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

Transcript

Ex 5.3, 9 Find 𝑑𝑦/𝑑π‘₯ in, y = sin^(βˆ’1) (2π‘₯/( 1 + 2π‘₯2 )) 𝑦 = sin^(βˆ’1) (2π‘₯/( 1 + 2π‘₯2 )) Putting x = tan ΞΈ 𝑦 = sin^(βˆ’1) (2π‘₯/( 1 + π‘₯2 )) 𝑦 = sin^(βˆ’1) ((2 π‘‘π‘Žπ‘›β‘πœƒ)/(1 + γ€–π‘‘π‘Žπ‘›γ€—^2 πœƒ)) 𝑦 = sin^(βˆ’1) ( sin⁑2ΞΈ) 𝑦 = 2πœƒ Putting value of ΞΈ = γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ 𝑦 = 2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ ("Since " 𝑠𝑖𝑛⁑2πœƒ" = " (2 π‘‘π‘Žπ‘›β‘πœƒ)/(1 + γ€–π‘‘π‘Žπ‘›γ€—^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 (2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯" ) " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 (𝑑 (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯" ) " )/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 (1/(1+ π‘₯^2 )) π’…π’š/𝒅𝒙 = 𝟐/(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.