# Misc 19 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Jan. 3, 2020 by Teachoo

Last updated at Jan. 3, 2020 by Teachoo

Transcript

Misc 19 Using mathematical induction prove that ๐/๐๐ฅ(๐ฅ^๐) = ใ๐๐ฅใ^(๐โ1) for all positive integers ๐. Let P(๐) : ๐/๐๐ฅ (๐ฅ^๐) = ใ๐๐ฅใ^(๐โ1) For ๐ = 1 LHS = (๐(๐ฅ^1)" " )/๐๐ฅ = ๐๐ฅ/๐๐ฅ = 1 โด LHS = RHS Thus, ๐(๐) is true for ๐ = 1 Let us assume that Let ๐(๐) is true for ๐โ๐ต ๐(๐) : (๐(๐ฅ^๐ ") " )/๐๐ฅ = ใ๐ ๐ฅใ^(๐โ1) Now We have to prove that P(๐+1) is true ๐(๐+1) : (๐(๐ฅ^(๐ + 1) ") " )/๐๐ฅ = ใ(๐+1) ๐ฅใ^(๐ + 1 โ 1) (๐(๐ฅ^(๐ + 1) ") " )/๐๐ฅ = ใ(๐+1) ๐ฅใ^๐ Taking L.H.S (๐(๐ฅ^(๐ + 1) ") " )/๐๐ฅ = (๐(๐ฅ^(๐ ). ๐ฅ ") " )/๐๐ฅ โฆ(1) Using product rule As (๐ข๐ฃ)โ = ๐ขโ๐ฃ + ๐ฃโ๐ข where u = xk & v = x = (๐(๐ฅ^๐ ") " )/๐๐ฅ . ๐ฅ + ๐(๐ฅ )/๐๐ฅ . ๐ฅ^(๐ ) = (๐ (๐^๐ ") " )/๐ ๐ . ๐ฅ + 1 . ๐ฅ^(๐ ) = (ใ๐. ๐ใ^(๐โ๐) ) . ๐ฅ+๐ฅ^๐ = ใ๐. ๐ฅใ^(๐โ1 + 1) .+๐ฅ^๐ = ใ๐. ๐ฅใ^๐+๐ฅ^๐ = ๐ฅ^๐ (๐+1) = R.H.S Hence proved (From (1): (๐(๐ฅ^๐ ") " )/๐๐ฅ = ใ๐ ๐ฅใ^(๐โ1) ) Thus , ๐(๐+1) is true when ๐(๐) is true โด By principal of mathematical Induction ๐(๐) : ๐/๐๐ฅ (๐ฅ^๐) = ใ๐๐ฅใ^(๐โ1) is true , ๐โ๐

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.