次の関数を微分せよ。 (1) $y = \frac{1}{(3x - 2)^2}$ (2) $y = \frac{1}{(x^2 + 2x + 3)^4}$ (3) $y = \sqrt{x^2 + 4x + 1}$ (4) $y = \frac{1}{\sqrt{x^2 + 1}}$ (5) $y = \sqrt[3]{(2x - 3)^2}$ (6) $y = \frac{4}{\sqrt[4]{4 - x^2}}$

はい、承知いたしました。

1. 問題の内容

次の関数を微分せよ。

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

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

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

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

(5) $y = \sqrt[3]{(2x - 3)^2}$

(6) $y = \frac{4}{\sqrt[4]{4 - x^2}}$

2. 解き方の手順

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

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

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

$y' = -4(x^2 + 2x + 3)^{-5} \cdot (2x + 2) = -8(x + 1)(x^2 + 2x + 3)^{-5} = -\frac{8(x + 1)}{(x^2 + 2x + 3)^5}$

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

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

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

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

(5) $y = \sqrt[3]{(2x - 3)^2} = (2x - 3)^{2/3}$

$y' = \frac{2}{3}(2x - 3)^{-1/3} \cdot 2 = \frac{4}{3}(2x - 3)^{-1/3} = \frac{4}{3\sqrt[3]{2x - 3}}$

(6) $y = \frac{4}{\sqrt[4]{4 - x^2}} = 4(4 - x^2)^{-1/4}$

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

3. 最終的な答え

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

(2) $y' = -\frac{8(x + 1)}{(x^2 + 2x + 3)^5}$

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

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

(5) $y' = \frac{4}{3\sqrt[3]{2x - 3}}$

(6) $y' = \frac{2x}{(4 - x^2)^{5/4}}$