Example 27 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
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Transcript
Example 27 Differentiate โ(((๐ฅโ3) (๐ฅ2+4))/( 3๐ฅ2+ 4๐ฅ + 5)) ๐ค.๐.๐ก. ๐ฅ. Let y =โ(((๐ฅ โ 3) (๐ฅ^2 + 4))/( 3๐ฅ2+ 4๐ฅ + 5)) Taking log on both sides ๐๐๐โก๐ = ๐๐๐ โ(((๐ โ ๐) (๐๐ + ๐))/( ๐๐๐+ ๐๐ + ๐)) logโก๐ฆ = logโก (((๐ฅ โ 3) (๐ฅ2 + 4))/( 3๐ฅ2+ 4๐ฅ + 5))^(1/2) ๐๐๐โก๐ = ๐/๐ log ((๐ฅ โ 3) (๐ฅ2 + 4))/((3๐ฅ2+ 4๐ฅ + 5) ) (๐๐ ๐๐๐ logโกใ๐^๐ ใ=๐ logโก๐) ๐๐๐โก๐ = ๐/๐ log ((๐ฅ โ 3) (๐ฅ2 + 4))/((3๐ฅ2+ 4๐ฅ + 5) ) ๐๐๐โก๐ฆ = 1/2 (logโกใ (๐ฅโ3)ใ+ใlog ใโก(๐ฅ2 + 4)โlogโกใ (3๐ฅ2 + 4๐ฅ + 5)ใ ) Differentiating ๐ค.๐.๐ก.๐ฅ (๐ (logโก๐ฆ))/๐๐ฅ = 1/2 ((๐(logโก(๐ฅโ3)+ใlog ใโกใ(๐ฅ^2+4)โlogโกใ (3๐ฅ^2+4๐ฅ+5)ใ ใ ) )/๐๐ฅ) (๐ (logโก๐ฆ))/๐๐ฅ = 1/2 ((๐(logโก(๐ฅโ3)))/๐๐ฅ " + " ๐(logโก(๐ฅ^2+4) )/๐๐ฅ " โ " ๐(logโก(3๐ฅ^2+4๐ฅ+5) )/๐๐ฅ) 1/๐ฆ (๐๐ฆ/๐๐ฅ) = 1/2 [1/(๐ฅ โ 3) " . " ๐(๐ฅ โ3)/๐๐ฅ " + " 1/(๐ฅ^2 + 4) " . " (๐ (๐ฅ^2 + 4))/๐๐ฅ " โ " 1/((3๐ฅ^2+ 4๐ฅ+ 5)) " . " (๐ (3๐ฅ^2+ 4๐ฅ+ 5))/(๐๐ฅ )] ๐๐ ๐๐๐ ๐๐๐โก๐๐=๐๐๐โก๐+๐๐๐ โก๐ &๐๐๐โกใ๐/๐ใ=๐๐๐โกใ๐โ๐๐๐โก๐ ใ 1/๐ฆ (๐๐ฆ/๐๐ฅ) = 1/2 [1/((๐ฅ โ 3) ) " " (1โ0) "+ " 1/(๐ฅ^2+ 4) " " (2๐ฅ + 0)" โ " 1/(3๐ฅ^2+ 4๐ฅ + 5) " " (6๐ฅ +4+0)" " ] 1/๐ฆ (๐๐ฆ/๐๐ฅ) = 1/2 (1/((๐ฅ โ 3) )+2๐ฅ/(๐ฅ^2+ 4)โ(6๐ฅ + 4)/(3๐ฅ^2 + 4๐ฅ + 5)) ๐๐ฆ/๐๐ฅ = ๐ฆ/2 (1/((๐ฅ โ 3) )+2๐ฅ/(๐ฅ^2+ 4)โ(6๐ฅ + 4)/(3๐ฅ^2 + 4๐ฅ + 5)) ๐ ๐/๐ ๐ = ๐/๐ โ(((๐ โ ๐)(๐^๐+ ๐))/(๐๐^๐+ ๐๐ + ๐)) (๐/((๐ โ ๐) )+๐๐/(๐^๐+ ๐)โ(๐๐ + ๐)/(๐๐^๐ + ๐๐ + ๐))
