次の関数を微分する問題です。 (1) $y = (x^2 + x - 2)^6$ (2) $y = (4 - x^2)^3$ (3) $y = e^{x^2}$ (4) $y = e^{\sin x}$ (5) $y = \log(x^2 + 1)$ (6) $y = \log|\sin x|$ (7) $y = \sqrt[3]{x^2 + 1}$ (8) $y = \frac{1}{\sqrt{x^2 + 1}}$

はい、承知いたしました。与えられた関数の微分を計算します。

1. 問題の内容

次の関数を微分する問題です。

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

(2) $y = (4 - x^2)^3$

(3) $y = e^{x^2}$

(4) $y = e^{\sin x}$

(5) $y = \log(x^2 + 1)$

(6) $y = \log|\sin x|$

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

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

2. 解き方の手順

各関数について、微分を計算します。合成関数の微分法（連鎖律）を必要に応じて使用します。

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

$y' = 6(x^2 + x - 2)^5 \cdot (2x + 1)$

$y' = 6(2x+1)(x^2 + x - 2)^5$

(2) $y = (4 - x^2)^3$

$y' = 3(4 - x^2)^2 \cdot (-2x)$

$y' = -6x(4 - x^2)^2$

(3) $y = e^{x^2}$

$y' = e^{x^2} \cdot (2x)$

$y' = 2xe^{x^2}$

(4) $y = e^{\sin x}$

$y' = e^{\sin x} \cdot (\cos x)$

$y' = \cos x e^{\sin x}$

(5) $y = \log(x^2 + 1)$

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

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

(6) $y = \log|\sin x|$

$y' = \frac{1}{\sin x} \cdot (\cos x)$

$y' = \frac{\cos x}{\sin x} = \cot x$

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

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

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

(8) $y = \frac{1}{\sqrt{x^2 + 1}} = (x^2 + 1)^{-1/2}$

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

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

3. 最終的な答え

(1) $y' = 6(2x+1)(x^2 + x - 2)^5$

(2) $y' = -6x(4 - x^2)^2$

(3) $y' = 2xe^{x^2}$

(4) $y' = \cos x e^{\sin x}$

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

(6) $y' = \cot x$

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

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