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

Transcript

Misc 5 Differentiate ๐‘ค.๐‘Ÿ.๐‘ก. ๐‘ฅ the function, (ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2)/โˆš(2๐‘ฅ+7 ) , โ€“ 2 < ๐‘ฅ < 2 Let ๐‘ฆ= (ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2)/โˆš(2๐‘ฅ+7 ) Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘‘/๐‘‘๐‘ฅ ((ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2)/โˆš(2๐‘ฅ+7 )) Using quotient rule As (๐‘ข/๐‘ฃ)^โ€ฒ = (๐‘ข^โ€ฒ ๐‘ฃ โˆ’ ๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 where u = cosโˆ’1 ๐‘ฅ/2 & v = โˆš(2๐‘ฅ+7) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (๐‘‘(ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2)/๐‘‘๐‘ฅ . โˆš(2๐‘ฅ + 7 ) โˆ’ ๐‘‘(โˆš(2๐‘ฅ + 7 ))/๐‘‘๐‘ฅ . ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 )/(โˆš(2๐‘ฅ+7 ))^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((โˆ’1)/โˆš(1 โˆ’(๐‘ฅ/2)^2 ) . ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ/2) โˆš(2๐‘ฅ + 7 ) โˆ’ 1/(2โˆš(2๐‘ฅ + 7)) . ๐‘‘(2๐‘ฅ + 7)/๐‘‘๐‘ฅ . ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 )/(โˆš(2๐‘ฅ + 7 ))^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((โˆ’1)/โˆš(ใ€–(4 โˆ’ ๐‘ฅ)/4ใ€—^2 ) . 1/2 โˆš(2๐‘ฅ + 7 ) โˆ’ 1/(2โˆš(2๐‘ฅ + 7)) . (2 + 0) . ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 )/((2๐‘ฅ + 7) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((โˆ’2)/โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 ) . 1/2 โˆš(2๐‘ฅ + 7 ) โˆ’ 1/(2โˆš(2๐‘ฅ + 7)) . 2 . ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 )/((2๐‘ฅ + 7) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((โˆ’ โˆš(2๐‘ฅ + 7 ))/( โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 )) โˆ’ (. ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2)/โˆš(2๐‘ฅ + 7) )/((2๐‘ฅ + 7) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’ โˆš(2๐‘ฅ + 7 ) ( โˆš(2๐‘ฅ + 7 )) โˆ’ ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 ))/((2๐‘ฅ + 7) (โˆš(2๐‘ฅ + 7 )) (โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 )) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ใ€–โˆ’(โˆš(2๐‘ฅ + 7 ))ใ€—^2/((2๐‘ฅ + 7) ( โˆš(2๐‘ฅ + 7 )) (โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 )) ) "+ " (ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 ))/((2๐‘ฅ + 7) ( โˆš(2๐‘ฅ + 7 )) (โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 )) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’(2๐‘ฅ + 7))/((2๐‘ฅ + 7) โˆš(2๐‘ฅ + 7 ) โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 )) "+ " (ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 )/((2๐‘ฅ+7) โˆš(2๐‘ฅ + 7 ) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’(2๐‘ฅ + 7))/((2๐‘ฅ + 7) โˆš(2๐‘ฅ + 7 ) โˆš(ใ€–4 โˆ’ ๐‘ฅใ€—^2 )) "+ " (ใ€–๐‘๐‘œ๐‘ ใ€—^(โˆ’1 ) ๐‘ฅ/2 )/((2๐‘ฅ+7) (2๐‘ฅ + 7)^(1/2) ) ๐’…๐’š/๐’…๐’™ = (โˆ’๐Ÿ)/(โˆš(๐Ÿ๐’™ + ๐Ÿ•) โˆš(ใ€–๐Ÿ’ โˆ’ ๐’™ใ€—^๐Ÿ ))โˆ’(ใ€–๐’„๐’๐’”ใ€—^(โˆ’๐Ÿ ) ๐’™/๐Ÿ )/((๐Ÿ๐’™ + ๐Ÿ•)^(๐Ÿ‘/๐Ÿ) )

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.