①微分
log
(
x
+
x
2
+
A
)
{
log
|
x
+
x
2
+
A
|
}
′
=
1
x
+
x
2
+
A
(
x
+
x
2
+
A
)
′
=
1
x
+
x
2
+
A
{
1
+
(
x
2
+
A
)
−
1
2
2
⋅
2
x
}
=
1
x
+
x
2
+
A
(
1
+
x
x
2
+
A
)
=
1
x
+
x
2
+
A
⋅
x
+
x
2
+
A
x
2
+
A
=
1
x
2
+
A
戻る
[
ひ
]
[
は行
]
[
索引トップ
]
