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

[题目]设 (x-y,y|x)=(x)^2-(y)^2 求f(x,y)

6.讨论下列函数的连续性:-|||-(1) (x,y)=dfrac ({x)^2-(y)^2}({x)^2+(y)^2}-|||-(2) (x,y)=dfrac

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

计算 =(int )_(L)dfrac ((x-y)dx+(x+y)dy)({x)^2+(y)^2}-|||-f[(x-y)(x+(x+y)dy)/(x^2+x

(x,y)=(x)^2-(y)^2-x-|||-__ B . (x,y)=(x)^2-(y)^2-x-|||-__C . (x,y)=(x)^2-(y)^2-x

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

2)y=(x)/(y)+(y)/(x),y|_(x=1)=2;2)$y'=\frac{x}{y}+\frac{y}{x},y|_{x=1}=2;$

证明u(x,y)=(x-y)(x^2+4 xy+y^2)u(x,y)=(x-y)(x^2+4 xy+y^2)u(x,y)=(x-y)(x^2+4 xy+y^2)

已知(x,y)=(e)^x+(y-2)arcsin sqrt ({x)^2+(y)^2}，求(x,y)=(e)^x+(y-2)arcsin sqrt ({x)^

[单选题]设(x-y)(x+2+y)-15=0，则x+y的值是()。A.-5或3B.-3或5C.3D.5