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1. Chapter 13 Class 11 Limits and Derivatives
2. Serial order wise
3. Miscellaneous

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

Miscellaneous 