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

Misc 23 If ๐‘ฆ=๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ) , โ€“ 1 โ‰ค ๐‘ฅ โ‰ค 1, show that (1โˆ’๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/ใ€–๐‘‘๐‘ฅใ€—^2 โˆ’ ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆ’๐‘Ž2 ๐‘ฆ 0 . ๐‘ฆ=๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ) Let ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ=๐‘ก ๐‘ฆ=๐‘’^๐‘ก Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ก/๐‘‘๐‘ก ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘(๐‘’^๐‘ก )/๐‘‘๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^๐‘ก ร— ๐‘‘๐‘ก/๐‘‘๐‘ฅ Putting value of ๐‘ก=ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) ร— ๐‘‘(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) ร— ๐‘Ž ๐‘‘(ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) ร— ๐‘Ž ((โˆ’1)/โˆš(1 โˆ’ ๐‘ฅ^2 )) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’๐‘Ž ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ))/โˆš(1 โˆ’ ๐‘ฅ^2 ) โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’๐‘Ž ๐‘’^(ใ€–๐‘Ž ๐‘๐‘œ๐‘ ใ€—^(โˆ’1) ๐‘ฅ" " ) โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’๐‘Ž ๐‘ฆ Differentiating again w.r.t x ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)/๐‘‘๐‘ฅ = d(โˆ’๐‘Ž๐‘ฆ)/๐‘‘๐‘ฅ ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ^2 ))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘( ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)/๐‘‘๐‘ฅ = โˆ’๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (โˆ’1)/(2โˆš(1 โˆ’ ๐‘ฅ^2 ))ร— (1โˆ’๐‘ฅ^2 )^โ€ฒ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+โˆš(1 โˆ’ ๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ (โˆ’1)/(2โˆš(1 โˆ’ ๐‘ฅ^2 ))ร— (โˆ’2๐‘ฅ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+โˆš(1 โˆ’ ๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฅ/โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+โˆš(1 โˆ’ ๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Multiplying โˆš(1 โˆ’ ๐‘ฅ^2 ) both sides ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+(โˆš(1 โˆ’ ๐‘ฅ^2 ))^2 (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+(1โˆ’๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘Ž ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+(1โˆ’๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = โˆ’โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘Ž ร—(โˆ’๐‘Ž๐‘ฆ)/โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+(1โˆ’๐‘ฅ^2 ) (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) = ๐‘Ž^2 ๐‘ฆ Hence proved From (1) โˆš(1 โˆ’ ๐‘ฅ^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = โˆ’๐‘Ž ๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’๐‘Ž๐‘ฆ)/โˆš(1 โˆ’ ๐‘ฅ^2 )

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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.