与えられた関数を微分する問題です。 (1) $y = 2\cos x + 3$ (2) $y = \sin(3x + 2)$ (3) $y = \sin^4 x$ (4) $y = \tan(\cos x)$ (5) $y = \frac{2}{\tan x}$ (6) $y = 2x \sin 2x$ (7) $y = x^3 \sin 4x$ (8) $y = \sin x \cos 2x$ (9) $y = \frac{x}{\cos^2 x}$ (10) $y = \frac{\sin x}{1 + \cos x}$

## 回答

1. 問題の内容

与えられた関数を微分する問題です。

(1) $y = 2\cos x + 3$

(2) $y = \sin(3x + 2)$

(3) $y = \sin^4 x$

(4) $y = \tan(\cos x)$

(5) $y = \frac{2}{\tan x}$

(6) $y = 2x \sin 2x$

(7) $y = x^3 \sin 4x$

(8) $y = \sin x \cos 2x$

(9) $y = \frac{x}{\cos^2 x}$

(10) $y = \frac{\sin x}{1 + \cos x}$

2. 解き方の手順

(1) $y = 2\cos x + 3$

$\frac{dy}{dx} = 2(-\sin x) + 0 = -2\sin x$

(2) $y = \sin(3x + 2)$

$\frac{dy}{dx} = \cos(3x + 2) \cdot 3 = 3\cos(3x + 2)$

(3) $y = \sin^4 x$

$\frac{dy}{dx} = 4\sin^3 x \cdot \cos x$

(4) $y = \tan(\cos x)$

$\frac{dy}{dx} = \frac{1}{\cos^2(\cos x)} \cdot (-\sin x) = -\frac{\sin x}{\cos^2(\cos x)}$

(5) $y = \frac{2}{\tan x} = 2\cot x$

$\frac{dy}{dx} = 2(-\frac{1}{\sin^2 x}) = -\frac{2}{\sin^2 x}$

または

$y = \frac{2}{\tan x}$ より、$\frac{dy}{dx} = 2 \cdot (-\frac{1}{\tan^2 x}) \cdot \frac{1}{\cos^2 x}= -\frac{2}{\tan^2x\cos^2x} = -\frac{2}{\sin^2 x}$.

(6) $y = 2x \sin 2x$

$\frac{dy}{dx} = 2\sin 2x + 2x \cdot (\cos 2x \cdot 2) = 2\sin 2x + 4x\cos 2x$

(7) $y = x^3 \sin 4x$

$\frac{dy}{dx} = 3x^2 \sin 4x + x^3 (\cos 4x \cdot 4) = 3x^2 \sin 4x + 4x^3 \cos 4x$

(8) $y = \sin x \cos 2x$

$\frac{dy}{dx} = \cos x \cos 2x + \sin x (-\sin 2x \cdot 2) = \cos x \cos 2x - 2\sin x \sin 2x$

(9) $y = \frac{x}{\cos^2 x}$

$\frac{dy}{dx} = \frac{\cos^2 x - x(2\cos x (-\sin x))}{\cos^4 x} = \frac{\cos^2 x + 2x\sin x\cos x}{\cos^4 x} = \frac{\cos x + 2x\sin x}{\cos^3 x}$

(10) $y = \frac{\sin x}{1 + \cos x}$

$\frac{dy}{dx} = \frac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2} = \frac{\cos x + 1}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$

3. 最終的な答え

(1) $\frac{dy}{dx} = -2\sin x$

(2) $\frac{dy}{dx} = 3\cos(3x + 2)$

(3) $\frac{dy}{dx} = 4\sin^3 x \cos x$

(4) $\frac{dy}{dx} = -\frac{\sin x}{\cos^2(\cos x)}$

(5) $\frac{dy}{dx} = -\frac{2}{\sin^2 x}$

(6) $\frac{dy}{dx} = 2\sin 2x + 4x\cos 2x$

(7) $\frac{dy}{dx} = 3x^2 \sin 4x + 4x^3 \cos 4x$

(8) $\frac{dy}{dx} = \cos x \cos 2x - 2\sin x \sin 2x$

(9) $\frac{dy}{dx} = \frac{\cos x + 2x\sin x}{\cos^3 x}$

(10) $\frac{dy}{dx} = \frac{1}{1 + \cos x}$