填空题
设二元函数f(u，v)具有二阶连续偏导数，且满足
又g(x，y)=f(x2-y2，2xy)，则
又g(x，y)=f(x2-y2，2xy)，则
- 1、
【正确答案】
1、0
【答案解析】[解析] 直接计算可得
dg=f'ud(x2-y2)+f'vd(2xy)=f'u·(2xdx-2ydy)+f'v·(2ydx+2xdy)
=(2xf'u+2yf'v)dx+(-2yf'u+2xf'v)dy，
从而g'x=2xf'u+2yf'v，g'y=-2yf'u+2xf'v
继续求偏导数又可得
g"xx=2f'u+2x(1f'u)'x+2y(f'v)'x
=2f'u+2x(2xf"uu+2yf"uv)+2y(2xf"vu+2yf"vv)
=2f'u+4x2f"vu+8xyf"uv+4y2f"vv，
g"yy=-2f'u-2y(-2yf"uu+2xf"vv)+2x(-2yf"vu+2xf"vv)
=-2f'u+4y2f"uu-8xyf"uv+4x2f"vv
由此即得 g"xx+g"yy=4(x2+y2)(f"uu+f"vv)=0
dg=f'ud(x2-y2)+f'vd(2xy)=f'u·(2xdx-2ydy)+f'v·(2ydx+2xdy)
=(2xf'u+2yf'v)dx+(-2yf'u+2xf'v)dy，
从而g'x=2xf'u+2yf'v，g'y=-2yf'u+2xf'v
继续求偏导数又可得
g"xx=2f'u+2x(1f'u)'x+2y(f'v)'x
=2f'u+2x(2xf"uu+2yf"uv)+2y(2xf"vu+2yf"vv)
=2f'u+4x2f"vu+8xyf"uv+4y2f"vv，
g"yy=-2f'u-2y(-2yf"uu+2xf"vv)+2x(-2yf"vu+2xf"vv)
=-2f'u+4y2f"uu-8xyf"uv+4x2f"vv
由此即得 g"xx+g"yy=4(x2+y2)(f"uu+f"vv)=0