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

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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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