（c）取 $u = \ln x$ 和 $v = x^2$ 得

$$\frac{du}{dx} = \frac{1}{x} \quad \text{and} \quad \frac{dv}{dx} = 2x \quad (x \neq 0)$$

因此由商法则

$$\begin{aligned} \frac{d}{dx}\left(\frac{\ln x}{x^2}\right) &= \frac{(1/x)x^2 - (\ln x)(2x)}{x^4} \ &= \frac{1 - 2 \ln x}{x^3} \end{aligned}$$

（d）引入中间变量 $z = x^2 + 1$ 则

$$y = \ln z \quad \text{and} \quad z = x^2 + 1$$

$$\frac{\text{dy}}{\text{dz}} = \frac{1}{z} \quad \text{and} \quad \frac{\text{dz}}{\text{dx}} = 2x$$

因此由链式法则

$$\frac{\text{dy}}{\text{dx}} = \frac{\text{dy}}{\text{dz}} \frac{\text{dz}}{\text{dx}} = \frac{1}{x^2 + 1}(2x) = \frac{2x}{x^2 + 1}$$

（e）取 $u = e^{-x}$ 和 $v = \sin x + \cos x$

$$\frac{du}{dx} = -e^{-x} \quad \text{and} \quad \frac{dv}{dx} = \cos x - \sin x$$

于是由乘积法则

$$\begin{aligned}\frac{dy}{dx} &= e^{-x}(\cos x - \sin x) + (\sin x + \cos x)(-e^{-x}) \ &= -2e^{-x}\sin x\end{aligned}$$

双曲函数在前面已经介绍过（见第 2.7.4 节），它们与指数函数密切相关，其导数很容易推导。根据它们的定义

$$\frac{d}{dx}(\sinh x) = \frac{d}{dx} \left[ \frac{e^x - e^{-x}}{2} \right] = \frac{1}{2}(e^x + e^{-x}) = \cosh x \quad (8.24a)$$

$$\frac{d}{dx}(\cosh x) = \frac{d}{dx} \left[ \frac{e^x + e^{-x}}{2} \right] = \frac{1}{2}(e^x - e^{-x}) = \sinh x \quad (8.24b)$$

$$\frac{d}{dx}(\tanh x) = \frac{d}{dx} \left[ \frac{\sinh x}{\cosh x} \right] = \frac{(\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{\cosh^2 x}$$

$$= \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x \quad (8.24c)$$