Misc 1 - Using differentials, find approximate value: 17/81

Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 4

Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 5 Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 6 Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 7 Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 8 Misc 1 - Chapter 6 Class 12 Application of Derivatives - Part 9

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

Transcript

Misc 1 Using differentials, find the approximate value of each of the following: (a) (17/81)^(1/4) (17/81)^(1/4) = (17)^(1/4)/(81)^(1/4) = (17)^(1/4)/3 Let ๐‘ฆ =๐‘ฅ^(1/4) Where ๐‘ฅ=16 and โ–ณ๐‘ฅ=1 Since ๐’š =๐’™^(๐Ÿ/๐Ÿ’) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^(1/4) )/๐‘‘๐‘ฅ = 1/4 ๐‘ฅ^(1/4 โˆ’ 1) = 1/4 ๐‘ฅ^((โˆ’3)/4) = 1/(4๐‘ฅ^(3/4) ) Now, โˆ†๐’š=๐’…๐’š/๐’…๐’™ โˆ†๐’™ โˆ†๐‘ฆ = (1/(4๐‘ฅ^(3/4) ))โˆ†๐‘ฅ Putting values โˆ†๐‘ฆ = 1/4 ร— 1/(16)^(3/4) ร— 1 = 1/4 ร— 1/(2^4 )^(3/4) = 1/4 ร— 1/2^3 = ๐Ÿ/๐Ÿ‘๐Ÿ Now, (17)^(1/4)=๐‘ฆ+โˆ†๐‘ฆ Putting values (17)^(1/4) = (16)^(1/4)+โˆ†๐‘ฆ (17)^(1/4) = (2^4 )^(1/4)+โˆ†๐‘ฆ (17)^(1/4) = 2+โˆ†๐‘ฆ (17)^(1/4) = 2 + 1/32 (17)^(1/4) = 2+ 0.03125 (๐Ÿ๐Ÿ•)^(๐Ÿ/๐Ÿ’) = 2.03125 Now, Approximate value of (๐Ÿ๐Ÿ•/๐Ÿ–๐Ÿ)^(๐Ÿ/๐Ÿ’) = (17)^(1/4)/3 = 2.03125/3 = 0.677 Hence, approximate value of (17/81)^(1/4) is 0.677 (As approximate value of (17)^(1/4) = 2.03125) Misc 1 Using differentials, find the approximate value of each of the following: (b) ใ€–(33)ใ€—^(โˆ’ 1/5)ใ€–(33)ใ€—^(โˆ’ 1/5) = 1/(33)^(1/5) Let ๐’š = ๐’™^(๐Ÿ/๐Ÿ“) where ๐‘ฅ = 32 & โˆ†๐‘ฅ = 1 Since ๐‘ฆ = ๐‘ฅ^(1/5) ๐’…๐’š/๐’…๐’™= (๐’… (๐’™^(๐Ÿ/๐Ÿ“)))/๐’…๐’™ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= 1/5 ๐‘ฅ^((1 )/5 โˆ’1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/5 ๐‘ฅ^((โˆ’4)/5) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ= 1/ใ€–5๐‘ฅใ€—^(4/5) Now, โˆ†๐’š = ๐’…๐’š/๐’…๐’™ โˆ†๐’™ โˆ†y = 1/ใ€–5๐‘ฅใ€—^(4/5) โˆ†๐‘ฅ Putting values โˆ†y = 1/ใ€–5(32)ใ€—^(4/5) ร— 1 โˆ†y = 1/5 "ร—" 1/ใ€–(32)ใ€—^(4/5) โˆ†y =1/5 "ร—" 1/ใ€–(2)ใ€—^(5 ร— 4/5) โˆ†y = 1/5 "ร—" 1/2^4 โˆ†y = 1/5 "ร—" 1/16 โˆ†y = ๐Ÿ/๐Ÿ–๐ŸŽ Now, (33)^(1/5)=๐‘ฆ+โˆ†๐‘ฆ Putting values (33)^(1/5) = (32)^(1/5)+โˆ†๐‘ฆ โˆ†y = 1/5 "ร—" 1/ใ€–(32)ใ€—^(4/5) โˆ†y =1/5 "ร—" 1/ใ€–(2)ใ€—^(5 ร— 4/5) โˆ†y = 1/5 "ร—" 1/2^4 โˆ†y = 1/5 "ร—" 1/16 โˆ†y = ๐Ÿ/๐Ÿ–๐ŸŽ Now, (33)^(1/5)=๐‘ฆ+โˆ†๐‘ฆ Putting values (33)^(1/5) = (32)^(1/5)+โˆ†๐‘ฆ ใ€–(33)ใ€—^(1/5) = ใ€–(2) ใ€—^(5 ร— 1/5)+1/80 ใ€–(33)ใ€—^(1/5) " =" 2+1/80 ใ€–(33)ใ€—^(1/5) " =" (1 + 160)/80 ใ€–(33)ใ€—^(1/5) = 161/80 But we need ๐Ÿ/(๐Ÿ‘๐Ÿ‘)^(๐Ÿ/๐Ÿ“) So, 1/(33)^(1/5) = 1/(161/80) 1/(33)^(1/5) = 80/161 1/(33)^(1/5) = 0.497 (33)^((โˆ’1)/5)= ๐ŸŽ.๐Ÿ’๐Ÿ—๐Ÿ• Thus, the approximate value of (33)^((โˆ’1)/5) ๐‘–๐‘  ๐ŸŽ.๐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 10 years. He provides courses for Maths and Science at Teachoo.