例 8.14求下列函数的导数 $x$ :

$$\begin{aligned} \text{(a) } \frac{3x+2}{2x^2+1} &\quad \text{(b) } \frac{2x+3}{x^2+x+1} \ \text{(c) } x^3 + 2x^2 - \frac{1}{x} + \frac{1}{x^2} + 3, \quad x \neq 0 \end{aligned}$$

解 (a) 取 $u = 3x + 2$ 和 $v = 2x^2 + 1$ 得

$$\frac{du}{dx} = 3 \quad \text{and} \quad \frac{dv}{dx} = 4x$$

因此，根据商法则，

$$\begin{aligned} \frac{d}{dx} \left[ \frac{3x+2}{2x^2+1} \right] &= \frac{v \left( \frac{du}{dx} \right) - u \left( \frac{dv}{dx} \right)}{v^2} \ &= \frac{(2x^2+1)3 - (3x+2)4x}{(2x^2+1)^2} \ &= \frac{-(6x^2+8x-3)}{(2x^2+1)^2} \end{aligned}$$

(b) 取 $u = 2x + 3$ 和 $v = x^2 + x + 1$

$$\frac{du}{dv} = 2 \quad \text{and} \quad \frac{dv}{dx} = 2x + 1$$

因此，根据商法则，

$$\begin{aligned} \frac{d}{dx} \left[ \frac{2x+3}{x^2+x+1} \right] &= \frac{(x^2+x+1)(2) - (2x+3)(2x+1)}{(x^2+x+1)^2} \ &= -\frac{(2x^2+6x+1)}{(x^2+x+1)^2} \end{aligned}$$

(c) 在这种情况下，我们可以将该函数表示为

$$y = x^3 + 2x^2 - x^{-1} + x^{-2} + 3, \quad x \neq 0$$

并逐项求导，得到

$$\begin{aligned} \frac{dy}{dx} &= 3x^2 + 2(2x) - (-x^{-2}) + (-2x^{-3}) + 0 = 3x^2 + 4x + x^{-2} - 2x^{-3} \ &= 3x^2 + 4x + \frac{1}{x^2} - \frac{2}{x^3}, \quad x \neq 0 \end{aligned}$$