Ex 5.7, 7 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Transcript

Ex 5.7, 7 Find the second order derivatives of the function 𝑒^6𝑥 cos⁡3𝑥 Let y = 𝑒^6𝑥 cos⁡3𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 . 𝑑𝑦/𝑑𝑥 = (𝑑(𝑒^6𝑥 " " cos⁡3𝑥))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(𝑒^6𝑥 )/𝑑𝑥 .cos⁡3𝑥 + (𝑑(cos⁡3𝑥))/𝑑𝑥 .〖 𝑒〗^6𝑥 𝑑𝑦/𝑑𝑥 =〖 𝑒〗^6𝑥 .𝑑(6𝑥)/𝑑𝑥 . cos⁡3𝑥 + (〖−sin〗⁡3𝑥) . (𝑑(3𝑥))/𝑑𝑥 . 〖 𝑒〗^6𝑥 Using product rule in 𝑒^6𝑥 𝑐𝑜𝑠⁡3𝑥 . As (𝑢𝑣)’= 𝑢’𝑣 + 𝑣’𝑢 where u = 𝑒^6𝑥 " & v =" cos⁡3𝑥 𝑑𝑦/𝑑𝑥 = 𝑒^6𝑥 . 6 . cos⁡3𝑥 − sin⁡3𝑥 . 3 . 𝑒^6𝑥 𝑑𝑦/𝑑𝑥 = 3〖 𝑒〗^6𝑥 (2〖 cos〗⁡3𝑥 − sin 3𝑥) Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑/𝑑𝑥 (𝑑𝑦/𝑑𝑥) = (𝑑(3 𝑒^6𝑥 (2 cos⁡3𝑥 − sin⁡3𝑥) ))/𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 3(𝑑(𝑒^6𝑥 (2 cos⁡3𝑥 − sin⁡3𝑥) ))/𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 3 (𝑑(〖 𝑒〗^6𝑥 )/𝑑𝑥 ."(2" cos⁡3𝑥 " − sin " 3𝑥") " +(𝑑(2 cos⁡3𝑥 − sin⁡3𝑥))/𝑑𝑥 ". " 𝑒^6𝑥 ) Using product Rule As (𝑢𝑣)’= 𝑢’𝑣 + 𝑣’𝑢 where v = 〖 𝑒〗^6𝑥 & v = 2 cos 3x − sin 3x (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 3 [6〖 𝑒〗^6𝑥 "(2" 〖 cos〗⁡3𝑥 " − sin " 3𝑥") + (" −"2" sin⁡〖3𝑥.3〗 " − " cos⁡3𝑥.3")" 𝑒^6𝑥 ] (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 3[12〖 𝑒〗^6𝑥 〖.cos〗⁡3𝑥 "− 6" 〖 𝑒〗^6𝑥 "sin " 3𝑥−〖6 𝑒〗^6𝑥 " sin " 3𝑥−3〖 𝑒〗^6𝑥 cos⁡3𝑥 ] (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 3[9𝑒^6𝑥 cos⁡3𝑥−12𝑒^6𝑥.sin⁡3𝑥 ] (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 9𝑒^6𝑥 [3 cos⁡3𝑥−4 sin⁡3𝑥 ] Hence , (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 ) = 𝟗𝒆^𝟔𝒙 [𝟑 𝒄𝒐𝒔⁡𝟑𝒙−𝟒 𝒔𝒊𝒏⁡𝟑𝒙 ]