# Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 11, 2021 by Teachoo

Last updated at March 11, 2021 by Teachoo

Transcript

Ex 5.5, 7 Differentiate the functions in, ใ(logโกใ๐ฅ)ใใ^๐ฅ + ๐ฅ^logโก๐ฅ Let ๐ฆ = ใ(logโกใ๐ฅ)ใใ^๐ฅ+ ๐ฅ^logโก๐ฅ Let ๐ข = ใ(logโกใ๐ฅ)ใใ^๐ฅ , ๐ฃ = ๐ฅ^logโก๐ฅ ๐ฆ = ๐ข+๐ฃ 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 = ใ(logโกใ๐ฅ)ใใ^๐ฅ Taking log both sides logโก๐ข = log ใ(logโกใ๐ฅ)ใใ^๐ฅ logโก๐ข = ๐ฅ . log (logโกใ๐ฅ)ใ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ข )/๐๐ข . ๐๐ข/๐๐ฅ = (๐(๐ฅ . log (logโกใ๐ฅ)ใ ) )/๐๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" ๐(๐ฅ)/๐๐ฅ . log (log" " ๐ฅ) + ๐(log (log" " ๐ฅ))/๐๐ฅ ร ๐ฅ Using product Rule As (๐ข๐ฃ)โ = ๐ขโ๐ฃ + ๐ฃโ๐ข 1/๐ข (๐๐ข/๐๐ฅ)" =" 1 . log (log" " ๐ฅ) + (1/(log" " ๐ฅ) .๐(log" " ๐ฅ)/๐๐ฅ) ร ๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" log (log" " ๐ฅ) + (1/(log" " ๐ฅ) . 1/๐ฅ) ร ๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" log (log" " ๐ฅ) + 1/logโก๐ฅ ร ๐ฅ/๐ฅ 1/๐ข (๐๐ข/๐๐ฅ)" =" log (log" " ๐ฅ) + 1/logโก๐ฅ ๐๐ข/๐๐ฅ " =" ๐ข (log (log" " ๐ฅ)" + " 1/logโก๐ฅ ) ๐๐ข/๐๐ฅ = (logโก๐ฅ )^๐ฅ (log (log" " ๐ฅ)" + " 1/logโก๐ฅ ) ๐๐ข/๐๐ฅ = (logโก๐ฅ )^๐ฅ ((logโกใ๐ฅ . ใlog ใโก(logโก๐ฅ ) +ใ 1)/logโก๐ฅ ) ๐๐ข/๐๐ฅ = (logโก๐ฅ )^๐ฅ/logโก๐ฅ (logโกใ๐ฅ . ใlog ใโก(logโก๐ฅ )+ใ 1) ๐๐ข/๐๐ฅ = (logโก๐ฅ )^(๐ฅ โ1) (logโกใ๐ฅ . ใlog ใโก(logโก๐ฅ )+ใ 1) Calculating ๐ ๐/๐ ๐ ๐ฃ = ๐ฅ^logโก๐ฅ Taking log both sides . logโก๐ฃ=logโกใ (๐ฅ^logโก๐ฅ )ใ logโก๐ฃ = log ๐ฅ . logโก๐ฅ logโก๐ฃ = (logโก๐ฅ )^2 (As ๐๐๐โก(๐^๐ )=๐ . ๐๐๐โก๐) Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ฃ )/๐๐ฅ = (๐(logโก๐ฅ )^2)/๐๐ฅ ๐(logโก๐ฃ )/๐๐ฅ . ๐๐ฃ/๐๐ฃ = (๐(logโก๐ฅ )^2)/๐๐ฅ ๐(logโก๐ฃ )/๐๐ฃ . ๐๐ฃ/๐๐ฅ = (๐(logโก๐ฅ )^2)/๐๐ฅ 1/๐ฃ . ๐๐ฃ/๐๐ฅ = (๐(logโก๐ฅ )^2)/๐๐ฅ 1/๐ฃ . ๐๐ฃ/๐๐ฅ = 2 logโก๐ฅ . ๐(logโก๐ฅ )/๐๐ฅ 1/๐ฃ . ๐๐ฃ/๐๐ฅ = 2 logโก๐ฅ . 1/๐ฅ ๐๐ฃ/๐๐ฅ = ๐ฃ ((2 logโก๐ฅ)/๐ฅ) ๐๐ฃ/๐๐ฅ = ๐ฅ^logโก๐ฅ ((2 logโก๐ฅ)/๐ฅ) ๐๐ฃ/๐๐ฅ = ๐ฅ^logโก๐ฅ /๐ฅ (2 logโก๐ฅ ) ๐๐ฃ/๐๐ฅ = ๐ฅ^(logโก๐ฅ โ 1) . 2 logโก๐ฅ ๐๐ฃ/๐๐ฅ = ใ2๐ฅใ^(logโก๐ฅ โ 1) . logโก๐ฅ Hence ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ Putting value of ๐๐ข/๐๐ฅ & ๐๐ฃ/๐๐ฅ ๐ ๐/๐ ๐ = (๐ฅ๐จ๐ โก๐ )^(๐ โ๐) (๐+๐ฅ๐จ๐ โก๐.๐ฅ๐จ๐ โก(๐ฅ๐จ๐ โก๐ ) ) + ใ๐๐ใ^(๐๐๐ ๐ โ๐). ๐๐๐โก๐

Ex 5.5

Ex 5.5, 1
Important

Ex 5.5, 2

Ex 5.5, 3 Important

Ex 5.5, 4

Ex 5.5, 5

Ex 5.5,6 Important

Ex 5.5, 7 Important You are here

Ex 5.5, 8

Ex 5.5, 9 Important

Ex 5.5, 10 Important

Ex 5.5, 11 Important

Ex 5.5, 12

Ex 5.5, 13

Ex 5.5, 14 Important

Ex 5.5, 15

Ex 5.5, 16 Important

Ex 5.5, 17 Important

Ex 5.5, 18

About the Author

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.