Davidee
已知 z=(3x2+y2)e-arctan
y
x ，求 ?z ?x ,?
2z
?x2 ,
?2z
?x?y .
解：对函数直接求偏导数有：
?z
?x=6xe
-arctanyx +(3x2+y2)e-arctanyx -y
1
x2
1+(yx)2
,
=6xe-arctan
y
x +(3x2+y2)e-arctan
y
x yx2+y2 ,
=e-arctan
y
x [6x+y(3x
2+y2)
x2+y2 ]。
?2z
?x2 =e
-arctanyx -y
1
x2
1+(yx)2
[6x+y(3x
2+y2)
x2+y2 ]+
e-arctan
y
x [6+6xy(x
2+y2)-y(3x2+y2)2x
(x2+y2)2 ],
=e-arctan
y
x { yx2+y2 [6x+y(3x
2+y2)
x2+y2 ]+[6+
4xy3
(x2+y2)2]},
=e-arctan
y
x [6+y
4+10xy3+3x2y2+6x3y
(x2+y2)2 ],
Davidee
由 ?z?x对 y 再求偏导数，得：
∵ ?z?x=e-arctan
y
x [6x+y(3x
2+y2)
x2+y2 ],
∴ ?
2z
?x?y
=e-arctan
y
x
-1x
1+(yx)2
[6x+y(3x
2+y2)
x2+y2 ]+
e-arctan
y
x [(3x
2+y2)+y2y](x2+y2)-y(3x2+y2)2y
(x2+y2)2 ,
=e-arctan
y
x xx2+y2 [6x+y(3x
2+y2)
x2+y2 ]+e
-arctanyx 3x4+y4
(x2+y2)2,
=e-arctan
y
x 9x
4+3x3y+6x2y2+xy3+y4
(x2+y2)3 。
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