Differential coefficient of sec (γ€–tanγ€—^(-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

Slide41.JPG

Slide42.JPG
Slide43.JPG
Slide44.JPG

  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Question 9 Differential coefficient of sec (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)x) w.r.t. x is (A) π‘₯/√(1 + π‘₯^2 ) (B) π‘₯/(1 + π‘₯^2 ) (C) x √(1+π‘₯^2 ) (D) 1/√(1 + π‘₯^2 ) Let y = sec (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) Differential coefficient sec (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)x) means π’…π’š/𝒅𝒙 Finding π’…π’š/𝒅𝒙 π’…π’š/𝒅𝒙 = 𝑑(sec⁑(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ = sec (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) tan (〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙) Γ— (𝑑(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ = sec (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) . 𝒙 . 1/(1 + π‘₯^2 ) = sec(〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙) Γ— 𝒙/(𝟏 + 𝒙^𝟐 ) Now, to find π’…π’š/𝒅𝒙 , we need to find value of sec(〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙) Finding value of sec(〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙) Let 𝜽 = 〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙 tanβ‘πœƒ=π‘₯ Now, let ABC be a triangle tanβ‘πœƒ=𝐴𝐡/𝐡𝐢 π‘₯=𝐴𝐡/𝐡𝐢 𝐴𝐡/𝐡𝐢=π‘₯/1 ∴ AB = 𝒙 & BC = 1 So, AC = √(〖𝐴𝐡〗^2+〖𝐡𝐢〗^2 ) = √(π‘₯^2+1^2 ) = √(𝒙^𝟐+𝟏) Now, cosβ‘γ€–πœƒ=𝐡𝐢/𝐴𝐢〗 =1/√(π‘₯^2 + 1) ∴ 𝐬𝐞𝐜⁑𝜽=1/cosβ‘πœƒ =1/(1/√(π‘₯^2+1)) =√(𝒙^𝟐+𝟏) Putting 𝛉 = 〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙 sec (〖𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙) =√(π‘₯^2+1) Putting value of sec (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) in (1) π’…π’š/𝒅𝒙 = sec(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) Γ— π‘₯/(1+ π‘₯^2 ) = √(𝒙^𝟐+𝟏) Γ— π‘₯/(1 + π‘₯^2 ) = √(π‘₯^2+1) Γ— π‘₯/(√(π‘₯^2 + 1))^2 = 𝒙/√(𝒙^𝟐 + 𝟏) So, the correct answer is (A)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.