Ex 5.5, 15 - 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, 15 Find ๐๐ฆ/๐๐ฅ of the functions in, ๐ฅ๐ฆ= ๐^((๐ฅ โ๐ฆ))Given ๐ฅ๐ฆ= ๐^((๐ฅ โ๐ฆ)) Taking log both sides log (๐ฅ๐ฆ) = log ๐^((๐ฅ โ๐ฆ)) log (๐ฅ๐ฆ) = (๐ฅ โ๐ฆ) log ๐ log ๐ฅ+logโก๐ฆ = (๐ฅ โ๐ฆ) (1) log ๐ฅ+logโก๐ฆ = (๐ฅ โ๐ฆ) (As ๐๐๐โก(๐^๐ )=๐ . ๐๐๐โก๐) Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(log ๐ฅ + logโก๐ฆ )/๐๐ฅ = (๐(๐ฅ โ ๐ฆ))/๐๐ฅ ๐(log ๐ฅ)/๐๐ฅ + ๐(logโก๐ฆ )/๐๐ฅ = ๐(๐ฅ)/๐๐ฅ โ ๐(๐ฆ)/๐๐ฅ 1/๐ฅ + ๐(logโก๐ฆ )/๐๐ฅ . ๐๐ฆ/๐๐ฆ = 1 โ ๐๐ฆ/๐๐ฅ 1/๐ฅ + ๐(logโก๐ฆ )/๐๐ฆ . ๐๐ฆ/๐๐ฅ = 1 โ ๐๐ฆ/๐๐ฅ 1/๐ฅ + 1/๐ฆ . ๐๐ฆ/๐๐ฅ = 1 โ ๐๐ฆ/๐๐ฅ 1/๐ฆ . ๐๐ฆ/๐๐ฅ + ๐๐ฆ/๐๐ฅ = 1 โ 1/๐ฅ ๐๐ฆ/๐๐ฅ (1/๐ฆ +1) = ("1 โ " 1/๐ฅ) ๐๐ฆ/๐๐ฅ ((1 + ๐ฆ)/๐ฆ) = ((๐ฅ โ 1)/๐ฅ) ๐๐ฆ/๐๐ฅ = ((๐ฅ โ 1)/๐ฆ) . (๐ฆ/(1 + ๐ฆ)) ๐ ๐/๐ ๐ = ๐(๐ โ ๐)/๐(๐ + ๐)
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo