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==微分法(数学Ⅱ/教科書レベル基本問題3)==
【導関数の公式Ⅰ】
nは正の整数,cは定数とする.
(1) y=cのとき,y'=0
(2) y=xnのとき,y'=nxn−1
• (2)が数学Ⅱで最も重要な微分公式です.
• 「導関数」と「微分」は同じものです.「導関数を求めよ」と「微分せよ」とは同じ意味です.
(公式の証明)
(1)←
y=c(定数)のとき,y=limh0cch=0
(2)←
y=xnのとき,
y=limh0(x+h)nxnh
次の分数式の分子は「二項定理」を使って展開したものであるが,実際には「約分したときにhが付いている項は消える」ので,初めの2項だけ見ればよい
=limh0xn+nxn1h+nC2xn2h2++hnxnh
=limh0(nxn1+nC2xn2h+)=nxn1
【例題1】
 導関数の公式を用いて,次の関数を微分してください.
(1) y=3
(2) y=x4
(解答)
(1)
y'=0
(2)
y'=4x3

• 空欄を「半角数字(1, 2, 3 など)」「半角英小文字」(a, b, c など)で埋めて,採点ボタンを押してください.
• 採点すれば採点結果と解説が出ます.見ているだけでは解説は出ません.
導関数の公式を用いて,次の関数を微分してください.
【問題1-1】
(1) y=4
y=
(2) y=x3
y=x
採点する
【問題1-2】
(1) y=−2
y=
(2) y=x5
y=x
採点する
【導関数の公式Ⅱ】
c, h, kは定数とする.
(3) y=cf(x)のとき,y'=cf'(x)
微分f'(x)を求めてから,定数cを掛けたらよいということ:c×f'(x)
(4) y=f(x)+g(x)のとき,y'=f'(x)+g'(x)
y=f(x)−g(x)のとき,y'=f'(x)−g'(x)
微分f'(x), g'(x)を求めてから,和や差を求めたらよいこと
※(3)(4)は定数倍,和差と微分とは,順序を変えても同じ結果になるということです.
 このような性質をもつ演算は線形であると言われ,数列,ベクトル,微分,積分など高校数学に登場する重要な演算は「線形性」があるので,とても計算しやすい.

 このことに関して,2つの関数の積や商f(x)g(x),f(x)g(x)が登場すると「線形性がない」ので,f(x)g(x),f(x)g(x)の微分は f(x)g(x),f(x)g(x)に等しくない ことに注意しましょう.線形性が言えるのは「定数倍」「和差」までです.
• (1)(2)を使って,定数項とn次式の微分を求め,(3)(4)を使って,その定数倍や和差を求めるというのが数学Ⅱでの微分計算になります.
(公式の証明)
(3)←
y=cf(x)(定数)のとき
y=limh0cf(x+h)cf(x)h=climh0f(x+h)f(x)h
=f(x)
(4)←
y=f(x)+g(x)のとき,
y=limh0{f(x+h)+g(x+h)}{f(x)+g(x)}h
=limh0{f(x+h)f(x)}+{g(x+h)g(x)}h
=limh0{f(x+h)f(x)h+g(x+h)g(x)h}
=f(x)+g(x)
y=f(x)−g(x)のときも同様に示せる
【例題2】
 導関数の公式を用いて,次の関数を微分してください.
(1) y=4x3
(2) y=3x4+2x−5
(解答)
(1)
y'=4×3x2=12x2
(2)
y'=3×4x3+2×1−0=12x3+2

導関数の公式を用いて,次の関数を微分してください.
【問題2-1】
y=3x2+4x+5
y=x+
採点する
導関数の公式を用いて,次の関数を微分してください.
【問題2-2】
y=x3+6x2−7
y=x+x
採点する
導関数の公式を用いて,次の関数を微分してください.
【問題2-3】
y=23x312x2
y=x−x
採点する
【関数の積の微分】
 導関数の公式Ⅱの箇所で説明したように,「関数の定数倍,関数の和差」については「線形性」が成り立つので,個別に微分した結果の「定数倍」「和差」で導関数が求められますが,2つの関数の積では「線形性は成り立たず」「1つずつ微分した結果の積は元の関数の導関数にはなりません」
【〇例1】
y=3x4y=3×4x3=12x3
【〇例2】
y=x2+3xy=2x+3
【×例3】
y=2x×3xy=2×3←これは間違い
y=6x2y=12x←これが正しい)
【×例4】
y=(x2+1)×(x2+2)y=2x×2x←これは間違い
y=x4+3x2+2y=4x3+6x←これが正しい)
 以上の結果から分かるように,2つの多項式の積で表される式を微分するには「まず初めに展開して,和差の形に直してから」微分します.
 数学Ⅲに進めば,2つ以上の関数の積や商の導関数を求める方法も習いますが,数学Ⅱでは「まず初めに展開」です.
(参考)
【問題2-3】のように,係数が分数になっている場合は,関数の割り算ではなく定数項の係数がたまたま分数になっているだけです.2/3=0.6666…など
【例題3】
 次の関数の導関数を求めてください.
(1) y=(2x+1)(3x+1)
(2) y=(2x+1)3
(解答)
「まず初めに展開して,和差の形に直してから」微分します.
(1)
y=6x2+5x+1だから
y=12x+5…(答)
(2)
y=8x3+12x2+6x+1だから
y=24x2+24x+6…(答)
 次の関数の導関数を求めてください.
【問題3-1】
y=x2(x+1)
y=x+x
採点する
 次の関数の導関数を求めてください.
【問題3-2】
y=(2x+1)(2x−3)
y=x
採点する
 次の関数の導関数を求めてください.
【問題3-3】
y=(x+1)3
y=x+x+
採点する
 ※次の公式は,数学Ⅲでは合成関数の微分法を用いて直ちに導けるものですが,数学Ⅱの範囲で「発展学習」として準公式にすることがあります.
(覚えたら計算が楽になり,間違いが減るので,先取りする・・・覚えなければダメだというものではない)
【1次関数の累乗の微分】
a, bは定数でa≠0nは正の整数とする.
y=(ax+b)nのとき,y=na(ax+b)n−1
n(ax+b)n−1に対してaが掛けてある点に注意
• この話は( )の中が1次式の場合だけで,2次式以上ではもっと複雑になります.
y=(ax2+bx+c)nのとき,yは数学Ⅲのお楽しみ
【例題4】
 次の関数の導関数を求めてください.
(1) y=(2x+3)4
(2) y=(3x−4)5
(解答)
(1)
y=8(2x+3)3…(答)
(2)
y=15(3x−4)4…(答)
 次の関数の導関数を求めてください.
【問題4-1】
y=(2x+1)3
y=(2x+1)
採点する
 次の関数の導関数を求めてください.
【問題4-2】
y=(−3x+2)4
y=(−3x+2)
採点する
【変数がx, y以外の文字の導関数】
 変数がx, y以外の文字の場合の導関数も,x, yの場合を参考にしながら求められる.
【例題5】
 次の関数の導関数を求めてください.(πは円周率を表す定数とする)
(1) L=2πrrを変数として微分してください.
(2) S=πr2rを変数として微分してください.
(3) y=at2+bt+ctを変数として微分してください.
(解答)
(1)
L=2π
次の書き方の方が,変数と関数がはっきりする
dLdr=2π
(2)
dSdr=2πr
(3)
dydt=2at+b
 次の関数を[ ]の文字について微分してください.(πは円周率を表す定数とする)
【問題5-1】
V=43πr3  ・・・[r]
dVdr=πr
採点する
 次の関数を[ ]の文字について微分してください.(a, bは定数とする)
【問題5-2】
s=at2+bt3  ・・・[t]
dsdt=at+bt
採点する
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