fx = y^2+2xz, fxx = 2z, fxx(0, 0, 1) = 2
fy = 2xy+z^2, fyz = 2z, fyz(0, -1, 0) = 0
2, ∂z/∂x = e^(xy)+ye^(xy) = (1+y)e^(xy)
∂^2z/∂x∂y = e^(xy) + x(1+y)e^(xy) = (1+x+xy)e^(xy)
第一题:
∵f(x,y,z)=xy^2+yz^2+zx^2
∴df=y^2dx+2xydy+z^2dy+2yzdz+x^2dz+2xzdx
=(y^2+2xz)dx+(z^2+2xy)dy+(x^2+2yz)dz
即:∂f/∂x=y^2+2xz;
∂f/∂y=z^2+2xy;
∂f/∂z=x^2+2yz.
进一步:
∂^2f/∂x^2=0+2z,代入数值有:∂^2f/∂x^2(0,0,1)=2*1=2;
∂^2f/∂yz=2z+0,代入数值有:∂^2f/∂yz(0,-1,0)=0。
第二题:
先对z求全微分,得:
dz=e^(xy)dx+x*e^(xy)*(ydx+xdy)
=e^(xy)*[dx+xydx+x^2dy]
=e^(xy)*[(1+xy)dx+x^2dy]
所以:
∂z/∂x=(1+xy)e^(xy);
∂z/∂y=x^2e^(xy);
进一步得:
∂^2z/∂xy=2xe^(xy)+x^2*e^(xy)*y
=e^(xy)(2x+yx^2)
=xe^(xy)(2+xy).