与えられた6つの関数をそれぞれ微分する問題です。 (1) $y = \sin^{-1}(4x)$ (2) $y = \cos^{-1}(\frac{x}{4})$ (3) $y = \tan^{-1}(\frac{3}{4}x)$ (4) $y = \sin^{-1}(\frac{2}{x})$ $(x \ge 2)$ (5) $y = \frac{\sin^{-1}x}{\cos^{-1}x}$ (6) $y = \sqrt{\tan^{-1}x}$

1. 問題の内容

与えられた6つの関数をそれぞれ微分する問題です。

(1) $y = \sin^{-1}(4x)$

(2) $y = \cos^{-1}(\frac{x}{4})$

(3) $y = \tan^{-1}(\frac{3}{4}x)$

(4) $y = \sin^{-1}(\frac{2}{x})$ $(x \ge 2)$

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

(6) $y = \sqrt{\tan^{-1}x}$

2. 解き方の手順

(1) $y = \sin^{-1}(4x)$

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (4x)^2}} \cdot 4 = \frac{4}{\sqrt{1 - 16x^2}}$

(2) $y = \cos^{-1}(\frac{x}{4})$

$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (\frac{x}{4})^2}} \cdot \frac{1}{4} = -\frac{1}{\sqrt{1 - \frac{x^2}{16}}} \cdot \frac{1}{4} = -\frac{1}{4\sqrt{\frac{16 - x^2}{16}}} = -\frac{1}{4 \cdot \frac{\sqrt{16-x^2}}{4}} = -\frac{1}{\sqrt{16 - x^2}}$

(3) $y = \tan^{-1}(\frac{3}{4}x)$

$\frac{dy}{dx} = \frac{1}{1 + (\frac{3}{4}x)^2} \cdot \frac{3}{4} = \frac{1}{1 + \frac{9}{16}x^2} \cdot \frac{3}{4} = \frac{1}{\frac{16 + 9x^2}{16}} \cdot \frac{3}{4} = \frac{16}{16 + 9x^2} \cdot \frac{3}{4} = \frac{12}{16 + 9x^2}$

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

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

$x \ge 2$ より $|x| = x$ なので、

$\frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 4}} \cdot (-\frac{2}{x^2}) = -\frac{2}{x\sqrt{x^2 - 4}}$

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

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

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ なので、

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

(6) $y = \sqrt{\tan^{-1}x} = (\tan^{-1}x)^{\frac{1}{2}}$

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

3. 最終的な答え

(1) $\frac{dy}{dx} = \frac{4}{\sqrt{1 - 16x^2}}$

(2) $\frac{dy}{dx} = -\frac{1}{\sqrt{16 - x^2}}$

(3) $\frac{dy}{dx} = \frac{12}{16 + 9x^2}$

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

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

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