Ex 5.4, 2 - Differentiate e sin-1 x - Class 12 CBSE NCERT - Ex 5.4

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.4, 2 (Method 1) Differentiate 𝑤.𝑟.𝑡. x in , 𝑒﷮ sin﷮−1﷯ 𝑥﷯ Let 𝑦 = 𝑒﷮ sin﷮−1﷯ 𝑥﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑 𝑦﷯﷮𝑑𝑥﷯ = 𝑑 𝑒﷮ sin﷮−1﷯ 𝑥﷯﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ sin﷮−1﷯ 𝑥﷯ . 𝑑 sin﷮−1﷯ 𝑥﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮ sin﷮−1﷯ 𝑥﷯ . 1﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝒅 𝒚﷯﷮𝒅𝒙﷯ = 𝒆﷮ 𝒔𝒊𝒏﷮−𝟏﷯ 𝒙﷯﷮ ﷮𝟏− 𝒙﷮𝟐﷯﷯﷯ Ex 5.4, 2 (Method 2) Differentiate 𝑤.𝑟.𝑡. x in , 𝑒﷮ sin﷮−1﷯ 𝑥﷯ Let 𝑦 = 𝑒﷮ sin﷮−1﷯ 𝑥﷯ Let sin﷮−1﷯ 𝑥=𝑡 𝑦 = 𝑒﷮𝑡﷯ Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑 𝑦﷯﷮𝑑𝑥﷯ = 𝑑 𝑒﷮𝑡﷯﷯﷮𝑑𝑥﷯ We need 𝑑𝑡 in denominator, so multiplying & Dividing by 𝑑𝑡 . 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑒﷮𝑡﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑒﷮𝑡﷯﷯﷮𝑑𝑥﷯ × 𝑑𝑡﷮𝑑𝑡﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑 𝑒﷮𝑡﷯﷯﷮𝑑𝑡﷯ × 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑒﷮𝑡﷯ × 𝑑𝑡﷮𝑑𝑥﷯ Putting value of 𝑡 𝑑𝑦﷮𝑑𝑥﷯= 𝑒﷮ sin﷮−1﷯ 𝑥﷯ × 𝑑 sin﷮−1﷯ 𝑥﷯﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑒﷮ sin﷮−1﷯ 𝑥﷯ × 1﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ 𝒅𝒚﷮𝒅𝒙﷯ = 𝒆﷮ 𝒔𝒊𝒏﷮−𝟏﷯ 𝒙﷯﷮ ﷮𝟏 − 𝒙﷮𝟐﷯﷯﷯

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