# Ex 5.7, 7

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 5.7, 7 Find the second order derivatives of the function 𝑒6𝑥 cos3𝑥 Let y = 𝑒6𝑥 cos3𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 . 𝑑𝑦𝑑𝑥 = 𝑑( 𝑒6𝑥 cos3𝑥)𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑 𝑒6𝑥𝑑𝑥 . cos3𝑥+ 𝑑( cos3𝑥)𝑑𝑥 . 𝑒6𝑥 𝑑𝑦𝑑𝑥 = 𝑒6𝑥 . 𝑑 6𝑥𝑑𝑥 . cos3𝑥 + ( −sin3𝑥) . 𝑑(3𝑥)𝑑𝑥 . 𝑒6𝑥 𝑑𝑦𝑑𝑥 = 𝑒6𝑥 . 6 . cos3𝑥 − sin3𝑥 . 3 . 𝑒6𝑥 𝑑𝑦𝑑𝑥 = 3 𝑒6𝑥 (2 cos3𝑥 − sin 3𝑥) Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑑𝑥 𝑑𝑦𝑑𝑥 = 𝑑(3 𝑒6𝑥(2 cos3𝑥 − sin3𝑥) )𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = 3 𝑑( 𝑒6𝑥(2 cos3𝑥 − sin3𝑥) )𝑑𝑥 𝑑2𝑦𝑑 𝑥2 = 3 𝑑 𝑒6𝑥𝑑𝑥 .(2 cos3𝑥 − sin 3𝑥) + 𝑑(2 cos3𝑥 − sin3𝑥)𝑑𝑥. 𝑒6𝑥 𝑑2𝑦𝑑 𝑥2 = 3 6 𝑒6𝑥(2 cos3𝑥 − sin 3𝑥) + (−2 sin3𝑥.3 − 𝑐𝑜𝑠3𝑥.3)e6x 𝑑2𝑦𝑑 𝑥2 = 3 12 𝑒6𝑥 .cos3𝑥− 6 𝑒6𝑥sin 3𝑥− 6 𝑒6𝑥6 sin 3𝑥−3 𝑒6𝑥 cos3𝑥 𝑑2𝑦𝑑 𝑥2 = 3 9 𝑒6𝑥 cos3𝑥−12 𝑒6𝑥. sin3𝑥 𝑑2𝑦𝑑 𝑥2 = 9 𝑒6𝑥 3 cos3𝑥−4 sin3𝑥 Hence , 𝒅𝟐𝒚𝒅 𝒙𝟐 = 𝟗 𝒆𝟔𝒙 𝟑 𝒄𝒐𝒔𝟑𝒙−𝟒 𝒔𝒊𝒏𝟑𝒙

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.