第7章 7.15

(1)

$y = x^4 + 2x^3$ $y' = 4x^3 + 6x^2$

$y'=0$ のとき $y' = 2x^2(2x+3) = 0$ $x = 0, -\dfrac{3}{2}$

(2)

$y = x^2 + \dfrac{1}{x}$ 定義域は $x \neq 0$ $y' = 2x - \dfrac{1}{x^2} = \dfrac{2x^3-1}{x^2}$

$y'=0$ のとき $2x^3 - 1 = 0$ $x = \dfrac{\sqrt[3]{4}}{2}$

(3)

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

$y'=0$ のとき $x = \pm 1$

(4)

$y = \frac{1}{4}x^2\sqrt{x+5} \qquad \text{定義域は } x \geq -5$

$y' = \frac{1}{4}\cdot 2x\sqrt{x+5} + \frac{1}{4}x^2\cdot\frac{1}{2\sqrt{x+5}} = \frac{1}{4}\left(\frac{4x(x+5)+x^2}{2\sqrt{x+5}}\right) = \frac{5x(x+4)}{8\sqrt{x+5}}$

$y'=0$ のとき $x=-4,0$

(5)

$y = x + \sin 2x \qquad y' = 1 + 2\cos 2x$

$y'=0$ のとき $\cos 2x = -\dfrac{1}{2}$ $x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}$

(6)

$y = \sqrt{1+\sin x} \qquad y' = \frac{\cos x}{2\sqrt{1+\sin x}}$

$y'=0$ のとき $\cos x = 0$ $x = \dfrac{\pi}{2}, \dfrac{3}{2}\pi$

(7)

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

$y' = 0$ のとき $x = \pm\dfrac{\pi}{3}, \pi$