次の6つの関数の導関数を求めよ。 (1) $y = x^x$ (2) $y = x^{\frac{1}{x}}$ (3) $y = (x+1)(x+2)(x+3)$ (4) $y = \log(x + \sqrt{x^2+1})$ (5) $y = a^{\log x}$ (6) $y = \sqrt{\frac{1-x}{1+x}}$

はい、承知いたしました。次の数学の問題を解きます。

1. 問題の内容

次の6つの関数の導関数を求めよ。

(1) $y = x^x$

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

(3) $y = (x+1)(x+2)(x+3)$

(4) $y = \log(x + \sqrt{x^2+1})$

(5) $y = a^{\log x}$

(6) $y = \sqrt{\frac{1-x}{1+x}}$

2. 解き方の手順

(1) $y = x^x$

両辺の自然対数を取ります。

$\log y = \log x^x = x \log x$

両辺を $x$ で微分します。

$\frac{1}{y} \frac{dy}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$

$\frac{dy}{dx} = y (\log x + 1) = x^x (\log x + 1)$

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

両辺の自然対数を取ります。

$\log y = \log x^{\frac{1}{x}} = \frac{1}{x} \log x$

両辺を $x$ で微分します。

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

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

(3) $y = (x+1)(x+2)(x+3)$

$y = (x^2 + 3x + 2)(x+3) = x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 = x^3 + 6x^2 + 11x + 6$

$\frac{dy}{dx} = 3x^2 + 12x + 11$

(4) $y = \log(x + \sqrt{x^2+1})$

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

(5) $y = a^{\log x}$

$\frac{dy}{dx} = a^{\log x} \cdot \log a \cdot \frac{1}{x} = \frac{a^{\log x} \log a}{x}$

(6) $y = \sqrt{\frac{1-x}{1+x}}$

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

$\log y = \frac{1}{2} \log (\frac{1-x}{1+x}) = \frac{1}{2} (\log(1-x) - \log(1+x))$

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

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

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

別の方法：

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

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

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

3. 最終的な答え

(1) $\frac{dy}{dx} = x^x (\log x + 1)$

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

(3) $\frac{dy}{dx} = 3x^2 + 12x + 11$

(4) $\frac{dy}{dx} = \frac{1}{\sqrt{x^2+1}}$

(5) $\frac{dy}{dx} = \frac{a^{\log x} \log a}{x}$

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