Ex 5.3, 13 - Find dy/dx in, y=cos-1 (2x/1+x2) - NCERT - Finding derivative of Inverse trigonometric functions

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.3, 13 Find 𝑑𝑦﷮𝑑𝑥﷯ in, y = cos–1 2𝑥﷮ 1+ 𝑥2 ﷯﷯ , −1 < x < 1 y = cos–1 2𝑥﷮ 1+ 𝑥2 ﷯﷯ Let 𝑥 = tan⁡𝜃 y = cos–1 2 tan﷮𝜃﷯﷮ 1+𝑡𝑎𝑛2𝜃 ﷯﷯ y = cos–1 ( sin 2 θ ) 𝑦 =cos–1 cos ﷮ 𝜋﷮2﷯ −2 𝜃﷯﷯﷯ 𝑦 = 𝜋﷮2﷯ − 2 𝜃 Putting value of θ = tan−1 x 𝑦 = 𝜋﷮2﷯ − 2 𝑡𝑎𝑛﷮−1﷯ 𝑥 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 . 𝑑(𝑦)﷮𝑑𝑥﷯ = 𝑑 𝜋﷮2﷯ − 2 𝑡𝑎𝑛﷮−1﷯ 𝑥 ﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 0 − 2 𝑑 (𝑡𝑎𝑛﷮−1﷯ 𝑥)﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 2 𝑑 (𝑡𝑎𝑛﷮−1﷯ 𝑥)﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = − 2 1﷮1 + 𝑥﷮2﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = −𝟐﷮𝟏 + 𝒙﷮𝟐﷯﷯

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.