次の関数を微分してください。$a \neq 0$ は定数です。 (1) $y = \log\frac{1 + \cos x}{\sin x}$ (2) $y = \frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|$ (3) $y = \log(\cos^2 x)$ (4) $y = \log\sqrt[3]{x^3 - 1}$ (5) $y = \log\frac{(x-2)^3}{(x+4)^4}$ (6) $y = \log(e^x + e^{-x})$ (7) $y = \frac{1}{e^{2x} + e^{-2x}}$ (8) $y = \frac{1}{e}(\log x)^e$ (9) $y = \frac{e^{3x} - e^{-3x}}{e^x - e^{-x}}$

はい、承知いたしました。それでは、与えられた問題を解いていきます。

1. 問題の内容

次の関数を微分してください。$a \neq 0$ は定数です。

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

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

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

(4) $y = \log\sqrt[3]{x^3 - 1}$

(5) $y = \log\frac{(x-2)^3}{(x+4)^4}$

(6) $y = \log(e^x + e^{-x})$

(7) $y = \frac{1}{e^{2x} + e^{-2x}}$

(8) $y = \frac{1}{e}(\log x)^e$

(9) $y = \frac{e^{3x} - e^{-3x}}{e^x - e^{-x}}$

2. 解き方の手順

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

$\frac{dy}{dx} = \frac{1}{\frac{1 + \cos x}{\sin x}} \cdot \frac{-\sin x \cdot \sin x - (1 + \cos x) \cdot \cos x}{\sin^2 x} = \frac{\sin x}{1 + \cos x} \cdot \frac{-\sin^2 x - \cos x - \cos^2 x}{\sin^2 x} = \frac{\sin x}{1 + \cos x} \cdot \frac{-1 - \cos x}{\sin^2 x} = \frac{-1}{\sin x} = -\csc x$

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

$\frac{dy}{dx} = \frac{1}{2a} \cdot \frac{1}{\frac{x-a}{x+a}} \cdot \frac{(x+a) - (x-a)}{(x+a)^2} = \frac{1}{2a} \cdot \frac{x+a}{x-a} \cdot \frac{2a}{(x+a)^2} = \frac{1}{2a} \cdot \frac{2a}{(x-a)(x+a)} = \frac{1}{x^2 - a^2}$

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

$\frac{dy}{dx} = \frac{1}{\cos^2 x} \cdot 2\cos x \cdot (-\sin x) = \frac{-2\cos x \sin x}{\cos^2 x} = -2\tan x$

(4) $y = \log\sqrt[3]{x^3 - 1} = \log(x^3 - 1)^{1/3} = \frac{1}{3}\log(x^3 - 1)$

$\frac{dy}{dx} = \frac{1}{3} \cdot \frac{1}{x^3 - 1} \cdot 3x^2 = \frac{x^2}{x^3 - 1}$

(5) $y = \log\frac{(x-2)^3}{(x+4)^4} = 3\log(x-2) - 4\log(x+4)$

$\frac{dy}{dx} = \frac{3}{x-2} - \frac{4}{x+4} = \frac{3(x+4) - 4(x-2)}{(x-2)(x+4)} = \frac{3x + 12 - 4x + 8}{x^2 + 2x - 8} = \frac{-x + 20}{x^2 + 2x - 8}$

(6) $y = \log(e^x + e^{-x})$

$\frac{dy}{dx} = \frac{1}{e^x + e^{-x}} \cdot (e^x - e^{-x}) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

(7) $y = \frac{1}{e^{2x} + e^{-2x}} = (e^{2x} + e^{-2x})^{-1}$

$\frac{dy}{dx} = -1(e^{2x} + e^{-2x})^{-2} \cdot (2e^{2x} - 2e^{-2x}) = \frac{-2(e^{2x} - e^{-2x})}{(e^{2x} + e^{-2x})^2}$

(8) $y = \frac{1}{e}(\log x)^e$

$\frac{dy}{dx} = \frac{1}{e} \cdot e(\log x)^{e-1} \cdot \frac{1}{x} = \frac{(\log x)^{e-1}}{x}$

(9) $y = \frac{e^{3x} - e^{-3x}}{e^x - e^{-x}}$

$\frac{dy}{dx} = \frac{(3e^{3x} + 3e^{-3x})(e^x - e^{-x}) - (e^{3x} - e^{-3x})(e^x + e^{-x})}{(e^x - e^{-x})^2}$

$= \frac{3e^{4x} - 3e^{2x} + 3e^{-2x} - 3e^{-4x} - (e^{4x} + e^{2x} - e^{-2x} - e^{-4x})}{(e^x - e^{-x})^2}$

$= \frac{2e^{4x} - 4e^{2x} + 4e^{-2x} - 2e^{-4x}}{(e^x - e^{-x})^2}$

$= \frac{2(e^{4x} - 2e^{2x} + 2e^{-2x} - e^{-4x})}{(e^x - e^{-x})^2}$

3. 最終的な答え

(1) $\frac{dy}{dx} = -\csc x$

(2) $\frac{dy}{dx} = \frac{1}{x^2 - a^2}$

(3) $\frac{dy}{dx} = -2\tan x$

(4) $\frac{dy}{dx} = \frac{x^2}{x^3 - 1}$

(5) $\frac{dy}{dx} = \frac{-x + 20}{x^2 + 2x - 8}$

(6) $\frac{dy}{dx} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

(7) $\frac{dy}{dx} = \frac{-2(e^{2x} - e^{-2x})}{(e^{2x} + e^{-2x})^2}$

(8) $\frac{dy}{dx} = \frac{(\log x)^{e-1}}{x}$

(9) $\frac{dy}{dx} = \frac{2(e^{4x} - 2e^{2x} + 2e^{-2x} - e^{-4x})}{(e^x - e^{-x})^2}$