次の関数を微分する問題です。 (1) $y = \tan^{-1}\sqrt{\frac{x-1}{2-x}}$ (2) $y = \log|\tan x + \sec x|$ (3) $y = \tan^{-1}(\cot x)$ (4) $y = (\tan^{-1}2x)^3$ (5) $y = \cos^{-1}\frac{1}{x}$ (6) $y = \sin^{-1}\frac{1}{x}$ (7) $y = \tan^{-1}(\frac{1}{\sqrt{2}}\tan\frac{x}{2})$ (8) $y = \sin^{-1}(2x)$ (9) $y = \cos^{-1}\sqrt{1-x^2}$

はい、承知いたしました。

1. 問題の内容

次の関数を微分する問題です。

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

(2) $y = \log|\tan x + \sec x|$

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

(4) $y = (\tan^{-1}2x)^3$

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

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

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

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

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

2. 解き方の手順

各関数の微分を計算します。

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

$u = \sqrt{\frac{x-1}{2-x}}$ と置くと、$y = \tan^{-1}u$。

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

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

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

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

$u^2 = \frac{x-1}{2-x}$

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

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

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

(2) $y = \log|\tan x + \sec x|$

$\frac{dy}{dx} = \frac{\sec^2 x + \sec x \tan x}{\tan x + \sec x} = \frac{\sec x(\sec x + \tan x)}{\tan x + \sec x} = \sec x$

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

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

(4) $y = (\tan^{-1}2x)^3$

$\frac{dy}{dx} = 3(\tan^{-1}2x)^2 \frac{2}{1+4x^2} = \frac{6(\tan^{-1}2x)^2}{1+4x^2}$

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

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

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

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

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

$\frac{dy}{dx} = \frac{1}{1 + \frac{1}{2}\tan^2\frac{x}{2}} \cdot \frac{1}{\sqrt{2}} \sec^2\frac{x}{2}\cdot \frac{1}{2} = \frac{1}{2}\frac{1}{1+\frac{1}{2}\tan^2\frac{x}{2}} \frac{1}{\sqrt{2}} \sec^2\frac{x}{2} = \frac{1}{\sqrt{2}} \cdot \frac{\sec^2(x/2)}{2+ \tan^2(x/2)} = \frac{1}{\sqrt{2}} \cdot \frac{1/\cos^2(x/2)}{2+\sin^2(x/2)/\cos^2(x/2)} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2\cos^2(x/2) + \sin^2(x/2)} = \frac{1}{\sqrt{2}} \cdot \frac{1}{\cos^2(x/2) + 1} = \frac{1}{2\sqrt{2}}\frac{2}{\cos x + 3}$

$\frac{1}{2\sqrt{2}} \frac{\frac{1}{\cos^2(x/2)}}{1 + \tan^2(x/2)/2} = \frac{1}{2\sqrt{2}} \cdot \frac{2}{2+\tan^2\frac{x}{2}}/\cos^2\frac{x}{2}$

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

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

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

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

3. 最終的な答え

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

(2) $\frac{dy}{dx} = \sec x$

(3) $\frac{dy}{dx} = -1$

(4) $\frac{dy}{dx} = \frac{6(\tan^{-1}2x)^2}{1+4x^2}$

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

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

(7) $\frac{dy}{dx} = \frac{1}{2\sqrt{2}}\frac{1}{3+ \cos(x)}$

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

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