设函数y=f（x）由方程sin（xy）=x+y确定，求y’和dy。

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对x求导，y是蚂源漏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'
所以闷烂y'=[ycos(xy)-1]/裂消[1-xcos(xy)]

所以dy={[ycos(xy)-1]/[1-xcos(xy)]}dx

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