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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.