Misc 12 - Find derivative: (ax + b)n - Chapter 13 Class 11 - Derivatives by formula - x^n formula

Misc 12  - Chapter 13 Class 11 Limits and Derivatives - Part 2
Misc 12  - Chapter 13 Class 11 Limits and Derivatives - Part 3
Misc 12  - Chapter 13 Class 11 Limits and Derivatives - Part 4

Misc 12  - Chapter 13 Class 11 Limits and Derivatives - Part 5
Misc 12  - Chapter 13 Class 11 Limits and Derivatives - Part 6

  1. Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
  2. Serial order wise

Transcript

Misc 12 (Method 1) Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax + b)n Let f(x) = (ax + b)n. We know that f’(x) = lim﷮h→0﷯ f x + h﷯ − f(x)﷮h﷯ Here, f(x) = (ax + b)n So, f(x + h) = (a(x + h) + b)n Putting values f’(x) = lim﷮h→0﷯ 𝑎 𝑥+ℎ﷯+𝑏﷯𝑛 − 𝑎𝑥 + 𝑏﷯𝑛﷮h﷯ f’(x) = lim﷮h→0﷯ 𝑎 𝑥+ℎ﷯+𝑏﷯𝑛 − 𝑎𝑥 + 𝑏﷯𝑛﷮h﷯ = lim﷮h→0﷯ 𝑎𝑥 + 𝑎ℎ + 𝑏﷯𝑛 − 𝑎𝑥 + 𝑏﷯𝑛﷮h﷯ = lim﷮h→0﷯ 𝑎𝑥 + 𝑏﷯ + 𝑎ℎ﷯﷮𝑛﷯ − 𝑎𝑥 + 𝑏﷯𝑛﷮h﷯ = lim﷮h→0﷯ 𝑎𝑥 + 𝑏﷯ 1+ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷯﷮𝑛﷯ − 𝑎𝑥 + 𝑏﷯𝑛﷮h﷯ = lim﷮h→0﷯ 𝑎𝑥 + 𝑏﷯﷮𝑛﷯ 𝟏+ 𝒂𝒉﷮𝒂𝒙 + 𝒃﷯﷯﷮𝒏﷯− 1﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 𝟏﷯﷮𝒏﷯ + 𝒏𝑪﷮𝟏﷯ 𝒂﷯﷮𝒏−𝟏﷯ 𝒂𝒉﷮𝒂𝒙 + 𝒃﷯﷯﷮𝟏﷯+ 𝒏𝑪﷮𝟐﷯ 𝟏﷯﷮𝒏−𝟐﷯ 𝒂𝒉﷮𝒂𝒙 + 𝒃﷯﷯﷮𝟐﷯+…. + 𝒂𝒉﷮𝒂𝒙 + 𝒃﷯﷯﷮𝒏﷯﷯ − 1﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 1+ 𝑛!﷮1! 𝑛 − 1﷯!﷯ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯−1﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 1 + 𝑛 . 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯−1﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 1 − 1﷯ + 𝑛 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 0 + 𝑛 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 𝑎ℎ𝑛﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + 𝑎ℎ﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n ℎ 𝑎𝑛﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ ℎ2 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + ℎ﷮𝑛﷯ 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n ℎ 𝑎𝑛﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ ℎ 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + ℎ﷮𝑛−1﷯ 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯﷮h﷯ = lim﷮h→0﷯(ax + b)n 𝑎𝑛﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ ℎ 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + ℎ﷮𝑛−1﷯ 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯ Putting h = 0 = lim﷮h→0﷯(ax + b)n 𝑎𝑛﷮𝑎𝑥 + 𝑏﷯﷯+ 𝑛𝐶﷮2﷯ (0) 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮2﷯+…. + (0)﷮𝑛−1﷯ 𝑎﷮𝑎𝑥 + 𝑏﷯﷯﷮𝑛﷯﷯ = (ax + b)n 𝑎𝑛﷮𝑎𝑥 + 𝑏﷯﷯+0+…+0﷯ = (ax + b)n . 𝑎𝑛﷮(𝑎𝑥 + 𝑏)﷯ = an (ax + b)n – 1 Hence f’ (x) = an (ax + b)n – 1 Misc12 (Method 2) Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax + b)n Let f(x) = (ax + b)n. Let p = (ax + b) So, f(x) = pn Now, f’(x) = (pn)’ f’(x) = npn–1 p’ Putting p = (ax + b) f’(x) = n (ax + b)n – 1 . (ax + b)’ f’(x) = n (ax + b)n – 1 . (ax + b)’ = n (ax + b)n – 1 [a . 1 . x1 – 1 + 0] = n (ax + b)n – 1 [a . 1 . x1 – 1 + 0] = n ( ax + b) n – 1 (ax0) = n (ax + b)n – 1 (a) = an (ax + b)n – 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.