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

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Ex 5.3, 15 Find 𝑑𝑦﷮𝑑𝑥﷯ in, y = sec–1 1﷮ 2𝑥2−1 ﷯﷯, 0 < x < 1﷮ ﷮2﷯﷯ y = sec–1 1﷮ 2𝑥2−1 ﷯﷯ 𝒔𝒆𝒄⁡𝒚 = 1﷮2𝑥2−1﷯ 𝟏﷮ 𝐜𝐨𝐬﷮𝒚﷯﷯ = 1﷮2 𝑥﷮2﷯−1﷯ cos⁡𝑦 = 2𝑥2−1 y = cos –1 (2𝑥2−1) Putting 𝑥 = 𝑐𝑜𝑠⁡𝜃 𝑦 = cos –1 (2 𝑐𝑜𝑠2𝜃−1) 𝑦 = cos –1 ( cos﷮2﷯𝜃) 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﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2 𝑑 𝑐𝑜𝑠﷮−1﷯ 𝑥 ﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 2 . −1﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = −𝟐﷮ ﷮𝟏 − 𝒙﷮𝟐﷯﷯﷯