Check sibling questions


Transcript

Question 1 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

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

About the Author

Davneet Singh

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