Example 31 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at June 11, 2021 by
Last updated at June 11, 2021 by
Transcript
Example 31 Differentiate ๐^๐ฅ ๐ค.๐.๐ก.๐ฅ, where a is a positive constant.Let y = ๐^๐ฅ Taking log on both sides logโก๐ฆ = logโก๐^๐ฅ ๐๐๐โก๐ = ๐ ๐๐๐โก ๐ Differentiating both sides ๐ค.๐.๐ก.๐ฅ (๐(logโก๐ฆ))/๐๐ฅ = ๐/๐๐ฅ(๐ฅ logโก๐) (๐(logโก๐ฆ))/๐๐ฅ = logโก๐ (๐๐ฅ/๐๐ฅ) (๐(logโก๐ฆ))/๐๐ฅ = ๐๐๐โก๐ (๐๐๐โกใ๐^๐=๐ ๐๐๐โก๐ ใ) (๐(logโก๐ฆ))/๐๐ฅ . ๐๐ฆ/๐๐ฆ = logโก๐ (๐(logโก๐ฆ))/๐๐ฆ . ๐๐ฆ/๐๐ฅ = logโก๐ 1/๐ฆ . ๐๐ฆ/๐๐ฅ = logโก๐ ๐๐ฆ/๐๐ฅ = ๐ฆ logโก๐ Putting back ๐ฆ = ๐^๐ฅ ๐๐ฆ/๐๐ฅ = ๐^๐ ๐๐๐โก๐
Logarithmic Differentiation - Type 1
Example 32 Important
Ex 5.5, 3 Important
Ex 5.5, 15
Ex 5.5, 5
Ex 5.5, 1 Important
Ex 5.5, 13
Ex 5.5, 14 Important
Ex 5.5, 16 Important
Ex 5.5, 17 Important
Ex 5.5, 18
Example 30 Important
Ex 5.5, 2
Misc 23 Important
Misc 3
Misc 7 Important
Example 46
Misc 9 Important
Example 45 (i)
Logarithmic Differentiation - Type 1
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