确定f(x,y)=4x+xy^2+y^2在圆形区域x^2+y^2
1个回答
二元函数求最值的方法
对x求偏导,并令其为0,4+y2=0,无解
说明此二元函数在R2内无驻点,故而在圆形区域x^2+y^2
相关问题
函数y=f(x)由方程x^2+y^2+e^xy=4确定,求y'
求函数z=x^2-xy+y^2在区域|x|+|y|
求函数x^2+y^2+2xy-2x在闭区域x^2+y^2
设y=f(x)由方程x^2+y^2=4xy确定,则dy/dx=
设F(xy,y^2/x)=x^2+y^2,求F(y^2/x,xy)
(x^2+4y^2-2xy)(x^2+4y^2+2xy)(x+2y)(x-2y)
f(xy,x-y)=x^2+3xy+y^2,则f(x,y)=?
(x^3+y^3)^2-4xy[x^4+x^2y^2+y^4-2xy(x^2-xy+y^2)]因式分解
x^2-2xy+y^2=14x^2-4xy+y^2=46x^2-5xy+y^2=0x^2-4xy+3y^2+4x-8y+
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x