与えられた13個の関数をそれぞれ微分する問題です。

1. 問題の内容

与えられた13個の関数をそれぞれ微分する問題です。

2. 解き方の手順

各関数の微分を順番に求めます。

(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) $y = \cos^3 x$

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

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

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

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

$y' = (\cos 3x) \cdot 3 \cdot \cos 5x + \sin 3x \cdot (-\sin 5x) \cdot 5 = 3\cos 3x \cos 5x - 5\sin 3x \sin 5x$

(8) $y = \log \left| \frac{2x-1}{2x+1} \right| = \log |2x-1| - \log |2x+1|$

$y' = \frac{2}{2x-1} - \frac{2}{2x+1} = \frac{2(2x+1)-2(2x-1)}{(2x-1)(2x+1)} = \frac{4x+2-4x+2}{4x^2-1} = \frac{4}{4x^2-1}$

(9) $y = e^{4x}$

$y' = e^{4x} \cdot 4 = 4e^{4x}$

(10) $y = (x+3)e^{-x}$

$y' = (1)e^{-x} + (x+3)(-e^{-x}) = e^{-x} - xe^{-x} - 3e^{-x} = -xe^{-x} - 2e^{-x} = -(x+2)e^{-x}$

(11) $y = e^x \cos x$

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

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

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

(13) $y = 3^{4x}$

$y' = 3^{4x} \cdot \log 3 \cdot 4 = 4\log 3 \cdot 3^{4x}$

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

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

(7) $y' = 3\cos 3x \cos 5x - 5\sin 3x \sin 5x$

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

(9) $y' = 4e^{4x}$

(10) $y' = -(x+2)e^{-x}$

(11) $y' = e^x (\cos x - \sin x)$

(12) $y' = \frac{1}{x}$

(13) $y' = 4\log 3 \cdot 3^{4x}$