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Ex 5.7, 17 - If y = (tan-1 x)2, show (x2 + 1) y2 + 2x(x2 + 1)

Ex 5.7, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.7, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 4
Ex 5.7, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 5
Ex 5.7, 17 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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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 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.