Check sibling questions

Differential coefficient of sec (〖tan〗^(-1)x) w.r.t. x is

(A) x/√(1 + x^2 ) 

(B) x/(1 + x^2 )

(C) x √(1+x^2 )  

(D) 1/√(1 + x^2 )

This question is similar to Ex 5.1, 21 - Chapter 5 Class 12 - Continuity and Differentiability

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Transcript

Question 9 Differential coefficient of sec (〖𝑡𝑎𝑛〗^(−1)x) w.r.t. x is (A) 𝑥/√(1 + 𝑥^2 ) (B) 𝑥/(1 + 𝑥^2 ) (C) x √(1+𝑥^2 ) (D) 1/√(1 + 𝑥^2 ) Let y = sec (〖𝑡𝑎𝑛〗^(−1) 𝑥) Differential coefficient sec (〖𝑡𝑎𝑛〗^(−1)x) means 𝒅𝒚/𝒅𝒙 Finding 𝒅𝒚/𝒅𝒙 𝒅𝒚/𝒅𝒙 = 𝑑(sec⁡(〖𝑡𝑎𝑛〗^(−1) 𝑥))/𝑑𝑥 = sec (〖𝑡𝑎𝑛〗^(−1) 𝑥) tan (〖𝒕𝒂𝒏〗^(−𝟏) 𝒙) × (𝑑(〖𝑡𝑎𝑛〗^(−1) 𝑥))/𝑑𝑥 = sec (〖𝑡𝑎𝑛〗^(−1) 𝑥) . 𝒙 . 1/(1 + 𝑥^2 ) = sec(〖𝒕𝒂𝒏〗^(−𝟏) 𝒙) × 𝒙/(𝟏 + 𝒙^𝟐 ) Now, to find 𝒅𝒚/𝒅𝒙 , we need to find value of sec(〖𝒕𝒂𝒏〗^(−𝟏) 𝒙) Finding value of sec(〖𝒕𝒂𝒏〗^(−𝟏) 𝒙) Let 𝜽 = 〖𝒕𝒂𝒏〗^(−𝟏) 𝒙 tan⁡𝜃=𝑥 Now, let ABC be a triangle tan⁡𝜃=𝐴𝐵/𝐵𝐶 𝑥=𝐴𝐵/𝐵𝐶 𝐴𝐵/𝐵𝐶=𝑥/1 ∴ AB = 𝒙 & BC = 1 So, AC = √(〖𝐴𝐵〗^2+〖𝐵𝐶〗^2 ) = √(𝑥^2+1^2 ) = √(𝒙^𝟐+𝟏) Now, cos⁡〖𝜃=𝐵𝐶/𝐴𝐶〗 =1/√(𝑥^2 + 1) ∴ 𝐬𝐞𝐜⁡𝜽=1/cos⁡𝜃 =1/(1/√(𝑥^2+1)) =√(𝒙^𝟐+𝟏) Putting 𝛉 = 〖𝒕𝒂𝒏〗^(−𝟏) 𝒙 sec (〖𝒕𝒂𝒏〗^(−𝟏) 𝒙) =√(𝑥^2+1) Putting value of sec (〖𝑡𝑎𝑛〗^(−1) 𝑥) in (1) 𝒅𝒚/𝒅𝒙 = sec(〖𝑡𝑎𝑛〗^(−1) 𝑥) × 𝑥/(1+ 𝑥^2 ) = √(𝒙^𝟐+𝟏) × 𝑥/(1 + 𝑥^2 ) = √(𝑥^2+1) × 𝑥/(√(𝑥^2 + 1))^2 = 𝒙/√(𝒙^𝟐 + 𝟏) So, the correct answer is (A)

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.