曲线ex+y一sin(xy)=e在点(0,1)处的切线方程为_________.

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问题 曲线ex+y一sin(xy)=e在点(0,1)处的切线方程为_________.

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答案y=([*]-1)x+1

解析 exy—sin(xy)=e两边对x求导得ex+y.(1+y)一cos(xy).(y+xy)=0,将x=0,y=1代入得y(0)=—1,所求的切线为y一1=(一1)x,即y=(一1)x+1.
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