设U(x,y)是由方程组U=yz+zx+xy,x^2+y^2+z^2=1所确定,试求Uxy

2025-05-14 00:26:47
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回答1:

x^2+y^2+z^2=1分别对x, 及y求导,得:
2x+2z z'(y)=0--> z'(x)= -x/z
2y+2z z'(y)=0--->z'(y)=-y/z
因此有:
Ux=yz'(x)+z'(x) x+z+y=-yx/z-x^2/z+z+y=y+z-x(y+x)/z
Uxy=1+z'(y)-x[ z-(y+x)z'(y)]/z^2=1-y/z-x[z+(y+x)y/z]/z^2=1-yz-x(z^2+y+x)/z^3