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

Transcript

Find the derivative of f given by f (x) = sec–1 π‘₯ assuming it exists. Let 𝑦 = sec^(–1) π‘₯ 𝑠𝑒𝑐⁑𝑦= π‘₯ π‘₯ =𝑠𝑒𝑐⁑𝑦 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑π‘₯/𝑑π‘₯ = (𝑑 (𝑠𝑒𝑐⁑𝑦 ))/𝑑π‘₯ 1 = (𝑑 (𝑠𝑒𝑐⁑𝑦 ))/𝑑π‘₯ We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = (𝑑 (sec⁑𝑦 ))/𝑑π‘₯ Γ— 𝑑𝑦/𝑑𝑦 1 = tan⁑𝑦 .sec⁑𝑦 . 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ tan⁑𝑦 .sec⁑𝑦= 1 𝑑𝑦/𝑑π‘₯ = 1/(π’•π’‚π’β‘π’š .γ€– sec〗⁑𝑦 ) 𝑑𝑦/𝑑π‘₯ = 1/((√(γ€–π¬πžπœγ€—^πŸβ‘π’š βˆ’ 𝟏)) .γ€– sec〗⁑𝑦 ) Putting value of 𝑠𝑒𝑐⁑𝑦 = π‘₯ 𝑑𝑦/𝑑π‘₯ = 1/((√(π‘₯^2 βˆ’ 1 ) ) . π‘₯) 𝑑𝑦/𝑑π‘₯ = 1/(π‘₯ √(π‘₯^2 βˆ’ 1 ) ) As 𝑦 = sec^(βˆ’1) π‘₯ So, 𝑠𝑒𝑐⁑𝑦 = π‘₯ Hence 𝒅(〖𝒔𝒆𝒄〗^(β€“πŸ) 𝒙)/𝒅𝒙 = 𝟏/(𝒙 √(𝒙^𝟐 βˆ’ 𝟏 ) ) As tan2 ΞΈ = sec2 ΞΈ – 1, tan ΞΈ = √("sec2 ΞΈ – 1" )

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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.