Misc 18 - Chapter 13 Class 11 Limits and Derivatives (Term 1 and Term 2)
Last updated at Nov. 30, 2019 by Teachoo
Miscellaneous (Term 1 and Term 2)
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
Misc 13
Misc 14 Important
Misc 15
Misc 16
Misc 17 Important
Misc 18 Important You are here
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
Miscellaneous (Term 1 and Term 2)
Last updated at Nov. 30, 2019 by Teachoo
Misc 18 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): sec〖x − 1〗/sec〖x + 1〗 Let f (x) = sec〖x − 1〗/sec〖x + 1〗 Let u = sec x – 1 & v = sec x + 1 ∴ f(x) = 𝑢/𝑣 So, f’(x) = (𝑢/𝑣)^′ Using quotient rule f’(x) = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 Finding u’ & v’ u = sec x – 1 u’ = (sec x – 1)’ = sec x tan x – 0 = sec x tan x & v = sec x + 1 v’= sec x tan x + 0 = sec x tan x Now, f’(x) = (𝑢/𝑣)^′ Derivative of sec x = sec x tan x Derivative of constant = 0 = (𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢)/𝑣^2 = ( (sec〖𝑥 tan〖𝑥)〗 (sec〖𝑥 + 1) − (sec〖𝑥 tan〖𝑥)〗 (sec〖𝑥 − 1)〗 〗 〗 〗)/〖(sec〖x + 1〗)〗^2 = ( sec〖𝑥 . tan𝑥 [(sec〖𝑥 + 1) − (sec〖𝑥 − 1)] 〗 〗 〗)/〖(sec〖x + 1〗)〗^2 = ( sec〖𝑥 . tan𝑥 (sec〖𝑥 + 1−〖 sec〗〖𝑥 + 1〗) 〗 〗)/〖(sec〖x + 1〗)〗^2 = sec〖𝑥 tan〖𝑥 (2 + 0)〗 〗/〖(sec〖𝑥 + 1〗)〗^2 = (𝟐 𝐬𝐞𝐜〖𝒙 𝐭𝐚𝐧𝒙 〗)/〖(𝒔𝒆𝒄〖𝒙 + 𝟏〗)〗^𝟐