If u = γ€–sinγ€—^(-1) (2x/(1 + x^2 )) and v = γ€–tanγ€—^(-1) (2x/(1 - x^2 )), then du/dv is

(A) 1/2 

(B) x

(C) (1 - x^2)/(1 + x^2 )   

(D) 1

This question is similar to Ex 5.3, 9 - Chapter 5 Class 12 - Continuity and Differentiability

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Question 10 If u = 〖𝑠𝑖𝑛〗^(βˆ’1) (2π‘₯/(1 + π‘₯^2 )) and v = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (2π‘₯/(1 βˆ’ π‘₯^2 )), then 𝑑𝑒/𝑑𝑣 is (A) 1/2 (B) π‘₯ (C) (1 βˆ’ π‘₯^2)/(1 + π‘₯^2 ) (D) 1 𝒅𝒖/𝒅𝒗=(𝒅𝒖⁄𝒅𝒙)/(𝒅𝒗⁄𝒅𝒙) So, we will find 𝑑𝑒/𝑑π‘₯ and 𝑑𝑣/𝑑π‘₯ Simplifying u u = 〖𝑠𝑖𝑛〗^(βˆ’1) (2π‘₯/(1 + π‘₯^2 )) Putting π‘₯=tanβ‘πœƒ u = 〖𝑠𝑖𝑛〗^(βˆ’1) ((2 π’•π’‚π’β‘πœ½)/(1 + 〖𝒕𝒂𝒏〗^𝟐 𝜽)) Using sin⁑〖2πœƒ=γ€— (2 tanβ‘πœƒ)/(1 + γ€–π‘‘π‘Žπ‘›γ€—^2 πœƒ) u = 〖𝑠𝑖𝑛〗^(βˆ’1) (sin⁑2πœƒ ) u = 2πœƒ Putting πœƒ=γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ u = πŸγ€–π’•π’‚π’γ€—^(βˆ’πŸ) 𝒙 Simplifying v v = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (2π‘₯/(1 βˆ’ π‘₯^2 )) Putting π‘₯=tanβ‘πœƒ v = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) ((2 π’•π’‚π’β‘πœ½)/(1 βˆ’ 〖𝒕𝒂𝒏〗^𝟐 𝜽)) Using tan⁑〖2πœƒ=γ€— (2 tanβ‘πœƒ)/(1 βˆ’ γ€–π‘‘π‘Žπ‘›γ€—^2 πœƒ) v = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (tan⁑2πœƒ ) v = 2πœƒ Putting πœƒ=γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ v = πŸγ€–π’•π’‚π’γ€—^(βˆ’πŸ) 𝒙 Finding 𝒅𝒖/𝒅𝒙 u = πŸγ€–π’•π’‚π’γ€—^(βˆ’πŸ) 𝒙 Differentiating w.r.t. π‘₯ 𝒅𝒖/𝒅𝒙=(𝑑〖(2π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ =2 Γ—(𝑑〖(π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ =2 Γ—1/γ€–1 + π‘₯γ€—^2 = 𝟐/γ€–πŸ + 𝒙〗^𝟐 Finding 𝒅𝒗/𝒅𝒙 v = πŸγ€–π’•π’‚π’γ€—^(βˆ’πŸ) 𝒙 Differentiating w.r.t. π‘₯ 𝒅𝒗/𝒅𝒙=(𝑑〖(2π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ =2 Γ—(𝑑〖(π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ =2 Γ—1/γ€–1 + π‘₯γ€—^2 = 𝟐/γ€–πŸ + 𝒙〗^𝟐 Now, 𝒅𝒖/𝒅𝒗=(𝑑𝑒⁄𝑑π‘₯)/(𝑑𝑣⁄𝑑π‘₯) =(2/γ€–1 + π‘₯γ€—^2 )/(1/γ€–1 + π‘₯γ€—^2 ) =𝟏 So, the correct answer is (D)

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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.