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函数F(x,y)=xysin1/(x^2+y^2)^1/2,(x,y)≠(0,0); 0,(x,y

函数f(x,y)=xy/√(x+y),(x,y)≠(0,0),=0,(x,y)=(0,0),求偏导数f'x(x,y)=y/[√(x+y)],(x,y)≠(0,0),=0,(x,y)=(0,0),而因lim(x→0,y=kx)f'x(x,y)=lim(x→0,y=kx)y/[√(x+y)]=lim(x→0)(kx)/{√[x+(kx)]}=k/[√(1+k)]与k有关,知极限lim(x→0,y→0)f'x(x,y)不存在,另一个同理.

f(x,y)={(x^2+y^2)sin[1(/x^2+y^2)],(x,y)≠(0,0) ;..{0,(x,y)=(0,0).f/x=2xsin[1/(x^2+y^2)]+(x^2+y^2)cos[1/(x^2+y^2)]*[-2x/(x^2+y^2)^2]=2xsin[1/(x^2+y^2)]-2x/(x^2+y^2)*cos[1/(x^2+y^2)],(x,y)≠(0,0); lim<△x→0>(△x)^2sin[1/(△x)^2]/△x=0.同理,f/y=

手画是画不出来的,它的 图像是马鞍型的.一三卦限向上翘的,二四卦限是向下的,需借助电脑才能画出的.

f(x,y)={(x^2+y^2)sin[1(/x^2+y^2)],(x,y)≠(0,0) ;..{0,(x,y)=(0,0).f/x=2xsin[1/(x^2+y^2)]+(x^2+y^2)cos[1/(x^2+y^2)]*[-2x/(x^2+y^2)^2]=2xsin[1/(x^2+y^2)]-2x/(x^2+y^2)*cos[1/(x^2+y^2)],(x,y)≠(0,0);lim<△x→0>(△x)^2sin[1/(△x)^2]/△x=0.同理,f/y=

函数 f(x,y) = xy/√(x+y),(x,y)≠(0,0), = 0, (x,y)=(0,0),求偏导数 f'x(x,y) = y/[√(x+y)],(x,y)≠(0,0), = 0,(x,y)=(0,0),而因 lim(x→0,y=kx)f'x(x,y) = lim(x→0,y=kx)y/[√(x+y)] = lim(x→0)(kx)/{√[x+(kx)]} = k/[√(1+k)] 与 k 有关,知极限 lim(x→0,y→0)f'x(x,y) 不存在,另一个同理.

f(x,y)=xsin[1/(x^2+y^2)]df(x.y)=sin[1/(x^2+y^2)]dx+xcos[1/(x^2+y^2)]*[(x^2+y^2)^(-1)]'df(x,y)=sin[1/(x^2+y^2)]dx-xcos[1/(x^2+y^2)]*(x^2+y^2)^(-2)*(2xdx+2ydy)所以:fx=sin[1/(x^2+y^2)]-2x^2cos[1/(x^2+y^

首先有确定x与y是两个变量,没有依赖关系,则fx=sin(y),fy=xcos(y),fxx=0,fxy=cos(y),fyx=cos(y),fyy=-xsin(y).

我觉得可以用夹逼法,因为-1 <= sinA <= 1所以-(x^2 + y^2) <= f(x,y) <= (x^2 + y^2)当x,y->0时,左右两边的极限都是0,所以f(x,y)->0

搜一下:f(x)=(x+y)sin(1/(x^2+y^2)),证明极限limf(x,y)=0

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