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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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Misc 19 Using mathematical induction prove that 𝑑/𝑑𝑥(𝑥^𝑛) = 〖𝑛𝑥〗^(𝑛−1) for all positive integers 𝑛. Let 𝐏(𝒏) : 𝑑/𝑑𝑥 (𝑥^𝑛) = 〖𝑛𝑥〗^(𝑛−1) For 𝒏 = 𝟏 Solving LHS (𝑑(𝑥^1)" " )/𝑑𝑥 = 𝑑𝑥/𝑑𝑥 = 1 = RHS Thus, 𝑷(𝒏) is true for 𝑛 = 1 Let us assume that 𝑷(𝒌) is true for 𝑘∈𝑵 𝑷(𝒌) : (𝑑 (𝑥^𝑘))/𝑑𝑥 = 〖𝑘 𝑥〗^(𝑘−1) Now We have to prove that P(𝒌+𝟏) is true 𝑃(𝑘+1) : (𝑑(𝑥^(𝑘 + 1))" " )/𝑑𝑥 = 〖(𝑘+1) 𝑥〗^(𝑘 + 1 − 1) (𝑑(𝑥^(𝑘 + 1)))/𝑑𝑥 = 〖(𝑘+1) 𝑥〗^𝑘 Taking L.H.S (𝑑(𝑥^(𝑘 + 1)))/𝑑𝑥 = (𝑑(𝑥^(𝑘 ). 𝑥))/𝑑𝑥 Using product rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 where u = xk & v = x = (𝑑(𝑥^𝑘)" " )/𝑑𝑥 . 𝑥 + 𝑑(𝑥 )/𝑑𝑥 . 𝑥^(𝑘 ) = (𝒅(𝒙^𝒌)" " )/𝒅𝒙 . 𝑥 + 1 . 𝑥^(𝑘 ) = (〖𝒌. 𝒙〗^(𝒌−𝟏) ) . 𝑥+𝑥^𝑘 = 〖𝑘. 𝑥〗^(𝑘−1 + 1) .+𝑥^𝑘 = 〖𝑘. 𝑥〗^𝑘+𝑥^𝑘 = 𝑥^𝑘 (𝑘+1) = R.H.S Hence proved (From (1): (𝑑(𝑥^𝑘 ") " )/𝑑𝑥 = 〖𝑘 𝑥〗^(𝑘−1) ) Thus , 𝑷(𝒌+𝟏) is true when 𝑷(𝒌) is true Therefore, By Principle of Mathematical Induction 𝑃(𝑛) : 𝑑/𝑑𝑥 (𝑥^𝑛) = 〖𝑛𝑥〗^(𝑛−1) is true for all 𝑛∈𝑵

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.