Question

设函数z=(x,y),由方程x+2y+xy-z-exp(z)=0确定,且Z(1,0)=0,Zx(1,0),Zy(1,0)

Answer 1

x+2y+xy-z-exp(z)=0.(1)

对(1)两边同时对x求偏导

1+y-Zx-(e^z)*Zx=0 .(2)

Zx=(1+y)/(e^z+1)

故Zx(1,0)=1/(e^0+1)=1/2

对(1)两边同时对y求偏导

2+x-Zy-(e^z)*Zy=0

Zy=(2+x)/(e^z+1)

故Zy(1,0)=(2+1)/(e^0+1)=3/2

对(2)两边同时对y求偏导

1-Zxy-(e^z)*Zy*Zx-(e^z)*Zxy=0

Zxy=[1-(e^z)*Zx*Zy]/(e^z+1)

故Zxy(1,0)=[1-e^0*(1/2)*(3/2)]/(e^0+1)=1/8