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

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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)

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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.