设 (x,y)=dfrac ({x)^2+(y)^2}({e)^xy+xysqrt ({x)^2+(y)^2}} ,则 (f')_(x)(1,0)= __ _.

设 (x+y,x-y)=2((x)^2+(y)^2)(e)^(x^2-{y)^2}, 则 _(x)(x,y)-(f)_(y)(x,y)= __

设 =(e)^xy+ln ((x)^2+(y)^2) 则（ ）A =(e)^xy+ln ((x)^2+(y)^2)B =(e)^xy+ln ((x)^2+(y)

证明:函数-|||-f(x,y)= ((x)^2+(y)^2)sin dfrac (1)(sqrt {{x)^2+(y)^2}}, ^2+(y)^2neq 0,

2.设 (xy,x+y)=(x)^2+(y)^2+xy (其中, =xy =x+y), 则 dfrac (partial f)(partial u)+dfrac

1.函数 (x+y,xy)=(x)^2+(y)^2-xy, 则 f(x,y)=

函数f(x,y)= dfrac (xy)({x)^2+(y)^2},(x)^2+(y)^2neq 0-|||-0, ^2+(y)^2=0在点（0，0）处（）。

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df

164 设f(x,y)有二阶连续偏导数, (x,y)=f((e)^xy,(x)^2+(y)^2), 且 f(x,y)=1-x-y+-|||-(sqrt ({(x

7.设 (x,y)=dfrac (1)(xy),r=sqrt ({x)^2+(y)^2} _(1)= (x,y)|(x,y)in {R)^2 dfrac (1)

设 f(x,y)= sin dfrac (1)({x)^2+(y)^2} , ^2+(y)^2neq 0,-|||-0, ^2+(y)^2=0,-|||-考察函