y=(e^x+1)/(e^x-1)
所以y'=[(e^x-1)*(e^x+1)'-(e^x-1)'*(e^x+1)]/(e^x-1)^2
=[(e^x-1)*e^x-e^x*(e^x+1)]/(e^x-1)^2
=(-2e^x)/(e^x-1)^2
还可以吧
y=(e^x+1)/(e^x-1)
所以y'=[(e^x-1)*(e^x+1)'-(e^x-1)'*(e^x+1)]/(e^x-1)^2
=[(e^x-1)*e^x-e^x*(e^x+1)]/(e^x-1)^2
=(-2e^x)/(e^x-1)^2
还可以吧