与えられた6つの逆三角関数または逆三角関数を含む関数を微分する問題です。 (1) $y = \sin^{-1} 4x$ (2) $y = \cos^{-1} \frac{x}{4}$ (3) $y = \tan^{-1} \frac{3}{4}x$ (4) $y = \sin^{-1} \frac{2}{x} \quad (x \ge 2)$ (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つの逆三角関数または逆三角関数を含む関数を微分する問題です。

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

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

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

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

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

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

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

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

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

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

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

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

ここで、$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$より、

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

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

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

3. 最終的な答え

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

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

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

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

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

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