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

Transcript

Misc 2 Show that the function given by f(x) = logโก๐‘ฅ/๐‘ฅ has maximum at x = e. Let f(๐‘ฅ) = logโก๐‘ฅ/๐‘ฅ Finding fโ€™(๐’™) fโ€™(๐‘ฅ) = ๐‘‘/๐‘‘๐‘ฅ (๐‘™๐‘œ๐‘”โก๐‘ฅ/๐‘ฅ) fโ€™(๐‘ฅ) = (๐‘‘(๐‘™๐‘œ๐‘”โก๐‘ฅ )/๐‘‘๐‘ฅ " " . ๐‘ฅ โˆ’๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ " . " ๐‘™๐‘œ๐‘”โก๐‘ฅ)/๐‘ฅ2 fโ€™(๐‘ฅ) = (1/๐‘ฅ ร— ๐‘ฅ โˆ’ ๐‘™๐‘œ๐‘”โก๐‘ฅ)/๐‘ฅ2 fโ€™(๐‘ฅ) = (1 โˆ’ ๐‘™๐‘œ๐‘”โก๐‘ฅ)/๐‘ฅ2 Using quotient rule as (๐‘ข/๐‘ฃ)^โ€ฒ = (๐‘ข^โ€ฒ ๐‘ฃโˆ’๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 Putting fโ€™(๐’™) = 0 (1 โˆ’ logโก๐‘ฅ)/๐‘ฅ2=0 1 โ€“ log ๐‘ฅ = 0 log ๐‘ฅ = 1 ๐‘ฅ = e Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ) = (1 โˆ’ logโก๐‘ฅ)/๐‘ฅ2 Diff w.r.t. ๐‘ฅ fโ€™โ€™(๐‘ฅ) = ๐‘‘/๐‘‘๐‘ฅ ((1 โˆ’ logโก๐‘ฅ)/๐‘ฅ2) log ๐‘ฅ=1 is only possible when ๐‘ฅ=๐‘’ Using quotient rule as (๐‘ข/๐‘ฃ)^โ€ฒ=(๐‘ข^โ€ฒ ๐‘ฃ โˆ’ ๐‘ฃ^โ€ฒ ๐‘ข)/(๐‘ฃ2 .) fโ€™โ€™(๐‘ฅ) = (๐‘‘(1 โˆ’ logโก๐‘ฅ )/๐‘‘๐‘ฅ . ๐‘ฅ2โˆ’ ๐‘‘(๐‘ฅ2)/๐‘‘๐‘ฅ . (1 โˆ’ logโก๐‘ฅ ))/(๐‘ฅ^2 )^2 = ((0 โˆ’ 1/๐‘ฅ) . ๐‘ฅ2 โˆ’ 2๐‘ฅ(1 โˆ’ logโก๐‘ฅ ))/๐‘ฅ4 = ((โˆ’1)/๐‘ฅ ร— ๐‘ฅ2 โˆ’ 2๐‘ฅ(1 โˆ’ logโก๐‘ฅ ))/๐‘ฅ^4 = (โˆ’๐‘ฅ โˆ’ 2๐‘ฅ(1 โˆ’ logโก๐‘ฅ ))/๐‘ฅ^4 = (โˆ’๐‘ฅ[1 + 2(1 โˆ’ logโก๐‘ฅ )])/๐‘ฅ4 = (โˆ’๐‘ฅ[1 + 2 โˆ’ 2 logโก๐‘ฅ ])/๐‘ฅ^4 = (โˆ’๐‘ฅ[3 โˆ’ 2 logโก๐‘ฅ ])/๐‘ฅ4 fโ€™โ€™(๐‘ฅ) = (โˆ’(3 โˆ’ 2 logโก๐‘ฅ ))/๐‘ฅ3 Putting ๐’™ = e fโ€™โ€™(๐‘’) = (โˆ’(3 โˆ’ 2 logโก๐‘’ ))/๐‘’3 = (โˆ’(3 โˆ’ 2))/๐‘’3 = (โˆ’1)/๐‘’3 = โ€“(1/๐‘’3) < 0 (๐‘Ž๐‘  logโกใ€–๐‘’=1ใ€— ) (โˆต1/๐‘’๐‘ฅ>0 ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’ ๐‘œ๐‘“ ๐‘ฅ โˆˆ๐‘น) Since fโ€™โ€™(๐‘ฅ) < 0 at ๐‘ฅ = e . โˆด ๐‘ฅ = e is point of maxima Hence, f(๐‘ฅ) is maximum at ๐’™ = e.

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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.