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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 ) ("As" 𝑑(γ€–π‘π‘œπ‘ γ€—^(βˆ’1) π‘₯)/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 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑/𝑑π‘₯ ((βˆ’1)/√(1 βˆ’ π‘₯^2 )) using Quotient Rule As, (𝑒/𝑣)^β€²= (𝑒’𝑣 βˆ’ 𝑣’𝑒)/𝑣^2 where u = 1 & v = √(1βˆ’π‘₯^2 ) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’ [ ((𝑑(1) )/𝑑π‘₯ . √(1 βˆ’ π‘₯^2 )βˆ’ (𝑑 (√(1 βˆ’ π‘₯^2 ) ))/𝑑π‘₯ . 1)/(√(1 βˆ’ π‘₯^2 ) )^2 ] (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = βˆ’[ (0 . √(1 βˆ’ π‘₯^2 )βˆ’ 1/(2 √(1 βˆ’ π‘₯^2 )) . 𝑑/𝑑π‘₯ (1 βˆ’ π‘₯^2) )/(√(1 βˆ’ π‘₯^2 ) )^2 ] = βˆ’[ ( 1/(2 √(1 βˆ’ π‘₯^2 )) . (0βˆ’2π‘₯))/((1 βˆ’ π‘₯^2 ) ) ] = βˆ’ (βˆ’1 (βˆ’ 2π‘₯))/(2 (√(1 βˆ’ π‘₯^2 ) ) (1 βˆ’ π‘₯^2 ) ) = ( βˆ’2π‘₯)/(2 (√(1 βˆ’ π‘₯^2 ) ) (1 βˆ’ π‘₯^2 ) ) = (βˆ’ π‘₯)/( (1 βˆ’ π‘₯^2 )^(1/2) (1 βˆ’ π‘₯^2 ) ) = (βˆ’ π‘₯)/( (1 βˆ’ π‘₯^2 )^(1/2 + 1) ) (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 ) ) = (βˆ’ cos⁑𝑦 " " )/( γ€–(sin⁑〖𝑦) γ€—γ€—^(2 Γ— 3/2 ) ) = (βˆ’ cos⁑𝑦 " " )/( γ€–(sin⁑〖𝑦) γ€—γ€—^(3 ) ) = (βˆ’cos⁑𝑦 " " )/( sin 𝑦 ) Γ— 1/sin2⁑𝑦 (𝒅^𝟐 π’š)/(𝒅𝒙^𝟐 ) = βˆ’π’„π’π’•β‘π’š. π’„π’π’”π’†π’„πŸ 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.