设曲线方程x3+y3+(x+1)cos(πy)+9=0,试求曲线在x=-1点处的切线和法线方程

x^3+y^3+(x+1)cos(πy)+9=0 x=-1时,-1+y^3-0+9=0 y^3=-8 y=-23x^2+3y^2y'+(x+1)π(-sin(πy))y'+cos(πy)=03x^2+(3y^2-π(x+1)sin(πy) )y'+cos(πy)=0y'=(3x^2+cos(πy))/[π(x+1)sin(πy)-3y^2] 代入(-1,-2)y'(-1)...