?z?x??z?u?u?x??z?v?v?x?2fu??ycosxfv?,
?z?y??z?u?u?y??z?v?v?y??fu??sinxfv?.
?z?x22???ycosxfuv??)?ysinxfv??ycosx(2fvu???ycosxfvv??) ?[2fu??ycosxfv?]?x?2(2fuu???4ycosxfuv???ysinxfv??y2cos2xfvv??, ?4fuu?z?x?y2???sinxfuv??)?cosxfv??ycosx(?fvu???sinxfvv??) ?[2fu??ycosxfv?]?y?2(?fuu???(2sinx?ycosx)fuv???ycosxsinxfvv???cosxfv?, ??2fuu?z?y22???2sinxfuv???sin2xfvv??. ???sinxfuv??)?sinx(?fvu???sinxfvv??)?fuu?[?fu??sinxfv?]?y??(?fuu五 证明题 (每小题6分,共18分) 1. 证明 limxyx?y623x?0y?0不存在.
证:当动点(x,y)沿着x轴(y?0)趋向于点(0,0)时,有 limx?0y?0xyx?y3362 ?0,
当动点(x,y)沿着曲线y?x趋向于点(0,0)时,有 lim33xyx?y6x?03y?x ? lim2xx63332x?0x?(x)?12,
故limxyx?y62x?0y?0不存在.
xyx?y0,xyx?y22??2. 证明:函数f(x,y)????22,x?yx?y222?0,?0在原点(0,0)连续且偏导数存在，但不可微.
2证:(1) 由f(x,y)?f(0,0)??x, 有
???0,?????0,?(x,y):x??,y??且(x,y)?(0,0),都有
f(x,y)?f(0,0)??,
即函数f(x,y)原点(0,0)连续.
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(2) fx?(0,0)?limf(?x,0)?f(0,0)?x?x?0?lim0?x?x?0?0,
fy?(0,0)?limf(0,?y)?f(0,0)?y?y?0?lim0?y?y?0?0.
(3)假设f(x,y)在原点(0,0)可微, 则有 df?fx?(0,0)?x?fy?(0,0)?y?0, 又?f?f(0??x,0??y)?f(0,0)??x?y(?x)?(?y)22, ??(?x)?(?y)，
22特别取?x??y,有lim?f?df?x?0?y??x??lim(?x)22?x?02(?x)22?12?0,与可微定义矛盾.
1?,?xysin223. 证明:函数f(x,y)??x?y?0,?x?yx?y2?0,?0在原点(0,0)可微.
2证: 由?f?f(0??x,0??y)?f(0,0)??x?ysin1(?x)?(?y)22, ??(?x)?(?y), 22有lim?f??0??lim?x?y(?x)?(?y)22?x?0sin1(?x)?(?y)22?0,
即?f?o(?)?0??x?0??y?o(?),故f(x,y)在原点(0,0)可微且df(0,0)?0.
注：函数f(x,y)在原点(0,0)的两个偏导数fx?(0,0)，fy?(0,0)存在，不一定有f(x,y)在(0,0)可
微，即全微分df(0,0)不一定存在.故不能由fx?(0,0)?0,fy?(0,0)?0得到df(0,0)?0.
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