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  • Using differentials, find the approximate value of each of the following up to 3 places of decimal 255 ^ 1/ 4

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(viii)  ( 255) ^{1/4}

Answers (1)

best_answer

Let's suppose y = (x)^{\frac{1}{4}} and let x = 256 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({256 - 1})^{\frac{1}{4}} - (256)^{\frac{1}{4}}
\Delta y = ({255})^{\frac{1}{4}} - 4
({255})^{\frac{1}{4}} = \Delta y + 4
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{4 (x)^{\frac{3}{4}}}.(-1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{4}} \ and \ \Delta x = -1)\\ dy = \frac{1}{4(256)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 64}.(-1)\\dy = \frac{1}{256}.(-1) \\dy = -0.003
Now,
(255)^{\frac{1}{4}} = \Delta y +4\\ (255)^{\frac{1}{4}} = (-0.003) + 4\\ (255)^{\frac{1}{4}} = 3.997
Hence, (255)^{\frac{1}{4}} is approximately equal to 3.997



        

Posted by

Gautam harsolia

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