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

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

(2) $y = \frac{1}{x \sqrt[4]{x}}$

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

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

(5) $y = \sqrt{1 + \frac{1}{\sqrt{x}}}$

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

2. 解き方の手順

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

$u = x$, $v = (1+x^3)^2$ とすると、

$u' = 1$, $v' = 2(1+x^3)(3x^2) = 6x^2(1+x^3)$

よって、

$y' = \frac{1 \cdot (1+x^3)^2 - x \cdot 6x^2(1+x^3)}{(1+x^3)^4} = \frac{(1+x^3)(1+x^3 - 6x^3)}{(1+x^3)^4} = \frac{1-5x^3}{(1+x^3)^3}$

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

$y' = -\frac{5}{4} x^{-\frac{5}{4} - 1} = -\frac{5}{4} x^{-\frac{9}{4}} = -\frac{5}{4x^{\frac{9}{4}}} = -\frac{5}{4x^2 \sqrt[4]{x}}$

(3) 積の微分公式 $(uv)' = u'v + uv'$ を使います。

$u = x$, $v = \sqrt{x^2+2}$ とすると、

$u' = 1$, $v' = \frac{1}{2\sqrt{x^2+2}} \cdot (2x) = \frac{x}{\sqrt{x^2+2}}$

よって、

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

(4) 商の微分公式を使います。

$u = x$, $v = \sqrt{1-x^2}$ とすると、

$u' = 1$, $v' = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}$

よって、

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

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

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

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

3. 最終的な答え

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

(2) $y' = -\frac{5}{4x^2 \sqrt[4]{x}}$

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

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

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

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