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

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) $y = x^4 - 3x^3 + 2x^2 - 4x + 1$

(2) $s = \frac{t^2 - 2t + 2}{2}$

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

(4) $s = \frac{2}{t^3} + \frac{2t-1}{t+1}$

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

(6) $s = \frac{1}{t\sqrt{t}}$

2. 解き方の手順

(1) $y = x^4 - 3x^3 + 2x^2 - 4x + 1$

各項を微分します。

$\frac{dy}{dx} = 4x^3 - 9x^2 + 4x - 4$

(2) $s = \frac{t^2 - 2t + 2}{2}$

$s = \frac{1}{2}t^2 - t + 1$

$\frac{ds}{dt} = t - 1$

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

積の微分公式 $(uv)' = u'v + uv'$ を用います。

$u = x^2 + 3$, $v = 2x - 1$ とすると、

$u' = 2x$, $v' = 2$

$\frac{dy}{dx} = 2x(2x - 1) + (x^2 + 3)(2) = 4x^2 - 2x + 2x^2 + 6 = 6x^2 - 2x + 6$

(4) $s = \frac{2}{t^3} + \frac{2t-1}{t+1}$

$s = 2t^{-3} + \frac{2t-1}{t+1}$

$\frac{ds}{dt} = -6t^{-4} + \frac{2(t+1) - (2t-1)}{(t+1)^2} = -\frac{6}{t^4} + \frac{2t+2-2t+1}{(t+1)^2} = -\frac{6}{t^4} + \frac{3}{(t+1)^2}$

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

$y = \frac{x^2 + 2x - 2}{x^{1/2}}$

$y = x^{3/2} + 2x^{1/2} - 2x^{-1/2}$

$\frac{dy}{dx} = \frac{3}{2}x^{1/2} + x^{-1/2} + x^{-3/2} = \frac{3}{2}\sqrt{x} + \frac{1}{\sqrt{x}} + \frac{1}{x\sqrt{x}}$

$\frac{dy}{dx} = \frac{3x^2 + 2x + 2}{2x\sqrt{x}}$

(6) $s = \frac{1}{t\sqrt{t}}$

$s = \frac{1}{t^{3/2}}$

$s = t^{-3/2}$

$\frac{ds}{dt} = -\frac{3}{2}t^{-5/2} = -\frac{3}{2t^{5/2}} = -\frac{3}{2t^2\sqrt{t}}$

3. 最終的な答え

(1) $\frac{dy}{dx} = 4x^3 - 9x^2 + 4x - 4$

(2) $\frac{ds}{dt} = t - 1$

(3) $\frac{dy}{dx} = 6x^2 - 2x + 6$

(4) $\frac{ds}{dt} = -\frac{6}{t^4} + \frac{3}{(t+1)^2}$

(5) $\frac{dy}{dx} = \frac{3x^2 + 2x + 2}{2x\sqrt{x}}$

(6) $\frac{ds}{dt} = -\frac{3}{2t^2\sqrt{t}}$