## 問題 2.1 の解答

## 問題 2.1 の解答

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1. 問題の内容

与えられた関数の導関数を求める問題です。全部で12個の関数について、それぞれ微分を実行します。

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2. 解き方の手順

以下に、それぞれの関数の導関数を求める手順と答えを示します。

**(1) $y = (x^2 + 1)^5 (x^3 - 2)^3$**

積の微分法と合成関数の微分法を用いる。

$y' = 5(x^2+1)^4(2x)(x^3-2)^3 + (x^2+1)^5 3(x^3-2)^2(3x^2)$

$= 10x(x^2+1)^4(x^3-2)^3 + 9x^2(x^2+1)^5(x^3-2)^2$

$= x(x^2+1)^4(x^3-2)^2[10(x^3-2) + 9x(x^2+1)]$

$= x(x^2+1)^4(x^3-2)^2[10x^3 - 20 + 9x^3 + 9x]$

$= x(x^2+1)^4(x^3-2)^2(19x^3+9x-20)$

**(2) $y = \log(\log x)$**

合成関数の微分法を用いる。

$y' = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$

**(3) $y = 2^x$**

$y' = 2^x \log 2$

**(4) $y = x^3 (x^2 + 1)^{3/2}$**

積の微分法と合成関数の微分法を用いる。

$y' = 3x^2 (x^2+1)^{3/2} + x^3 \frac{3}{2} (x^2+1)^{1/2} (2x)$

$= 3x^2 (x^2+1)^{3/2} + 3x^4 (x^2+1)^{1/2}$

$= 3x^2 (x^2+1)^{1/2} [x^2+1 + x^2]$

$= 3x^2 (x^2+1)^{1/2} (2x^2+1)$

**(5) $y = e^{x^x}$**

$u = x^x$ とおくと、$y = e^u$。

$\log u = x \log x$

$\frac{1}{u} \frac{du}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1$

$\frac{du}{dx} = u (\log x + 1) = x^x (\log x + 1)$

$y' = e^u \frac{du}{dx} = e^{x^x} x^x (\log x + 1)$

**(6) $y = (\sin x)^{\cos x}$**

両辺の対数をとる。

$\log y = \cos x \log (\sin x)$

$\frac{1}{y} y' = -\sin x \log(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x$

$y' = (\sin x)^{\cos x} \left[-\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x} \right]$

**(7) $y = \tan^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$**

$x = \tan \theta$ とおくと、$\frac{1-x^2}{1+x^2} = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta$

$y = \tan^{-1} (\cos 2\theta) = \tan^{-1} \left( \frac{1 - \tan^2(\pi/4 - \theta)}{1+\tan^2(\pi/4 - \theta)} \right) = \tan^{-1}\left( \tan(\pi/2 - 2\theta) \right)$

$y = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2 \tan^{-1} x$

$y' = - \frac{2}{1+x^2}$

**(8) $y = \sqrt{1 + 2\log x}$**

$y' = \frac{1}{2\sqrt{1+2\log x}} \cdot \frac{2}{x} = \frac{1}{x \sqrt{1+2\log x}}$

**(9) $y = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right)$**

$x = \tan \theta$ とおくと、$\frac{x}{\sqrt{1+x^2}} = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} = \frac{\tan \theta}{\sec \theta} = \sin \theta$

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

$y' = \frac{1}{1+x^2}$

**(10) $y = 2 \cos^{-1} \sqrt{\frac{x+1}{2}}$**

$x = \cos \theta$ とおくと、$\sqrt{\frac{x+1}{2}} = \sqrt{\frac{\cos \theta + 1}{2}} = \sqrt{\cos^2 \frac{\theta}{2}} = \cos \frac{\theta}{2}$

$y = 2 \cos^{-1} \left( \cos \frac{\theta}{2} \right) = 2 \cdot \frac{\theta}{2} = \theta = \cos^{-1} x$

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

**(11) $y = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}}$**

両辺の対数をとる。

$\log y = \frac{1}{2} \left[ \log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) \right]$

$\frac{1}{y} y' = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} \right]$

$y' = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \cdot \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} \right]$

$y' = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} \right]$

$= \frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \left[\frac{(x-2)(x-3)(x-4) + (x-1)(x-3)(x-4) - (x-1)(x-2)(x-4) - (x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} \right]$

分子 = $x^3-9x^2+26x-24+x^3-8x^2+19x-12-x^3+7x^2-14x+8-x^3+6x^2-11x+6 = -4x^2+20x-22 = -2(2x^2-10x+11)$

$y' = - \frac{2x^2-10x+11}{((x-1)(x-2)(x-3)(x-4))^{1/2}}$

**(12) $y = x \sqrt{a^2 - x^2} + a^2 \sin^{-1} \frac{x}{a}$**

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

$= \sqrt{a^2 - x^2} - \frac{x^2}{\sqrt{a^2 - x^2}} + \frac{a^2}{\sqrt{a^2 - x^2}}$

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

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3. 最終的な答え

(1) $y' = x(x^2+1)^4(x^3-2)^2(19x^3+9x-20)$

(2) $y' = \frac{1}{x \log x}$

(3) $y' = 2^x \log 2$

(4) $y' = 3x^2 (x^2+1)^{1/2} (2x^2+1)$

(5) $y' = e^{x^x} x^x (\log x + 1)$

(6) $y' = (\sin x)^{\cos x} \left[-\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x} \right]$

(7) $y' = - \frac{2}{1+x^2}$

(8) $y' = \frac{1}{x \sqrt{1+2\log x}}$

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

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

(11) $y' = - \frac{2x^2-10x+11}{\sqrt{(x-1)(x-2)(x-3)(x-4)}}$

(12) $y' = 2\sqrt{a^2 - x^2}$