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

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i) โ25.3Let y = โ๐ฅ where x = 25 & โณ x = 0.3 Since y = โ๐ฅ ๐๐ฆ/๐๐ฅ = (๐(โ๐ฅ))/๐๐ฅ = 1/(2โ๐ฅ) Now, โ๐ฆ = ๐๐ฆ/๐๐ฅ โณx = 1/(2โ๐ฅ) (0.3) = 1/(2โ25) (0.3) = 1/(2 ร 5) ร 0.3 = 0.3/10 = 0.03 Also, โ๐ฆ=๐(๐ฅ+โ๐ฅ)โ๐(๐ฅ) Putting values โ๐ฆ=โ(๐ฅ+โ๐ฅ)โโ๐ฅ 0. 03=โ(25+0. 3)โโ25 0. 03=โ(25. 3)โ5 0. 03+5=โ(25. 3) โ(25. 3)=5. 03 Hence, approximate value of โ25.3 is 5.03

Ex 6.4

Ex 6.4, 1 (i)
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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.