Logarithmic Differentiation - Type 2
Last updated at April 16, 2024 by Teachoo
Misc 11 Differentiate w.r.t. x the function, ๐ฅ^(๐ฅ^2โ 3)+(๐ฅโ3)๐ฅ^2, for ๐ฅ > 3 Let ๐ฆ=๐ฅ^(๐ฅ^2โ 3)+(๐ฅโ3)^(๐ฅ^2 ) And let ๐ข=๐ฅ^(๐ฅ^2โ 3) , ๐ฃ =(๐ฅโ3)^(๐ฅ^2 ) Now, ๐ = ๐+๐ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐๐ฆ/๐๐ฅ = (๐ (๐ข + ๐ฃ))/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ Calculating ๐ ๐/๐ ๐ ๐ข = ๐ฅ^(๐ฅ^2โ 3) Taking log on both sides log ๐ข=logโกใ๐ฅ^(๐ฅ^2โ 3) ใ log ๐ข=ใ(๐ฅใ^2โ 3). logโก๐ฅ Differentiating ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ข )/๐๐ฅ = ๐(ใ(๐ฅใ^2โ 3) logโก๐ฅ )/๐๐ฅ ๐(logโก๐ข )/๐๐ฅ . ๐๐ข/๐๐ข = ๐(ใ(๐ฅใ^2 โ 3) logโก๐ฅ )/๐๐ฅ " " ๐(logโก๐ข )/๐๐ข . ๐๐ข/๐๐ฅ = ๐(ใ(๐ฅใ^2โ 3) logโก๐ฅ )/๐๐ฅ " " 1/๐ข . ๐๐ข/๐๐ฅ = ๐(ใ(๐ฅใ^2โ 3) logโก๐ฅ )/๐๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = (๐ใ(๐ฅใ^2โ 3) )/๐๐ฅ . ใ logใโก๐ฅ + ๐(logโก๐ฅ )/๐๐ฅ . ใ(๐ฅใ^2โ 3) 1/๐ข . ๐๐ข/๐๐ฅ = (2๐ฅ โ0) ใ logใโก๐ฅ + 1/๐ฅ ร ใ(๐ฅใ^2โ 3) 1/๐ข . ๐๐ข/๐๐ฅ = 2๐ฅ . logโก๐ฅ + (๐ฅ^2โ 3)/๐ฅ ๐๐ข/๐๐ฅ = u (2๐ฅ "." logโก๐ฅ "+ " (๐ฅ^2โ 3)/๐ฅ) ๐ ๐/๐ ๐ = ๐^(๐^๐โ ๐) (๐๐ "." ๐๐๐โก๐ "+ " (๐^๐โ ๐)/๐) Calculating ๐ ๐/๐ ๐ ๐ฃ = (๐ฅโ3)๐ฅ^2 Taking log on both sides log ๐ฃ=logโกใ(๐ฅโ3)^(๐ฅ^2 ) ใ log ๐ฃ=ใ๐ฅ^2 . logใโกใ (๐ฅโ3)ใ Differentiating ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ฃ )/๐๐ฅ = (๐(ใ๐ฅ^2. logใโกใ (๐ฅ โ 3)ใ ) )/๐๐ฅ ๐(logโก๐ฃ )/๐๐ฅ . ๐๐ฃ/๐๐ฃ = (๐(ใ๐ฅ^2. logใโกใ (๐ฅโ3)ใ ) )/๐๐ฅ ๐(logโก๐ฃ )/๐๐ฃ . ๐๐ฃ/๐๐ฅ = (๐(ใ๐ฅ^2. logใโกใ (๐ฅโ3)ใ ) )/๐๐ฅ 1/๐ฃ . ๐๐ฃ/๐๐ฅ = (๐(ใ๐ฅ^2. logใโกใ (๐ฅโ3)ใ ) )/๐๐ฅ 1/๐ฃ . ๐๐ฃ/๐๐ฅ = ๐(๐ฅ^2 )/๐๐ฅ . log (๐ฅโ3) + ๐(log" " (๐ฅ โ 3))/๐๐ฅ . ๐ฅ^2 1/๐ฃ . ๐๐ฃ/๐๐ฅ = 2๐ฅ . log (๐ฅโ3) + 1/((๐ฅ โ 3) ). (๐(๐ฅ โ 3)" " )/๐๐ฅ . ๐ฅ^2 1/๐ฃ . ๐๐ฃ/๐๐ฅ = 2๐ฅ . log (๐ฅโ3) + 1/((๐ฅ โ 3) ) . ๐ฅ^2 1/๐ฃ . ๐๐ฃ/๐๐ฅ = 2๐ฅ. log (๐ฅโ3) + ๐ฅ^2/(๐ฅ โ3) ๐๐ฃ/๐๐ฅ = ๐ฃ (2๐ฅ". " log" " (๐ฅโ3)" + " ๐ฅ^2/(๐ฅ โ3)) ๐ ๐/๐ ๐ = (๐โ๐)๐^๐ (๐๐". " ๐ฅ๐จ๐ " " (๐โ๐)" + " ๐^๐/(๐ โ๐)) Now, ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ = ๐^(๐^๐โ ๐) ((๐^๐โ ๐)/๐+๐๐ ๐ฅ๐จ๐ โก๐ ) + (๐โ๐)๐^๐ (๐^๐/(๐ โ๐)+๐๐ .๐ฅ๐จ๐ โก(๐ โ๐) )