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

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

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

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

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

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

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

(4) $y = \tan 3x$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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