解:∵把x=1代入曲线方程得1+e^(2y)=4-2e^y
==>e^(2y)+2e^y-3=0
==>(e^y+3)(e^y-1)=0
又e^y>0
∴e^y-1=0 ==>e^y=1 ==>y=0
∴切点的坐标是(1,0)
∵对原方程两端求关于x导数
得1+2y'e^(2y)=-2(y+xy')e^(xy)
==>1+2y'e^(2y)=-2ye^(xy)-2xy'e^(xy)
==>2(e^(2y)+xe^(xy))y'=-1-2ye^(xy)
==>y'=[-1-2ye^(xy)]/[2(e^(2y)+xe^(xy))]
把切点的坐标(1,0)代入y',得y'=-1/4
∴过点(1,0),且斜率是-1/4的直线方程是y=-(x-1)/4
故此曲线在x=1处的切线方程是y=-(x-1)/4
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