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

Transcript

Ex 5.3, 10 Find 𝑑𝑦/𝑑π‘₯ in, 𝑦 = tan–1 ((3π‘₯βˆ’ π‘₯^3)/( 1βˆ’ 3π‘₯2 )) , βˆ’ 1/√3 < π‘₯ < 1/√3 𝑦 = tan–1 ((3π‘₯βˆ’ π‘₯^3)/( 1βˆ’ 3π‘₯2 )) Putting x = tan ΞΈ y = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) ((3 tanβ‘γ€–πœƒ βˆ’ tan⁑3 πœƒγ€—)/(1 βˆ’ 3 tan⁑2 πœƒ)) y = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (tan⁑〖3 πœƒγ€—) 𝑦 = 3πœƒ Putting value of ΞΈ = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ 𝑦 = 3γ€– (π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) (π‘‘π‘Žπ‘›β‘3ΞΈ " = " (3 π‘‘π‘Žπ‘›β‘γ€–πœƒβˆ’π‘‘π‘Žπ‘›β‘3 πœƒγ€—)/(1βˆ’3 π‘‘π‘Žπ‘›β‘2 πœƒ)) 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 3γ€–(π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯") " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ =3 (𝑑〖(π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯") " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 3 (1/(1 + π‘₯^2 )) π’…π’š/𝒅𝒙 = πŸ‘/(𝟏 +γ€– 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.