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Ex 6.4, 1 (ii) - Using differentials, find approximate value of √49.5

Ex 6.4, 1 (ii) - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.4, 1 (ii) - Chapter 6 Class 12 Application of Derivatives - Part 3


Transcript

Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ii) √49.5Let y = √𝑥 where x = 49 & △ x = 0.5 Since y = √𝑥 𝑑𝑦/𝑑𝑥 = (𝑑(√𝑥))/𝑑𝑥 = 1/(2√𝑥) Now, ∆𝑦 = 𝑑𝑦/𝑑𝑥 △x = 1/(2√𝑥) (0.5) = 1/(2 × √49) × 0.5 = 1/(2 × 7) × 0.5 = 0.5/14 = 0.036 Also, ∆𝑦=𝑓(𝑥+∆𝑥)−𝑓(𝑥) Putting values ∆𝑦=√(𝑥+∆𝑥)−√𝑥 0. 036=√(49+0. 5)−√49 0. 036=√49.5−7 0. 036+7=√49.5 √49.5=7. 036 Hence, approximate value of √49.5 is 7.036

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.