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

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Question 1 Using mathematical induction prove that š‘‘/š‘‘š‘„(š‘„^š‘›) = ć€–š‘›š‘„ć€—^(š‘›āˆ’1) for all positive integers š‘›. Let š(š’) : š‘‘/š‘‘š‘„ (š‘„^š‘›) = ć€–š‘›š‘„ć€—^(š‘›āˆ’1) For š’ = šŸ Solving LHS (š‘‘(š‘„^1)" " )/š‘‘š‘„ = š‘‘š‘„/š‘‘š‘„ = 1 = RHS Thus, š‘·(š’) is true for š‘› = 1 Let us assume that š‘·(š’Œ) is true for š‘˜āˆˆš‘µ š‘·(š’Œ) : (š‘‘ (š‘„^š‘˜))/š‘‘š‘„ = ć€–š‘˜ š‘„ć€—^(š‘˜āˆ’1) Now We have to prove that P(š’Œ+šŸ) is true š‘ƒ(š‘˜+1) : (š‘‘(š‘„^(š‘˜ + 1))" " )/š‘‘š‘„ = 怖(š‘˜+1) š‘„ć€—^(š‘˜ + 1 āˆ’ 1) (š‘‘(š‘„^(š‘˜ + 1)))/š‘‘š‘„ = 怖(š‘˜+1) š‘„ć€—^š‘˜ Taking L.H.S (š‘‘(š‘„^(š‘˜ + 1)))/š‘‘š‘„ = (š‘‘(š‘„^(š‘˜ ). š‘„))/š‘‘š‘„ Using product rule As (š‘¢š‘£)’ = š‘¢ā€™š‘£ + š‘£ā€™š‘¢ where u = xk & v = x = (š‘‘(š‘„^š‘˜)" " )/š‘‘š‘„ . š‘„ + š‘‘(š‘„ )/š‘‘š‘„ . š‘„^(š‘˜ ) = (š’…(š’™^š’Œ)" " )/š’…š’™ . š‘„ + 1 . š‘„^(š‘˜ ) = (ć€–š’Œ. š’™ć€—^(š’Œāˆ’šŸ) ) . š‘„+š‘„^š‘˜ = ć€–š‘˜. š‘„ć€—^(š‘˜āˆ’1 + 1) .+š‘„^š‘˜ = ć€–š‘˜. š‘„ć€—^š‘˜+š‘„^š‘˜ = š‘„^š‘˜ (š‘˜+1) = R.H.S Hence proved (From (1): (š‘‘(š‘„^š‘˜ ") " )/š‘‘š‘„ = ć€–š‘˜ š‘„ć€—^(š‘˜āˆ’1) ) Thus , š‘·(š’Œ+šŸ) is true when š‘·(š’Œ) is true Therefore, By Principle of Mathematical Induction š‘ƒ(š‘›) : š‘‘/š‘‘š‘„ (š‘„^š‘›) = ć€–š‘›š‘„ć€—^(š‘›āˆ’1) is true for all š‘›āˆˆš‘µ

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