Question

设y=y(x)由方程x3+y3-2xy=1所确定求该曲线过点(0.1)的切线方程

Answer 1

对x^3+y^3-2xy=1求导得

3x^2+3y^2*y'-2y-2xy'=0,

整理得(3y^2-2x)y'=2y-3x^2,

∴y'=(2y-3x^2)/(3y^2-2x),

把x=0,y=1代入上式得y'(0)=2/3,

∴该曲线在点(0,1)处的切线方程是y-1=2x/3,即2x-3y+3=0.