与えられた複数の関数について、それぞれの導関数を求める。

1. 問題の内容

与えられた複数の関数について、それぞれの導関数を求める。

2. 解き方の手順

各関数の導関数を求める手順は以下の通りです。

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

積の微分法と合成関数の微分法を用いる。

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

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

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

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

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

合成関数の微分法を用いる。

$f'(x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x\log x}$

(3) $f(x) = 2^x$

$f'(x) = 2^x \log 2$

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

積の微分法と合成関数の微分法を用いる。

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

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

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

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

$y = x^x$とおくと $\log y = x \log x$

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

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

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

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

両辺の対数をとると $\log f(x) = \cos x \log(\sin x)$

両辺を微分すると

$\frac{f'(x)}{f(x)} = -\sin x \log(\sin x) + \cos x \frac{\cos x}{\sin x}$

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

(7) $f(x) = \sin^{-1} (x^3+1)$

定義域に注意。$-1 \le x^3 + 1 \le 1$ より $-2 \le x^3 \le 0$, $-\sqrt[3]{2} \le x \le 0$

$f'(x) = \frac{1}{\sqrt{1 - (x^3+1)^2}} (3x^2) = \frac{3x^2}{\sqrt{-x^6 - 2x^3}}$

(8) $f(x) = \tan^{-1} (\frac{1-x^2}{1+x^2})$

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

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

(9) $f(x) = \sqrt{1+2\log x}$

$f'(x) = \frac{1}{2\sqrt{1+2\log x}} \cdot \frac{2}{x} = \frac{1}{x\sqrt{1+2\log x}}$

(10) $f(x) = \exp(\tan^{-1} x) = e^{\tan^{-1} x}$

$f'(x) = e^{\tan^{-1} x} \cdot \frac{1}{1+x^2} = \frac{e^{\tan^{-1} x}}{1+x^2}$

(11) $f(x) = x\sqrt{a^2 - x^2} + a^2 \sin^{-1} \frac{x}{a}$ ($a > 0$)

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

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

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

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

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

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

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

(14) $f(x) = \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}}$

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

$\frac{f'(x)}{f(x)} = \frac{1}{2} \left( \frac{1}{x-1} + \frac{1}{x-2} - \frac{1}{x-3} - \frac{1}{x-4} \right)$

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

$= \frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \cdot \frac{(x-2)(x-3)(x-4)+(x-1)(x-3)(x-4)-(x-1)(x-2)(x-4)-(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)}$

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

分子は $-2x^2+10x-11$なので

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

(15) $f(x) = \sqrt[3]{\frac{x^2+1}{(x-1)^2}}$

$\log f(x) = \frac{1}{3} [\log(x^2+1) - 2\log(x-1)]$

$\frac{f'(x)}{f(x)} = \frac{1}{3} \left( \frac{2x}{x^2+1} - \frac{2}{x-1} \right) = \frac{2}{3} \left( \frac{x}{x^2+1} - \frac{1}{x-1} \right)$

$f'(x) = \sqrt[3]{\frac{x^2+1}{(x-1)^2}} \cdot \frac{2}{3} \left( \frac{x(x-1) - (x^2+1)}{(x^2+1)(x-1)} \right) = \sqrt[3]{\frac{x^2+1}{(x-1)^2}} \cdot \frac{2}{3} \frac{-x-1}{(x^2+1)(x-1)} = \frac{-2(x+1)}{3\sqrt[3]{(x^2+1)^2 (x-1)^5}}$

(16) $f(x) = \sinh x$

$f'(x) = \cosh x$

(17) $f(x) = \cosh x$

$f'(x) = \sinh x$

(18) $f(x) = \tanh x$

$f'(x) = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x$

3. 最終的な答え

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

(2) $\frac{1}{x\log x}$

(3) $2^x \log 2$

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

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

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

(7) $\frac{3x^2}{\sqrt{-x^6 - 2x^3}}$

(8) $\frac{-2x}{1+x^4}$

(9) $\frac{1}{x\sqrt{1+2\log x}}$

(10) $\frac{e^{\tan^{-1} x}}{1+x^2}$

(11) $2\sqrt{a^2-x^2}$

(12) $\frac{1}{1+x^2}$

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

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

(15) $\frac{-2(x+1)}{3\sqrt[3]{(x^2+1)^2 (x-1)^5}}$

(16) $\cosh x$

(17) $\sinh x$

(18) $\frac{1}{\cosh^2 x} = \operatorname{sech}^2 x$