画像に示された関数を微分する問題です。具体的には、以下の関数について微分を求めます。 10.(1) $y = 5x^3$ 10.(2) $y = x^2 - \sqrt{2}$ 10.(3) $y = \frac{1}{4}(3x^4 - 4x^2)$ 10.(4) $y = \frac{x^4 - x^2}{2}$ 11.(1) $y = (x+3)(2x-3)$ 11.(2) $y = (2x-1)(x^2+3x-1)$ 11.(3) $s = (t^2+2)(t^3+1)$ 11.(4) $y = \frac{3x}{x-2}$ 11.(5) $s = \frac{1}{t+2}$ 11.(6) $y = x^3 + \frac{2}{x-1}$ 11.(7) $y = \frac{2x-3}{x+1}$ 11.(8) $y = \frac{x^2}{x-1}$

はい、承知いたしました。画像にある問題の微分を計算します。

1. 問題の内容

画像に示された関数を微分する問題です。具体的には、以下の関数について微分を求めます。

10.(1) $y = 5x^3$

10.(2) $y = x^2 - \sqrt{2}$

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

10.(4) $y = \frac{x^4 - x^2}{2}$

11.(1) $y = (x+3)(2x-3)$

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

11.(3) $s = (t^2+2)(t^3+1)$

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

11.(5) $s = \frac{1}{t+2}$

11.(6) $y = x^3 + \frac{2}{x-1}$

11.(7) $y = \frac{2x-3}{x+1}$

11.(8) $y = \frac{x^2}{x-1}$

2. 解き方の手順

それぞれの関数について、以下の微分公式や計算法則を用いて微分を計算します。

* 定数倍の微分: $(cf(x))' = cf'(x)$

* 和と差の微分: $(f(x) \pm g(x))' = f'(x) \pm g'(x)$

* べき関数の微分: $(x^n)' = nx^{n-1}$

* 積の微分: $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$

* 商の微分: $(\frac{f(x)}{g(x)})' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$

* 合成関数の微分: $\frac{d}{dx} f(g(x)) = f'(g(x))g'(x)$

それぞれの問題について、微分を計算する手順を以下に示します。

10.(1) $y = 5x^3$

$y' = 5 \cdot 3x^2 = 15x^2$

10.(2) $y = x^2 - \sqrt{2}$

$y' = 2x - 0 = 2x$

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

$y = \frac{3}{4}x^4 - x^2$

$y' = \frac{3}{4} \cdot 4x^3 - 2x = 3x^3 - 2x$

10.(4) $y = \frac{x^4 - x^2}{2}$

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

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

11.(1) $y = (x+3)(2x-3)$

$y = 2x^2 - 3x + 6x - 9 = 2x^2 + 3x - 9$

$y' = 4x + 3$

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

$y = 2x^3 + 6x^2 - 2x - x^2 - 3x + 1 = 2x^3 + 5x^2 - 5x + 1$

$y' = 6x^2 + 10x - 5$

11.(3) $s = (t^2+2)(t^3+1)$

$s = t^5 + t^2 + 2t^3 + 2$

$s' = 5t^4 + 2t + 6t^2$

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

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

11.(5) $s = \frac{1}{t+2}$

$s = (t+2)^{-1}$

$s' = -1(t+2)^{-2} \cdot 1 = -\frac{1}{(t+2)^2}$

11.(6) $y = x^3 + \frac{2}{x-1}$

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

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

11.(7) $y = \frac{2x-3}{x+1}$

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

11.(8) $y = \frac{x^2}{x-1}$

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

3. 最終的な答え

10.(1) $y' = 15x^2$

10.(2) $y' = 2x$

10.(3) $y' = 3x^3 - 2x$

10.(4) $y' = 2x^3 - x$

11.(1) $y' = 4x + 3$

11.(2) $y' = 6x^2 + 10x - 5$

11.(3) $s' = 5t^4 + 6t^2 + 2t$

11.(4) $y' = -\frac{6}{(x-2)^2}$

11.(5) $s' = -\frac{1}{(t+2)^2}$

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

11.(7) $y' = \frac{5}{(x+1)^2}$

11.(8) $y' = \frac{x(x-2)}{(x-1)^2}$