練習16：以下の関数を微分してください。 (1) $y = \cos 2x$ (2) $y = \sqrt{2} \sin(3x + \frac{\pi}{4})$ (3) $y = \sin^2 x$ (4) $y = \tan^2 x$ (5) $y = \frac{1}{\sin x}$ (6) $y = \cos^2 3x$ 練習17：以下の関数を微分してください。 (1) $y = x \sin x + \cos x$ (2) $y = x \cos x - \sin x$

はい、承知いたしました。練習16のすべての問題と、練習17の問題1と2を解きます。

1. 問題の内容

練習16：以下の関数を微分してください。

(1) $y = \cos 2x$

(2) $y = \sqrt{2} \sin(3x + \frac{\pi}{4})$

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

(4) $y = \tan^2 x$

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

(6) $y = \cos^2 3x$

練習17：以下の関数を微分してください。

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

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

2. 解き方の手順

練習16

(1) $y = \cos 2x$

$y' = -\sin 2x \cdot (2x)' = -2 \sin 2x$

(2) $y = \sqrt{2} \sin(3x + \frac{\pi}{4})$

$y' = \sqrt{2} \cos(3x + \frac{\pi}{4}) \cdot (3x + \frac{\pi}{4})' = 3\sqrt{2} \cos(3x + \frac{\pi}{4})$

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

$y' = 2 \sin x \cdot (\sin x)' = 2 \sin x \cos x = \sin 2x$

(4) $y = \tan^2 x$

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

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

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

(6) $y = \cos^2 3x$

$y' = 2 \cos 3x \cdot (\cos 3x)' = 2 \cos 3x \cdot (-\sin 3x) \cdot (3x)' = -6 \cos 3x \sin 3x = -3 \sin 6x$

練習17

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

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

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

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

3. 最終的な答え

練習16

(1) $y' = -2 \sin 2x$

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

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

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

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

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

練習17

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

(2) $y' = -x \sin x$