设函数y=y(x)由方程x^2y^2+y=1(y>0)所确定 求dy/dx|x=0,d^2y/dx^2|x=0.

∵x²y²+y=1(y>0)
==>2xy²+2x²yy'+y'=0 (等式两端对x求导数).(1)
==>2y²+4xyy'+4xyy'+2x²(yy''+y'²)+y''=0 (等式两端对x求导数).(2)
∴由(1)得y'=-2xy²/(2x²y+1)
由(2)得y''=-(2y²+8xyy'+2x²y'²)/(2x²+1)
∵当x=0时,y=1
∴dy/dx|(x=0)=y'|(x=0)=-2*0*1²/(2*0²*1+1)=0
d²y/dx²|(x=0)=y''|(x=0)=-(2*1²+8*0*1*0+2*0²*0²)/(2*1²+1)=-2/3.