sin(xy)=x²y²+e^xy,两边求导得到:cos(xy)(ydx+xdy)=2xy^2dx+2x^2ydy+e^(xy)(ydx+xdy)y[cos(xy)-e^(xy)]dx+x[cos(xy)-e^(xy)]dy=2xy^2dx+2x^2ydyx[cos(xy)-e^(xy)-2xy]dy=y[2xy-cos(xy)+e^(xy)]dx所以:dy/dx=y[2xy-cos(xy)+e^(xy)]/{x[cos(xy)-e^(xy)-2xy]}.