f(x,y)=(x+y∧2)i+(2xy-1)j 在此力场中,场力做的功与路径无关

f(x+y)=f(x)+f(y)+2xy f(xy,x-y)=x^2+y^2+xy;f(x,y)/∂x;f(x,y)/∂y f(x+y,x-y)=(x^2-y^2)/2xy,求f(x,y)! f(x+y,x-y)=2xy(x-y),求f(x,y) f(xy,x-y)=x^2+3xy+y^2,则f(x,y)=? f(xy,x-y)=x^2-xy+y^2.求f(x,y) y'=-2xy;求f(x). 函数f(xy,x+y)=x^2+xy+y^2,则df(x,y)=? 设f(x+y,x-y)=x^2+xy,求f(x,y) f(x^2-y^2,x-y)=x+y+x^2-xy,求f'(x,y)对y求导 设f(x+y,xy)=x^2+y^2-xy,则（∂f(x,y)）/∂y= f(2x+y,2y+x)=xy,求f(x,y) 设f（x+y,xy）=x^2+y^2,则f（x,y） f(x+y)+f(xy-1)=f(x)f(y)+2f(n)表达式 设函数f(xy,x+y)=x^2+y^2+xy,求∂f(x,y)／∂x和∂f(x,y)／∂y 已知f(x+y)=f(x)+f(y)+2xy,求f(0) y'=-2xy;求f(x).y'=2y/x+2;求y 高数题!f(x,y)=x^3-2xy+3y^2,求f(x+y,x-y).要借题步骤