y=2x/(x-1)²
y'=[2(x-1)²-2x*2(x-1)]/(x-1)^4
=(2x²-4x+2-4x²+2x)/(x-1)^4
=(-2x²-2x+2)/(x-1)^4
=-2(x²+x-1)/(x-1)^4
y=2/x²+1
k=y'=2*(-2)x^(-3)+1
當(dāng)x=1時,k=4+1=5
設(shè)切線方程為：y=5x+m
切線過（1,1）,則m=-4
所以曲線y=2/x²+1在點(diǎn)P（1,1）處的切線方程是：y=5x-4