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

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

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

3.单选题-|||-二元函数极限 lim _(xarrow 0)(dfrac (1-cos sqrt {{x)^2+(y)^2}}({e)^2+(y)^2-1}

1.设 sin y+(e)^x-x(y)^2=0, 求 dfrac (dy)(dx).-|||-2.设 ln sqrt ({x)^2+(y)^2}=arctan

设D: leqslant (x)^2+(y)^2, =dfrac (dxdy)(sqrt {{x)^2+(y)^2}}=()leqslant yleqslant

(6) arctan dfrac (y)(x)=ln sqrt ({x)^2+(y)^2}

已知函数 z=f(x,y) 连续且满足 lim _(xarrow 1)dfrac (f(x,y)-x+2y+2)(sqrt {{(x-1))^2+(y)^2}}

12.设方程 +sqrt ({x)^2+(y)^2+(z)^2}=sqrt (2) 确定了函数 =z(x,y), 则z(x,y)-|||-在点 (1,0,-1)

(9)设 f(x,y)= dfrac (1)({({x)^2+(y)^2)}^2},1leqslant xleqslant 3, dfrac (sqrt {3)

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