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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Concept wise

Transcript

Ex 5.3, 13 Find 𝑑𝑦/𝑑π‘₯ in, y = cos–1 (2π‘₯/( 1+ π‘₯2 )) , βˆ’1 < x < 1 𝑦 = cos–1 (2π‘₯/( 1+ π‘₯2 )) Let π‘₯ = tanβ‘πœƒ 𝑦 = cos–1 ((2 tanβ‘πœƒ)/( 1 + π‘‘π‘Žπ‘›2πœƒ )) 𝑦 = cos–1 (sin 2ΞΈ) 𝑦 ="cos–1" (γ€–cos 〗⁑(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‘₯ Since x = tan ΞΈ ∴ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) x = ΞΈ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (𝑑(𝑦))/𝑑π‘₯ = (𝑑 (" " πœ‹/2 " βˆ’ " γ€–2π‘‘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 ) ((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.