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Ex 5.7, 7 - Find second order derivatives of e6x cos 3x - Ex 5.7

Ex 5.7, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 7 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Ex 5.7, 7 Find the second order derivatives of the function 𝑒^6π‘₯ cos⁑3π‘₯ Let y = 𝑒^6π‘₯ cos⁑3π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ . 𝑑𝑦/𝑑π‘₯ = (𝑑(𝑒^6π‘₯ " " cos⁑3π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑑(𝑒^6π‘₯ )/𝑑π‘₯ .cos⁑3π‘₯ + (𝑑(cos⁑3π‘₯))/𝑑π‘₯ .γ€– 𝑒〗^6π‘₯ 𝑑𝑦/𝑑π‘₯ =γ€– 𝑒〗^6π‘₯ .𝑑(6π‘₯)/𝑑π‘₯ . cos⁑3π‘₯ + (γ€–βˆ’sin〗⁑3π‘₯) . (𝑑(3π‘₯))/𝑑π‘₯ . γ€– 𝑒〗^6π‘₯ Using product rule in 𝑒^6π‘₯ π‘π‘œπ‘ β‘3π‘₯ . As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 where u = 𝑒^6π‘₯ " & v =" cos⁑3π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑒^6π‘₯ . 6 . cos⁑3π‘₯ βˆ’ sin⁑3π‘₯ . 3 . 𝑒^6π‘₯ 𝑑𝑦/𝑑π‘₯ = 3γ€– 𝑒〗^6π‘₯ (2γ€– cos〗⁑3π‘₯ βˆ’ sin 3π‘₯) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = (𝑑(3 𝑒^6π‘₯ (2 cos⁑3π‘₯ βˆ’ sin⁑3π‘₯) ))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 3(𝑑(𝑒^6π‘₯ (2 cos⁑3π‘₯ βˆ’ sin⁑3π‘₯) ))/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 3 (𝑑(γ€– 𝑒〗^6π‘₯ )/𝑑π‘₯ ."(2" cos⁑3π‘₯ " βˆ’ sin " 3π‘₯") " +(𝑑(2 cos⁑3π‘₯ βˆ’ sin⁑3π‘₯))/𝑑π‘₯ ". " 𝑒^6π‘₯ ) Using product Rule As (𝑒𝑣)’= 𝑒’𝑣 + 𝑣’𝑒 where v = γ€– 𝑒〗^6π‘₯ & v = 2 cos 3x βˆ’ sin 3x (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 3 [6γ€– 𝑒〗^6π‘₯ "(2" γ€– cos〗⁑3π‘₯ " βˆ’ sin " 3π‘₯") + (" βˆ’"2" sin⁑〖3π‘₯.3γ€— " βˆ’ " cos⁑3π‘₯.3")" 𝑒^6π‘₯ ] (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 3[12γ€– 𝑒〗^6π‘₯ γ€–.cos〗⁑3π‘₯ "βˆ’ 6" γ€– 𝑒〗^6π‘₯ "sin " 3π‘₯βˆ’γ€–6 𝑒〗^6π‘₯ " sin " 3π‘₯βˆ’3γ€– 𝑒〗^6π‘₯ cos⁑3π‘₯ ] (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 3[9𝑒^6π‘₯ cos⁑3π‘₯βˆ’12𝑒^6π‘₯.sin⁑3π‘₯ ] (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 9𝑒^6π‘₯ [3 cos⁑3π‘₯βˆ’4 sin⁑3π‘₯ ] Hence , (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = πŸ—π’†^πŸ”π’™ [πŸ‘ π’„π’π’”β‘πŸ‘π’™βˆ’πŸ’ π’”π’Šπ’β‘πŸ‘π’™ ]

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.