第5章 5.8

(1)

$y' = \frac{0-2x}{(x^2)^2} = \frac{-2}{x^3}$

(2)

$s' = \frac{0-3\cdot1}{t^2} - \frac{0-2\cdot3t^2}{(t^3)^2} = -\frac{3}{t^2}+\frac{6}{t^4}$

(3)

$w = (1-z^2)\left(1+\frac{1}{z^2}\right) = 1-z^2+\frac{1}{z^2}-1 = -z^2+\frac{1}{z^2}$

$w' = -2z - \frac{2}{z^3}$

(4)

$y' = 3(2x-3)^2 \cdot 2 = 6(2x-3)^2$

(5)

$y' = 4(x^2+1)^3 \cdot 2x = 8x(x^2+1)^3$

(6)

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

(7)

$y' = \frac{0-2\cdot(3-4x)\cdot(-4)}{(3-4x)^4} = \frac{8}{(3-4x)^3}$

(8)

$y' = \frac{0-4(x^2-1)^3\cdot 2x}{(x^2-1)^8} = -\frac{8x}{(x^2-1)^5}$