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# Solution for Using Differential, Find the Approximate Value of the Following: √ 25 . 02 - CBSE (Science) Class 12 - Mathematics

#### Question

1 Using differential, find the approximate value of the following:

$\sqrt{25 . 02}$

#### Solution

$\text { Consider the function y } = f\left( x \right) = \sqrt{x} .$

$\text { Let }:$

$x = 25$

$x + ∆ x = 25 . 02$

$\text { Then, }$

$∆ x = 0 . 02$

$\text { For} x = 25,$

$y = \sqrt{25} = 5$

$\text { Let }:$

$dx = ∆ x = 0 . 02$

$\text { Now,} y = \sqrt{x}$

$\Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}$

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 25} = \frac{1}{10}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{10} \times 0 . 02 = 0 . 002$

$\Rightarrow ∆ y = 0 . 002$

$\therefore \sqrt{25 . 02} = y + ∆ y = 5 . 002$

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Solution Using Differential, Find the Approximate Value of the Following: √ 25 . 02 Concept: Approximations.
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