与えられた関数を微分する問題です。ここでは、(1) $y = \tan 2x \cos 2x$、(2) $y = \cot \frac{x}{2} \sin \frac{x}{2}$、(3) $y = \sin^2 3x \csc 3x$、(4) $y = \sec \frac{x}{3} \cos^2 \frac{x}{3}$、(5) $y = \tan^2 \frac{x}{4} \cot \frac{x}{4}$、(6) $y = \frac{\tan 4x}{\sec 4x}$、(7) $y = \frac{\cot 5x}{\csc 5x}$、(8) $y = \frac{\sec \frac{x}{5}}{\csc \frac{x}{5}}$をそれぞれ微分します。

はい、承知いたしました。以下の形式で問題の解答を作成します。

1. 問題の内容

与えられた関数を微分する問題です。ここでは、(1) $y = \tan 2x \cos 2x$、(2) $y = \cot \frac{x}{2} \sin \frac{x}{2}$、(3) $y = \sin^2 3x \csc 3x$、(4) $y = \sec \frac{x}{3} \cos^2 \frac{x}{3}$、(5) $y = \tan^2 \frac{x}{4} \cot \frac{x}{4}$、(6) $y = \frac{\tan 4x}{\sec 4x}$、(7) $y = \frac{\cot 5x}{\csc 5x}$、(8) $y = \frac{\sec \frac{x}{5}}{\csc \frac{x}{5}}$をそれぞれ微分します。

2. 解き方の手順

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

$y' = 2\cos 2x$

(2) $y = \cot \frac{x}{2} \sin \frac{x}{2} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \sin \frac{x}{2} = \cos \frac{x}{2}$

$y' = -\frac{1}{2} \sin \frac{x}{2}$

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

$y' = 3\cos 3x$

(4) $y = \sec \frac{x}{3} \cos^2 \frac{x}{3} = \frac{1}{\cos \frac{x}{3}} \cos^2 \frac{x}{3} = \cos \frac{x}{3}$

$y' = -\frac{1}{3} \sin \frac{x}{3}$

(5) $y = \tan^2 \frac{x}{4} \cot \frac{x}{4} = \tan^2 \frac{x}{4} \frac{1}{\tan \frac{x}{4}} = \tan \frac{x}{4}$

$y' = \frac{1}{4} \sec^2 \frac{x}{4}$

(6) $y = \frac{\tan 4x}{\sec 4x} = \frac{\frac{\sin 4x}{\cos 4x}}{\frac{1}{\cos 4x}} = \sin 4x$

$y' = 4\cos 4x$

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

$y' = -5\sin 5x$

(8) $y = \frac{\sec \frac{x}{5}}{\csc \frac{x}{5}} = \frac{\frac{1}{\cos \frac{x}{5}}}{\frac{1}{\sin \frac{x}{5}}} = \frac{\sin \frac{x}{5}}{\cos \frac{x}{5}} = \tan \frac{x}{5}$

$y' = \frac{1}{5} \sec^2 \frac{x}{5}$

3. 最終的な答え

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

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

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

(4) $y' = -\frac{1}{3} \sin \frac{x}{3}$

(5) $y' = \frac{1}{4} \sec^2 \frac{x}{4}$

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

(7) $y' = -5\sin 5x$

(8) $y' = \frac{1}{5} \sec^2 \frac{x}{5}$