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Ex 5.3, 15 - Find dy/dx in y= sec-1 (1 / 2x2 - 1) - Chapter 5

Ex 5.3, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.3, 15 Find 𝑑𝑦/𝑑π‘₯ in, y = sec–1 (1/( 2π‘₯2βˆ’1 )), 0 < x < 1/√2 y = sec–1 (1/( 2π‘₯^2 βˆ’ 1 )) π’”π’†π’„β‘π’š = 1/(2π‘₯^2 βˆ’ 1) 𝟏/πœπ¨π¬β‘π’š = 1/(2π‘₯^2 βˆ’ 1) cos⁑𝑦 = 2π‘₯2βˆ’1 y = cos –1 (2π‘₯2βˆ’1) Putting π‘₯ = cos⁑θ 𝑦 = cos –1 (2π‘π‘œπ‘ 2πœƒβˆ’1) 𝑦 = cos –1 (cos⁑2 πœƒ) 𝑦 = 2πœƒ Putting value of ΞΈ = cosβˆ’1 x 𝑦 = 2 γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ . (𝑑(𝑦))/𝑑π‘₯ = (𝑑 (2γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 (𝑑 (γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯" " ))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 . ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) π’…π’š/𝒅𝒙 = (βˆ’πŸ)/√(𝟏 βˆ’ 𝒙^𝟐 ) (π‘π‘œπ‘ β‘2πœƒ " = 2 " γ€–π‘π‘œπ‘ γ€—^2 πœƒβˆ’1) Since x = cos ΞΈ ∴ γ€–π‘π‘œπ‘ γ€—^(βˆ’1) x = ΞΈ ((γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯")β€˜ = " (βˆ’1)/√(1 βˆ’ π‘₯^2 ))

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.