与えられた6つの関数をそれぞれ微分する問題です。 (1) $y = \frac{2x}{x^2+1}$ (2) $y = \frac{1}{6}\sqrt{2x^2+1}$ (3) $y = \sqrt{\frac{2x+1}{x^2+1}}$ (4) $y = \frac{\sin x}{\sin^2 x + 1}$ (5) $y = \sin\{\cos(3x)\}$ (6) $y = \log\{\tan(x^2)\}$

1. 問題の内容

与えられた6つの関数をそれぞれ微分する問題です。

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

(2) $y = \frac{1}{6}\sqrt{2x^2+1}$

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

(4) $y = \frac{\sin x}{\sin^2 x + 1}$

(5) $y = \sin\{\cos(3x)\}$

(6) $y = \log\{\tan(x^2)\}$

2. 解き方の手順

(1) 商の微分公式を用います。$y = \frac{u}{v}$のとき、$y' = \frac{u'v - uv'}{v^2}$

$u = 2x$, $v = x^2+1$とすると、$u' = 2$, $v' = 2x$なので、

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

(2) 合成関数の微分を用います。$y = \frac{1}{6}(2x^2+1)^{1/2}$なので、

$y' = \frac{1}{6} \cdot \frac{1}{2} (2x^2+1)^{-1/2} \cdot 4x = \frac{4x}{12\sqrt{2x^2+1}} = \frac{x}{3\sqrt{2x^2+1}}$

(3) 合成関数の微分と商の微分公式を用います。

$y = \sqrt{\frac{2x+1}{x^2+1}}$なので、

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

(4) 商の微分公式を用います。

$y = \frac{\sin x}{\sin^2 x + 1}$なので、

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

(5) 合成関数の微分を用います。

$y = \sin\{\cos(3x)\}$なので、

$y' = \cos\{\cos(3x)\} \cdot (-\sin(3x)) \cdot 3 = -3\sin(3x)\cos(\cos(3x))$

(6) 合成関数の微分を用います。

$y = \log\{\tan(x^2)\}$なので、

$y' = \frac{1}{\tan(x^2)} \cdot \frac{1}{\cos^2(x^2)} \cdot 2x = \frac{2x}{\tan(x^2)\cos^2(x^2)} = \frac{2x}{\frac{\sin(x^2)}{\cos(x^2)} \cos^2(x^2)} = \frac{2x}{\sin(x^2)\cos(x^2)} = \frac{4x}{2\sin(x^2)\cos(x^2)} = \frac{4x}{\sin(2x^2)}$

3. 最終的な答え

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

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

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

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

(5) $y' = -3\sin(3x)\cos(\cos(3x))$

(6) $y' = \frac{4x}{\sin(2x^2)}$