Question

已知y=y(x)由方程cos(x+y)+e^(x-y)=2 y' y'|(x,y)=(0,0)

Answer 1

对方程两边同时求导得,﹣﹙y＋xy′﹚sin﹙xy﹚＋e^y＋﹙x＋1﹚y′e^y＝0

令x=0则方程cos(xy)+(x+1)*e^y=2为1＋e^y=2,得y=0,即切点坐标为﹙0,0﹚

将﹙0,0﹚带入﹣﹙y＋xy′﹚sin﹙xy﹚＋e^y＋﹙x＋1﹚y′e^y＝0得y′=﹣1

则过切点﹙0,0﹚的切线方程是y=﹣x,法线方程是y=x.