次の関数を微分せよ。 (1) $y = \sin^2 3x$ (2) $y = 2\sin x \cos x (1 - 2\sin^2 x)$ (3) $y = \sin^3 x \cos^3 x$ (4) $y = (\sin x + \cos x)^2$ (5) $y = \frac{1}{\sin x}$ (6) $y = \frac{1}{\tan 2x}$ (7) $y = \frac{\sin x}{1 + \cos x}$

1. 問題の内容

次の関数を微分せよ。

(1) $y = \sin^2 3x$

(2) $y = 2\sin x \cos x (1 - 2\sin^2 x)$

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

(4) $y = (\sin x + \cos x)^2$

(5) $y = \frac{1}{\sin x}$

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

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

2. 解き方の手順

(1) $y = \sin^2 3x$

連鎖律を用いて微分する。

$\frac{dy}{dx} = 2\sin 3x \cdot (\sin 3x)' = 2\sin 3x \cdot \cos 3x \cdot (3x)' = 2\sin 3x \cos 3x \cdot 3 = 6\sin 3x \cos 3x = 3\sin 6x$

(2) $y = 2\sin x \cos x (1 - 2\sin^2 x)$

三角関数の恒等式を用いる。$2\sin x \cos x = \sin 2x$、$1 - 2\sin^2 x = \cos 2x$

$y = \sin 2x \cos 2x = \frac{1}{2} \sin 4x$

$\frac{dy}{dx} = \frac{1}{2} (\sin 4x)' = \frac{1}{2} \cos 4x \cdot (4x)' = \frac{1}{2} \cos 4x \cdot 4 = 2\cos 4x$

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

$y = (\sin x \cos x)^3 = \left(\frac{1}{2} \sin 2x\right)^3 = \frac{1}{8} \sin^3 2x$

$\frac{dy}{dx} = \frac{1}{8} \cdot 3 \sin^2 2x \cdot (\sin 2x)' = \frac{3}{8} \sin^2 2x \cdot \cos 2x \cdot (2x)' = \frac{3}{8} \sin^2 2x \cdot \cos 2x \cdot 2 = \frac{3}{4} \sin^2 2x \cos 2x$

(4) $y = (\sin x + \cos x)^2$

$y = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + \sin 2x$

$\frac{dy}{dx} = (1 + \sin 2x)' = 0 + \cos 2x \cdot (2x)' = \cos 2x \cdot 2 = 2\cos 2x$

(5) $y = \frac{1}{\sin x} = \csc x$

$\frac{dy}{dx} = -\csc x \cot x = -\frac{\cos x}{\sin^2 x}$

(6) $y = \frac{1}{\tan 2x} = \cot 2x$

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

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

商の微分法を用いる。

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

3. 最終的な答え

(1) $y' = 3\sin 6x$

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

(3) $y' = \frac{3}{4} \sin^2 2x \cos 2x$

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

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

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

(7) $y' = \frac{1}{1+\cos x}$