If y = log ((1 - x 2 )/(1 + x 2 )),then dy/dx is equal to
(A) ใ4x 3 /(1-x 4 )
(B) (-4x)/(1-x 4 )
(C) 1/(4-x 4 )
(D) (-4x 3 )/(1-x 4 )
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NCERT Exemplar - MCQs
Last updated at Dec. 16, 2024 by Teachoo
Question 19 If y = log ((1 โ ๐ฅ^2)/(1 + ๐ฅ^2 )),then ๐๐ฆ/๐๐ฅ is equal to (A) ใ4๐ฅใ^3/(1โ๐ฅ^4 ) (B) (โ4๐ฅ)/(1โ๐ฅ^4 ) (C) 1/(4โ๐ฅ^4 ) (D) (โ4๐ฅ^3)/(1โ๐ฅ^4 ) y=log((1 โ ๐ฅ^2)/(1 +ใ ๐ฅใ^2 )) ๐ฒ=๐ฅ๐จ๐ (๐โ๐^๐ )โ๐ฅ๐จ๐ โกใ(๐+๐^๐)ใ Differentiating wrt ๐ฅ ๐ ๐/๐ ๐=๐(log(1 โ ๐ฅ^2 ) โ logโกใ(1 + ๐ฅ^2)ใ )/๐๐ฅ =๐(log(1 โ ๐ฅ^2 ))/๐๐ฅโ๐(log(1 + ๐ฅ^2 ))/๐๐ฅ =๐/((๐ โ ๐^๐ ) ) ๐ (๐ โ ๐^๐ )/๐ ๐โ๐/((๐ + ๐^๐ ) ) ๐ (๐ + ๐^๐ )/๐ ๐ =1/((1 โ ๐ฅ^2 ) )(0โ2๐ฅ)โ1/((1 + ๐ฅ^2 ) ) (0+2๐ฅ) =(โ2๐ฅ)/((1 โ ๐ฅ^2 ) )โ2๐ฅ/((1 + ๐ฅ^2 ) ) =โ2๐ฅ(1/((1 โใ ๐ฅใ^2 ) )+1/((1 +ใ ๐ฅใ^2 ) )) =โ2๐ฅ((1 + ๐ฅ^2 + 1 โใ ๐ฅใ^2)/(1 โ ๐ฅ^2 )(1 + ๐ฅ^2 ) ) =โ2๐ฅ(2/(1 โใ ๐ฅใ^2 )(1 + ๐ฅ^2 ) ) =(โ๐๐)/(๐ โ ๐^๐ ) So, the correct answer is (B)