Check Full Chapter Explained - Continuity and Differentiability - Continuity and Differentiability Class 12

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

Transcript

Example 41 (Method 1) If y = ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ, show that (1 โ€“ ๐‘ฅ2) ๐‘‘2๐‘ฆ/๐‘‘๐‘ฅ2 โˆ’ ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 . We have ๐‘ฆ = ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/โˆš(ใ€–1 โˆ’ ๐‘ฅใ€—^2 ) โˆš((๐Ÿโˆ’๐’™^๐Ÿ ) ) ๐’š^โ€ฒ = ๐Ÿ Squaring both sides ("As " ๐‘‘(ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ " = " 1/โˆš(ใ€–1 โˆ’ ๐‘ฅใ€—^2 )) (โˆš((1โˆ’๐‘ฅ^2 ) ) ๐‘ฆ^โ€ฒ )^2 = 1^2 (1โˆ’๐‘ฅ^2 )(๐‘ฆ^โ€ฒ )^2 = 1 Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ ((1โˆ’๐‘ฅ^2 )(๐‘ฆ^โ€ฒ )^2 ) = (๐‘‘(1))/๐‘‘๐‘ฅ d(1 โˆ’ x^2 )/๐‘‘๐‘ฅ (๐‘ฆ^โ€ฒ )^2+(1โˆ’๐‘ฅ^2 ) ๐‘‘((๐‘ฆ^โ€ฒ )^2 )/๐‘‘๐‘ฅ = 0 โˆ’2๐‘ฅ(๐‘ฆ^โ€ฒ )^2+(1โˆ’๐‘ฅ^2 ) 2๐‘ฆ^โ€ฒ ร— ๐‘ฆ^โ€ฒโ€ฒ = 0 ใ€–2yใ€—^โ€ฒ [โˆ’๐’™๐’š^โ€ฒ+(๐Ÿโˆ’๐’™^๐Ÿ ) ๐’š^โ€ฒโ€ฒ ] = 0 โˆ’๐‘ฅ๐‘ฆ^โ€ฒ+(1โˆ’๐‘ฅ^2 ) ๐‘ฆ^โ€ฒโ€ฒ=0 (ใ€–๐Ÿโˆ’๐’™ใ€—^๐Ÿ ) (๐’…^๐Ÿ ๐’š)/ใ€–๐’…๐’™ใ€—^๐Ÿ โˆ’ ๐’™ . ๐’…๐’š/๐’…๐’™ = 0 (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 = (โˆ’1)/( 2) (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’3)/2 ). (0โˆ’2๐‘ฅ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 = (โˆ’1)/( 2) (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’3)/2 ). (โˆ’2๐‘ฅ) (๐’…^๐Ÿ ๐’š)/ใ€–๐’…๐’™ใ€—^๐Ÿ = ๐’™(ใ€–๐Ÿโˆ’๐’™ใ€—^๐Ÿ )^((โˆ’๐Ÿ‘)/๐Ÿ ) Example 41 (Method 2) If y = ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ, show that (1 โ€“ ๐‘ฅ2) ๐‘‘2๐‘ฆ/๐‘‘๐‘ฅ2 โˆ’ ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 . We have ๐‘ฆ = ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1/โˆš(ใ€–1 โˆ’ ๐‘ฅใ€—^2 ) ๐’…๐’š/๐’…๐’™ = (ใ€–๐Ÿโˆ’๐’™ใ€—^๐Ÿ )^((โˆ’๐Ÿ)/( ๐Ÿ)) ("As " ๐‘‘(ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ " = " 1/โˆš(ใ€–1 โˆ’ ๐‘ฅใ€—^2 )) Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = (๐‘‘(ใ€–1 โˆ’ ๐‘ฅใ€—^2 )^((โˆ’1)/( 2)))/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 = (โˆ’1)/( 2) (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’1)/( 2) โˆ’1) . ๐‘‘(ใ€–1 โˆ’ ๐‘ฅใ€—^2 )/๐‘‘๐‘ฅ (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 = (โˆ’1)/( 2) (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’3)/2 ). (0โˆ’2๐‘ฅ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 = (โˆ’1)/( 2) (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’3)/2 ). (โˆ’2๐‘ฅ) (๐’…^๐Ÿ ๐’š)/ใ€–๐’…๐’™ใ€—^๐Ÿ = ๐’™(ใ€–๐Ÿโˆ’๐’™ใ€—^๐Ÿ )^((โˆ’๐Ÿ‘)/๐Ÿ ) Now, We need to prove (ใ€–1โˆ’๐‘ฅใ€—^2 ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 โˆ’ ๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 Solving LHS (ใ€–1โˆ’๐‘ฅใ€—^2 ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 โˆ’ ๐‘ฅ . ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (ใ€–1โˆ’๐‘ฅใ€—^2 ) . (๐‘ฅใ€– (ใ€–1โˆ’๐‘ฅใ€—^2 )ใ€—^((โˆ’3)/2 ) ) โˆ’ ๐‘ฅ (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’1)/( 2)) = ๐‘ฅใ€– (ใ€–1โˆ’๐‘ฅใ€—^2 )ใ€—^(๐Ÿ + ((โˆ’๐Ÿ‘)/๐Ÿ) )โˆ’๐‘ฅ (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’1)/( 2)) = ๐‘ฅใ€– (ใ€–1โˆ’๐‘ฅใ€—^2 )ใ€—^((โˆ’1)/( 2))โˆ’๐‘ฅ (ใ€–1โˆ’๐‘ฅใ€—^2 )^((โˆ’1)/( 2)) = 0 = RHS 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 10 years. He provides courses for Maths and Science at Teachoo.