How do you differentiate #y=((x+1)/(x-1))^2#?

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Answer 1 · 2015-05-16T01:04:12

I would use the Chain and Quotient Rule:
#y'=2((x+1)/(x-1))((x-1)-(x+1))/(x-1)^2=#
#=2((x+1)/(x-1))(-2)/(x-1)^2=#
#=-4(x+1)/(x-1)^3#