与えられた9個の関数をそれぞれ微分せよ。

わかりました。画像の問題を解きます。

1. 問題の内容

与えられた9個の関数をそれぞれ微分せよ。

2. 解き方の手順

各関数について、微分の公式と合成関数の微分法を適用して解きます。

(1) $y = \cos(3x + \frac{\pi}{6})$

$y' = -\sin(3x + \frac{\pi}{6}) \cdot 3 = -3\sin(3x + \frac{\pi}{6})$

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

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

(3) $y = \sqrt{1 + \sin x}$

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

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

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

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

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

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

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

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

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

(8) $y = \sin x \cdot \cos^2 x$

$y' = \cos x \cdot \cos^2 x + \sin x \cdot (2\cos x (-\sin x)) = \cos^3 x - 2\sin^2 x \cos x = \cos x(\cos^2 x - 2\sin^2 x) = \cos x(\cos^2 x - 2(1 - \cos^2 x)) = \cos x(3\cos^2 x - 2)$

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

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

3. 最終的な答え

(1) $y' = -3\sin(3x + \frac{\pi}{6})$

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

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

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

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

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

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

(8) $y' = \cos x(3\cos^2 x - 2)$

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