Ex 5.5, 13 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 5.5, 13 Find ๐๐ฆ/๐๐ฅ of the functions in, ๐ฆ^๐ฅ = ๐ฅ^๐ฆ Given, ๐ฆ^๐ฅ = ๐ฅ^๐ฆ Taking log both sides log (๐ฆ^๐ฅ ) = log (๐ฅ^๐ฆ ) ๐ฅ . log ๐ฆ=๐ฆ.logโก๐ฅ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. (๐(๐ฅ . log ๐ฆ))/๐๐ฅ = ๐(๐ฆ.ใ logใโก๐ฅ )/๐๐ฅ (As ๐๐๐โก(๐^๐ )=๐ . ๐๐๐โก๐) Using product Rule As (๐ข๐ฃ)โ = ๐ขโ๐ฃ + ๐ฃโ๐ข ๐(๐ฅ)/๐๐ฅ . log ๐ฆ+ ๐(logโก๐ฆ )/๐๐ฅ . ๐ฅ =" " ๐(๐ฆ)/๐๐ฅ " ". log ๐ฅ + ๐(logโก๐ฅ )/๐๐ฅ . ๐ฆ log ๐ฆ+๐ฅ . ๐(logโก๐ฆ )/๐๐ฅ . ๐ฅ = ๐๐ฆ/๐๐ฅ log ๐ฅ + 1/๐ฅ . ๐ฆ log ๐ฆ+๐ฅ . ๐(logโก๐ฆ )/๐๐ฅ . ๐๐ฆ/๐๐ฆ = ๐๐ฆ/๐๐ฅ . log ๐ฅ + ๐ฆ/๐ฅ log ๐ฆ+๐ฅ . ๐(logโก๐ฆ )/๐๐ฆ . ๐๐ฆ/๐๐ฅ = ๐๐ฆ/๐๐ฅ . log ๐ฅ + ๐ฆ/๐ฅ log ๐ฆ+๐ฅ . 1/๐ฆ . ๐๐ฆ/๐๐ฅ = ๐๐ฆ/๐๐ฅ . log ๐ฅ + ๐ฆ/๐ฅ log ๐ฆ+ ๐ฅ/๐ฆ . ๐๐ฆ/๐๐ฅ = ๐๐ฆ/๐๐ฅ . log ๐ฅ + ๐ฆ/๐ฅ ๐ฅ/๐ฆ . ๐๐ฆ/๐๐ฅ โ ๐๐ฆ/๐๐ฅ . log ๐ฅ = ๐ฆ/๐ฅ โ log ๐ฆ ๐๐ฆ/๐๐ฅ (๐ฅ/๐ฆ โ log ๐ฅ) = ๐ฆ/๐ฅ โ log ๐ฆ ๐๐ฆ/๐๐ฅ ((๐ฅ โ ๐ฆ logโก๐ฅ)/๐ฆ) = (๐ฆ โ ๐ฅ logโก๐ฆ)/๐ฅ ๐๐ฆ/๐๐ฅ = (๐ฆ โ ๐ฅ logโก๐ฆ)/๐ฅ . ๐ฆ/(๐ฅ โ ๐ฆ logโก๐ฅ ) ๐ ๐/๐ ๐ = ๐(๐ โ ๐ ๐๐๐โก๐ )/๐(๐ โ ๐ ๐๐๐โก๐ )
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Ex 5.5, 13 You are here
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