与えられた9つの関数をそれぞれ微分する問題です。 (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) $y = \cos^{-1}\sqrt{1-x^2}$

1. 問題の内容

与えられた9つの関数をそれぞれ微分する問題です。

(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

各関数について、微分を計算します。

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

$u = \frac{x-1}{2-x}$ とおくと、 $\frac{du}{dx} = \frac{(2-x) - (x-1)(-1)}{(2-x)^2} = \frac{2-x+x-1}{(2-x)^2} = \frac{1}{(2-x)^2}$

$v = \sqrt{u} = \sqrt{\frac{x-1}{2-x}}$ とおくと、 $\frac{dv}{du} = \frac{1}{2\sqrt{u}} = \frac{1}{2\sqrt{\frac{x-1}{2-x}}}$

$y = \tan^{-1}v$ なので、 $\frac{dy}{dv} = \frac{1}{1+v^2} = \frac{1}{1+\frac{x-1}{2-x}} = \frac{2-x}{2-x+x-1} = 2-x$

したがって、 $\frac{dy}{dx} = \frac{dy}{dv}\frac{dv}{du}\frac{du}{dx} = (2-x) \cdot \frac{1}{2\sqrt{\frac{x-1}{2-x}}} \cdot \frac{1}{(2-x)^2} = \frac{1}{2\sqrt{\frac{x-1}{2-x}}(2-x)} = \frac{1}{2\sqrt{(x-1)(2-x)^2/(2-x)}} = \frac{1}{2\sqrt{(x-1)(2-x)}}$

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

$u = \tan x + \sec x$ とおくと、 $\frac{du}{dx} = \sec^2 x + \sec x \tan x = \sec x (\sec x + \tan x)$

$y = \log|u|$ なので、 $\frac{dy}{du} = \frac{1}{u} = \frac{1}{\tan x + \sec x}$

したがって、 $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{\tan x + \sec x} \cdot \sec x(\sec x + \tan x) = \sec x$

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

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

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

したがって、 $\frac{dy}{dx} = \frac{1}{1+\cot^2 x} (-\csc^2 x) = \frac{-\csc^2 x}{\csc^2 x} = -1$

あるいは、 $\cot x = \tan (\frac{\pi}{2} - x)$ なので $y = \tan^{-1}(\tan (\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$ よって $y' = -1$.

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

$u = \tan^{-1} 2x$ とおくと、 $\frac{du}{dx} = \frac{1}{1+(2x)^2} \cdot 2 = \frac{2}{1+4x^2}$

$y = u^3$ なので、 $\frac{dy}{du} = 3u^2 = 3(\tan^{-1} 2x)^2$

したがって、 $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = 3(\tan^{-1} 2x)^2 \cdot \frac{2}{1+4x^2} = \frac{6(\tan^{-1} 2x)^2}{1+4x^2}$

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

$\frac{d}{dx} \cos^{-1} x = \frac{-1}{\sqrt{1-x^2}}$

$\frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}$

したがって、 $\frac{dy}{dx} = \frac{-1}{\sqrt{1-(\frac{1}{x})^2}} \cdot (-\frac{1}{x^2}) = \frac{1}{x^2\sqrt{1-\frac{1}{x^2}}} = \frac{1}{x^2\sqrt{\frac{x^2-1}{x^2}}} = \frac{1}{x^2 \frac{\sqrt{x^2-1}}{|x|}} = \frac{1}{|x|\sqrt{x^2-1}}$

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

$\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}$

したがって、 $\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{1}{x})^2}} \cdot (-\frac{1}{x^2}) = \frac{-1}{x^2\sqrt{1-\frac{1}{x^2}}} = \frac{-1}{x^2\sqrt{\frac{x^2-1}{x^2}}} = \frac{-1}{x^2 \frac{\sqrt{x^2-1}}{|x|}} = \frac{-1}{|x|\sqrt{x^2-1}}$

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

$u = \frac{1}{\sqrt{2}}\tan\frac{x}{2}$ とおくと、 $\frac{du}{dx} = \frac{1}{\sqrt{2}}\sec^2(\frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2\sqrt{2}}\sec^2(\frac{x}{2})$

$y = \tan^{-1}u$ なので、 $\frac{dy}{du} = \frac{1}{1+u^2} = \frac{1}{1+\frac{1}{2}\tan^2(\frac{x}{2})}$

したがって、 $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{1+\frac{1}{2}\tan^2(\frac{x}{2})} \cdot \frac{1}{2\sqrt{2}}\sec^2(\frac{x}{2}) = \frac{\sec^2(\frac{x}{2})}{2\sqrt{2} + \sqrt{2}\tan^2(\frac{x}{2})} = \frac{1/\cos^2(x/2)}{2\sqrt{2}(1+1/2\tan^2(x/2))} = \frac{1}{2\sqrt{2}[\cos^2(x/2)+\sin^2(x/2)]} =\frac{1}{2\sqrt{2} [\cos^2(\frac{x}{2}) + \frac{1}{2}\sin^2(\frac{x}{2})]} = \frac{1}{2\sqrt{2} (\cos^2 \frac{x}{2}+\frac{1}{2} \sin^2\frac{x}{2})}$.

$\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$より，$2\sin^2(\frac{x}{2}) = 1-\cos x$より$\sin^2(x/2)= \frac{1-\cos x}{2}$、$\cos x = 2\cos^2(\frac{x}{2})-1$より$\cos^2(\frac{x}{2}) = \frac{1+\cos x}{2}$.

ゆえに$\cos^2(\frac{x}{2})+\frac{1}{2}\sin^2(\frac{x}{2}) = \frac{1+\cos x}{2} + \frac{1-\cos x}{4} = \frac{2+2\cos x+1-\cos x}{4} = \frac{3+\cos x}{4}$.

$\frac{dy}{dx} = \frac{1}{2\sqrt{2}[\frac{3+\cos x}{4}]} = \frac{2}{\sqrt{2}(3+\cos x)} = \frac{\sqrt{2}}{3+\cos x}$

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

$u=2x$, $\frac{du}{dx} = 2$.

$y = \sin^{-1} u$. $\frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} = \frac{1}{\sqrt{1-4x^2}}$.

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

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

$u = \sqrt{1-x^2}$ とおくと $\frac{du}{dx} = \frac{1}{2\sqrt{1-x^2}}(-2x) = \frac{-x}{\sqrt{1-x^2}}$

$y = \cos^{-1}u$なので、 $\frac{dy}{du} = \frac{-1}{\sqrt{1-u^2}} = \frac{-1}{\sqrt{1-(1-x^2)}} = \frac{-1}{\sqrt{x^2}} = \frac{-1}{|x|}$

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

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

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

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