Misc 19 - Using mathematical induction prove d/dx (xn) = nxn-1

Misc 19 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Misc 19 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Misc 19 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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

Transcript

Misc 19 Using mathematical induction prove that ๐‘‘/๐‘‘๐‘ฅ(๐‘ฅ^๐‘›) = ใ€–๐‘›๐‘ฅใ€—^(๐‘›โˆ’1) for all positive integers ๐‘›. Let ๐(๐’) : ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^๐‘›) = ใ€–๐‘›๐‘ฅใ€—^(๐‘›โˆ’1) For ๐’ = ๐Ÿ Solving LHS (๐‘‘(๐‘ฅ^1)" " )/๐‘‘๐‘ฅ = ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = 1 = RHS Thus, ๐‘ท(๐’) is true for ๐‘› = 1 Let us assume that ๐‘ท(๐’Œ) is true for ๐‘˜โˆˆ๐‘ต ๐‘ท(๐’Œ) : (๐‘‘ (๐‘ฅ^๐‘˜))/๐‘‘๐‘ฅ = ใ€–๐‘˜ ๐‘ฅใ€—^(๐‘˜โˆ’1) Now We have to prove that P(๐’Œ+๐Ÿ) is true ๐‘ƒ(๐‘˜+1) : (๐‘‘(๐‘ฅ^(๐‘˜ + 1))" " )/๐‘‘๐‘ฅ = ใ€–(๐‘˜+1) ๐‘ฅใ€—^(๐‘˜ + 1 โˆ’ 1) (๐‘‘(๐‘ฅ^(๐‘˜ + 1)))/๐‘‘๐‘ฅ = ใ€–(๐‘˜+1) ๐‘ฅใ€—^๐‘˜ Taking L.H.S (๐‘‘(๐‘ฅ^(๐‘˜ + 1)))/๐‘‘๐‘ฅ = (๐‘‘(๐‘ฅ^(๐‘˜ ). ๐‘ฅ))/๐‘‘๐‘ฅ Using product rule As (๐‘ข๐‘ฃ)โ€™ = ๐‘ขโ€™๐‘ฃ + ๐‘ฃโ€™๐‘ข where u = xk & v = x = (๐‘‘(๐‘ฅ^๐‘˜)" " )/๐‘‘๐‘ฅ . ๐‘ฅ + ๐‘‘(๐‘ฅ )/๐‘‘๐‘ฅ . ๐‘ฅ^(๐‘˜ ) = (๐’…(๐’™^๐’Œ)" " )/๐’…๐’™ . ๐‘ฅ + 1 . ๐‘ฅ^(๐‘˜ ) = (ใ€–๐’Œ. ๐’™ใ€—^(๐’Œโˆ’๐Ÿ) ) . ๐‘ฅ+๐‘ฅ^๐‘˜ = ใ€–๐‘˜. ๐‘ฅใ€—^(๐‘˜โˆ’1 + 1) .+๐‘ฅ^๐‘˜ = ใ€–๐‘˜. ๐‘ฅใ€—^๐‘˜+๐‘ฅ^๐‘˜ = ๐‘ฅ^๐‘˜ (๐‘˜+1) = R.H.S Hence proved (From (1): (๐‘‘(๐‘ฅ^๐‘˜ ") " )/๐‘‘๐‘ฅ = ใ€–๐‘˜ ๐‘ฅใ€—^(๐‘˜โˆ’1) ) Thus , ๐‘ท(๐’Œ+๐Ÿ) is true when ๐‘ท(๐’Œ) is true Therefore, By Principle of Mathematical Induction ๐‘ƒ(๐‘›) : ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^๐‘›) = ใ€–๐‘›๐‘ฅใ€—^(๐‘›โˆ’1) is true for all ๐‘›โˆˆ๐‘ต

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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.