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Ex 6.4, 1 (i) - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at Dec. 8, 2021 by

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Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) √25.3Let y = √𝒙 Thus, √(𝟐𝟓.𝟑) = y + ∆𝒚 Here, ∆𝒚 = 𝒅𝒚/𝒅𝒙 △x where x = 25 & △x = 0.3 Since y = √𝑥 𝒅𝒚/𝒅𝒙 = (𝑑(√𝑥))/𝑑𝑥 = 𝟏/(𝟐√𝒙) Now, ∆𝒚 = 𝒅𝒚/𝒅𝒙 △x = 1/(2√𝑥) △x Putting x = 25 & △x = 0.3 = 1/(2√25) (0.3) = 1/(2 × 5) × 0.3 = 0.3/10 = 0.03 Therefore, √25.3 = y + ∆𝑦 Putting values √25.3 =√25+0.03 √(𝟐𝟓. 𝟑)=𝟓. 𝟎𝟑 Hence, approximate value of √25.3 is 5.03

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