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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.4, 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (xv) ใ€–(32.15)ใ€—^(1/5) Let ๐‘ฆ=(๐‘ฅ)^( 1/5) where ๐‘ฅ=32 & โˆ†๐‘ฅ=0. 15 Now, ๐‘ฆ=(๐‘ฅ)^( 1/5) Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^( 1/5) )/๐‘‘๐‘ฅ=1/5 ๐‘ฅ^( 1/5 โˆ’ 1) =1/5 ๐‘ฅ^( (โˆ’ 4)/( 5) )=1/(5ใ€– ๐‘ฅใ€—^( 4/5 ) ) Using โˆ†๐‘ฆ=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ†๐‘ฅ โˆ†๐‘ฆ=1/(5 ใ€–๐‘ฅ ใ€—^(4/5) ) โˆ†๐‘ฅ Putting Values โˆ†๐‘ฆ=1/(5(32)^( 4/5) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5(2)^( 5 ร— 4/5) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5(2)^( 4) ) ร— (0. 15) โˆ†๐‘ฆ=1/(5 ร— 16) ร— 0. 15 โˆ†๐‘ฆ=(0. 15" " )/80 โˆ†๐‘ฆ=0. 00187 We know that โˆ†๐‘ฆ=๐‘“(๐‘ฅ+โˆ†๐‘ฅ)โˆ’๐‘“(๐‘ฅ) So, โˆ†๐‘ฆ=(๐‘ฅ+โˆ†๐‘ฅ)^( 1/5)โˆ’ใ€–๐‘ฅ ใ€—^(1/5) Putting Values 0. 00187=(32+0. 15)^( 1/5)โˆ’(32)^( 1/5) 0. 00187=(32. 15)^( 1/5)โˆ’(2)^(5 ร— 1/5 ) 0. 00187=(32. 15)^( 1/5)โˆ’2 0. 00187+2=(32. 15)^( 1/5) 2. 00187=(32. 15)^( 1/5) (32. 15)^( 1/5)=2. 00187 Hence, Approximate Values of (32. 15)^( 1/5) is ๐Ÿ. ๐ŸŽ๐ŸŽ๐Ÿ๐Ÿ–๐Ÿ•

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.