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# Using differentials, find the approximate value of each of the following up to 3 places of decimal 401^ 1 / 2

1. Using differentials, find the approximate value of each of the following up to 3
places of decimal.
(x) $( 401 ) ^{1/2 }$

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Let's suppose $y = (x)^{\frac{1}{2}}$ and let x = 400 and $\Delta x = 1$
Then,
$\Delta y = ({x+\Delta x})^{\frac{1}{2}} - (x)^{\frac{1}{2}}$
$\Delta y = ({400 + 1})^{\frac{1}{2}} - (400)^{\frac{1}{2}}$
$\Delta y = ({401})^{\frac{1}{2}} - 20$
$({401})^{\frac{1}{2}} = \Delta y + 20$
Now, we can say that $\Delta y$  is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2 (x)^{\frac{1}{2}}}.(1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = (x)^{\frac{1}{2}} \ and \ \Delta x = 1)\\ dy = \frac{1}{2(400)^{\frac{1}{2}}}.(1)\\ dy = \frac{1}{2\times 20}.(1)\\dy = \frac{1}{40}.(1) \\dy = 0.025$
Now,
$(401)^{\frac{1}{2}} = \Delta y +20\\ (401)^{\frac{1}{2}} = (0.025) + 20\\ (401)^{\frac{1}{2}} = 20.025$
Hence, $(401)^{\frac{1}{2}}$ is approximately equal to 20.025

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