与えられた6つの関数を微分し、その答えを求めます。 (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つの関数を微分し、その答えを求めます。

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

商の微分公式 $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ を用います。

$u = 2x$, $v = x^2 + 1$ とすると、$u' = 2$, $v' = 2x$ です。

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

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

$y = \frac{1}{6}(2x^2 + 1)^{\frac{1}{2}}$ と書き換えます。

合成関数の微分公式を用いると、$y' = \frac{1}{6} \cdot \frac{1}{2} (2x^2 + 1)^{-\frac{1}{2}} \cdot (4x) = \frac{4x}{12\sqrt{2x^2 + 1}} = \frac{x}{3\sqrt{2x^2 + 1}}$

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

$y = \left(\frac{2x+1}{x^2+1}\right)^{\frac{1}{2}}$ と書き換えます。

合成関数の微分公式と商の微分公式を用いると、$y' = \frac{1}{2}\left(\frac{2x+1}{x^2+1}\right)^{-\frac{1}{2}} \cdot \frac{2(x^2+1) - (2x+1)(2x)}{(x^2+1)^2} = \frac{1}{2}\sqrt{\frac{x^2+1}{2x+1}} \cdot \frac{2x^2+2-4x^2-2x}{(x^2+1)^2} = \frac{1}{2}\sqrt{\frac{x^2+1}{2x+1}} \cdot \frac{-2x^2-2x+2}{(x^2+1)^2} = \frac{-x^2-x+1}{(x^2+1)^2}\sqrt{\frac{x^2+1}{2x+1}} = \frac{-x^2-x+1}{(x^2+1)^{3/2}\sqrt{2x+1}}$

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

商の微分公式 $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ を用います。

$u = \sin x$, $v = \sin^2 x + 1$ とすると、$u' = \cos x$, $v' = 2\sin x \cos x$ です。

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

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

合成関数の微分公式を用いると、$y' = \cos\{\cos(3x)\} \cdot (-\sin(3x)) \cdot 3 = -3\sin(3x)\cos\{\cos(3x)\}$

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

合成関数の微分公式を用いると、$y' = \frac{1}{\tan(x^2)} \cdot \frac{1}{\cos^2(x^2)} \cdot 2x = \frac{2x}{\tan(x^2) \cos^2(x^2)} = \frac{2x \cos(x^2)}{\sin(x^2) \cos^2(x^2)} = \frac{2x}{\sin(x^2) \cos(x^2)} = \frac{4x}{2\sin(x^2)\cos(x^2)} = \frac{4x}{\sin(2x^2)}$

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

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

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

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