e^y+xy+e^x=0
兩邊同時(shí)對(duì)x求導(dǎo)得：
e^y·y '+y+xy '+e^x=0
得y '=-(y+e^x)/(x+e^y)
y ''=-[(y '+e^x)(x+e^y)-(y+e^x)(1+e^y·y ')]/(x+e^y)²
當(dāng)x=0時(shí),e^y+1=0,題目應(yīng)該有問題,求不出y啊錯(cuò)了。。是e^y+xy-e^x=0e^y+xy-e^x=0兩邊同時(shí)對(duì)x求導(dǎo)得：e^y·y '+y+xy '-e^x=0得y '=-(y-e^x)/(x+e^y)y ''=-[(y '-e^x)(x+e^y)-(y-e^x)(1+e^y·y ')]/(x+e^y)²當(dāng)x=0時(shí)，e^y-1=0，得y=0，y '=-(y-e^x)/(x+e^y)=-(0-1)/(0+1)=1故y ''=-[(1-1)(0+1)-(0-1)(1+1)]/(0+1)²=-2