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

Last updated at Sept. 17, 2019 by Teachoo

Last updated at Sept. 17, 2019 by Teachoo

Transcript

Ex 5.3, 15 Find 𝑑𝑦𝑑𝑥 in, y = sec–1 1 2𝑥2−1 , 0 < x < 1 2 y = sec–1 1 2𝑥2−1 𝒔𝒆𝒄𝒚 = 12𝑥2−1 𝟏 𝐜𝐨𝐬𝒚 = 12 𝑥2−1 cos𝑦 = 2𝑥2−1 y = cos –1 (2𝑥2−1) Putting 𝑥 = 𝑐𝑜𝑠𝜃 𝑦 = cos –1 (2 𝑐𝑜𝑠2𝜃−1) 𝑦 = cos –1 ( cos2𝜃) 𝑦 = 2𝜃 Putting value of θ = cos−1 x 𝑦 = 2 𝑐𝑜𝑠−1 𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . 𝑑(𝑦)𝑑𝑥 = 𝑑 2 𝑐𝑜𝑠−1 𝑥 𝑑𝑥 𝑑𝑦𝑑𝑥 = 2 𝑑 𝑐𝑜𝑠−1 𝑥 𝑑𝑥 𝑑𝑦𝑑𝑥 = 2 . −1 1 − 𝑥2 𝒅𝒚𝒅𝒙 = −𝟐 𝟏 − 𝒙𝟐

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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.