PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Using Differential, Find the Approximate Value of the 25 1 3 ? - PUC Karnataka Science Class 12 - Mathematics

#### Question

Using differential, find the approximate value of the ${25}^\frac{1}{3}$ ?

#### Solution

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

$\text {Let }:$

$x = 27$

$x + ∆ x = 25$

$\text { Then,}$

$\bigtriangleup x = - 2$

$\text { For } x = 27,$

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

$\text { Let }:$

$dx = ∆ x = - 2$

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

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

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

$\therefore ∆ y = dy = \frac{dy}{dx}dx = \frac{1}{27} \times \left( - 2 \right) = - 0 . 07407$

$\Rightarrow ∆ y = - 0 . 07407$

$\therefore \left( 25 \right)^\frac{1}{3} = y + ∆ y = 2 . 9259$

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Solution Using Differential, Find the Approximate Value of the 25 1 3 ? Concept: Approximations.
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