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

Transcript

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
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.