Use differentials to approximate sqrt{36.6}. | vedclass

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Question

(A) Let $y = f(x) = \sqrt{x}$.We choose $x = 36$ and $\Delta x = 0.6$ such that $x + \Delta x = 36.6$.We know that $dy \approx \Delta y$,where $dy = f

Vedclass · Accepted Answer

$6.05$

Vedclass · Answer

(A) Let $y = f(x) = \sqrt{x}$.We choose $x = 36$ and $\Delta x = 0.6$ such that $x + \Delta x = 36.6$.We know that $dy \approx \Delta y$,where $dy = f'(x) \Delta x$.First,find the derivative: $f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$.At $x = 36$,$f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{2 	imes 6} = \frac{1}{12}$.Now,calculate the differential $dy$:$dy = f'(36) \Delta x = \frac{1}{12} 	imes 0.6 = \frac{0.6}{12} = 0.05$.Since $f(x + \Delta x) \approx f(x) + dy$,we have:$\sqrt{36.6} \approx \sqrt{36} + 0.05 = 6 + 0.05 = 6.05$.

Use differentials to approximate $\sqrt{36.6}$.

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