与えられた関数を微分せよという問題です。以下の関数について、それぞれ微分を計算します。 (1) $y = 2x - \cos x$ (2) $y = \sin x - \tan x$ (3) $y = \cos(2x - 1)$ (4) $y = \tan 3x$ (5) $y = \cos(\sin x)$ (6) $y = \sin x^2$ (7) $y = \tan x^2$ (8) $y = \cos^3 x$ (9) $y = \tan^3 x$ (10) $y = \frac{1}{\cos x}$ (11) $y = \frac{1}{\sin^2 x}$ (12) $y = x \sin 2x$ (13) $y = \sin x \cos x$ (14) $y = \sin 3x \cos 5x$

はい、承知いたしました。画像の数学の問題を解いていきます。

1. 問題の内容

与えられた関数を微分せよという問題です。以下の関数について、それぞれ微分を計算します。

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

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

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

(4) $y = \tan 3x$

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

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

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

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

(9) $y = \tan^3 x$

(10) $y = \frac{1}{\cos x}$

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

(12) $y = x \sin 2x$

(13) $y = \sin x \cos x$

(14) $y = \sin 3x \cos 5x$

2. 解き方の手順

各関数の微分を計算します。

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

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

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

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

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

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

(4) $y = \tan 3x$

$y' = \frac{1}{\cos^2 3x} \cdot 3 = 3\sec^2 3x$

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

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

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

$y' = \cos x^2 \cdot 2x = 2x\cos x^2$

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

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

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

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

(9) $y = \tan^3 x$

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

(10) $y = \frac{1}{\cos x} = \sec x$

$y' = \frac{\sin x}{\cos^2 x} = \sec x \tan x$

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

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

(12) $y = x \sin 2x$

$y' = \sin 2x + x \cdot \cos 2x \cdot 2 = \sin 2x + 2x\cos 2x$

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

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

または

$y' = \frac{1}{2}\cos 2x \cdot 2 = \cos 2x$

(14) $y = \sin 3x \cos 5x$

積和の公式を使う: $\sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B))$

$y = \frac{1}{2}(\sin(3x+5x) + \sin(3x-5x)) = \frac{1}{2}(\sin 8x + \sin(-2x)) = \frac{1}{2}(\sin 8x - \sin 2x)$

$y' = \frac{1}{2}(8\cos 8x - 2\cos 2x) = 4\cos 8x - \cos 2x$

3. 最終的な答え

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

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

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

(4) $y' = 3\sec^2 3x$

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

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

(7) $y' = 2x \sec^2 x^2$

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

(9) $y' = 3\tan^2 x \sec^2 x$

(10) $y' = \sec x \tan x$

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

(12) $y' = \sin 2x + 2x\cos 2x$

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

(14) $y' = 4\cos 8x - \cos 2x$