设函数y=f（x）由方程sin（xy）=x+y确定,求y’和dy.

对x求导,y是x的函数
所以cos(xy)*(xy)'=1+y'
cos(xy)*(x'*y+x*y')=1+y'
cos(xy)*(y+x*y')=1+y'
ycos(xy)+xcos(xy)*y'=1+y'
所以y'=[ycos(xy)-1]/[1-xcos(xy)]
所以dy={[ycos(xy)-1]/[1-xcos(xy)]}dx