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

Transcript

Ex 5.7, 17 (Method 1) If 𝑦= γ€–(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)γ€—^(2 ), show that γ€–(π‘₯^2+1)γ€—^(2 ) 𝑦2 + 2π‘₯ γ€–(π‘₯^2+1)γ€—^ 𝑦1 = 2 We have y = γ€–(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)γ€—^(2 ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ y’ = 2 tanβˆ’1 π‘₯ Γ— 1/(1 + π‘₯^2 ) (1 + π‘₯^2) y’ = 2 tanβˆ’1 π‘₯ Again differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ [𝑦^β€² (1+π‘₯^2 )]^β€² = 2 Γ— 1/(1 +γ€– π‘₯γ€—^2 ) [𝑦^β€² (1+π‘₯^2 )]^β€² = 2/(1 + π‘₯^2 ) (tan^(βˆ’1)⁑π‘₯ )^β€²=1/(1+π‘₯^2 ) Using product rule (1+π‘₯^2 )^β€² + 𝑦^β€²β€² (1 + π‘₯^2) = 2/(1 + π‘₯^2 ) 2π‘₯ 𝑦^β€²+𝑦^β€²β€² (1 +π‘₯^2 ) = 2/(1 + π‘₯^2 ) 2π‘₯ 𝑦^β€² (1 +π‘₯^2 )+𝑦^β€²β€² (1 +π‘₯^2 )Γ—(1 +π‘₯^2 ) = 2 2π‘₯(1 +π‘₯^2 ) 𝑦^β€²+𝑦^β€²β€² (1 +π‘₯^2 )^2 = 2 π’š^β€²β€² (𝟏 +𝒙^𝟐 )^𝟐+πŸπ’™(𝟏 +𝒙^𝟐 ) π’š^β€² = 2 Hence Proved Ex 5.7, 17 (Method 2) If 𝑦= γ€–(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)γ€—^(2 ), show that γ€–(π‘₯^2+1)γ€—^(2 ) 𝑦2 + 2π‘₯ γ€–(π‘₯^2+1)γ€—^ 𝑦1 = 2 We have y = γ€–(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)γ€—^(2 ) Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑 (γ€–(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)γ€—^(2 )))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ . 𝑑(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = 2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ . 1/(1 +γ€– π‘₯γ€—^2 ) Hence, 𝑦1 = 𝑑𝑦/𝑑π‘₯ = (2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 ) Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑/𝑑π‘₯ ((2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 2 𝑑/𝑑π‘₯ ((γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )) (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 2 [(𝑑(γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/𝑑π‘₯ . (1 +γ€– π‘₯γ€—^2 ) βˆ’ 𝑑(1 +γ€– π‘₯γ€—^2 )/𝑑π‘₯ .γ€– π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )^2 ] (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 2 [(1/(1 +γ€– π‘₯γ€—^2 ) . (1 +γ€– π‘₯γ€—^2 ) βˆ’ (𝑑(1)/𝑑π‘₯ + 𝑑(π‘₯^2 )/𝑑π‘₯) . γ€– π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )^2 ] Using quotient Rule As, (𝑒/𝑣)^β€²= (𝑒’𝑣 βˆ’ 𝑣’𝑒)/𝑣^2 where u = tan-1 x & v = 1 + x2 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 2 [( 1 βˆ’ (0 + 2π‘₯) γ€– π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )^2 ] (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 2 [( 1 βˆ’ 2π‘₯ .γ€– π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )^2 ] Thus, π’š2 = 2 [( 𝟏 βˆ’ πŸπ’™ .γ€– 𝒕𝒂𝒏〗^(βˆ’πŸ) 𝒙)/(𝟏 +γ€– 𝒙〗^𝟐 )^𝟐 ] We need to show γ€–(π‘₯^2+1)γ€—^(2 ) 𝑦2 + 2π‘₯ γ€–(π‘₯^2+1)γ€—^ 𝑦1 = 2 Solving LHS γ€–(π‘₯^2+1)γ€—^(2 ) 𝑦2 + 2π‘₯ γ€–(π‘₯^2+1)γ€—^ 𝑦1 = γ€–(π‘₯^2+1)γ€—^(2 ). 2 [( 1 βˆ’ 2π‘₯ .γ€– π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )^2 ] + 2π‘₯ (π‘₯2+1) . ((2 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯)/(1 +γ€– π‘₯γ€—^2 )) = 2 (1βˆ’2π‘₯ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯) + 4π‘₯ γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) π‘₯ = 2 – 4x 〖𝒕𝒂𝒏〗^(βˆ’πŸ) x + 4x 〖𝒕𝒂𝒏〗^(βˆ’πŸ) x = 2 Hence proved .

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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.