Last updated at May 29, 2018 by Teachoo

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) = 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

Misc 1
Important

Misc 2

Misc 3

Misc 4

Misc 5

Misc 6 Important

Misc 7

Misc 8

Misc 9 Important

Misc 10

Misc 11

Misc 12 You are here

Misc 13

Misc 14

Misc 15

Misc 16

Misc 17

Misc 18

Misc 19

Misc 20

Misc 21

Misc 22

Misc 23

Misc 24 Important

Misc 25

Misc 26

Misc 27 Important

Misc 28 Important

Misc 29

Misc 30 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.