Example 30 - Differentiate root (x-3) (x^2+4) / 3x^2 + 4x + 5

Example 30 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Example 30 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Example 30 Differentiate โˆš(((๐‘ฅโˆ’3) (๐‘ฅ2+4))/( 3๐‘ฅ2+ 4๐‘ฅ + 5)) ๐‘ค.๐‘Ÿ.๐‘ก. ๐‘ฅ.Let y =โˆš(((๐‘ฅ โˆ’ 3) (๐‘ฅ^2 + 4))/( 3๐‘ฅ2+ 4๐‘ฅ + 5)) Taking log on both sides ๐’๐’๐’ˆโก๐’š = ๐’๐’๐’ˆ โˆš(((๐’™ โˆ’ ๐Ÿ‘) (๐’™๐Ÿ + ๐Ÿ’))/( ๐Ÿ‘๐’™๐Ÿ+ ๐Ÿ’๐’™ + ๐Ÿ“)) logโก๐‘ฆ = logโก (((๐‘ฅ โˆ’ 3) (๐‘ฅ2 + 4))/( 3๐‘ฅ2+ 4๐‘ฅ + 5))^(1/2) ๐’๐’๐’ˆโก๐’š = ๐Ÿ/๐Ÿ log ((๐‘ฅ โˆ’ 3) (๐‘ฅ2 + 4))/((3๐‘ฅ2+ 4๐‘ฅ + 5) ) (๐‘ˆ๐‘ ๐‘–๐‘›๐‘” logโกใ€–๐‘Ž^๐‘ ใ€—=๐‘ logโก๐‘Ž) ๐’๐’๐’ˆโก๐’š = ๐Ÿ/๐Ÿ log ((๐‘ฅ โˆ’ 3) (๐‘ฅ2 + 4))/((3๐‘ฅ2+ 4๐‘ฅ + 5) ) ๐‘™๐‘œ๐‘”โก๐‘ฆ = 1/2 (logโกใ€– (๐‘ฅโˆ’3)ใ€—+ใ€–log ใ€—โก(๐‘ฅ2 + 4)โˆ’logโกใ€– (3๐‘ฅ2 + 4๐‘ฅ + 5)ใ€— ) Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ (๐‘‘ (logโก๐‘ฆ))/๐‘‘๐‘ฅ = 1/2 ((๐‘‘(logโก(๐‘ฅโˆ’3)+ใ€–log ใ€—โกใ€–(๐‘ฅ^2+4)โˆ’logโกใ€– (3๐‘ฅ^2+4๐‘ฅ+5)ใ€— ใ€— ) )/๐‘‘๐‘ฅ) (๐‘‘ (logโก๐‘ฆ))/๐‘‘๐‘ฅ = 1/2 ((๐‘‘(logโก(๐‘ฅโˆ’3)))/๐‘‘๐‘ฅ " + " ๐‘‘(logโก(๐‘ฅ^2+4) )/๐‘‘๐‘ฅ " โˆ’ " ๐‘‘(logโก(3๐‘ฅ^2+4๐‘ฅ+5) )/๐‘‘๐‘ฅ) 1/๐‘ฆ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = 1/2 [1/(๐‘ฅ โˆ’ 3) " . " ๐‘‘(๐‘ฅ โˆ’3)/๐‘‘๐‘ฅ " + " 1/(๐‘ฅ^2 + 4) " . " (๐‘‘ (๐‘ฅ^2 + 4))/๐‘‘๐‘ฅ " โ€“ " 1/((3๐‘ฅ^2+ 4๐‘ฅ+ 5)) " . " (๐‘‘ (3๐‘ฅ^2+ 4๐‘ฅ+ 5))/(๐‘‘๐‘ฅ )] ๐‘ˆ๐‘ ๐‘–๐‘›๐‘” ๐‘™๐‘œ๐‘”โก๐‘Ž๐‘=๐‘™๐‘œ๐‘”โก๐‘Ž+๐‘๐‘œ๐‘ โก๐‘ &๐‘™๐‘œ๐‘”โกใ€–๐‘Ž/๐‘ใ€—=๐‘™๐‘œ๐‘”โกใ€–๐‘Žโˆ’๐‘™๐‘œ๐‘”โก๐‘ ใ€— 1/๐‘ฆ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = 1/2 [1/((๐‘ฅ โˆ’ 3) ) " " (1โˆ’0) "+ " 1/(๐‘ฅ^2+ 4) " " (2๐‘ฅ + 0)" โˆ’ " 1/(3๐‘ฅ^2+ 4๐‘ฅ + 5) " " (6๐‘ฅ +4+0)" " ] 1/๐‘ฆ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = 1/2 (1/((๐‘ฅ โˆ’ 3) )+2๐‘ฅ/(๐‘ฅ^2+ 4)โˆ’(6๐‘ฅ + 4)/(3๐‘ฅ^2 + 4๐‘ฅ + 5)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฆ/2 (1/((๐‘ฅ โˆ’ 3) )+2๐‘ฅ/(๐‘ฅ^2+ 4)โˆ’(6๐‘ฅ + 4)/(3๐‘ฅ^2 + 4๐‘ฅ + 5)) ๐’…๐’š/๐’…๐’™ = ๐Ÿ/๐Ÿ โˆš(((๐’™ โˆ’ ๐Ÿ‘)(๐’™^๐Ÿ+ ๐Ÿ’))/(๐Ÿ‘๐’™^๐Ÿ+ ๐Ÿ’๐’™ + ๐Ÿ“)) (๐Ÿ/((๐’™ โˆ’ ๐Ÿ‘) )+๐Ÿ๐’™/(๐’™^๐Ÿ+ ๐Ÿ’)โˆ’(๐Ÿ”๐’™ + ๐Ÿ’)/(๐Ÿ‘๐’™^๐Ÿ + ๐Ÿ’๐’™ + 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.