Ex 5.3, 14 - Find dy/dx in, y= sin-1 (2x root 1-x2) - CBSE

Ex 5.3, 14 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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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‘₯ Since x = sin ΞΈ ∴ 〖𝑠𝑖𝑛〗^(βˆ’1) x = ΞΈ 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 ) ((sin^(βˆ’1)⁑π‘₯ )^β€²= 1/√(1 βˆ’ π‘₯^2 ))

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.