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两边同事求导 对1+sin(x+y)=e^(xy)求导得
cos(x+y)(1+y')=e^(xy)*(y+xy'),
∴cos(x+y)-ye^(xy)=[xe^(xy)-cos(x+y)]y',
∴y'=[cos(x+y)-ye^(xy)]/[xe^(xy)-cos(x+y)].