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# Solution for Using Differential, Find the Approximate Value of the √ 401 ? - CBSE (Science) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the $\sqrt{401}$ ?

#### Solution

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

$\text { Let }:$

$x = 400$

$x + ∆ x = 401$

$\text { Then },$

$∆ x = 1$

$\text { For } x = 400,$

$y = \sqrt{400} = 20$

$\text { Let }:$

$dx = ∆ x = 1$

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

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

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 400} = \frac{1}{40}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{40} \times 1 = \frac{1}{40}$

$\Rightarrow ∆ y = \frac{1}{40} = 0 . 025$

$\therefore \sqrt{401} = y + ∆ y = 20 . 025$

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Solution for question: Using Differential, Find the Approximate Value of the √ 401 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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