# 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ii)$\sqrt { 49.5 }$

Lets suppose $y = \sqrt x$ and let x = 49 and $\Delta x = 0.5$
Then,
$\Delta y = \sqrt{x+\Delta x} - \sqrt x$
$\Delta y = \sqrt{49+0.5} - \sqrt 49$
$\Delta y = \sqrt{49.5} - 7$
$\sqrt{49.5} = \Delta y +7$
Now, we can say that $\Delta y$  is approximately equal to dy
$dy = \frac{dy}{dx}\Delta x\\ dy = \frac{1}{2\sqrt x}.(0.5) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because y = \sqrt x \ and \ \Delta x = 0.5)\\ dy = \frac{1}{2\sqrt 49}.(0.5)\\ dy = \frac{1}{14}.(0.5)\\ dy = 0.035$
Now,
$\sqrt{49.5} = \Delta y +7\\ \sqrt {49.5} = 0.035 + 7\\ \sqrt{49.5} = 7.035$
Hence, $\sqrt{49.5}$ is approximately equal to 7.035

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