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

Transcript

Ex 5.3, 12 Find 𝑑𝑦/𝑑π‘₯ in, y = sin–1 ((1βˆ’ π‘₯^2)/( 1+ π‘₯2 )) , 0 < x < 1 y = sin–1 ((1βˆ’ π‘₯^2)/( 1+ π‘₯2 )) Putting x = tan ΞΈ y = 〖𝑠𝑖𝑛〗^(βˆ’1) (γ€–1 βˆ’ tan^2γ€—β‘πœƒ/γ€–1 + tan^2γ€—β‘πœƒ ) y = 〖𝑠𝑖𝑛〗^(βˆ’1) (cos⁑2πœƒ) 𝑦 =〖𝑠𝑖𝑛〗^(βˆ’1) (γ€–sin 〗⁑(πœ‹/2 βˆ’2πœƒ) ) 𝑦 = πœ‹/2 βˆ’ 2πœƒ Putting value of ΞΈ = tanβˆ’1 x 𝑦 = πœ‹/2 βˆ’ 2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . (𝑑(𝑦))/𝑑π‘₯ = (𝑑 (" " πœ‹/2 " βˆ’ " 2γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯" " ))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 0 βˆ’ 2 (𝑑〖 (π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = βˆ’2 (𝑑〖 (π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = βˆ’2 (1/(1 + π‘₯^2 )) π’…π’š/𝒅𝒙 = (βˆ’πŸ)/(𝟏 + 𝒙^𝟐 ) Since x = tan ΞΈ ∴ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) x = ΞΈ ((γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯") β€˜ = " 1/(1 + π‘₯^2 ))

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.