次の関数を微分する。 (1) $y = \sin 5x \cos 5x$ (2) $y = x - 2x\cos^2 x$ (3) $y = \sin^3 x \cos^3 x$

1. 問題の内容

次の関数を微分する。

(1) $y = \sin 5x \cos 5x$

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

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

2. 解き方の手順

(1)

$y = \sin 5x \cos 5x = \frac{1}{2} \sin 10x$ と変形できる。

$\frac{dy}{dx} = \frac{1}{2} (\sin 10x)' = \frac{1}{2} (\cos 10x)(10x)' = \frac{1}{2} (\cos 10x) \cdot 10 = 5 \cos 10x$

(2)

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

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

(3)

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

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

3. 最終的な答え

(1) $\frac{dy}{dx} = 5 \cos 10x$

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

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