If  y = log ((1 - x 2 )/(1 + x 2 )),then dy/dx is equal to

(A) ใ€–4x 3 /(1-x 4

(B) (-4x)/(1-x 4 )

(C) 1/(4-x 4

(D) (-4x 3 )/(1-x 4 )

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Transcript

Question 19 If y = log ((1 โˆ’ ๐‘ฅ^2)/(1 + ๐‘ฅ^2 )),then ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ is equal to (A) ใ€–4๐‘ฅใ€—^3/(1โˆ’๐‘ฅ^4 ) (B) (โˆ’4๐‘ฅ)/(1โˆ’๐‘ฅ^4 ) (C) 1/(4โˆ’๐‘ฅ^4 ) (D) (โˆ’4๐‘ฅ^3)/(1โˆ’๐‘ฅ^4 ) y=log((1 โˆ’ ๐‘ฅ^2)/(1 +ใ€– ๐‘ฅใ€—^2 )) ๐ฒ=๐ฅ๐จ๐ (๐Ÿโˆ’๐’™^๐Ÿ )โˆ’๐ฅ๐จ๐ โกใ€–(๐Ÿ+๐’™^๐Ÿ)ใ€— Differentiating wrt ๐‘ฅ ๐’…๐’š/๐’…๐’™=๐‘‘(log(1 โˆ’ ๐‘ฅ^2 ) โˆ’ logโกใ€–(1 + ๐‘ฅ^2)ใ€— )/๐‘‘๐‘ฅ =๐‘‘(log(1 โˆ’ ๐‘ฅ^2 ))/๐‘‘๐‘ฅโˆ’๐‘‘(log(1 + ๐‘ฅ^2 ))/๐‘‘๐‘ฅ =๐Ÿ/((๐Ÿ โˆ’ ๐’™^๐Ÿ ) ) ๐’…(๐Ÿ โˆ’ ๐’™^๐Ÿ )/๐’…๐’™โˆ’๐Ÿ/((๐Ÿ + ๐’™^๐Ÿ ) ) ๐’…(๐Ÿ + ๐’™^๐Ÿ )/๐’…๐’™ =1/((1 โˆ’ ๐‘ฅ^2 ) )(0โˆ’2๐‘ฅ)โˆ’1/((1 + ๐‘ฅ^2 ) ) (0+2๐‘ฅ) =(โˆ’2๐‘ฅ)/((1 โˆ’ ๐‘ฅ^2 ) )โˆ’2๐‘ฅ/((1 + ๐‘ฅ^2 ) ) =โˆ’2๐‘ฅ(1/((1 โˆ’ใ€– ๐‘ฅใ€—^2 ) )+1/((1 +ใ€– ๐‘ฅใ€—^2 ) )) =โˆ’2๐‘ฅ((1 + ๐‘ฅ^2 + 1 โˆ’ใ€– ๐‘ฅใ€—^2)/(1 โˆ’ ๐‘ฅ^2 )(1 + ๐‘ฅ^2 ) ) =โˆ’2๐‘ฅ(2/(1 โˆ’ใ€– ๐‘ฅใ€—^2 )(1 + ๐‘ฅ^2 ) ) =(โˆ’๐Ÿ’๐’™)/(๐Ÿ โˆ’ ๐’™^๐Ÿ’ ) So, the correct answer is (B)

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.