Ex 5.5, 7 - Chapter 5 Class 12 Continuity and Differentiability (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 5 Class 12 Continuity and Differentiability
Ex 5.1, 13
Ex 5.1, 16
Ex 5.1, 18 Important
Ex 5.1, 28 Important
Ex 5.1, 30 Important
Ex 5.1, 34 Important
Ex 5.2, 5
Ex 5.2, 9 Important
Ex 5.2, 10 Important
Ex 5.3, 10 Important
Ex 5.3, 14
Example 29 Important
Example 30 Important
Ex 5.5,6 Important
Ex 5.5, 7 Important You are here
Ex 5.5, 11 Important
Ex 5.5, 16 Important
Ex 5.6, 7 Important
Ex 5.6, 11 Important
Example 38 Important
Ex 5.7, 14 Important
Question 4 Important Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Example 39 (i)
Example 40 (i)
Example 42 Important
Misc 6 Important
Misc 15 Important
Misc 16 Important
Misc 22 Important
Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 16, 2024 by Teachoo
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