与えられた関数 $y$ を $x$ で微分する問題です。 (1) $y = \log 3x$ (2) $y = \log_{10}(-4x)$ (3) $y = \log |x^2 - 1|$ (4) $y = (\log x)^3$ (5) $y = \log_2 |\cos x|$ (6) $y = \log(\log x)$ (7) $y = \log \frac{1 + \sin x}{1 - \sin x}$ (8) $y = e^{6x}$

はい、承知いたしました。画像にある問題について、微分を計算します。

1. 問題の内容

与えられた関数 $y$ を $x$ で微分する問題です。

(1) $y = \log 3x$

(2) $y = \log_{10}(-4x)$

(3) $y = \log |x^2 - 1|$

(4) $y = (\log x)^3$

(5) $y = \log_2 |\cos x|$

(6) $y = \log(\log x)$

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

(8) $y = e^{6x}$

2. 解き方の手順

微分公式：

* $(\log x)' = \frac{1}{x}$

* $(\log_a x)' = \frac{1}{x \ln a}$

* $(\sin x)' = \cos x$

* $(\cos x)' = -\sin x$

* $(e^x)' = e^x$

* 合成関数の微分: $\{f(g(x))\}' = f'(g(x)) g'(x)$

* 商の微分: $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$

(1) $y = \log 3x$

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

(2) $y = \log_{10}(-4x)$

$y' = \frac{1}{-4x \ln 10} \cdot (-4) = \frac{1}{x \ln 10}$

(3) $y = \log |x^2 - 1|$

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

(4) $y = (\log x)^3$

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

(5) $y = \log_2 |\cos x|$

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

(6) $y = \log(\log x)$

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

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

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

(8) $y = e^{6x}$

$y' = e^{6x} \cdot 6 = 6 e^{6x}$

3. 最終的な答え

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

(2) $y' = \frac{1}{x \ln 10}$

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

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

(5) $y' = -\frac{\tan x}{\ln 2}$

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

(7) $y' = \frac{2}{\cos x} = 2 \sec x$

(8) $y' = 6 e^{6x}$