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Question 21 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at March 16, 2023 by

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

(A) γ€–4x 3 /(1-x 4 )Β 

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

(C) 1/(4-x 4 )Β 

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

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Transcript

Question 21 If y = log ((1 βˆ’ π‘₯^2)/(1 + π‘₯^2 )),then 𝑑𝑦/𝑑π‘₯ is equal to (A) γ€–4π‘₯γ€—^3/(1βˆ’π‘₯^4 ) (B) (βˆ’4π‘₯)/(1βˆ’π‘₯^4 ) (C) 1/(4βˆ’π‘₯^4 ) (D) (βˆ’4π‘₯^3)/(1βˆ’π‘₯^4 ) y=log((1 βˆ’ π‘₯^2)/(1 +γ€– π‘₯γ€—^2 )) 𝐲=π₯𝐨𝐠(πŸβˆ’π’™^𝟐 )βˆ’π₯𝐨𝐠⁑〖(𝟏+𝒙^𝟐)γ€— Differentiating wrt π‘₯ π’…π’š/𝒅𝒙=𝑑(log(1 βˆ’ π‘₯^2 ) βˆ’ log⁑〖(1 + π‘₯^2)γ€— )/𝑑π‘₯ =𝑑(log(1 βˆ’ π‘₯^2 ))/𝑑π‘₯βˆ’π‘‘(log(1 + π‘₯^2 ))/𝑑π‘₯ =𝟏/((𝟏 βˆ’ 𝒙^𝟐 ) ) 𝒅(𝟏 βˆ’ 𝒙^𝟐 )/π’…π’™βˆ’πŸ/((𝟏 + 𝒙^𝟐 ) ) 𝒅(𝟏 + 𝒙^𝟐 )/𝒅𝒙 =1/((1 βˆ’ π‘₯^2 ) )(0βˆ’2π‘₯)βˆ’1/((1 + π‘₯^2 ) ) (0+2π‘₯) =(βˆ’2π‘₯)/((1 βˆ’ π‘₯^2 ) )βˆ’2π‘₯/((1 + π‘₯^2 ) ) =βˆ’2π‘₯(1/((1 βˆ’γ€– π‘₯γ€—^2 ) )+1/((1 +γ€– π‘₯γ€—^2 ) )) =βˆ’2π‘₯((1 + π‘₯^2 + 1 βˆ’γ€– π‘₯γ€—^2)/(1 βˆ’ π‘₯^2 )(1 + π‘₯^2 ) ) =βˆ’2π‘₯(2/(1 βˆ’γ€– π‘₯γ€—^2 )(1 + π‘₯^2 ) ) =(βˆ’πŸ’π’™)/(𝟏 βˆ’ 𝒙^πŸ’ ) So, the correct answer is (B)

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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.