x + 2 - 佐藤(邦)

平成 26 年度微積分解法演習４
（ H26.
学籍番号
氏名
担当佐藤邦）
（本日の講義終了時に回収します。
解以外の途中の計算等も余白に書くこと。）
問１
次の関数を微分せよ。
1
⃝
1 √
x+4
⃝
1
⃝
2
⃝
2
(
)√
x + 2 2x − 1
⃝
3
2
√
3
x+2
⃝
4
{


)− 1
)− 3
1(
1
d
1
d(
)√
√
=
x+4 2 =− x+4 2 =− (
dx x + 4
dx
2
2 x+4 x+4
)√
d ( 2
x + 2 2x − 1 =
dx
√
5
(2x − 3)4
=− √ 1

(
)3
2
x+4
}
{
}
) √
( 2
) d√
d( 2
x +2
2x − 1 + x + 2
2x − 1
dx
dx
√
( 2
) 1
4x2 − 2x + x2 + 2
5x2 − 2x + 2
1
√
√
·2=
= √
= 2x 2x − 1 + x + 2 ·
2 2x − 1
2x − 1
2x − 1
⃝
3
)1
) 1 −1 1 (
)− 2
d √
d(
1(
1
3
x+2=
x+2 3 = x+2 3 = x+2 3 = √
3
dx
dx
3
3
3 (x + 2)2
⃝
4
)4
) 4 −1
)− 1
d √
4(
8(
8
d(
5
(2x − 3)4 =
2x − 3 5 = 2x − 3 5 · 2 = 2x − 3 5 = √
5
dx
dx
5
5
5 2x − 3
問２
⃝
1
⃝
2
⃝
3
⃝
4
1
)√
− (
2 x+4 x+4
5x2 − 2x + 2
√
2x − 1
1
√
3 3 (x + 2)2
8
√
5
5 2x − 3
次の値を求めよ。
√
3
54
(i) √
3
2
(ii) log4 8
(iii) loge−3 e
(iv) 3log3 27
(v) log3
√ )
√ )
(√
(√
11 + 2 + log3
11 − 2
√
√
√
√
3
3
3
√
( )1
54
27 × 2
27 × 3 2 √
3
3
3 = 33 3 = 3
√
√
(i) √
=
=
=
27
=
3
3
3
3
2
2
2
(ii) log4 8 =
log2 8
log2 23
3 log2 2
3
=
=
=
2
log2 4
log2 2
2 log2 2
2
loge e
loge e
1
1
=
=
=−
−3
loge e
−3 loge e
−3
3
(
)
3
= 3log3 3 = 33 log3 3 = 33 = 27
公式 aloga A = A より 27 と書いても OK
(iii) loge−3 e =
(iv) 3log3 27
(v) log3
√ )
√ )
(√
(√
(
)
11 + 2 + log3 11 − 2 = log3 11 − 2 = log3 32 = 2 log3 3 = 2
(i)
(ii)
3
3
2
(iii)
(iv)
(v)
1
3
27
2
−
問３
次の関数を微分せよ。
2x3 −5
(1) e
(2) log3
(
)
4x + 1
√ (
)
(4)
log x2 − 4
1
(3) 2x
e + e−2x
u = 2x3 − 5 →
(2)
u = 4x + 1 →
(3)
( 2x
)
−2x
( 2x
)
2
e
−
e
d
1
1
−2x
= −(
=−(
)2 · 2e − 2e
)2
dx e2x + e−2x
e2x + e−2x
e2x + e−2x
(4)
u = log x − 4
(5) 2
(
)
d
d loge u
1 d
du
1 1
4
)
log3 4x + 1 =
=
log u ·
=
·4= (
dx
dx loge 3
log 3 du
dx
log 3 u
4x + 1 log 3
)
d
→
dx
√
(
)
(
)
d√
d √ du
d 1 d
log x2 − 4 =
u=
u·
=
u2 ·
log x2 − 4
dx
du
dx
du
dx
1
x
1 −1
√ (
u 2· 2
· 2x = (
)
)
2
x −4
x2 − 4
log x2 − 4
=
(5)
2
)
d u
d u du
d 2x3 −5
3
e
=
e =
e ·
= eu · 6x2 = 6x2 e2x −5
dx
dx
du
dx
(1)
(
(
log 3x+4
u = 2log
(
)
3x+4
→ log u = log 2log
(
)
3x+4
(
)
= log 3x + 4 log 2
(
)
1 du
d
d
du
1
3 log 2
d
) · 3 · log 2 →
)
log u =
log 3x + 4 log 2 →
log u
=(
=(
dx
dx
du
dx
u dx
3x + 4
3x + 4
→
→
du
3 log 2
d log
) →
=u (
2
dx
dx
3x + 4
(1)
6x2 e2x
3 −5
(
(
)
3x+4
= 2log
(
)
3x+4
(
)
log 3x+4
3 log 2
3 log 2 2
(
)=
(
)
3x + 4
3x + 4
(2)
(3)
4
)
4x + 1 log 3
)
(
2 e2x − e−2x
−(
)2
e2x + e−2x
(4)
(5)
(
(
x
)√ (
)
x2 − 4
log x2 − 4
)
log 3x+4
3 log 2 2
(
)
3x + 4

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