1. Using differentials, find the approximate value of each of the following:

a)   ( 17/81) ^{1/4 }

Answers (1)

Let y = x^\frac{1}{4}  and  x = \frac{16}{81} \ and \ \Delta x = \frac{1}{81}
\Delta y = (x+\Delta x)^\frac{1}{4}-x^\frac{1}{4}
         = (\frac{16}{81}+\frac{1}{81})^\frac{1}{4}-(\frac{16}{81})^\frac{1}{4}
(\frac{17}{81})^\frac{1}{4} = \Delta y + \frac{2}{3}
Now, we know that \Delta y is approximate equals to dy
So,
dy = \frac{dy}{dx}.\Delta x \\ = \frac{1}{4x^\frac{3}{4}}.\frac{1}{81} \ \ \ \ \ \ \ (\because y = x^\frac{1}{4} \ and \ \Delta x = \frac{1}{81})\\ = \frac{1}{4(\frac{16}{81})^\frac{3}{4}}.\frac{1}{81} = \frac{27}{4\times 8}.\frac{1}{81} = \frac{1}{96}
Now,
(\frac{17}{81})^\frac{1}{4} = \Delta y + \frac{2}{3} = \frac{1}{96}+\frac{2}{3} = \frac{65}{96} = 0.677
Hence, (\frac{17}{81})^\frac{1}{4} is approximately equal to 0.677

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