数学演習1 No.2 2005. 4.20
1.2 微分係数と導関数（解答）
担当：市原問題4 次の関数の与えられたxの値における微分係数を,定義に基づいて計算しなさい. (1)y=x3 [x= 0]
h→0lim
(0 +h)3−03
h = lim
h→0
h3 h = lim
h→0h2= 0
(2)y= 1
x−2 [x=−1]
h→0lim
1
(−1 +h)−2 − 1 (−1)−2
h = lim
h→0
1 h−3 +1
3
h = lim
h→0
3 + (h−3) 3(h−3)
h = lim
h→0
h 3(h−3)
h
= lim
h→0
h
3(h−3) ×(3(h−3)) h×(3(h−3)) = lim
h→0
h
h(3(h−3)) = lim
h→0
1
3(h−3) =−1 9
(3)y= 1
√x [x= 3]
h→0lim
√ 1
3 +h− 1
√3
h = lim
h→0
√3−√ 3 +h
√3√ 3 +h
h = lim
h→0
√3−√ 3 +h
√3√
3 +h ×√ 3√
3 +h h×√
3√ 3 +h
= lim
h→0
√3−√ 3 +h h√
3√
3 +h = lim
h→0
(√ 3−√
3 +h)×(√ 3 +√
3 +h) h√
3√
3 +h×(√ 3 +√
3 +h)
= lim
h→0
3−(3 +h) h√
3√
3 +h(√ 3 +√
3 +h) = lim
h→0
−h h√
3√
3 +h(√ 3 +√
3 +h)
= lim
h→0
√ −1 3√
3 +h(√ 3 +√
3 +h) =− 1
√3√ 3(√
3 +√
3) =− 1 3×2√
3 =−
√3 18
(4)y=√3
x [x= 1]
h→0lim
√3
1 +h−√3 1
h = lim
h→0
¡√3
1 +h−√3 1¢
ס (√3
1 +h)2+ (√3
1 +h)(√3
1) + (√3 1)2¢ hס
(√3
1 +h)2+ (√3
1 +h)(√3
1) + (√3 1)2¢
= lim
h→0
¡√3 1 +h¢3
−¡√3 1¢3
h¡ (√3
1 +h)2+ (√3
1 +h)(√3 1) + (√3
1)2¢ = lim
h→0
(1 +h)−1 h¡
(√3
1 +h)2+ (√3
1 +h)(√3
1) + (√3 1)2¢
= lim
h→0
h h¡
(√3
1 +h)2+ (√3
1 +h)(√3 1) + (√3
1)2¢ = lim
h→0
1 (√3
1 +h)2+ (√3
1 +h)(√3
1) + (√3 1)2
= 1
(√3
1)2+ (√3 1)(√3
1) + (√3
1)2 = 1
1 + 1 + 1 = 1 3
問題5 次の関数を定義に基づいて(limを使って)微分しなさい. (1)y=−x2+ 3
y= lim
h→0
(−(x+h)2+ 3)−(−x2+ 3)
h = lim
h→0
(−x2−2xh−h2+ 3)−(−x2+ 3) h
= lim
h→0
−2xh−h2
h = lim
h→0(−2x−h) =−2x
(2)y=x+2 x
y= lim
h→0
µ
(x+h) + 2 x+h
¶
− µ
x+2 x
¶
h = lim
h→0
x+h+ 2
x+h−x− 2 x h
= lim
h→0
h+ 2 x+h− 2
x
h = lim
h→0
h+2x−2(x+h) (x+h)x
h = lim
h→0
h+ −2h (x+h)x
h
= lim
h→0
h µ
1 + −2 (x+h)x
¶
h = lim
h→0
µ
1 + −2 (x+h)x
¶
= 1− 2 x2
(3)y= 1
√x+ 1
y= lim
h→0
p 1
(x+h) + 1− 1
√x+ 1
h = lim
h→0
√x+ 1−p
(x+h) + 1 p(x+h) + 1√
x+ 1 h
= lim
h→0
√x+ 1−p
(x+h) + 1 p(x+h) + 1√
x+ 1 ×p
(x+h) + 1√ x+ 1 h×p
(x+h) + 1√
x+ 1 = lim
h→0
√x+ 1−p
(x+h) + 1 hp
(x+h) + 1√ x+ 1
= lim
h→0
³√
x+ 1−p
(x+h) + 1
´
×
³√
x+ 1 +p
(x+h) + 1
´ hp
(x+h) + 1√ x+ 1×
³√
x+ 1 +p
(x+h) + 1
´
= lim
h→0
(x+ 1)−((x+h) + 1) hp
(x+h) + 1√ x+ 1
³√
x+ 1 +p
(x+h) + 1
´
= lim
h→0
−h hp
(x+h) + 1√
x+ 1³√
x+ 1 +p
(x+h) + 1´
= lim
h→0
p −1
(x+h) + 1√
x+ 1³√
x+ 1 +p
(x+h) + 1´
= −1
√x+ 1√
x+ 1¡√
x+ 1 +√ x+ 1¢
=− 1
2(x+ 1)√ x+ 1
問題6 次の関数を微分しなさい.
(1)x10−99x4+ 99 導関数は, 10x9−396x3
(2) 4 3x− 1
(−x)5 4 3x− 1
(−x)5 =4
3x−1−(−x)−5= 4
3x−1+x−5 より,導関数は,−4
3x−2−5x−6
(3)−3x32 +√6
x −3x32 +√6
x=−3x32 +x16 より,導関数は,−9 2x12 +1
6x−56
(4) (x+ 1)3 (x+ 1)3=x3+ 3x2+ 3x+ 1より,導関数は, 3x2+ 6x+ 3
(5) µ
3− 1 x3
¶2
µ 3− 1
x3
¶2
= 32−6· 1 x3 + 1
x6 = 9−6x−3+x−6より,導関数は, 18x−4−6x−7
(6) (x−2+ 1)(4−√4 x)
(x−2+ 1)(4−√4
x) = (x−2+ 1)(4−x14) = 4x−2−x−74 + 4−x14 より, 導関数は,−8x−3+7
4x−114 −1 4x−34
