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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 𝑥 assuming it exists. Let 𝑦 = sec^(–1) 𝑥 𝑠𝑒𝑐⁡𝑦= 𝑥 𝑥 =𝑠𝑒𝑐⁡𝑦 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (𝑠𝑒𝑐⁡𝑦 ))/𝑑𝑥 1 = (𝑑 (𝑠𝑒𝑐⁡𝑦 ))/𝑑𝑥 We need 𝑑𝑦 in denominator, so multiplying & Dividing by 𝑑𝑦. 1 = (𝑑 (sec⁡𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = tan⁡𝑦 .sec⁡𝑦 . 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 tan⁡𝑦 .sec⁡𝑦= 1 𝑑𝑦/𝑑𝑥 = 1/(𝒕𝒂𝒏⁡𝒚 .〖 sec〗⁡𝑦 ) 𝑑𝑦/𝑑𝑥 = 1/((√(〖𝐬𝐞𝐜〗^𝟐⁡𝒚 − 𝟏)) .〖 sec〗⁡𝑦 ) Putting value of 𝑠𝑒𝑐⁡𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = 1/((√(𝑥^2 − 1 ) ) . 𝑥) 𝑑𝑦/𝑑𝑥 = 1/(𝑥 √(𝑥^2 − 1 ) ) As 𝑦 = sec^(−1) 𝑥 So, 𝑠𝑒𝑐⁡𝑦 = 𝑥 Hence 𝒅(〖𝒔𝒆𝒄〗^(–𝟏) 𝒙)/𝒅𝒙 = 𝟏/(𝒙 √(𝒙^𝟐 − 𝟏 ) ) As tan2 θ = sec2 θ – 1, tan θ = √("sec2 θ – 1" )

About the Author

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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.