设函数y=y(x)是由方程cos(xy)=x+y所以确定的隐函数,求函数曲线y=y(x),过点（0,1）的切线方程

cos(xy)=x+y两边微分,得dx+dy-sin(xy)*(x*dy+y*dx)=0
dx(1-ysin(xy))+dy(1-xsin(xy))=0
dy/dx=(ysin(xy)-1)/(1-xsin(xy))即为切线斜率
代入数值,斜率为-1,过点（0,1）,所以切线为x+y=1