与えられた8つの関数をそれぞれ微分する問題です。 (1) $y = x^2 + \sin x$ (2) $y = x - 5\cos x$ (3) $y = x \sin x$ (4) $y = \sin x (\cos x - 1)$ (5) $y = \frac{\cos x}{\sin x}$ (6) $y = \frac{\sin x}{\sin x + \cos x}$ (7) $y = \sin 2x$ (8) $y = \cos 3x$

1. 問題の内容

与えられた8つの関数をそれぞれ微分する問題です。

(1) $y = x^2 + \sin x$

(2) $y = x - 5\cos x$

(3) $y = x \sin x$

(4) $y = \sin x (\cos x - 1)$

(5) $y = \frac{\cos x}{\sin x}$

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

(7) $y = \sin 2x$

(8) $y = \cos 3x$

2. 解き方の手順

(1) $y = x^2 + \sin x$

$y' = \frac{d}{dx}(x^2) + \frac{d}{dx}(\sin x) = 2x + \cos x$

(2) $y = x - 5\cos x$

$y' = \frac{d}{dx}(x) - 5\frac{d}{dx}(\cos x) = 1 - 5(-\sin x) = 1 + 5\sin x$

(3) $y = x \sin x$

積の微分法を用いる: $(uv)' = u'v + uv'$

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

(4) $y = \sin x (\cos x - 1) = \sin x \cos x - \sin x$

$y' = \frac{d}{dx}(\sin x \cos x) - \frac{d}{dx}(\sin x)$

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

$y' = \cos 2x - \cos x$

(5) $y = \frac{\cos x}{\sin x}$

商の微分法を用いる: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$

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

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

商の微分法を用いる

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

(7) $y = \sin 2x$

合成関数の微分法を用いる: $\frac{d}{dx} \sin(f(x)) = \cos(f(x)) \cdot f'(x)$

$y' = \cos(2x) \cdot \frac{d}{dx}(2x) = \cos(2x) \cdot 2 = 2\cos 2x$

(8) $y = \cos 3x$

合成関数の微分法を用いる: $\frac{d}{dx} \cos(f(x)) = -\sin(f(x)) \cdot f'(x)$

$y' = -\sin(3x) \cdot \frac{d}{dx}(3x) = -\sin(3x) \cdot 3 = -3\sin 3x$

3. 最終的な答え

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

(2) $y' = 1 + 5\sin x$

(3) $y' = \sin x + x\cos x$

(4) $y' = \cos 2x - \cos x$

(5) $y' = -\csc^2 x = -\frac{1}{\sin^2 x}$

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

(7) $y' = 2\cos 2x$

(8) $y' = -3\sin 3x$