximate value: $f(33) \approx 0.5 + (-\frac{1}{320})(1)$.$f(33) \approx 0.5 - 0.003125 = 0.496875$.Rounding to three decimal places,we get $0.497$.

Using differentials,find the approximate value of $(33)^{-\frac{1}{5}}$.

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