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Last updated at May 29, 2018 by Teachoo

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Ex 5.7, 9 Find the second order derivatives of the function log ( log𝑥) Let y = log ( log𝑥) Differentiating 𝑤.𝑟.𝑡.𝑥 . 𝑑𝑦𝑑𝑥 = 𝑑( log ( log𝑥))𝑑𝑥 𝑑𝑦𝑑𝑥 = 1 log𝑥 . 𝑑( log𝑥)𝑑𝑥 𝑑𝑦𝑑𝑥 = 1 log𝑥 . 1𝑥 𝑑𝑦𝑑𝑥 = 1 𝑥 . log𝑥 Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 1 𝑥 . log𝑥 𝑑2𝑦𝑑 𝑥2 = 𝑑(1)𝑑𝑥 𝑥 . log𝑥 − 𝑑 𝑥 . log𝑥𝑑𝑥 . 1 𝑥 . log𝑥2 𝑑2𝑦𝑑 𝑥2 = 0 . 𝑥 . log𝑥 − 𝑑 𝑥 . log𝑥𝑑𝑥 . 1 𝑥 . log𝑥2 𝑑2𝑦𝑑 𝑥2 = − 𝑑 𝑥 . log𝑥𝑑𝑥 𝑥 . log𝑥2 𝑑2𝑦𝑑 𝑥2 = − 𝑑(𝑥)𝑑𝑥 . log𝑥 + 𝑑( log𝑥)𝑑𝑥 . 𝑥 𝑥 . log𝑥2 𝑑2𝑦𝑑 𝑥2 = 1. log𝑥 + 1𝑥 × 𝑥 𝑥. log𝑥2 𝑑2𝑦𝑑 𝑥2 = − log𝑥 +1 𝑥 . log𝑥2 Thus , 𝒅𝟐𝒚𝒅 𝒙𝟐 = − 𝒍𝒐𝒈𝒙 +𝟏 𝒙 . 𝒍𝒐𝒈𝒙𝟐

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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.