与えられた関数について、導関数の定義に従って導関数を求める問題です。 (1) $ax^2 + bx + c$ (2) $x^4$ (3) $\frac{1}{x}$ (4) $\frac{1}{x^2}$ (5) $\sqrt[3]{x}$

1. 問題の内容

与えられた関数について、導関数の定義に従って導関数を求める問題です。

(1) $ax^2 + bx + c$

(2) $x^4$

(3) $\frac{1}{x}$

(4) $\frac{1}{x^2}$

(5) $\sqrt[3]{x}$

2. 解き方の手順

導関数の定義は、

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

です。各関数について、この定義を用いて導関数を求めます。

(1) $f(x) = ax^2 + bx + c$ の場合

$f(x+h) = a(x+h)^2 + b(x+h) + c = a(x^2 + 2xh + h^2) + bx + bh + c = ax^2 + 2axh + ah^2 + bx + bh + c$

$f(x+h) - f(x) = (ax^2 + 2axh + ah^2 + bx + bh + c) - (ax^2 + bx + c) = 2axh + ah^2 + bh$

$\frac{f(x+h) - f(x)}{h} = \frac{2axh + ah^2 + bh}{h} = 2ax + ah + b$

$f'(x) = \lim_{h \to 0} (2ax + ah + b) = 2ax + b$

(2) $f(x) = x^4$ の場合

$f(x+h) = (x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$

$f(x+h) - f(x) = (x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) - x^4 = 4x^3h + 6x^2h^2 + 4xh^3 + h^4$

$\frac{f(x+h) - f(x)}{h} = \frac{4x^3h + 6x^2h^2 + 4xh^3 + h^4}{h} = 4x^3 + 6x^2h + 4xh^2 + h^3$

$f'(x) = \lim_{h \to 0} (4x^3 + 6x^2h + 4xh^2 + h^3) = 4x^3$

(3) $f(x) = \frac{1}{x}$ の場合

$f(x+h) = \frac{1}{x+h}$

$f(x+h) - f(x) = \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}$

$\frac{f(x+h) - f(x)}{h} = \frac{-h}{x(x+h)h} = \frac{-1}{x(x+h)}$

$f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x^2}$

(4) $f(x) = \frac{1}{x^2}$ の場合

$f(x+h) = \frac{1}{(x+h)^2}$

$f(x+h) - f(x) = \frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2(x+h)^2} = \frac{x^2 - (x^2 + 2xh + h^2)}{x^2(x+h)^2} = \frac{-2xh - h^2}{x^2(x+h)^2}$

$\frac{f(x+h) - f(x)}{h} = \frac{-2xh - h^2}{x^2(x+h)^2h} = \frac{-2x - h}{x^2(x+h)^2}$

$f'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2(x+h)^2} = \frac{-2x}{x^2x^2} = \frac{-2}{x^3}$

(5) $f(x) = \sqrt[3]{x} = x^{\frac{1}{3}}$ の場合

$f(x+h) = (x+h)^{\frac{1}{3}}$

$f(x+h) - f(x) = (x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}$

$\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h}$

ここで、$a^3-b^3 = (a-b)(a^2+ab+b^2)$を利用して分子を有理化します。

$\frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h} = \frac{((x+h)^{\frac{1}{3}} - x^{\frac{1}{3}})((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}})}{h((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}})} = \frac{(x+h)-x}{h((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}})} = \frac{h}{h((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}})} = \frac{1}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}$

$f'(x) = \lim_{h \to 0} \frac{1}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}} = \frac{1}{x^{\frac{2}{3}} + x^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}} = \frac{1}{3x^{\frac{2}{3}}} = \frac{1}{3\sqrt[3]{x^2}}$

3. 最終的な答え

(1) $2ax + b$

(2) $4x^3$

(3) $-\frac{1}{x^2}$

(4) $-\frac{2}{x^3}$

(5) $\frac{1}{3\sqrt[3]{x^2}}$