# Ex 5.7, 12

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.7, 12 If y= 𝑐𝑜𝑠−1 𝑥 , Find 𝑑2𝑦𝑑𝑥2 in terms of 𝑦 alone. Let y = 𝑐𝑜𝑠−1 𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦𝑑𝑥 = 𝑑( 𝑐𝑜𝑠−1 𝑥)𝑑𝑥 𝑑𝑦𝑑𝑥 = −1 1 − 𝑥2 Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝑑𝑥 −1 1 − 𝑥2 𝑑2𝑦𝑑 𝑥2 = 𝑑𝑑𝑥 −1 1 − 𝑥2 𝑑2𝑦𝑑 𝑥2 = − 𝑑 1 𝑑𝑥 . 1 − 𝑥2 − 𝑑 1 − 𝑥2 𝑑𝑥 . 1 1 − 𝑥2 2 𝑑2𝑦𝑑 𝑥2 = − 0 . 1 − 𝑥2 − 12 1 − 𝑥2 . 𝑑𝑑𝑥 (1 − 𝑥2) 1 − 𝑥2 2 = − 12 1 − 𝑥2 . 0−2𝑥 1 − 𝑥2 = − −1 − 2𝑥2 1 − 𝑥2 1 − 𝑥2 = −2𝑥2 1 − 𝑥2 1 − 𝑥2 = − 𝑥 1 − 𝑥2 12 1 − 𝑥2 = − 𝑥 1 − 𝑥2 12 + 1 𝒅𝟐𝒚𝒅 𝒙𝟐 = − 𝒙 𝟏 − 𝒙𝟐 𝟑𝟐 But we need to calculate 𝑑2𝑦𝑑 𝑥2 in terms of y . y = 𝑐𝑜𝑠−1 𝑥 cos𝑦 = 𝑥 𝑥 = cos𝑦 Putting 𝑥 = cos𝑦 in equation 𝑑2𝑦𝑑 𝑥2 = − 𝑥 1 − 𝑥2 32 𝑑2𝑦𝑑 𝑥2 = −cos𝑦 1 − ( cos𝑦)2 32 = − cos𝑦 ( sin2𝑦) 32 = − cos𝑦 ( sin𝑦) 2 × 32 = − cos𝑦 ( sin𝑦) 3 = cos𝑦 sin 𝑦 × 1 sin2𝑦 𝒅𝟐𝒚𝒅 𝒙𝟐 = 𝒄𝒐𝒕𝒚. 𝒄𝒐𝒔𝒆𝒄𝟐 𝒚

About the Author

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.