第1章 1.15

(1)

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

$y' = 2(x^2 + x + 1) \cdot (2x + 1) = 4x^3 + 6x^2 + 6x + 2$

$y'' = 12x^2 + 12x + 6$

(2)

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

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

$y'' = 3(2x+1)^2 \cdot 2 + 6\{(2x+1)^2 + x(2(2x+1)\cdot 2)\} = 6(2x+1)^2 + 6(2x+1)^2 + 24x(2x+1)$

$= 12(2x+1)^2 + 24x(2x+1) = 12(2x+1)(2x+1+2x) = 12(2x+1)(4x+1)$

(3)

$y = \frac{x^2}{a^2 + x^2}$

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

$y'' = \frac{2a^2(a^2+x^2)^2 - 2a^2x \cdot 2(a^2+x^2) \cdot 2x}{(a^2+x^2)^4} = \frac{2a^2(a^2+x^2)^2 - 8a^2x^2(a^2+x^2)}{(a^2+x^2)^4}$

$= \frac{(a^2+x^2)\{2a^2((a^2+x^2) - 8a^2x^2\}}{(a^2+x^2)^4} = \frac{2a^2(a^2-3x^2)}{(a^2+x^2)^3}$

(4)

$y = \sqrt{2x - x^2}$

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

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

(5)

$y = a\sin(cx) + b\cos(cx)$

$y' = ac\cos(cx) + b\{-c\sin(cx)\} = ac\cos(cx) - bc\sin(cx)$

$y'' = -ac^2\sin(cx) - bc^2\cos(cx)$

(6)

$y = \log\left|\tan\frac{x}{2}\right|$

$y' = \dfrac{\dfrac{1}{2}}{\cos^2\dfrac{x}{2}\cdot\tan\dfrac{x}{2}} = \dfrac{1}{2\cos^2\dfrac{x}{2}\cdot\tan\dfrac{x}{2}} = \dfrac{1}{2\cos\dfrac{x}{2}\cdot\sin\dfrac{x}{2}} = \dfrac{1}{\sin x}$

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