次の関数を微分せよ。 (1) $y = (1 + \log x)^2$ (2) $y = \log(x^3 - 3x + 5)$ (3) $y = \log(\sin^2 x)$ (4) $y = \log(\log x)$ (5) $y = \log(x + \sqrt{x^2 + 1})$

1. 問題の内容

次の関数を微分せよ。

(1) $y = (1 + \log x)^2$

(2) $y = \log(x^3 - 3x + 5)$

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

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

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

2. 解き方の手順

(1) $y = (1 + \log x)^2$

合成関数の微分を用いる。$u = 1 + \log x$ とおくと、$y = u^2$。

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = 2u = 2(1 + \log x)$

$\frac{du}{dx} = \frac{1}{x}$

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

(2) $y = \log(x^3 - 3x + 5)$

合成関数の微分を用いる。$u = x^3 - 3x + 5$ とおくと、$y = \log u$。

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{x^3 - 3x + 5}$

$\frac{du}{dx} = 3x^2 - 3$

$\frac{dy}{dx} = \frac{1}{x^3 - 3x + 5} \cdot (3x^2 - 3) = \frac{3x^2 - 3}{x^3 - 3x + 5}$

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

$y = \log(\sin^2 x) = \log((\sin x)^2) = 2\log(\sin x)$

$u = \sin x$ とおくと、$y = 2\log u$。

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = \frac{2}{u} = \frac{2}{\sin x}$

$\frac{du}{dx} = \cos x$

$\frac{dy}{dx} = \frac{2}{\sin x} \cdot \cos x = \frac{2\cos x}{\sin x} = 2\cot x$

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

$u = \log x$ とおくと、$y = \log u$。

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{\log x}$

$\frac{du}{dx} = \frac{1}{x}$

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

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

$u = x + \sqrt{x^2 + 1}$ とおくと、$y = \log u$。

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{du} = \frac{1}{u} = \frac{1}{x + \sqrt{x^2 + 1}}$

$\frac{du}{dx} = 1 + \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = 1 + \frac{x}{\sqrt{x^2 + 1}} = \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}}$

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

3. 最終的な答え

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

(2) $\frac{3x^2 - 3}{x^3 - 3x + 5}$

(3) $2\cot x$

(4) $\frac{1}{x\log x}$

(5) $\frac{1}{\sqrt{x^2 + 1}}$