Misc 18 - Find derivative: sec x - 1 / sec x + 1 - CBSE - Miscellaneou

Misc 18 - Chapter 13 Class 11 Limits and Derivatives - Part 2
Misc 18 - Chapter 13 Class 11 Limits and Derivatives - Part 3

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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 . tan⁡𝑥 [(sec⁡〖𝑥 + 1) − (sec⁡〖𝑥 − 1)] 〗 〗 〗)/〖(sec⁡〖x + 1〗)〗^2 = ( sec⁡〖𝑥 . tan⁡𝑥 (sec⁡〖𝑥 + 1−〖 sec〗⁡〖𝑥 + 1〗) 〗 〗)/〖(sec⁡〖x + 1〗)〗^2 = sec⁡〖𝑥 tan⁡〖𝑥 (2 + 0)〗 〗/〖(sec⁡〖𝑥 + 1〗)〗^2 = (𝟐 𝐬𝐞𝐜⁡〖𝒙 𝐭𝐚𝐧⁡𝒙 〗)/〖(𝒔𝒆𝒄⁡〖𝒙 + 𝟏〗)〗^𝟐

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.