Y=(1+x/1-x) ^2 ? find dy y=(1+x/1-x) ^2

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Answer 1 · 2018-04-01T16:31:40

#y=((1+x)/(1-x)) ^2#

differentiating, using quotient rule.

#y=((1+x)^2)/(1-x) ^2#

#dy/dx = ([(1-x) ^2][(1+x) ^2]' -[(1+x) ^2][(1-x) ^2]')/[(1-x) ^2]^2#

#dy/dx = ([(1-x) ^2][2(1+x)] -[(1+x) ^2][-2(1-x)])/[(1-x) ^4]#

#dy/dx = (2(1+x)(1-x) ^2 +2(1-x)(1+x) ^2)/[(1-x) ^4]#

#dy/dx = (2(1+x))/(1-x)^2 +(2(1+x) ^2)/[(1-x) ^3]#