# Ex 5.7, 17 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.7, 17 If 𝑦= ( 𝑡𝑎𝑛−1 𝑥)2 , show that ( 𝑥2+1)2 𝑦2 + 2𝑥 ( 𝑥2+1) 𝑦1 = 2 We have y = ( 𝑡𝑎𝑛−1 𝑥)2 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦𝑑𝑥 = 𝑑 ( ( 𝑡𝑎𝑛−1 𝑥)2 )𝑑𝑥 𝑑𝑦𝑑𝑥 = 2 𝑡𝑎𝑛−1 𝑥 . 𝑑 𝑡𝑎𝑛−1 𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 = 2 𝑡𝑎𝑛−1 𝑥 . 11 + 𝑥2 Hence, 𝑦1 = 𝑑𝑦𝑑𝑥 = 2 𝑡𝑎𝑛−1 𝑥1 + 𝑥2 Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 2 𝑡𝑎𝑛−1 𝑥1 + 𝑥2 𝑑2𝑦𝑑 𝑥2 = − 2 𝑑𝑑𝑥 2 𝑡𝑎𝑛−1 𝑥1 + 𝑥2 𝑑2𝑦𝑑 𝑥2 = 2 𝑑 2 𝑡𝑎𝑛−1 𝑥𝑑𝑥 . 1 + 𝑥2 − 𝑑 1 + 𝑥2𝑑𝑥 . 𝑡𝑎𝑛−1 𝑥 1 + 𝑥22 𝑑2𝑦𝑑 𝑥2 = 2 11 + 𝑥2 . 1 + 𝑥2 − 𝑑 1𝑑𝑥 + 𝑑 𝑥2𝑑𝑥 . 𝑡𝑎𝑛−1 𝑥 1 + 𝑥22 𝑑2𝑦𝑑 𝑥2 = 2 1− 0 + 2𝑥 𝑡𝑎𝑛−1 𝑥 1 + 𝑥22 𝑑2𝑦𝑑 𝑥2 = 2 1−2𝑥 . 𝑡𝑎𝑛−1 𝑥 1 + 𝑥22 Thus, 𝒚2 = 2 𝟏−𝟐𝒙 . 𝒕𝒂𝒏−𝟏 𝒙 𝟏 + 𝒙𝟐𝟐 We need to show ( 𝑥2+1)2 𝑦2 + 2𝑥 ( 𝑥2+1) 𝑦1 = 2 Solving LHS ( 𝑥2+1)2 𝑦2 + 2𝑥 ( 𝑥2+1) 𝑦1 = ( 𝑥2+1)2 . 2 1−2𝑥 . 𝑡𝑎𝑛−1 𝑥 1 + 𝑥22 + 2𝑥 (𝑥2+1) . 2 𝑡𝑎𝑛−1 𝑥1 + 𝑥2 = 2 (1−2𝑥 𝑡𝑎𝑛−1 𝑥) + 4𝑥 𝑡𝑎𝑛−1 𝑥 = 2 – 4x 𝑡𝑎𝑛−1 x + 4x 𝑡𝑎𝑛−1 x = 2 Hence proved .

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Davneet Singh

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.