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(x
2
+y
2
，xy)，则
，又g(x，y)=f(x
2
+y
2
，xy)，则
【正确答案】
【答案解析】x
2
-y
2
[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" u d(x 2 +y 2 )+f" v d(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 +4x 2 f" uu +4xyf" uv +y 2 f" 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 +4y 2 f" uu +4xyf" uv +x 2 f" vu ．
故g" xx -g" yy =4(x 2 -y 2 )f" uu -(x 2 -y 2 )f" vv ．
=(x 2 -y 2 )(4f" uu -f" vv )=x 2 -y 2 ．
dg=f" u d(x 2 +y 2 )+f" v d(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 +4x 2 f" uu +4xyf" uv +y 2 f" 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 +4y 2 f" uu +4xyf" uv +x 2 f" vu ．
故g" xx -g" yy =4(x 2 -y 2 )f" uu -(x 2 -y 2 )f" vv ．
=(x 2 -y 2 )(4f" uu -f" vv )=x 2 -y 2 ．