Ex 5.3, 14 (i).jpg

Ex 5.3 (ii).jpg

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

Transcript

Ex 5.3, 14 Find 𝑑𝑦/𝑑π‘₯ in, y = sin–1 (2π‘₯ √(1βˆ’π‘₯^2 )) , βˆ’ 1/√2 < x < 1/√2 y = sin–1 (2π‘₯ √(1βˆ’π‘₯^2 )) Putting π‘₯ =π‘ π‘–π‘›β‘πœƒ 𝑦 = sin–1 (2 sinβ‘πœƒ √(1βˆ’γ€–π‘ π‘–π‘›γ€—^2 πœƒ)) 𝑦 = sin–1 ( 2 sin ΞΈ √(γ€–π‘π‘œπ‘ γ€—^2 πœƒ) ) 𝑦 ="sin–1 " (γ€–"2 sin ΞΈ" 〗⁑cosβ‘πœƒ ) 𝑦 = sin–1 (sin⁑〖2 πœƒ)γ€— 𝑦 = 2ΞΈ Putting value of ΞΈ = sinβˆ’1 x 𝑦 = 2 〖𝑠𝑖𝑛〗^(βˆ’1) π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . (𝑑(𝑦))/𝑑π‘₯ = (𝑑 (γ€–2 sin^(βˆ’1)〗⁑π‘₯ ))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 (𝑑〖 (𝑠𝑖𝑛〗^(βˆ’1) π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 (1/√(1 βˆ’γ€– π‘₯γ€—^2 )) π’…π’š/𝒅𝒙 = 𝟐/√(𝟏 βˆ’ 𝒙^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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.