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# Solution for Using Differential, Find the Approximate Value of the ( 29 ) 1 3 ? - CBSE (Science) Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the $\left( 29 \right)^\frac{1}{3}$ ?

#### Solution

$\text {Consider the function }y = f\left( x \right) = \left( x \right)^\frac{1}{3} .$

$\text{Let }:$

$x = 27$

$x + ∆ x = 29$

$\text { Then,}$

$∆ x = 2$

$\text { For } x = 27,$

$y = \left( 27 \right)^\frac{1}{3} = 3$

$\text { Let }:$

$dx = ∆ x = 2$

$\text { Now }, y = \left( x \right)^\frac{1}{3}$

$\Rightarrow \frac{dy}{dx} = \frac{1}{3 \left( x \right)^\frac{2}{3}}$

$\Rightarrow \left( \frac{dy}{dx} \right)_{x = 27} = \frac{1}{27}$

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{27} \times 2 = 0 . 074$

$\Rightarrow ∆ y = 0 . 074$

$\therefore \left( 29 \right)^\frac{1}{3} = y + ∆ y = 3 . 074$

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: 9.16 | Page no. 9

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Solution for question: Using Differential, Find the Approximate Value of the ( 29 ) 1 3 ? concept: Approximations. For the courses CBSE (Science), CBSE (Commerce), PUC Karnataka Science, CBSE (Arts)
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