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Misc 10 Differentiate w.r.t. x the function, ๐ฅ๐ฅ + ๐ฅ๐ + ๐^๐ฅ+ ๐๐, for some fixed ๐ >0 and ๐ฅ> 0Let ๐ฆ= ๐ฅ๐ฅ + ๐ฅ๐ + ๐^๐ฅ+ ๐๐ And let u=๐ฅ๐ฅ , ๐ฃ=๐ฅ๐ , ๐ค=๐^๐ฅ Now, ๐=๐+๐+๐+๐๐ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐๐ฆ/๐๐ฅ = ๐(๐ข + ๐ฃ + ๐ค + ๐๐)/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐(๐ข)/๐๐ฅ +๐(๐ฃ)/๐๐ฅ+๐(๐ค)/๐๐ฅ + ๐(๐๐)/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ +๐๐ฃ/๐๐ฅ+๐๐ค/๐๐ฅ + 0 ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ + ๐๐ค/๐๐ฅ Calculating ๐ ๐/๐ ๐ ๐ข =๐ฅ^๐ฅ Taking log on both sides logโก๐ข=logโกใ๐ฅ^๐ฅ ใ logโก๐ข=๐ฅ .logโก๐ฅ (๐^๐ ๐๐ ๐๐๐๐ ๐ก๐๐๐ก) Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐(logโก๐ข )/๐๐ฅ = ๐(๐ฅ .ใ logใโก๐ฅ )/๐๐ฅ ๐(logโก๐ข )/๐๐ฅ . ๐๐ข/๐๐ข = ๐(๐ฅ .ใ logใโก๐ฅ )/๐๐ฅ ๐(logโก๐ข )/๐๐ข . ๐๐ข/๐๐ฅ = ๐๐ฅ/๐๐ฅ . logโก๐ฅ + (๐(ใ logใโก๐ฅ))/๐๐ฅ. ๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = logโก๐ฅ + 1/๐ฅ . ๐ฅ 1/๐ข . ๐๐ข/๐๐ฅ = logโก๐ฅ + 1 ๐๐ข/๐๐ฅ = u (1+ log ๐ฅ)โก ๐ ๐/๐ ๐ = ๐^๐ (๐+ ๐๐๐ ๐)โก Calculating ๐ ๐/๐ ๐ ๐ฃ=๐ฅ^๐ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐๐ฃ/๐๐ฅ= ๐(๐ฅ^๐ )/๐๐ฅ ๐ ๐/๐ ๐= ๐๐^(๐ โ๐) Calculating ๐ ๐/๐ ๐ ๐ค=๐^๐ฅ Differentiating both sides ๐ค.๐.๐ก.๐ฅ. ๐๐ค/๐๐ฅ = ๐(๐^๐ฅ )/๐๐ฅ ๐๐ค/๐๐ฅ = ๐^๐ฅ .logโก๐ Therefore, ๐๐ฆ/๐๐ฅ = ๐๐ข/๐๐ฅ + ๐๐ฃ/๐๐ฅ + ๐๐ค/๐๐ฅ = ๐^๐ (๐+ ๐๐๐ ๐) + ๐๐^(๐ โ๐) + ๐^๐ .๐๐๐โก๐

Miscellaneous

Misc 1

Misc 2

Misc 3

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7 Important

Misc 8

Misc 9 Important

Misc 10 You are here

Misc 11 Important

Misc 12

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18

Misc 19 Important

Misc 20

Misc 21

Misc 22

Misc 23 Important

Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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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.