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Miscellaneous

Misc 1 (i)

Misc 1 (ii) Important

Misc 1 (iii)

Misc 1 (iv) Important

Misc 2

Misc 3 Important

Misc 4 Important

Misc 5

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11

Misc 12 Important You are here

Misc 13

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Misc 15

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19

Misc 20 Important

Misc 21

Misc 22 Important

Misc 23

Misc 24 Important

Misc 25

Misc 26

Misc 27 Important

Misc 28 Important

Misc 29 Important

Misc 30 Important

Last updated at Sept. 6, 2021 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