与えられた12個の関数を微分する問題です。これらの関数は、逆三角関数（$\sin^{-1} x$, $\cos^{-1} x$, $\tan^{-1} x$）を含んでいます。

1. 問題の内容

与えられた12個の関数を微分する問題です。これらの関数は、逆三角関数（$\sin^{-1} x$, $\cos^{-1} x$, $\tan^{-1} x$）を含んでいます。

2. 解き方の手順

各関数の微分を以下に示します。

(1) $y = \sin^{-1}(3x)$

$\frac{dy}{dx} = \frac{1}{\sqrt{1 - (3x)^2}} \cdot 3 = \frac{3}{\sqrt{1 - 9x^2}}$

(2) $y = \sin^{-1}(ax)$ (aは定数)

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

(3) $y = \cos^{-1}(3x)$

$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot 3 = -\frac{3}{\sqrt{1 - 9x^2}}$

(4) $y = \cos^{-1}(ax)$ (aは定数)

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

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

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

(6) $y = \tan^{-1}(ax)$ (aは定数)

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

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

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

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

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

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

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

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

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

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

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

(12) $y = (\tan^{-1}(x))^2$

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

3. 最終的な答え

(1) $\frac{3}{\sqrt{1 - 9x^2}}$

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

(3) $-\frac{3}{\sqrt{1 - 9x^2}}$

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

(5) $\frac{3}{1 + 9x^2}$

(6) $\frac{a}{1 + a^2x^2}$

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

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

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

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

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

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