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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 38 (Method 1) If y = 怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„, show that (1 ā€“ š‘„2) š‘‘2š‘¦/š‘‘š‘„2 āˆ’ š‘„ š‘‘š‘¦/š‘‘š‘„ = 0 . We have š‘¦ = 怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„ Differentiating š‘¤.š‘Ÿ.š‘”.š‘„ š‘‘š‘¦/š‘‘š‘„ = š‘‘(怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„)/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„ = 1/āˆš(怖1 āˆ’ š‘„怗^2 ) āˆš((šŸāˆ’š’™^šŸ ) ) š’š^ā€² = šŸ Squaring both sides ("As " š‘‘(怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„)/š‘‘š‘„ " = " 1/āˆš(怖1 āˆ’ š‘„怗^2 )) (āˆš((1āˆ’š‘„^2 ) ) š‘¦^ā€² )^2 = 1^2 (1āˆ’š‘„^2 )(š‘¦^ā€² )^2 = 1 Again Differentiating š‘¤.š‘Ÿ.š‘”.š‘„ š‘‘/š‘‘š‘„ ((1āˆ’š‘„^2 )(š‘¦^ā€² )^2 ) = (š‘‘(1))/š‘‘š‘„ d(1 āˆ’ x^2 )/š‘‘š‘„ (š‘¦^ā€² )^2+(1āˆ’š‘„^2 ) š‘‘((š‘¦^ā€² )^2 )/š‘‘š‘„ = 0 āˆ’2š‘„(š‘¦^ā€² )^2+(1āˆ’š‘„^2 ) 2š‘¦^ā€² Ɨ š‘¦^ā€²ā€² = 0 怖2y怗^ā€² [āˆ’š’™š’š^ā€²+(šŸāˆ’š’™^šŸ ) š’š^ā€²ā€² ] = 0 āˆ’š‘„š‘¦^ā€²+(1āˆ’š‘„^2 ) š‘¦^ā€²ā€²=0 (怖šŸāˆ’š’™ć€—^šŸ ) (š’…^šŸ š’š)/怖š’…š’™ć€—^šŸ āˆ’ š’™ . š’…š’š/š’…š’™ = 0 Example 38 (Method 2) If y = 怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„, show that (1 ā€“ š‘„2) š‘‘2š‘¦/š‘‘š‘„2 āˆ’ š‘„ š‘‘š‘¦/š‘‘š‘„ = 0 . We have š‘¦ = 怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„ Differentiating š‘¤.š‘Ÿ.š‘”.š‘„ š‘‘š‘¦/š‘‘š‘„ = š‘‘(怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„)/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„ = 1/āˆš(怖1 āˆ’ š‘„怗^2 ) š’…š’š/š’…š’™ = (怖šŸāˆ’š’™ć€—^šŸ )^((āˆ’šŸ)/( šŸ)) ("As " š‘‘(怖š‘ š‘–š‘›ć€—^(āˆ’1) š‘„)/š‘‘š‘„ " = " 1/āˆš(怖1 āˆ’ š‘„怗^2 )) Again Differentiating š‘¤.š‘Ÿ.š‘”.š‘„ š‘‘/š‘‘š‘„ (š‘‘š‘¦/š‘‘š‘„) = (š‘‘(怖1 āˆ’ š‘„怗^2 )^((āˆ’1)/( 2)))/š‘‘š‘„ (š‘‘^2 š‘¦)/怖š‘‘š‘„怗^2 = (āˆ’1)/( 2) (怖1āˆ’š‘„怗^2 )^((āˆ’1)/( 2) āˆ’1) . š‘‘(怖1 āˆ’ š‘„怗^2 )/š‘‘š‘„ (š‘‘^2 š‘¦)/怖š‘‘š‘„怗^2 = (āˆ’1)/( 2) (怖1āˆ’š‘„怗^2 )^((āˆ’3)/2 ). (0āˆ’2š‘„) (š‘‘^2 š‘¦)/怖š‘‘š‘„怗^2 = (āˆ’1)/( 2) (怖1āˆ’š‘„怗^2 )^((āˆ’3)/2 ). (āˆ’2š‘„) (š’…^šŸ š’š)/怖š’…š’™ć€—^šŸ = š’™(怖šŸāˆ’š’™ć€—^šŸ )^((āˆ’šŸ‘)/šŸ ) Now, We need to prove (怖1āˆ’š‘„怗^2 ) (š‘‘^2 š‘¦)/怖š‘‘š‘„怗^2 āˆ’ š‘„ . š‘‘š‘¦/š‘‘š‘„ = 0 Solving LHS (怖1āˆ’š‘„怗^2 ) (š‘‘^2 š‘¦)/怖š‘‘š‘„怗^2 āˆ’ š‘„ . š‘‘š‘¦/š‘‘š‘„ = (怖1āˆ’š‘„怗^2 ) . (š‘„怖 (怖1āˆ’š‘„怗^2 )怗^((āˆ’3)/2 ) ) āˆ’ š‘„ (怖1āˆ’š‘„怗^2 )^((āˆ’1)/( 2)) = š‘„怖 (怖1āˆ’š‘„怗^2 )怗^(šŸ + ((āˆ’šŸ‘)/šŸ) )āˆ’š‘„ (怖1āˆ’š‘„怗^2 )^((āˆ’1)/( 2)) = š‘„怖 (怖1āˆ’š‘„怗^2 )怗^((āˆ’1)/( 2))āˆ’š‘„ (怖1āˆ’š‘„怗^2 )^((āˆ’1)/( 2)) = 0 = RHS Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.