第5章 5.14

(1)

$y = x^2(4x^2+12x+9) = 4x^4+12x^3+9x^2$

$y' = 16x^3+36x^2+18x = 2x(8x^2+18x+9) = 2x(2x+3)(4x+3)$

(2)

$y' = 2(x-1)(2x+1)^3+(x-1)^2 \cdot 3(2x+1)^2 \cdot 2 = (x-1)(2x+1)^2\{2(2x+1)+6(x-1)\} = (x-1)(2x+1)^2(10x-4)$

$= 2(x-1)(2x+1)^2(5x-2)$

(3)

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

$= \dfrac{(4+x^2)\{4x(4-x^2)+2x(4+x^2)\}}{(4-x^2)^2} = \dfrac{(4+x^2)(16x-4x^3+8x+2x^3)}{(4-x^2)^2} = \dfrac{2x(4+x^2)(12-x^2)}{(4-x^2)^2}$

(4)

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

$= \dfrac{(2x+1)^2(-2x^2-4x-18)}{(x^2-3)^3} = -\dfrac{2(2x+1)^2(x^2+2x+9)}{(x^2-3)^3}$