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Last updated at May 29, 2023 by Teachoo

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) = limh→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) = limh→0 𝑎 𝑥+ℎ+𝑏𝑛 − 𝑎𝑥 + 𝑏𝑛h f’(x) = limh→0 𝑎 𝑥+ℎ+𝑏𝑛 − 𝑎𝑥 + 𝑏𝑛h = limh→0 𝑎𝑥 + 𝑎ℎ + 𝑏𝑛 − 𝑎𝑥 + 𝑏𝑛h = limh→0 𝑎𝑥 + 𝑏 + 𝑎ℎ𝑛 − 𝑎𝑥 + 𝑏𝑛h = limh→0 𝑎𝑥 + 𝑏 1+ 𝑎ℎ𝑎𝑥 + 𝑏𝑛 − 𝑎𝑥 + 𝑏𝑛h = limh→0 𝑎𝑥 + 𝑏𝑛 𝟏+ 𝒂𝒉𝒂𝒙 + 𝒃𝒏− 1h = limh→0(ax + b)n 𝟏𝒏 + 𝒏𝑪𝟏 𝒂𝒏−𝟏 𝒂𝒉𝒂𝒙 + 𝒃𝟏+ 𝒏𝑪𝟐 𝟏𝒏−𝟐 𝒂𝒉𝒂𝒙 + 𝒃𝟐+…. + 𝒂𝒉𝒂𝒙 + 𝒃𝒏 − 1h = limh→0(ax + b)n 1+ 𝑛!1! 𝑛 − 1! 𝑎ℎ𝑎𝑥 + 𝑏+ 𝑛𝐶2 𝑎ℎ𝑎𝑥 + 𝑏2+…. + 𝑎ℎ𝑎𝑥 + 𝑏𝑛−1h = limh→0(ax + b)n 1 + 𝑛 . 𝑎ℎ𝑎𝑥 + 𝑏+ 𝑛𝐶2 𝑎ℎ𝑎𝑥 + 𝑏2+…. + 𝑎ℎ𝑎𝑥 + 𝑏𝑛−1h = limh→0(ax + b)n 1 − 1 + 𝑛 𝑎ℎ𝑎𝑥 + 𝑏+ 𝑛𝐶2 𝑎ℎ𝑎𝑥 + 𝑏2+…. + 𝑎ℎ𝑎𝑥 + 𝑏𝑛h = limh→0(ax + b)n 0 + 𝑛 𝑎ℎ𝑎𝑥 + 𝑏+ 𝑛𝐶2 𝑎ℎ𝑎𝑥 + 𝑏2+…. + 𝑎ℎ𝑎𝑥 + 𝑏𝑛h = limh→0(ax + b)n 𝑎ℎ𝑛𝑎𝑥 + 𝑏+ 𝑛𝐶2 𝑎ℎ𝑎𝑥 + 𝑏2+…. + 𝑎ℎ𝑎𝑥 + 𝑏𝑛h = limh→0(ax + b)n ℎ 𝑎𝑛𝑎𝑥 + 𝑏+ 𝑛𝐶2 ℎ2 𝑎𝑎𝑥 + 𝑏2+…. + ℎ𝑛 𝑎𝑎𝑥 + 𝑏𝑛h = limh→0(ax + b)n ℎ 𝑎𝑛𝑎𝑥 + 𝑏+ 𝑛𝐶2 ℎ 𝑎𝑎𝑥 + 𝑏2+…. + ℎ𝑛−1 𝑎𝑎𝑥 + 𝑏𝑛h = limh→0(ax + b)n 𝑎𝑛𝑎𝑥 + 𝑏+ 𝑛𝐶2 ℎ 𝑎𝑎𝑥 + 𝑏2+…. + ℎ𝑛−1 𝑎𝑎𝑥 + 𝑏𝑛 Putting h = 0 = limh→0(ax + b)n 𝑎𝑛𝑎𝑥 + 𝑏+ 𝑛𝐶2 (0) 𝑎𝑎𝑥 + 𝑏2+…. + (0)𝑛−1 𝑎𝑎𝑥 + 𝑏𝑛 = (ax + b)n 𝑎𝑛𝑎𝑥 + 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