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

Transcript

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 LHS = (𝑑(𝑥^1)" " )/𝑑𝑥 = 𝑑𝑥/𝑑𝑥 = 1 ∴ LHS = RHS Thus, 𝑃(𝑛) is true for 𝑛 = 1 Let us assume that Let 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 …(1) 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 ∴ By principal of mathematical Induction 𝑃(𝑛) : 𝑑/𝑑𝑥 (𝑥^𝑛) = 〖𝑛𝑥〗^(𝑛−1) is true , 𝑛∈𝑁

Miscellaneous 