以下の関数の2次導関数 $y^{(2)}$ を求めよ。 (1) $y = x^4 + 3x^2 + 2x + 7$ (2) $y = (x^3 + 1)^2$ (3) $y = \sqrt{x+1}$ (4) $y = \sqrt[3]{x^2 + 3}$ (5) $y = \frac{1}{x^2 + 1}$ (6) $y = \frac{x}{x^2 + 1}$

はい、承知いたしました。画像にある6つの関数の2次導関数を求めます。

1. 問題の内容

以下の関数の2次導関数 $y^{(2)}$ を求めよ。

(1) $y = x^4 + 3x^2 + 2x + 7$

(2) $y = (x^3 + 1)^2$

(3) $y = \sqrt{x+1}$

(4) $y = \sqrt[3]{x^2 + 3}$

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

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

2. 解き方の手順

各関数について、まず1次導関数 $y'$ を求め、次にその導関数 $y'$ を微分して2次導関数 $y^{(2)}$ を求めます。

(1) $y = x^4 + 3x^2 + 2x + 7$

$y' = 4x^3 + 6x + 2$

$y'' = 12x^2 + 6$

(2) $y = (x^3 + 1)^2 = x^6 + 2x^3 + 1$

$y' = 6x^5 + 6x^2$

$y'' = 30x^4 + 12x$

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

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

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

(4) $y = \sqrt[3]{x^2 + 3} = (x^2 + 3)^{\frac{1}{3}}$

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

$y'' = \frac{2}{3} \cdot \frac{(x^2+3)^{\frac{2}{3}} - x \cdot \frac{2}{3}(x^2+3)^{-\frac{1}{3}} \cdot 2x}{(x^2+3)^{\frac{4}{3}}} = \frac{2}{3} \cdot \frac{(x^2+3) - \frac{4}{3}x^2}{(x^2+3)^{\frac{5}{3}}} = \frac{2}{3} \cdot \frac{3x^2 + 9 - 4x^2}{3(x^2+3)^{\frac{5}{3}}} = \frac{2(9-x^2)}{9(x^2+3)^{\frac{5}{3}}}$

(5) $y = \frac{1}{x^2 + 1} = (x^2 + 1)^{-1}$

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

$y'' = -2 \cdot \frac{(x^2+1)^2 - x \cdot 2(x^2+1) \cdot 2x}{(x^2+1)^4} = -2 \cdot \frac{(x^2+1) - 4x^2}{(x^2+1)^3} = -2 \cdot \frac{1-3x^2}{(x^2+1)^3} = \frac{6x^2 - 2}{(x^2+1)^3}$

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

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

$y'' = \frac{-2x(x^2+1)^2 - (1-x^2) \cdot 2(x^2+1) \cdot 2x}{(x^2+1)^4} = \frac{-2x(x^2+1) - 4x(1-x^2)}{(x^2+1)^3} = \frac{-2x^3 - 2x - 4x + 4x^3}{(x^2+1)^3} = \frac{2x^3 - 6x}{(x^2+1)^3}$

3. 最終的な答え

(1) $y'' = 12x^2 + 6$

(2) $y'' = 30x^4 + 12x$

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

(4) $y'' = \frac{2(9-x^2)}{9(x^2+3)^{\frac{5}{3}}}$

(5) $y'' = \frac{6x^2 - 2}{(x^2+1)^3}$

(6) $y'' = \frac{2x^3 - 6x}{(x^2+1)^3}$