曲线ex+y-sin(xy)=e在点(0,1)处的切线方程为______.

admin2020-03-18  33

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

选项

答案[*]

解析 ex+y-sin(xy)=e两边对x求导得ex+y.(1+y’)-cos(xy).(y+xy’)=0,
将x=0,y=1代入得y’(0)=
故所求的切线方程为
转载请注明原文地址:https://kaotiyun.com/show/LcD4777K
0

最新回复(0)