与えられた12個の関数について、それぞれの導関数を求める問題です。

はい、承知しました。問題の導関数を求めます。

1. 問題の内容

与えられた12個の関数について、それぞれの導関数を求める問題です。

2. 解き方の手順

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

積の微分公式: $(uv)' = u'v + uv'$を使う。

$u = (x^2+1)^5$とすると、$u' = 5(x^2+1)^4(2x) = 10x(x^2+1)^4$

$v = (x^3-2)^3$とすると、$v' = 3(x^3-2)^2(3x^2) = 9x^2(x^3-2)^2$

$y' = u'v + uv' = 10x(x^2+1)^4(x^3-2)^3 + (x^2+1)^5 \cdot 9x^2(x^3-2)^2$

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

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

合成関数の微分: $y' = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x}$

(3) $y = 2^x$

$y' = 2^x \log 2$

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

積の微分公式: $y' = 3x^2(x^2+1)^{3/2} + x^3 \cdot \frac{3}{2}(x^2+1)^{1/2}(2x) = 3x^2(x^2+1)^{3/2} + 3x^4(x^2+1)^{1/2}$

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

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

$u = x^x$とすると、$\log u = x \log x$

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

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

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

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

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

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

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

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

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

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

(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) $y = \arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)$

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

(10) $y = 2\arccos\sqrt{\frac{x+1}{2}}$

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

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

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

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

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

$y' = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \left[ \frac{(x-2)+(x-1)}{(x-1)(x-2)} - \frac{(x-4)+(x-3)}{(x-3)(x-4)} \right] = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \left[ \frac{2x-3}{(x-1)(x-2)} - \frac{2x-7}{(x-3)(x-4)} \right]$

$y' = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \frac{(2x-3)(x-3)(x-4) - (2x-7)(x-1)(x-2)}{(x-1)(x-2)(x-3)(x-4)}$

$y' = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \frac{(2x-3)(x^2-7x+12) - (2x-7)(x^2-3x+2)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \frac{(2x^3-14x^2+24x-3x^2+21x-36) - (2x^3-6x^2+4x-7x^2+21x-14)}{(x-1)(x-2)(x-3)(x-4)}$

$y' = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \frac{2x^3-17x^2+45x-36 - (2x^3-13x^2+25x-14)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{2} \sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)}} \frac{-4x^2+20x-22}{(x-1)(x-2)(x-3)(x-4)}$

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

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

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

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

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

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

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

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

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

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

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

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