Question

曲线sin(xy)+ln(y-x)=x在点(0，1)处的切线方程为______． 详细步骤

Answer 1

对x求导cos(xy)*(xy)'-1/(y-x)*(y-x)'=1cos(xy)*(y+x*y')-(y'-1)/(y-x)=1y*cos(xy)+xcos(xy)*y'-y'/(y-x)+1/(y-x)=1y'=[1-1/(y-x)-y*cos(xy)]/(xcos(xy)-1/(y-x)]x=0,y=1所以斜率k=y'=-1/(-1)=1所以x-y+1=0