NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

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

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### Transcript

Question 21 If y = log ((1 β π₯^2)/(1 + π₯^2 )),then ππ¦/ππ₯ is equal to (A) γ4π₯γ^3/(1βπ₯^4 ) (B) (β4π₯)/(1βπ₯^4 ) (C) 1/(4βπ₯^4 ) (D) (β4π₯^3)/(1βπ₯^4 ) y=log((1 β π₯^2)/(1 +γ π₯γ^2 )) π²=π₯π¨π (πβπ^π )βπ₯π¨π β‘γ(π+π^π)γ Differentiating wrt π₯ ππ/ππ=π(log(1 β π₯^2 ) β logβ‘γ(1 + π₯^2)γ )/ππ₯ =π(log(1 β π₯^2 ))/ππ₯βπ(log(1 + π₯^2 ))/ππ₯ =π/((π β π^π ) ) π(π β π^π )/ππβπ/((π + π^π ) ) π(π + π^π )/ππ =1/((1 β π₯^2 ) )(0β2π₯)β1/((1 + π₯^2 ) ) (0+2π₯) =(β2π₯)/((1 β π₯^2 ) )β2π₯/((1 + π₯^2 ) ) =β2π₯(1/((1 βγ π₯γ^2 ) )+1/((1 +γ π₯γ^2 ) )) =β2π₯((1 + π₯^2 + 1 βγ π₯γ^2)/(1 β π₯^2 )(1 + π₯^2 ) ) =β2π₯(2/(1 βγ π₯γ^2 )(1 + π₯^2 ) ) =(βππ)/(π β π^π ) So, the correct answer is (B)