与えられた12個の関数の導関数をそれぞれ求める問題です。

1. 問題の内容

与えられた12個の関数の導関数をそれぞれ求める問題です。

2. 解き方の手順

各問題ごとに解き方を説明します。

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

積の微分公式 $(uv)' = u'v + uv'$を使います。

$u = (x^2+1)^5$, $v = (x^3-2)^3$とします。

$u' = 5(x^2+1)^4 (2x) = 10x(x^2+1)^4$

$v' = 3(x^3-2)^2 (3x^2) = 9x^2(x^3-2)^2$

よって、

$y' = 10x(x^2+1)^4 (x^3-2)^3 + (x^2+1)^5 9x^2(x^3-2)^2 = x(x^2+1)^4 (x^3-2)^2 [10(x^3-2) + 9x(x^2+1)] = x(x^2+1)^4 (x^3-2)^2 (19x^3 + 9x - 20)$

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

合成関数の微分を使います。

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

(3) $y = 2^x$

$y' = 2^x \log 2$

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

積の微分公式を使います。

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

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

$x^x = e^{x \log x}$であるから$y=e^{e^{x \log x}}$.

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

または、 $x^x = f(x)$とおくと、$\log f(x) = x \log x$となり、$\frac{f'(x)}{f(x)} = \log x + 1$。よって、$f'(x) = x^x (\log x + 1)$。

$y = e^{f(x)}$なので、$y' = e^{f(x)} f'(x) = e^{x^x} x^x (\log x + 1)$.

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

両辺の対数を取ると、$\log y = \cos x \log (\sin x)$

両辺を$x$で微分すると、

$\frac{y'}{y} = -\sin x \log(\sin x) + \cos x \cdot \frac{\cos x}{\sin x} = -\sin x \log(\sin x) + \frac{\cos^2 x}{\sin x}$

よって、$y' = (\sin x)^{\cos x} \left( \frac{\cos^2 x}{\sin x} - \sin x \log(\sin x) \right)$

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

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

$y = \tan^{-1} (\cos 2\theta)$.

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

合成関数の微分より

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

(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、$\frac{x}{\sqrt{1+x^2}} = \frac{\tan \theta}{\sqrt{1+\tan^2 \theta}} = \frac{\tan \theta}{\sec \theta} = \sin \theta$.

よって、$y = \sin^{-1} (\sin \theta) = \theta = \tan^{-1} x$.

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

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

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

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

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

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

両辺の対数を取ると、$\log y = \frac{1}{2} \left[ \log(x-1) + \log(x-2) - \log(x-3) - \log(x-4) \right]$.

両辺を$x$で微分すると、$\frac{y'}{y} = \frac{1}{2} \left[ \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} \right] = \frac{1}{2} \frac{(x-2)(x-3)(x-4)+(x-1)(x-3)(x-4)-(x-1)(x-2)(x-4)-(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{2} \frac{-4x^2+10x-8}{(x-1)(x-2)(x-3)(x-4)}$

$y' = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \frac{1}{2} \frac{-4x^2+10x-8}{(x-1)(x-2)(x-3)(x-4)} = \frac{-2x^2+5x-4}{\sqrt{(x-1)(x-2)(x-3)(x-4)^3}}$

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

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

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

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

(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) $y' = \frac{-2x^2+5x-4}{\sqrt{(x-1)(x-2)(x-3)(x-4)^3}}$

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