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

Transcript

Ex 5.7, 12 If y= γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ , Find 𝑑2𝑦/𝑑π‘₯2 in terms of 𝑦 alone. Let y = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/√(1 βˆ’ π‘₯^2 ) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑/𝑑π‘₯ ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) ("As" 𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯)/𝑑π‘₯=(βˆ’1)/√(1 βˆ’ π‘₯^2 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑/𝑑π‘₯ ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’π‘‘/𝑑π‘₯ (1βˆ’π‘₯^2 )^((βˆ’1)/2) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’((βˆ’1)/2) (1βˆ’π‘₯^2 )^((βˆ’1)/2 βˆ’ 1)Γ— (1βˆ’π‘₯^2 )^β€² (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 1/2 (1βˆ’π‘₯^2 )^((βˆ’3)/2) Γ— (βˆ’2π‘₯) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’π‘₯(1βˆ’π‘₯^2 )^((βˆ’3)/2) (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = (βˆ’ 𝒙)/( (𝟏 βˆ’ 𝒙^𝟐 )^(πŸ‘/𝟐 ) ) But we need to calculate (𝑑^2 𝑦)/(𝑑π‘₯^2 ) in terms of y . y = γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯ cos⁑𝑦 = π‘₯ π‘₯ = cos⁑𝑦 Putting π‘₯ = cos⁑𝑦 in equation (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’ π‘₯)/( (1 βˆ’ π‘₯^2 )^(3/2 ) ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = (βˆ’cos⁑𝑦 " " )/( (1 βˆ’ γ€–(cos⁑𝑦)γ€—^2 )^(3/2 ) ) = (βˆ’ cos⁑𝑦 " " )/( γ€–(sin2⁑〖𝑦) γ€—γ€—^(3/2 ) ) (As 1βˆ’cos⁑〖2 γ€— πœƒ = sin⁑2 πœƒ) = (βˆ’ cos⁑𝑦 " " )/( γ€–(sin⁑〖𝑦) γ€—γ€—^(2 Γ— 3/2 ) ) = (βˆ’ π‘π‘œπ‘ β‘π‘¦)/( γ€–(𝑠𝑖𝑛⁑〖𝑦) γ€—γ€—^3 ) = (βˆ’ π‘π‘œπ‘ β‘π‘¦)/sin⁑𝑦 Γ—1/( γ€–(𝑠𝑖𝑛⁑𝑦)γ€—^2 ) = βˆ’πœπ¨π­β‘π’š 𝒄𝒐𝒔𝒆𝒄^𝟐 π’š

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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.