Misc 10 - Differentiate xx + xa + ax + aa - Class 12 NCERT

Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Misc  10 - Chapter 5 Class 12 Continuity and Differentiability - Part 5

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Misc 10 Differentiate w.r.t. x the function, ๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘Ž + ๐‘Ž^๐‘ฅ+ ๐‘Ž๐‘Ž, for some fixed ๐‘Ž >0 and ๐‘ฅ> 0Let ๐‘ฆ= ๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘Ž + ๐‘Ž^๐‘ฅ+ ๐‘Ž๐‘Ž And let u=๐‘ฅ๐‘ฅ , ๐‘ฃ=๐‘ฅ๐‘Ž , ๐‘ค=๐‘Ž^๐‘ฅ Now, ๐’š=๐’–+๐’—+๐’˜+๐’‚๐’‚ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘ข + ๐‘ฃ + ๐‘ค + ๐‘Ž๐‘Ž)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘ข)/๐‘‘๐‘ฅ +๐‘‘(๐‘ฃ)/๐‘‘๐‘ฅ+๐‘‘(๐‘ค)/๐‘‘๐‘ฅ + ๐‘‘(๐‘Ž๐‘Ž)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ +๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘‘๐‘ค/๐‘‘๐‘ฅ + 0 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘‘๐‘ค/๐‘‘๐‘ฅ Calculating ๐’…๐’–/๐’…๐’™ ๐‘ข =๐‘ฅ^๐‘ฅ Taking log on both sides logโก๐‘ข=logโกใ€–๐‘ฅ^๐‘ฅ ใ€— logโก๐‘ข=๐‘ฅ .logโก๐‘ฅ (๐‘Ž^๐‘Ž ๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘ ๐‘ก๐‘Ž๐‘›๐‘ก) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ฅ = ๐‘‘(๐‘ฅ .ใ€– logใ€—โก๐‘ฅ )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ฅ . ๐‘‘๐‘ข/๐‘‘๐‘ข = ๐‘‘(๐‘ฅ .ใ€– logใ€—โก๐‘ฅ )/๐‘‘๐‘ฅ ๐‘‘(logโก๐‘ข )/๐‘‘๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ . logโก๐‘ฅ + (๐‘‘(ใ€– logใ€—โก๐‘ฅ))/๐‘‘๐‘ฅ. ๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = logโก๐‘ฅ + 1/๐‘ฅ . ๐‘ฅ 1/๐‘ข . ๐‘‘๐‘ข/๐‘‘๐‘ฅ = logโก๐‘ฅ + 1 ๐‘‘๐‘ข/๐‘‘๐‘ฅ = u (1+ log ๐‘ฅ)โก ๐’…๐’–/๐’…๐’™ = ๐’™^๐’™ (๐Ÿ+ ๐’๐’๐’ˆ ๐’™)โก Calculating ๐’…๐’—/๐’…๐’™ ๐‘ฃ=๐‘ฅ^๐‘Ž Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ= ๐‘‘(๐‘ฅ^๐‘Ž )/๐‘‘๐‘ฅ ๐’…๐’—/๐’…๐’™= ๐’‚๐’™^(๐’‚ โˆ’๐Ÿ) Calculating ๐’…๐’˜/๐’…๐’™ ๐‘ค=๐‘Ž^๐‘ฅ Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘‘(๐‘Ž^๐‘ฅ )/๐‘‘๐‘ฅ ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐‘Ž^๐‘ฅ .logโก๐‘Ž Therefore, ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘๐‘ข/๐‘‘๐‘ฅ + ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘‘๐‘ค/๐‘‘๐‘ฅ = ๐’™^๐’™ (๐Ÿ+ ๐’๐’๐’ˆ ๐’™) + ๐’‚๐’™^(๐’‚ โˆ’๐Ÿ) + ๐’‚^๐’™ .๐’๐’๐’ˆโก๐’‚

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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.