设函数y=f(x)由方程(y)^2+(y)^2ln x-4=0所确定,则(y)^2+(y)^2ln x-4=0= ( )(y)^2+(y)^2ln x-4=0(y)^2+(y)^2ln x-4=0(y)^2+(y)^2ln x-4=0(y)^2+(y)^2ln x-4=0

讨论下列函数的连续性:-|||-^2+(y)^2neq 0,-|||-f(x,y)= ) (y)^2ln ((x)^2+(y)^2) 0,=0;

设f(x,y)= ^4+{y)^2},(x)^2+(y)^2neq 0 0,(x)^2+(y)^2=0 .处是否连续？设,试问在点处是否连续？

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

设函数 y=y(x) 由方程 ^3+x(y)^2+(x)^2y+6=0 确定,设函数 y=y(x) 由方程 ^3+x(y)^2+(x)^2y+6=0 确定,

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

27 判断题 y=ln x^2和y=2ln x是同一函数。A. 正确B. 错误

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

[题目]设函数 y=f(x) 由方程 +2ln x=(y)^4 所确定,-|||-则曲线 y=f(x) 在点(1,1)处的切线方程是 __-|||-__

球面^2+(y)^2+(z)^2+4x+6y+2z+10=0的球心坐标为A.^2+(y)^2+(z)^2+4x+6y+2z+10=0B.^2+(y)^2+(z)

殳 =(u)^2ln v =dfrac (y)(x), =2x-3y,-|||-则 dfrac (partial z)(partial y)= ()-|||-A