与えられた6つの関数 $f(x)$ に対して、それぞれの導関数を求める問題です。 (1) $f(x) = \frac{1}{x^2+1}$ (2) $f(x) = \frac{x}{x^2+1}$ (3) $f(x) = \frac{x+2}{x^2+1}$ (4) $f(x) = \frac{2x+1}{x^2+x+1}$ (5) $f(x) = x^2(5x^3+3)$ (6) $f(x) = x^2(5x^3+3)^5$

1. 問題の内容

与えられた6つの関数 $f(x)$ に対して、それぞれの導関数を求める問題です。

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

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

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

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

(5) $f(x) = x^2(5x^3+3)$

(6) $f(x) = x^2(5x^3+3)^5$

2. 解き方の手順

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

$f'(x) = -\frac{1}{(x^2+1)^2} \cdot (2x) = -\frac{2x}{(x^2+1)^2}$

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

$f'(x) = \frac{(x^2+1) \cdot 1 - x \cdot (2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$

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

$f'(x) = \frac{(x^2+1) \cdot 1 - (x+2) \cdot (2x)}{(x^2+1)^2} = \frac{x^2+1 - 2x^2 - 4x}{(x^2+1)^2} = \frac{-x^2 - 4x + 1}{(x^2+1)^2}$

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

$f'(x) = \frac{(x^2+x+1) \cdot 2 - (2x+1) \cdot (2x+1)}{(x^2+x+1)^2} = \frac{2x^2+2x+2 - (4x^2+4x+1)}{(x^2+x+1)^2} = \frac{-2x^2-2x+1}{(x^2+x+1)^2}$

(5) $f(x) = x^2(5x^3+3) = 5x^5+3x^2$

$f'(x) = 25x^4 + 6x$

(6) $f(x) = x^2(5x^3+3)^5$

$f'(x) = 2x(5x^3+3)^5 + x^2 \cdot 5(5x^3+3)^4 \cdot (15x^2) = 2x(5x^3+3)^5 + 75x^4(5x^3+3)^4 = x(5x^3+3)^4 [2(5x^3+3) + 75x^3] = x(5x^3+3)^4 (10x^3 + 6 + 75x^3) = x(5x^3+3)^4 (85x^3+6)$

3. 最終的な答え

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

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

(3) $f'(x) = \frac{-x^2 - 4x + 1}{(x^2+1)^2}$

(4) $f'(x) = \frac{-2x^2-2x+1}{(x^2+x+1)^2}$

(5) $f'(x) = 25x^4 + 6x$

(6) $f'(x) = x(5x^3+3)^4 (85x^3+6)$