Miscellaneous

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Misc 5 Important

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Misc 10 You are here

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Misc 22 Important

Question 1 Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

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