2. 関数のグラフと微分 2.1 微分係数と導関数（解答）

数学演習1 No.3 2004.10. 6

2. 関数のグラフと微分 2.1 微分係数と導関数（解答）

問題5 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6 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→01 + −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´

׳√

x+ 1 +p

(x+h) + 1´ hp

(x+h) + 1√

x+ 1׳√

x+ 1 +p

(x+h) + 1´

= lim

h→0

(x+ 1)−((x+h) + 1) hp

(x+h) + 1√

x+ 1³√

x+ 1 +p

(x+h) + 1´

= lim

h→0

−h hp

(x+h) + 1√ x+ 1

³√

x+ 1 +p

(x+h) + 1

´

= lim

h→0

p −1

(x+h) + 1√ x+ 1

³√

x+ 1 +p

(x+h) + 1

´

= −1

√x+ 1√

x+ 1¡√

x+ 1 +√ x+ 1¢

=− 1

2(x+ 1)√ x+ 1

問題7 次の関数を微分しなさい. (1)x¹⁰−99x⁴+ 99

導関数はy= (x¹⁰−99x⁴+ 99)⁰= 10x⁹−396x³

(2) 4 3x− 1

(−x)⁵ 4

3x− 1 (−x)⁵ = 4

3 1 x− 1

−x⁵ = 4

3x⁻¹+x⁻⁵ より,導関数はy=−4

3x⁻²−5x⁻⁶

(3)−3x³²+√⁶ x

−3x³² +√⁶

x=−3x³² +x¹⁶ より,導関数はy=−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

=¡

3−x⁻³¢₂

= 3²−6x⁻³+(x⁻³)²= 9−6x⁻³+x⁻⁶より,導関数はy= 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⁻²x¹⁴ + 4−x¹⁴ = 4x⁻²−x⁻⁷⁴ + 4−x¹⁴

よって, 導関数はy=−8x⁻³+7

4x⁻¹¹⁴ −1 4x⁻³⁴