求由方程cos（xy）=x²y²所确定的函数y的微分

两边对x求导：
-(y+xy')sin(xy)=2xy^2+2x^2yy'
解得：y'=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]
所以dy=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]dx可是答案是dy=（-y/x）dx哦，是喔，可以继续化简：
y'=-[ysin(xy)+2xy^2]/[2x^2y+xsin(xy)]
=-y[sin(xy)+2xy]/x[2xy+sin(xy)]
=-y/x
故答案是对的