设曲线方程是sin（xy）-e^2x+y^3=0 求它在x=0处的切线方程和法线方程

x=0代入方程,得0-1+y^3=0,得y=1,即切点为(0,1)方程两边对x求导：cos(xy)(y+xy')-2e^2x+3y'y^2=0得：y'=[2e^2x-ycos(xy)]/[xcos(xy)+3y^2]将点(0,1)代入,得y'=[2-1]/[0+3]=1/3因此,由点斜式,得：切线方程为：y=x/3+1...