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# Solution for Using Differential, Find the Approximate Value of the √ 36 . 6 ? - CBSE (Commerce) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the $\sqrt{36 . 6}$ ?

#### Solution

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

$\text { Let }:$

$x = 36$

$x + ∆ x = 36 . 6$

$\text { Then},$

$∆ x = 0 . 6$

$\text { For } x = 36,$

$y = \sqrt{36} = 6$

$\text { Let }:$

$dx = ∆ x = 0 . 6$

$\text { Now,} y = \left( x \right)^\frac{1}{2}$

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

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 36} = \frac{1}{12}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{12} \times 0 . 6 = 0 . 05$

$\Rightarrow ∆ y = 0 . 05$

$\therefore \sqrt{36 . 6} = y + ∆ y = 6 . 05$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 12 by R D Sharma (Set of 2 Volume) (2018-19 Session) (with solutions)
Chapter 14: Differentials, Errors and Approximations
Q: 9.24 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the √ 36 . 6 ? concept: Approximations. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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