与えられた関数を微分する問題です。 (1) $y = \sin^{-1} 3x$ (2) $y = \sin^{-1} \frac{x}{3}$ (3) $y = \cos^{-1} 3x$ (4) $y = \cos^{-1} \frac{x}{3}$ (5) $y = \tan^{-1} 2x$ (6) $y = \tan^{-1} x^2$

1. 問題の内容

与えられた関数を微分する問題です。

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

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

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

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

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

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

2. 解き方の手順

(1) $y = \sin^{-1} 3x$ の微分

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

(2) $y = \sin^{-1} \frac{x}{3}$ の微分

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

(3) $y = \cos^{-1} 3x$ の微分

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

(4) $y = \cos^{-1} \frac{x}{3}$ の微分

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

(5) $y = \tan^{-1} 2x$ の微分

$y' = \frac{1}{1+(2x)^2} \cdot (2x)' = \frac{1}{1+4x^2} \cdot 2 = \frac{2}{1+4x^2}$

(6) $y = \tan^{-1} x^2$ の微分

$y' = \frac{1}{1+(x^2)^2} \cdot (x^2)' = \frac{1}{1+x^4} \cdot 2x = \frac{2x}{1+x^4}$

3. 最終的な答え

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

(2) $y' = \frac{1}{\sqrt{9-x^2}}$

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

(4) $y' = -\frac{1}{\sqrt{9-x^2}}$

(5) $y' = \frac{2}{1+4x^2}$

(6) $y' = \frac{2x}{1+x^4}$