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  • Using differentials, find the approximate value of each of the following up to 3 places of decimal 15^1 / 4

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(vi)  (15 )^{1/4}

Answers (1)

best_answer

Let's suppose y = (x)^{\frac{1}{4}} and let x = 16 and \Delta x = -1
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{4}} - (x)^{\frac{1}{4}}
\Delta y = ({16 - 1})^{\frac{1}{4}} - (16)^{\frac{1}{4}}
\Delta y = ({15})^{\frac{1}{4}} - 2
({15})^{\frac{1}{4}} = \Delta y + 2
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(16)^{\frac{3}{4}}}.(-1)\\ dy = \frac{1}{4\times 8}.(-1)\\dy = \frac{1}{32}.(-1) \\dy = -0.031
Now,
(15)^{\frac{1}{4}} = \Delta y +2\\ (15)^{\frac{1}{4}} = (-0.031) + 2\\ (15)^{\frac{1}{4}} = 1.969
Hence, (15)^{\frac{1}{4}} is approximately equal to 1.969  



        

Posted by

Gautam harsolia

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