微分積分学１ No.2 2005. 4.20
1.2 微分係数と導関数 担当：市原
問題4 次の関数のx= 3における微分係数を,定義に基づいてもとめなさい. (1)y=x2+ 1
h→0lim
((3 +h)2+ 1)−(32+ 1)
h = lim
h→0
9 + 6h+h2+ 1−10
h = lim
h→0
6h+h2 h = lim
h→06 +h= 6
(2)y=√ x
h→0lim
√3 +h−√ 3
h = lim
h→0
(√
3 +h−√ 3)(√
3 +h+√ 3) h(√
3 +h+√
3) = lim
h→0
(3 +h)−3 h(√
3 +h+√ 3)
= lim
h→0
h h(√
3 +h+√
3) = lim
h→0
√ 1
3 +h+√ 3 = 1
2√ 3 =
√3 6
(3)y= 1 x
h→0lim 1 3 +h−1
3 h = lim
h→0
3−(3 +h) 3(3 +h)
h = lim
h→0
−h 3(3 +h)
h = lim
h→0
−h
3(3 +h)×(3(3 +h)) h×(3(3 +h))
= lim
h→0
−h
3h(3 +h)= lim
h→0
−1
3(3 +h) =−1 9
問題5 次の関数を微分しなさい. (1)y=x−4+x3
導関数は,y=¡
x−4+x3¢0
=−4x−5+ 3x2
(2)y= 2√
x+x5= 2x12 +x5
よって導関数は,y=
³
2x12 +x5
´0
= 2×1
2 ×x−12 + 5x4=x−12 + 5x4
(3)y= (2√
x+x)(3x2+2x) = (2x12+x)(3x2+2x) = 6x12x2+4x12x+3x3+2x2= 6x52+4x32+3x3+2x2
よって導関数は,y=
³
6x52 + 4x32 + 3x3+ 2x2
´0
= 15x32 + 6x12 + 9x2+ 4x
