填空题
设函数f(u，v)具有二阶连续偏导数，且满足
，又g(x，y)=f(x2+y2，xy)，则
，又g(x，y)=f(x2+y2，xy)，则
- 1、
【正确答案】
1、x2-y2
【答案解析】 利用一阶全微分形式不变性可得复合函数g(x，y)的一阶全微分
dg=f'ud(x2+y2)+f'vd(x，y)
=2(xdx+ydy)f'u+(ydx+xdy)f'v
=(2xf'u+yf'v)dx+(2yf'u+xf'u)dy．
从而g'x=2xf'u+yf'v，g'y=2yf'u+xf'v．继续求g(x，y)的二阶偏导数，又有
g"xx=2f'u+2x(f'u)'x+y(f'v)'x
=2f'u+2x(2xf"uu+yf"uv)+y(2xf"vu+yf"vv)
=2f'u+4x2f"uu+4xyf"uv+y2f"vv，
g"yy=2f'u+2y(f'u)'y+x(f'v)'y
=2f'u+2y(2yf"uu+xf"uv)+x(2yf"vu+xf"vv)
=2f'u+4y2f"uu+4xyf"uv+x2f"vu．
故g"xx-g"yy=4(x2-y2)f"uu-(x2-y2)f"vv．
=(x2-y2)(4f"uu-f"vv)=x2-y2．
dg=f'ud(x2+y2)+f'vd(x，y)
=2(xdx+ydy)f'u+(ydx+xdy)f'v
=(2xf'u+yf'v)dx+(2yf'u+xf'u)dy．
从而g'x=2xf'u+yf'v，g'y=2yf'u+xf'v．继续求g(x，y)的二阶偏导数，又有
g"xx=2f'u+2x(f'u)'x+y(f'v)'x
=2f'u+2x(2xf"uu+yf"uv)+y(2xf"vu+yf"vv)
=2f'u+4x2f"uu+4xyf"uv+y2f"vv，
g"yy=2f'u+2y(f'u)'y+x(f'v)'y
=2f'u+2y(2yf"uu+xf"uv)+x(2yf"vu+xf"vv)
=2f'u+4y2f"uu+4xyf"uv+x2f"vu．
故g"xx-g"yy=4(x2-y2)f"uu-(x2-y2)f"vv．
=(x2-y2)(4f"uu-f"vv)=x2-y2．