【答案解析】解析:由题知 f(e,0)=1,f
x
'(x,y)=yx
y-1
,f
x
'(e,0)=0,f
y
'(x,y)=x
y
lnx,f
y
'(e,0)=1, f
xx
''(x,y)=y(y-1)x
y-2
,f
xx
''(e,0)=0,f
xy
''(x,y)=x
y-1
+yx
y-1
lnx,f
xy
''(e,0)=e
-1
, f
yy
''(x,y)=x
y
(lnx)
2
,f
yy
''(e,0)=1. 因此f(x,y)在点(e,0)处展开的二阶泰勒公式为 f(x,y)=f(e,0)+(x-e)f
x
'(e,0)+(y-0)f
y
'(e,0)+

[(x-e)
2
f
xx
''(e,0)+ 2(x-e)(y-0)f
xy
''(e,0)+(y-0)
2
f
yy
''(e,0)]+R
3
