# Ex 5.3, 13 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Nov. 19, 2019 by Teachoo

Last updated at Nov. 19, 2019 by Teachoo

Transcript

Ex 5.3, 13 Find 𝑑𝑦𝑑𝑥 in, y = cos–1 2𝑥 1+ 𝑥2 , −1 < x < 1 y = cos–1 2𝑥 1+ 𝑥2 Let 𝑥 = tan𝜃 y = cos–1 2 tan𝜃 1+𝑡𝑎𝑛2𝜃 y = cos–1 ( sin 2 θ ) 𝑦 =cos–1 cos 𝜋2 −2 𝜃 𝑦 = 𝜋2 − 2 𝜃 Putting value of θ = tan−1 x 𝑦 = 𝜋2 − 2 𝑡𝑎𝑛−1 𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . 𝑑(𝑦)𝑑𝑥 = 𝑑 𝜋2 − 2 𝑡𝑎𝑛−1 𝑥 𝑑𝑥 𝑑𝑦𝑑𝑥 = 0 − 2 𝑑 (𝑡𝑎𝑛−1 𝑥)𝑑𝑥 𝑑𝑦𝑑𝑥 = − 2 𝑑 (𝑡𝑎𝑛−1 𝑥)𝑑𝑥 𝑑𝑦𝑑𝑥 = − 2 11 + 𝑥2 𝒅𝒚𝒅𝒙 = −𝟐𝟏 + 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.