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Using differentials, find the approximate value of each of the following up to 3 places of decimal 0 0037^1/ 2

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(xi)  ( 0.0037 ) ^{1/2 }

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Lets suppose y = (x)^{\frac{1}{2}} and let x = 0.0036 and \Delta x = 0.0001
Then,
\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}
\Delta y = ({0.0036 + 0.0001})^{\frac{1}{2}} - (0.0036)^{\frac{1}{2}}
\Delta y = ({0.0037})^{\frac{1}{2}} - 0.06
({0.0037})^{\frac{1}{2}} = \Delta y + 0.06
Now, we can say that \Delta y  is approximately equal to dy
dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(0.0001) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 0.0001)\\ dy = \frac{1}{2(0.0036)^{\frac{1}{2}}}.(0.0001)\\ dy = \frac{1}{2\times 0..06}.(0.0001)\\dy = \frac{1}{0.12}.(0.0001) \\dy = 0.0008
Now,
(0.0037)^{\frac{1}{2}} = \Delta y +0.06\\ (0.0037)^{\frac{1}{2}} = (0.0008) + 0.06\\ (0.0037)^{\frac{1}{2}} = 0.0608
Hence, (0.0037)^{\frac{1}{2}} is approximately equal to 0.060  (because we need to take up to three decimal places)



        

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