Using differentials, find the approximate value of the following up to 3 places of decimal.
$\sqrt{49.5}$

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Solution

$\sqrt{49.5}$
Consider $f (x) = \sqrt{x} \Rightarrow f^{'} (x) = \frac{1}{2 \sqrt{x}}$
Let x = 49 and $D e l t a x = 0.5$
Now $f (x + Δ x) ≃ f (x) + Δ x f^{'} (x)$
$∴ \sqrt{x + Δ x} ≃ \sqrt{x} + \frac{1}{2 \sqrt{x}} \times Δ x$
$\Rightarrow \sqrt{49.5} ≃ \sqrt{49} + \frac{0.5}{2 \sqrt{49}} = 7 + \frac{05}{140} . = 7 + 0.036 \Rightarrow \sqrt{49.5} ≃ 7.036$

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