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

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)'=1
cos(xy)*(y+x*y')-(y'-1)/(y-x)=1
y*cos(xy)+xcos(xy)*y'-y'/(y-x)+1/(y-x)=1
y'=[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