既然我们已经对极限有了概念性的理解，并且具备了计算极限的实际能力，我们便为微积分的学习奠定了基础；微积分是数学的一个分支，在其中我们计算导数与积分。大多数数学家与史学家都认为，微积分是由英国人 Isaac Newton $\text{(1643–1727)}$ 与德国人 Gottfried Leibniz $\text{(1646–1716)}$ 各自独立发展起来的，二人的画像见图 3.2。当我们把微积分的发展归功于 Newton 与 Leibniz 时，实际上是指 Newton 与 Leibniz 最先理解了导数与积分之间的关系。两位数学家都受益于前人的工作，如 Barrow、Fermat 与 Cavalieri。这两位数学家最初的关系似乎友善；然而，在后来的岁月里，关于谁的工作居先爆发了一场激烈的争论。尽管 Newton 似乎确实更早提出了微积分背后的思想，但我们今天普遍使用的记号却要归功于 Leibniz。

我们通过重温割线与切线的概念来开始微积分的学习。回想一下，我们用函数上一点 $(a,f(a))$ 处的割线斜率来估计变化率，即一个变量相对于另一个变量的变化快慢。我们可以通过选取一个靠近 $a$ 的 $x$ 值，并过点 $(a,f(a))$ 与 $\left( {x,f(x)} \right)$ 作一条直线（如图 3.3 所示）来得到割线的斜率。这条直线的斜率由一个差商形式的方程给出：

$$m_{\text{sec}} = \frac{f(x) - f(a)}{x - a}.$$

我们也可以利用这个方程，把 $x$ 换成 $a + h$（其中 $h$ 是一个接近 0 的值），来计算函数在某点 a 处的割线斜率。接着，我们可以计算过点 $(a,f(a))$ 与 $(a + h,f\left( {a + h} \right))$ 的直线的斜率。此时，我们看到该割线的斜率由以下带增量 $h$ 的差商给出：

$$m_{\text{sec}} = \frac{f\left( {a + h} \right) - f(a)}{a + h - a} = \frac{f\left( {a + h} \right) - f(a)}{h}.$$

设 $f$ 为定义在区间 $I$ 上的函数，且 $I$ 含有 $a.$ 若 $x \neq a$ 属于 $I,$ 则

$$Q = \frac{f(x) - f(a)}{x - a}$$ (3.1)

是一个差商。

此外，若选取 $h \neq 0$ 使得 $a + h$ 属于 $I,$ 则

$$Q = \frac{f\left( {a + h} \right) - f(a)}{h}$$ (3.2)

是带增量 $h$ 的差商。

通过此小程序查看导数的发展过程。

计算割线斜率的这两个表达式在图 3.3 中均有说明。我们将看到，求割线斜率的这两种方法各有价值。视具体情形，我们可以选用其中一种。我们选取时的主要考量通常取决于计算的简便程度。

在图 3.4(a) 中我们看到，随着 $x$ 的值趋近于 $a,$ 割线的斜率对函数在 $a$ 处的变化率给出了更好的估计。此外，割线本身趋近于函数在 $a$ 处的切线，该切线代表了割线的极限。类似地，图 3.4(b) 显示，随着 $h$ 的值趋近于 $0,$ 割线也趋近于切线。切线在 $a$ 处的斜率即函数在 $a$ 处的变化率，如图 3.4(c) 所示。

你可以利用此网站来探究图像，看它们在某点处是否有切线。

在图 3.5 中，我们在一系列关于 $x = 1$ 的逐渐缩小的区间内展示了 $f(x) = \sqrt{x}$ 的图像及其在点 $(1,1)$ 处的切线。随着区间变窄，函数的图像与其切线趋于重合，使得切线上的值成为函数在 $x$ 接近 $1$ 时的良好近似。事实上，函数 $f(x)$ 的图像在 $x = 1$ 的紧邻邻域内本身似乎是局部线性的。

我们可以形式化地将函数的切线定义如下。

设 $f(x)$ 为定义在含有 $a$ 的开区间内的函数。$f(x)$ 在 $a$ 处的切线是过点 $\left( {a,f(a)} \right)$、且具有如下斜率的直线

$$m_{\text{tan}} = \underset{x\rightarrow a}{\text{lim}}\frac{f(x) - f(a)}{x - a}$$ (3.3)

当该极限存在时。

等价地，我们可以将 $f(x)$ 在 $a$ 处的切线定义为过点 $\left( {a,f(a)} \right)$、且具有如下斜率的直线

$$m_{\text{tan}} = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {a + h} \right) - f(a)}{h}$$ (3.4)

当该极限存在时。

正如我们用两种不同的表达式来定义割线的斜率，我们也用两种不同的形式来定义切线的斜率。本书中我们同时使用这两种定义形式。与前面一样，选用哪种定义取决于具体情形。既然我们已经形式化地定义了函数在一点处的切线，便可以用这一定义来求切线方程。

求曲线 $f(x) = x^{2}$ 在 $x = 3$ 处的切线方程。

首先求切线的斜率。在本例中，使用公式 3.3。

$$\begin{array}{clccl} m_{\text{tan}} & {= \underset{x\rightarrow 3}{\text{lim}}\frac{f(x) - f(3)}{x - 3}} & & & \text{Apply the definition.} \\ & {= \underset{x\rightarrow 3}{\text{lim}}\frac{x^{2} - 9}{x - 3}} & & & {\text{Substitute}\ f(x) = x^{2}\ \text{and}\ f(3) = 9.} \\ & {= \underset{x\rightarrow 3}{\text{lim}}\frac{\left( {x - 3} \right)\left( {x + 3} \right)}{x - 3} = \underset{x\rightarrow 3}{\text{lim}}\left( {x + 3} \right) = 6} & & & \text{Factor the numerator to evaluate the limit.} \end{array}$$

接下来，求切线上的一点。由于该直线与 $f(x)$ 的图像在 $x = 3$ 处相切，它经过点 $\left( {3,f(3)} \right).$ 我们有 $f(3) = 9,$ 故切线经过点 $\left( {3,9} \right).$

利用点斜式直线方程，取斜率 $m = 6$ 与点 $\left( {3,9} \right),$ 我们得到直线 $y - 9 = 6\left( {x - 3} \right).$ 化简后得 $y = 6x - 9.$ 函数 $f(x) = x^{2}$ 的图像及其在 $3$ 处的切线如图 3.6 所示。

用公式 3.4 求曲线 $f(x) = x^{2}$ 在 $x = 3$ 处的切线斜率。

步骤与示例 3.1 非常相似。定义见公式 3.4。

$$\begin{array}{clccl} m_{\text{tan}} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {3 + h} \right) - f(3)}{h}} & & & \text{Apply the definition.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\left( {3 + h} \right)^{2} - 9}{h}} & & & {\text{Substitute}\ f\left( {3 + h} \right) = {(3 + h)}^{2}\ \text{and}\ f(3) = 9.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{9 + 6h + h^{2} - 9}{h}} & & & \text{Expand and simplify to evaluate the limit.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{h\left( {6 + h} \right)}{h} = \underset{h\rightarrow 0}{\text{lim}}\left( {6 + h} \right) = 6} & & & \end{array}$$

用另一种定义我们得到了相同的切线斜率值，这表明两个公式可以互换使用。

求曲线 $f(x) = 1\text{/}x$ 在 $x = 2$ 处的切线方程。

我们可以使用公式 3.3，但如我们所见，使用公式 3.4 结果相同。

$$\begin{array}{clccl} m_{\text{tan}} & {= \underset{x\rightarrow 2}{\text{lim}}\frac{f(x) - f(2)}{x - 2}} & & & \text{Apply the definition.} \\ & {= \underset{x\rightarrow 2}{\text{lim}}\frac{\frac{1}{x} - \frac{1}{2}}{x - 2}} & & & {\text{Substitute}\ f(x) = \frac{1}{x}\ \text{and}\mspace{2mu} f(2) = \frac{1}{2}.} \\ & {= \underset{x\rightarrow 2}{\text{lim}}\frac{\frac{1}{x} - \frac{1}{2}}{x - 2} \cdot \frac{2x}{2x}} & & & \begin{array}{l} {\text{Multiply numerator and denominator by}\ 2x\ \text{to}} \\ \text{simplify fractions.} \end{array} \\ & {= \underset{x\rightarrow 2}{\text{lim}}\frac{\left( {2 - x} \right)}{\left( {x - 2} \right)\left( {2x} \right)}} & & & \text{Simplify.} \\ & {= \underset{x\rightarrow 2}{\text{lim}}\frac{-1}{2x}} & & & {\text{Simplify using}\ \frac{2 - x}{x - 2} = -1,\text{for}\ x \neq 2.} \\ & {= - \frac{1}{4}} & & & \text{Evaluate the limit.} \end{array}$$

现在已知切线斜率为 $- \frac{1}{4}.$ 为求切线方程，我们还需要直线上的一点。我们知道 $f(2) = \frac{1}{2}.$ 由于切线经过点 $(2,\frac{1}{2})$，我们可以用点斜式直线方程来求切线方程。于是切线方程为 $y = - \frac{1}{4}x + 1.$ 函数 $f(x) = \frac{1}{x}$ 与 $y = - \frac{1}{4}x + 1$ 的图像如图 3.7 所示。

求曲线 $f(x) = \sqrt{x}$ 在 $x = 4$ 处的切线斜率。

为了求函数在一点处的切线斜率而要计算的那种极限，在许多学科的大量应用中都会出现。这些应用包括物理学中的速度与加速度、商业中的边际利润函数，以及生物学中的增长率。这种极限出现得如此频繁，以至于我们给这个值一个专门的名称：导数。求导数的过程称为微分（法）。

设 $f(x)$ 为定义在含有 $a$ 的开区间内的函数。函数 $f(x)$ 在 $a$ 处的导数，记作 $f^{\prime}(a),$ 定义为

$$f^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{f(x) - f(a)}{x - a}$$ (3.5)

当该极限存在时。

或者，我们也可以将 $f(x)$ 在 $a$ 处的导数定义为

$$f^{\prime}(a) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {a + h} \right) - f(a)}{h}.$$ (3.6)

对于 $f(x) = x^{2},$ 用表格借助公式 3.5 估计 $f^{\prime}(3)$。

取略小于 $3$ 与略大于 $3$ 的 $x$ 值制作表格。

观察该表，我们认为一个良好的估计是 $f^{\prime}(3) = 6.$

对于 $f(x) = x^{2},$ 用表格借助公式 3.6 估计 $f^{\prime}(3)$。

对于 $f(x) = 3x^{2} - 4x + 1,$ 用公式 3.5 求 $f^{\prime}(2)$。

将给定的函数与数值直接代入公式。

$$\begin{array}{clccl} {f^{\prime}(2)} & {= \underset{x\rightarrow 2}{\text{lim}}\frac{f(x) - f(2)}{x - 2}} & & & \text{Apply the definition.} \\ & {= \underset{x\rightarrow 2}{\text{lim}}\frac{\left( {3x^{2} - 4x + 1} \right) - 5}{x - 2}} & & & {\text{Substitute}\ f(x) = 3x^{2} - 4x + 1\ \text{and}\ f(2) = 5.} \\ & {= \underset{x\rightarrow 2}{\text{lim}}\frac{(x - 2)(3x + 2)}{x - 2}} & & & \text{Simplify and factor the numerator.} \\ & {= \underset{x\rightarrow 2}{\text{lim}}(3x + 2)} & & & \text{Cancel the common factor.} \\ & {= 8} & & & \text{Evaluate the limit.} \end{array}$$

对于 $f(x) = 3x^{2} - 4x + 1,$ 用公式 3.6 求 $f^{\prime}(2)$。

利用这个方程，我们可以将函数的两个值代入公式，应当得到与示例 3.5 相同的结果。

$$\begin{array}{clccl} {f^{\prime}(2)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {2 + h} \right) - f(2)}{h}} & & & \text{Apply the definition.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{(3{(2 + h)}^{2} - 4\left( {2 + h} \right) + 1) - 5}{h}} & & & \begin{array}{l} {\text{Substitute}\ f(2) = 5\ \text{and}} \\ {f\left( {2 + h} \right) = 3{(2 + h)}^{2} - 4\left( {2 + h} \right) + 1.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{3h^{2} + 8h}{h}} & & & \text{Simplify the numerator.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{h(3h + 8)}{h}} & & & \text{Factor the numerator.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}(3h + 8)} & & & \text{Cancel the common factor.} \\ & {= 8} & & & \text{Evaluate the limit.} \end{array}$$

无论使用公式 3.5 还是公式 3.6，结果都相同。

对于 $f(x) = x^{2} + 3x + 2,$ 求 $f^{\prime}(1)$。

既然我们能够计算导数，便可以将其用于速度问题。回想一下，若 $s(t)$ 是沿坐标轴运动的物体的位置，则该物体在时间间隔 $\left\lbrack {a,t} \right\rbrack$（当 $t > a$ 时）或 $\left\lbrack {t,a} \right\rbrack$（当 $t < a$ 时）内的平均速度由差商给出

$$v_{\text{ave}} = \frac{s(t) - s(a)}{t - a}.$$ (3.7)

当 $t$ 的值趋近于 $a$ 时，$v_{\text{ave}}$ 的值趋近于我们称为 $a$ 处瞬时速度的那个值。也就是说，$a$ 处的瞬时速度，记作 $v(a),$ 由下式给出

$$v(a) = s^{\prime}(a) = \underset{t\rightarrow a}{\text{lim}}\frac{s(t) - s(a)}{t - a}.$$ (3.8)

为了更好地理解平均速度与瞬时速度之间的关系，见图 3.8。在该图中，切线（以红色显示）的斜率是物体在时刻 $t = a$ 的瞬时速度，其位置由函数 $s(t)$ 给出。割线（以绿色显示）的斜率是物体在时间间隔 $\left\lbrack {a,t} \right\rbrack$ 内的平均速度。

我们可以用公式 3.5 计算瞬时速度，也可以利用数值表估计运动物体的速度。随后可用公式 3.7 验证该估计。

弹簧上悬挂的铅垂正在上下振荡。它在时刻 $t$ 相对于一条固定水平线的位置由 $s(t) = \text{sin}\ t$（图 3.9）给出。用数值表估计 $v(0)$，并用公式 3.5 检验该估计。

我们可以通过取趋近于 $0$ 的 $t$ 值计算平均速度表，来估计 $t = 0$ 处的瞬时速度，如表 3.1 所示。

表 3.1 利用趋近于 0 的 t 值所得的平均速度

由表可见，时间间隔 $\left\lbrack {-0.1,0} \right\rbrack$ 内的平均速度为 $0.998334166,$ 时间间隔 $\left\lbrack {-0.01,0} \right\rbrack$ 内的平均速度为 $0.9999833333,$ 依此类推。利用此数值表，看来一个良好的估计是 $v(0) = 1.$

利用公式 3.5，我们可以看到

$$v(0) = s^{\prime}(0) = \underset{t\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} t - \text{sin}\mspace{2mu} 0}{t - 0} = \underset{t\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} t}{t} = 1.$$

因此，实际上 $v(0) = 1.$

一块石头从 $64$ 英尺高处落下。它在 $t$ 秒后的离地高度由 $s(t) = -16t^{2} + 64,0 \leq t \leq 2$ 给出。用公式 3.5 求其落下 $1$ 秒后的瞬时速度。

如本节所见，函数切线斜率与瞬时速度是相关的概念。二者都通过求导数得到，且都度量函数的瞬时变化率，即函数沿其图像任一点处的变化率。

函数 $f(x)$ 在 $a$ 处的瞬时变化率即其导数 $f^{\prime}(a)$。

Hennessey Venom GT 的最高速度可达 $270.49$ mph，是世界上最快的汽车之一。在测试中，它从 $0$ 加速到 $60$ mph 用了 $3.05$ 秒，从 $0\ \text{to}\ 100$ mph 用了 $5.88$ 秒，从 $0\ \text{to}\ 200$ mph 用了 $14.51$ 秒，从 $0\ \text{to}\ 229.9$ mph 用了 $19.96$ 秒。利用这些数据，就速度的变化率（即加速度）在趋近于 $229.9$ mph 时的情形得出结论。该车加速的快慢看来是在增大、减小，还是保持不变？

首先注意到 $60$ mph = $88$ ft/s，$100$ mph $\approx 146.67$ ft/s，$200$ mph $\approx 293.33$ ft/s，$229.9$ mph $\approx 337.19$ ft/s。我们可以将这些信息汇总于表中。

表 3.2 不同 t 值处的 $v(t)$

现在计算该车以英尺每二次秒为单位的、在区间 $\left\lbrack {t,19.96} \right\rbrack$（当 $t$ 趋近于 $19.96$ 时）上的平均加速度，如下表所示。

表 3.3 平均加速度

随着速度趋近于 $229.9$ mph $\text{(}337.19$ ft/s)，该车加速的快慢在减小。

房主设定恒温器，使室内温度从晚上 $9$ 点的 $70\text{°}\text{F}$ 开始下降，夜间降至最低 $60\text{°}$，并在次日早晨 $7$ 点回升到 $70\text{°}$。设室内温度由 $T(t) = 0.4t^{2} - 4t + 70$ 给出（$0 \leq t \leq 10$），其中 $t$ 为晚上 $9$ 点之后经过的小时数。求午夜时温度的瞬时变化率。

由于午夜是晚上 $9$ 点之后 $3$ 小时，我们要求 $T^{\prime}(3)$。参见公式 3.5。

$$\begin{array}{clccl} {T^{\prime}(3)} & {= \underset{t\rightarrow 3}{\text{lim}}\frac{T(t) - T(3)}{t - 3}} & & & \text{Apply the definition.} \\ & {= \underset{t\rightarrow 3}{\text{lim}}\frac{0.4t^{2} - 4t + 70 - 61.6}{t - 3}} & & & \begin{array}{l} {\text{Substitute}\ T(t) = 0.4t^{2} - 4t + 70\ \text{and}} \\ {T(3) = 61.6.} \end{array} \\ & {= \underset{t\rightarrow 3}{\text{lim}}\frac{0.4t^{2} - 4t + 8.4}{t - 3}} & & & \text{Simplify.} \\ & {= \underset{t\rightarrow 3}{\text{lim}}\frac{0.4(t - 3)(t - 7)}{t - 3}} & & & {= \underset{t\rightarrow 3}{\text{lim}}\frac{0.4(t - 3)(t - 7)}{t - 3}} \\ & {= \underset{t\rightarrow 3}{\text{lim}}0.4(t - 7)} & & & \text{Cancel.} \\ & {= -1.6} & & & \text{Evaluate the limit.} \end{array}$$

午夜时温度的瞬时变化率为 $-1.6\text{°}\text{F}$ 每小时。

某玩具公司可以按每套 $p = -0.01x + 400$ 美元的价格售出 $x$ 套电子游戏机。生产 $x$ 套的成本由 $C(x) = 100x + 10,000$ 美元给出。求当生产 $10,000$ 套时利润的变化率。该玩具公司应增产还是减产？

生产 $x$ 套游戏机所获得的利润 $P(x)$ 为 $R(x) - C(x)$，其中 $R(x)$ 为售出 $x$ 套的销售收入。由于公司可以按每套 $p = -0.01x + 400$ 的价格售出 $x$ 套，

$$R(x) = xp = x\left( {-0.01x + 400} \right) = -0.01x^{2} + 400x.$$

于是，

$$P(x) = -0.01x^{2} + 300x - 10,000.$$

因此，计算利润的变化率得到

$$\begin{array}{cl} {P^{\prime}(10000)} & {= \underset{x\rightarrow 10000}{\text{lim}}\frac{P(x) - P(10000)}{x - 10000}} \\ & {= \underset{x\rightarrow 10000}{\text{lim}}\frac{-0.01x^{2} + 300x - 10000 - 1990000}{x - 10000}} \\ & {= \underset{x\rightarrow 10000}{\text{lim}}\frac{-0.01x^{2} + 300x - 2000000}{x - 10000}} \\ & {= 100.} \end{array}$$

由于利润的变化率 $P^{\prime}(10,000) > 0$ 且 $P(10,000) > 0$，该公司应当增产。

某咖啡店确定，按每个司康饼 $s$ 美元收费时，每日利润由 $P(s) = -20s^{2} + 150s - 10$ 给出。该店目前每个司康饼收费 $\text{\$}3.25$。求 $P^{\prime}(3.25)$，即价格为 $\text{\$}3.25$ 时利润的变化率，并判断该店是否应考虑提高或降低司康饼的价格。

在以下习题中，使用公式 3.1 求每个函数 $y = f(x)$ 在 $x_{1}$ 与 $x_{2}$ 之间的割线斜率。

1.

$f(x) = 4x + 7;x_{1} = 2,x_{2} = 5$

2.

$f(x) = 8x - 3;x_{1} = -1,x_{2} = 3$

3.

$f(x) = x^{2} + 2x + 1;x_{1} = 3,x_{2} = 3.5$

4.

$f(x) = \text{−}x^{2} + x + 2;x_{1} = 0.5,x_{2} = 1.5$

5.

$f(x) = \frac{4}{3x - 1};x_{1} = 1,x_{2} = 3$

6.

$f(x) = \frac{x - 7}{2x + 1};x_{1} = 0,x_{2} = 2$

7.

$f(x) = \sqrt{x};x_{1} = 1,x_{2} = 16$

8.

$f(x) = \sqrt{x - 9};x_{1} = 10,x_{2} = 13$

9.

$f(x) = x^{1\text{/}3} + 1;x_{1} = 0,x_{2} = 8$

10.

$f(x) = 6x^{2\text{/}3} + 2x^{1\text{/}3};x_{1} = 1,x_{2} = 27$

对于以下函数，

1. 用公式 3.4 求切线斜率 $m_{\text{tan}} = f^{\prime}(a)$，并

2. 求 $f$ 在 $x = a$ 处的切线方程。

11.

$f(x) = 3 - 4x,a = 2$

12.

$f(x) = \frac{x}{5} + 6,a = -1$

13.

$f(x) = x^{2} + x,a = 1$

14.

$f(x) = 1 - x - x^{2},a = 0$

15.

$f(x) = \frac{7}{x},a = 3$

16.

$f(x) = \sqrt{x + 8},a = 1$

17.

$f(x) = 2 - 3x^{2},a = -2$

18.

$f(x) = \frac{-3}{x - 1},a = 4$

19.

$f(x) = \frac{2}{x + 3},a = -4$

20.

$f(x) = \frac{3}{x^{2}},a = 3$

对于以下函数 $y = f(x),$ 用公式 3.5 求 $f^{\prime}(a)$。

21.

$f(x) = 5x + 4,a = -1$

22.

$f(x) = -7x + 1,a = 3$

23.

$f(x) = x^{2} + 9x,a = 2$

24.

$f(x) = 3x^{2} - x + 2,a = 1$

25.

$f(x) = \sqrt{x},a = 4$

26.

$f(x) = \sqrt{x - 2},a = 6$

27.

$f(x) = \frac{1}{x},a = 2$

28.

$f(x) = \frac{1}{x - 3},a = -1$

29.

$f(x) = \frac{1}{x^{3}},a = 1$

30.

$f(x) = \frac{1}{\sqrt{x}},a = 4$

在以下习题中，给定函数 $y = f(x)$，

1. 对表格中给出的每个 $x$ 值，求过点 $Q\left( {x,f(x)} \right)$ 的割线 $PQ$ 的斜率。

2. 利用 a. 的答案估计点 $P$ 处切线斜率的值。

3. 利用 b. 的答案求 $f$ 在点 $P$ 处的切线方程。

31.

\[T\] $f(x) = x^{2} + 3x + 4,P\left( {1,8} \right)$（保留 $6$ 位小数。）

32.

\[T\] $f(x) = \frac{x + 1}{x^{2} - 1},P\left( {0,-1} \right)$

33.

\[T\] $f(x) = 10e^{0.5x},P\left( {0,10} \right)$（保留 $4$ 位小数。）

34.

\[T\] $f(x) = \text{tan}\mspace{2mu}(x),P\left( {\pi,0} \right)$

\[T\] 对于以下位置函数 $y = s(t)$，一物体沿直线运动，其中 $t$ 的单位为秒，$s$ 的单位为米。求

1. 从 $t = 2$ 到 $t = 2 + h$ 的平均速度的简化表达式；

2. 在 $t = 2$ 与 $t = 2 + h$ 之间的平均速度，其中 $\text{(i)}\ h = 0.1,$ $\text{(ii)}\ h = 0.01,$ $\text{(iii)}\ h = 0.001,$ $\text{(iv)}\ h = 0.0001$；以及

3. 利用 a. 的答案估计 $t = 2$ 秒时的瞬时速度。

35.

$s(t) = \frac{1}{3}t + 5$

36.

$s(t) = t^{2} - 2t$

37.

$s(t) = 2t^{3} + 3$

38.

$s(t) = \frac{16}{t^{2}} - \frac{4}{t}$

39.

利用以下图像求 a. $f^{\prime}(1)$ 与 b. $f^{\prime}(6)$。

40.

利用以下图像求 a. $f^{\prime}(-3)$ 与 b. $f^{\prime}(1.5)$。

在以下习题中，用导数的极限定义证明对每个给定函数，导数在 $x = a$ 处不存在。

41.

$f(x) = x^{1\text{/}3},x = 0$

42.

$f(x) = x^{2\text{/}3},x = 0$

43.

$f(x) = \left\{ \begin{matrix}

{1,x < 1} \\

{x,x \geq 1}

\end{matrix} \right.,x = 1$

44.

$f(x) = \frac{|x|}{x},x = 0$

45.

\[T\] 一辆赛车沿直线赛道行驶 $t$ 秒后的位置（单位：英尺）由函数 $s(t) = 8t^{2} - \frac{1}{16}t^{3}$ 建模。

1. 求车辆在下列时间间隔内的平均速度，保留四位小数：

1. \[4, 4.1\]

2. \[4, 4.01\]

3. \[4, 4.001\]

4. \[4, 4.0001\]

2. 利用 a. 的结论，就车辆在 $t = 4$ 秒时的瞬时速度给出判断。

46.

\[T\] 一个球沿斜面滚下的距离（单位：英尺）由函数 $s(t) = 14t^{2}$ 建模，其中 t 为球开始滚动后经过的秒数。

1. 求球在下列时间间隔内的平均速度：

1. \[5, 5.1\]

2. \[5, 5.01\]

3. \[5, 5.001\]

4. \[5, 5.0001\]

2. 利用 a. 的答案就球在 $t = 5$ 秒时的瞬时速度给出判断。

47.

两辆车并排沿一条直路出发。如图所示，它们的位置函数由 $s = f(t)$ 与 $s = g(t)$ 给出，其中 $s$ 的单位为英尺，$t$ 的单位为秒。

1. 在 $t = 2$ 秒时哪辆车行驶得更远？

2. 在 $t = 3$ 秒时每辆车的大致速度是多少？

3. 在 $t = 4$ 秒时哪辆车行驶得更快？

4. 关于 $t = 4$ 秒时两车的位置可以得出什么结论？

48.

\[T\] 生产 $x$ 千罐蛋黄酱的总成本 $C(x)$（单位：百美元）由 $C(x) = 0.000003x^{3} + 4x + 300$ 给出。

1. 计算在下列区间内每罐的平均成本：

1. \[100, 100.1\]

2. \[100, 100.01\]

3. \[100, 100.001\]

4. \[100, 100.0001\]

2. 利用 a. 的答案估计生产 $100,000$ 罐蛋黄酱的平均成本。

49.

\[T\] 对于函数 $f(x) = x^{3} - 2x^{2} - 11x + 12$，完成下列事项。

1. 用图形计算器在合适的视窗中绘制 f 的图像。

2. 使用计算器上的 ZOOM 功能，近似求出使 $m_{\text{tan}} = f^{\prime}(a) = 0$ 的两个 $x = a$ 值。

50.

\[T\] 对于函数 $f(x) = \frac{x}{1 + x^{2}}$，完成下列事项。

1. 用图形计算器在合适的视窗中绘制 $f$ 的图像。

2. 使用计算器上的 ZOOM 功能，近似求出使 $m_{\text{tan}} = f^{\prime}(a) = 0$ 的 $x = a$ 值。

51.

设 $N(x)$ 计算一辆行驶 $x$ 英里的车辆所消耗的汽油加仑数。假设该车油耗为 $30$ mpg。

1. 求 $N(x)$ 的数学表达式。

2. $N(100\text{)}$ 是多少？解释其物理意义。

3. $N^{\prime}(100)$ 是多少？解释其物理意义。

52.

\[T\] 对于函数 $f(x) = x^{4} - 5x^{2} + 4$，完成下列事项。

1. 用图形计算器在合适的视窗中绘制 $f$ 的图像。

2. 在图形计算器上使用数值求导数的 $\text{nDeriv}$ 函数，估计 $f^{\prime}(-2),f^{\prime}(-0.5),f^{\prime}(1.7)$ 与 $f^{\prime}(2.718)$。

53.

\[T\] 对于函数 $f(x) = \frac{x^{2}}{x^{2} + 1}$，完成下列事项。

1. 用图形计算器在合适的视窗中绘制 $f$ 的图像。

2. 在图形计算器上使用 $\text{nDeriv}$ 函数求 $f^{\prime}(-4),f^{\prime}(-2),f^{\prime}(2)$ 与 $f^{\prime}(4)$。

如我们此前所见，函数在某点的导数给出了该点处函数切线（即割线极限）的斜率或变化率。若在给定时刻对位置函数求导，我们就得到该时刻的速度。由此可以合理地推断：若知道函数在每一点处的导数，就能获得关于函数行为的有价值信息。然而，即便只利用上一节的方法求少数几个点的导数，这一过程也会很快变得相当繁琐。本节中，我们定义导函数，并学习求其导数的一般方法。

导函数给出原函数在定义域内每一可导点处的导数。我们可以如下形式化地定义导函数。

设 $f$ 为一个函数。导函数（记作 $f^{\prime}$）是一个函数，其定义域由满足下列极限存在的那些 $x$ 值组成：

$$f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}.$$ (3.9)

若存在 $f^{\prime}(a)$，则称函数 $f(x)$ 在点 $a$ 处可导。更一般地，若函数在开集 $S$ 中每一点都可导，则称该函数在区间 $S$ 上可导；而可导函数指的是在其定义域上 $f^{\prime}(x)$ 存在的函数。

在接下来的几个示例中，我们将利用方程 3.9 求函数的导数。

求 $f(x) = \sqrt{x}$ 的导数。

直接从导函数的定义出发。使用方程 3.1。

$$\begin{array}{clccl} {f^{\prime}(x)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\sqrt{x + h} - \sqrt{x}}{h}} & & & \begin{array}{l} {\text{Substitute}\ f\left( {x + h} \right) = \sqrt{x + h}\ \text{and}\ f(x) = \sqrt{x}} \\ {\text{into}\ f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}} & & & \begin{array}{l} \text{Multiply numerator and denominator by} \\ {\sqrt{x + h} + \sqrt{x}\ \text{without distributing in the}} \\ \text{denominator.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{h}{h(\sqrt{x + h} + \sqrt{x})}} & & & \text{Multiply the numerators and simplify.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{1}{(\sqrt{x + h} + \sqrt{x})}} & & & {\text{Cancel the}\ h.} \\ & {= \frac{1}{2\sqrt{x}}} & & & \text{Evaluate the limit.} \end{array}$$

求函数 $f(x) = x^{2} - 2x$ 的导数。

此处采用同样的步骤，但无需乘以共轭因式。

$$\begin{array}{clccl} {f^{\prime}(x)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{({(x + h)}^{2} - 2\left( {x + h} \right)) - (x^{2} - 2x)}{h}} & & & \begin{array}{l} {\text{Substitute}\ f\left( {x + h} \right) = {(x + h)}^{2} - 2(x + h)\ \text{and}} \\ {f(x) = x^{2} - 2x\ \text{into}} \\ {f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{x^{2} + 2xh + h^{2} - 2x - 2h - x^{2} + 2x}{h}} & & & {\text{Expand}\ {(x + h)}^{2} - 2(x + h).} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{2xh - 2h + h^{2}}{h}} & & & \text{Simplify.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{h(2x - 2 + h)}{h}} & & & {\text{Factor out}\ h\ \text{from the numerator.}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}(2x - 2 + h)} & & & {\text{Cancel the common factor of}\ h.} \\ & {= 2x - 2} & & & \text{Evaluate the limit.} \end{array}$$

求 $f(x) = x^{2}$ 的导数。

我们用多种不同的记号来表示函数的导数。在示例 3.12 中我们已表明，若 $f(x) = x^{2} - 2x,$ 则 $f^{\prime}(x) = 2x - 2.$ 若将这个函写作 $y = x^{2} - 2x$ 的形式，则可将导数写作 $y^{\prime} = 2x - 2$ 或 $\frac{dy}{dx} = 2x - 2.$ 通过写出 $\frac{d}{dx}\left( {x^{2} - 2x} \right) = 2x - 2$ 也可传达同样的信息。因此，对于函数 $y = f(x)$，下列每种记号都表示 $f(x)$ 的导数：

$$f^{\prime}(x),\ \frac{dy}{dx},\ y^{\prime},\ \frac{d}{dx}\left( {f(x)} \right).$$

作为 $f^{\prime}(a)$ 的替代，我们也可使用 $\frac{dy}{dx}\left| \begin{array}{l}

\\

{}_{x = a}

\end{array} \right.$ 这种 $\frac{dy}{dx}$ 记号（称为 Leibniz 记号）在工程与物理中十分常见。为了更好地理解这一记号，回想一下：函数在某点的导数是当割线趋近于切线时割线斜率的极限。这些割线斜率常表示为 $\frac{\text{Δ}y}{\text{Δ}x}$ 的形式，其中 $\text{Δ}y$ 是与 $x$ 的差所对应的 $y$ 值的差，记作 $\text{Δ}x$（图 3.11）。因此，导数可看作 $y$ 关于 $x$ 的瞬时变化率，可表示为

$$\frac{dy}{dx} = \underset{\text{Δ}x\rightarrow 0}{\text{lim}}\frac{\text{Δ}y}{\text{Δ}x}.$$

我们已经讨论过如何绘制函数图像，因此，给定一个函数或其导函数的方程，我们都可以绘制其图像。若两者都已知，我们预期会在这两幅函数图像之间观察到对应关系，因为 $f^{\prime}(x)$ 给出了函数 $f(x)$ 的变化率（即 $f(x)$ 处切线的斜率）$f(x)\operatorname{)).}$

在示例 3.11 中我们发现，对于 $f(x) = \sqrt{x}$，有 $f^{'}(x) = \frac{1}{2\sqrt{x}}.$ 若在同一坐标系中绘制这两个函数的图像（如图 3.12），我们便可以利用图像来理解这两个函数之间的关系。首先，我们注意到 $f(x)$ 在其整个定义域上递增，这意味着它在各点处切线的斜率都为正。因此，我们预期在其定义域上对所有 $x$ 都有 $f^{\prime}(x) > 0$。此外，随着 $x$ 增大，$f(x)$ 处切线的斜率不断减小，我们预期会看到 $f^{\prime}(x)$ 相应地减小。我们还观察到 $f\prime(0)$ 无定义，且 $\underset{x\rightarrow 0^{+}}{\text{lim}}f^{\prime}(x) = \text{+}\infty，$ 这对应于 $f(x)$ 在 $0$ 处的垂直切线。

在示例 3.12 中我们发现，对于 $f(x) = x^{2} - 2x$，有 $f^{\prime}(x) = 2x - 2.$ 这些函数的图像如图 3.13 所示。可以观察到，当 $x < 1$ 时 $f(x)$ 递减；对于这些相同的 $x$ 值，有 $f^{\prime}(x) < 0.$ 当 $x > 1$ 时，$f(x)$ 递增且 $f^{\prime}(x) > 0.$ 此外，$f(x)$ 在 $x = 1$ 处有水平切线，且 $f^{\prime}(1) = 0.$

利用 $f(x)$ 的如下图像，描绘 $f^{\prime}(x)$ 的图像。

解答如下图所示。可以观察到，在 $\left( {–2,3} \right)$ 上 $f(x)$ 递增且 $f^{\prime}(x) > 0$。此外，在 $\left( {\text{−}\infty,-2} \right)$ 与 $\left( {3,\text{+}\infty} \right)$ 上 $f(x)$ 递减且 $f^{\prime}(x) < 0$。还需注意，$f(x)$ 在 $–2$ 和 $3$ 处有水平切线，且 $f^{\prime}(-2) = 0$、$f^{\prime}(3) = 0$。

描绘 $f(x) = x^{2} - 4$ 的图像。在哪些区间上 $f^{\prime}(x)$ 的图像位于 $x$ 轴上方？

既然我们已经会画导数图像，接下来考察图像的行为。首先，我们考虑可导性与连续性之间的关系。我们将看到，若函数在一点可导，则它在该点必连续；然而，在一点连续的函数未必在该点可导。事实上，函数在一点连续却因多种原因而不可导的情况是可能发生的。

设 $f(x)$ 为一个函数，$a$ 在其定义域内。若 $f(x)$ 在点 $a$ 可导，则 $f$ 在点 $a$ 连续。

若 $f(x)$ 在点 $a$ 可导，则 $f^{\prime}(a)$ 存在，且

$$f^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{f(x) - f(a)}{x - a}.$$

我们想通过证明 $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a)$ 来说明 $f(x)$ 在点 $a$ 连续。因此，

$$\begin{array}{clccc} {\underset{x\rightarrow a}{\text{lim}}f(x)} & {= \underset{x\rightarrow a}{\text{lim}}\left( {f(x) - f(a) + f(a)} \right)} & & & \\ & {= \underset{x\rightarrow a}{\text{lim}}\left( {\frac{f(x) - f(a)}{x - a} \cdot \left( {x - a} \right) + f(a)} \right)} & & & {\text{Multiply and divide}\ f(x) - f(a)\ \text{by}\ x - a.} \\ & {= \left( {\underset{x\rightarrow a}{\text{lim}}\frac{f(x) - f(a)}{x - a}} \right) \cdot \left( {\underset{x\rightarrow a}{\text{lim}}\left( {x - a} \right)} \right) + \underset{x\rightarrow a}{\text{lim}}f(a)} & & & \\ & {= f\prime(a) \cdot 0 + f(a)} & & & \\ & {= f(a\operatorname{).}} & & & \end{array}$$

因此，由于 $f(a)$ 有定义且 $\underset{x\rightarrow a}{\text{lim}}f(x) = f(a)$，我们得出结论：$f$ 在点 $a$ 连续。

□

我们刚刚证明了可导必连续，现在来考察连续性是否蕴含可导。为了回答这个问题，我们考察函数 $f(x) = |x|$。这个函数在处处连续；然而 $f^{\prime}(0)$ 无定义。这一观察使我们相信连续性并不蕴含可导。进一步探讨。对于 $f(x) = |x|$，

$$f^{\prime}(0) = \underset{x\rightarrow 0}{\text{lim}}\frac{f(x) - f(0)}{x - 0} = \underset{x\rightarrow 0}{\text{lim}}\frac{|x| - |0|}{x - 0} = \underset{x\rightarrow 0}{\text{lim}}\frac{|x|}{x}.$$

该极限不存在，因为

$$\underset{x\rightarrow 0^{-}}{\text{lim}}\frac{|x|}{x} = -1\ \text{and}\ \underset{x\rightarrow 0^{+}}{\text{lim}}\frac{|x|}{x} = 1.$$

见图 3.14。

我们再考虑几个连续函数不可导的情形。考虑函数 $f(x) = \sqrt[3]{x}$：

$$f^{\prime}(0) = \underset{x\rightarrow 0}{\text{lim}}\frac{\sqrt[3]{x} - 0}{x - 0} = \underset{x\rightarrow 0}{\text{lim}}\frac{1}{\sqrt[3]{x^{2}}} = \text{+}\infty.$$

因此 $f^{\prime}(0)$ 不存在。观察 $f(x) = \sqrt[3]{x}$ 的图像即可明了。该函数在 $0$ 处有垂直切线（图 3.15）。

函数 $f(x) = \left\{ \begin{array}{l}

{x\mspace{2mu}\text{sin}\mspace{2mu}\left( \frac{1}{x} \right)\ \text{if}\ x \neq 0} \\

{0\ \text{if}\ x = 0}

\end{array} \right.$ 在 $0.$ 处也表现出有趣的不可导行为。我们看到

$$f^{\prime}(0) = \underset{x\rightarrow 0}{\text{lim}}\frac{x\mspace{2mu}\text{sin}\mspace{2mu}\left( {1\text{/}x} \right) - 0}{x - 0} = \underset{x\rightarrow 0}{\text{lim}}\text{sin}\mspace{2mu}\left( \frac{1}{x} \right).$$

该极限不存在，本质上是因为当割线趋近于零时，其斜率不断改变方向（图 3.16）。

综上所述：

1. 我们观察到，若函数不连续，则它必不可导，因为每一个可导函数都必连续。然而，若函数连续，它仍可能不可导。

2. 我们看到，$f(x) = |x|$ 在 $0$ 处不可导，因为左右两侧切线斜率的极限不相同。从图像上看，这导致函数图像在 $0$ 处出现一个尖角。由此我们得出结论：要使函数在一点可导，它在该点必须是“光滑”的。

3. 正如我们在 $f(x) = \sqrt[3]{x}$ 的例子中看到的，函数在出现垂直切线的点处不可导。

4. 正如我们在 $f(x) = \left\{ \begin{array}{l}

{x\mspace{2mu}\text{sin}\mspace{2mu}\left( \frac{1}{x} \right)\ \text{if}\ x \neq 0} \\

{0\ \text{if}\ x = 0}

\end{array} \right.$ 函数也可能以更复杂的方式在一点处不可导。

一家玩具公司想为玩具车设计一条轨道，它先沿抛物线，然后转为直线（图 3.17）。描述该轨道的函数应具有形式 $f(x) = \left\{ \begin{matrix}

{\frac{1}{10}x^{2} + bx + c\mspace{2mu}\text{if}\mspace{2mu} x < -10} \\

{- \frac{1}{4}x + \frac{5}{2}\mspace{2mu}\text{if}\mspace{2mu} x \geq -10}

\end{matrix} \right.$ 其中 $x$ 与 $f(x)$ 的单位为英寸。为使小车沿轨道平稳运动，函数 $f(x)$ 必须在 $-10$ 处既连续又可导。求使 $f(x)$ 既连续又可导的 $b$ 和 $c$ 的值。

要使函数在 $x = -10$ 处连续，需满足 $\underset{x\rightarrow-10^{-}}{\text{lim}}f(x) = f(-10)$。因此，由于

$$\underset{x\rightarrow\text{−}10^{-}}{\text{lim}}f(x) = \frac{1}{10}{(-10)}^{2} - 10b + c = 10 - 10b + c$$

且 $f(-10) = 5$，我们必有 $10 - 10b + c = 5$。等价地，有 $c = 10b - 5$。

要使函数在 $-10$ 处可导，

$$f^{\prime}(-10) = \underset{x\rightarrow\text{−}10}{\text{lim}}\frac{f(x) - f(-10)}{x + 10}$$

必须存在。由于 $f(x)$ 在左右两侧由不同的规则定义，我们必须分别从右侧与左侧求该极限，然后令它们相等：

$$\begin{array}{clccc} {\underset{x\rightarrow\text{−}10^{-}}{\text{lim}}\frac{f(x) - f(-10)}{x + 10}} & {= \underset{x\rightarrow\text{−}10^{-}}{\text{lim}}\frac{\frac{1}{10}x^{2} + bx + c - 5}{x + 10}} & & & \\ & {= \underset{x\rightarrow\text{−}10^{-}}{\text{lim}}\frac{\frac{1}{10}x^{2} + bx + \left( {10b - 5} \right) - 5}{x + 10}} & & & {\text{Substitute}\ c = 10b - 5.} \\ & {= \underset{x\rightarrow\text{−}10^{-}}{\text{lim}}\frac{x^{2} - 100 + 10bx + 100b}{10(x + 10)}} & & & \text{Multiply numerator and denominator by 10.} \\ & {= \underset{x\rightarrow\text{−}10^{-}}{\text{lim}}\frac{(x + 10)(x - 10 + 10b)}{10(x + 10)}} & & & \text{Factor by grouping.} \\ & {= b - 2.} & & & \end{array}$$

我们还有

$$\begin{array}{cl} {\underset{x\rightarrow\text{−}10^{+}}{\text{lim}}\frac{f(x) - f(-10)}{x + 10}} & {= \underset{x\rightarrow\text{−}10^{+}}{\text{lim}}\frac{- \frac{1}{4}x + \frac{5}{2} - 5}{x + 10}} \\ & {= \underset{x\rightarrow\text{−}10^{+}}{\text{lim}}\frac{\text{−}(x + 10)}{4(x + 10)}} \\ & {= - \frac{1}{4}.} \end{array}$$

由此得到 $b - 2 = - \frac{1}{4}$。于是 $b = \frac{7}{4}$，且 $c = 10\left( \frac{7}{4} \right) - 5 = \frac{25}{2}$。

求使 $f(x) = \left\{ \begin{matrix}

{ax + b\ \text{if}\ x < 3} \\

{x^{2}\ \text{if}\ x \geq 3}

\end{matrix} \right.$ 在 $3$ 处既连续又可导的 $a$ 和 $b$ 的值。

函数的导数本身仍是一个函数，因此我们可以求导数的导数。例如，位置函数的导数是位置的变化率，即速度。速度的导数是速度的变化率，即加速度。对导数再求导所得到的新函数称为二阶导数。此外，我们可以继续求导，得到三阶导数、四阶导数，依此类推。这些统称为高阶导数。$y = f(x)$ 的高阶导数的记号可以用下列任意一种形式表示：

$$f^{''}(x),\ f\text{'''}(x),f^{(4)}(x)\text{,…},f^{(n)}(x)$$ $$y^{''}(x),y\text{'''}(x),y^{(4)}(x)\text{,…},y^{(n)}(x)$$ $$\frac{d^{2}y}{dx^{2}},\frac{d^{3}y}{dx^{3}},\frac{d^{4}y}{dx^{4}}\text{,…},\frac{d^{n}y}{dx^{n}}.$$

值得注意的是，$\frac{d^{2}y}{dx^{2}}$ 这一记号可以看作是为更紧凑地表达 $\frac{d}{dx}\left( \frac{dy}{dx} \right)$ 所作的尝试。类似地，$\frac{d}{dx}\left( {\frac{d}{dx}\left( \frac{dy}{dx} \right)} \right) = \frac{d}{dx}\left( \frac{d^{2}y}{dx^{2}} \right) = \frac{d^{3}y}{dx^{3}}$。

对于 $f(x) = 2x^{2} - 3x + 1$，求 $f^{''}(x)$。

先求 $f^{\prime}(x)$。

$$\begin{array}{clccl} {f^{\prime}(x)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\left( {2{(x + h)}^{2} - 3\left( {x + h} \right) + 1} \right) - (2x^{2} - 3x + 1)}{h}} & & & \begin{array}{l} {\text{Substitute}\ f(x) = 2x^{2} - 3x + 1} \\ \text{and} \\ {f\left( {x + h} \right) = 2{(x + h)}^{2} - 3\left( {x + h} \right) + 1} \\ {\text{into}\ f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{4xh + 2h^{2} - 3h}{h}} & & & \text{Simplify the numerator.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\left( {4x + 2h - 3} \right)} & & & \begin{array}{l} {\text{Factor out the}\ h\ \text{in the numerator}} \\ {\text{and cancel with the}\ h\ \text{in the}} \\ \text{denominator.} \end{array} \\ & {= 4x - 3} & & & \text{Take the limit.} \end{array}$$

接着，对 $f^{\prime}(x) = 4x - 3$ 求导以得到 $f^{''}(x)$。

$$\begin{array}{clccl} {f^{''}(x)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{f^{\prime}\left( {x + h} \right) - f^{\prime}(x)}{h}} & & & \begin{array}{l} {\text{Use}\ f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}\ \text{with}\ f^{\prime}(x)\ \text{in}} \\ {\text{place of}\ f(x).} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\left( {4\left( {x + h} \right) - 3} \right) - (4x - 3)}{h}} & & & \begin{array}{l} {\text{Substitute}\ f^{\prime}\left( {x + h} \right) = 4\left( {x + h} \right) - 3\ \text{and}} \\ {f^{\prime}(x) = 4x - 3.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}4} & & & \text{Simplify.} \\ & {= 4} & & & \text{Take the limit.} \end{array}$$

对 $f(x) = x^{2}$ 求 $f^{''}(x)$。

某粒子沿坐标轴的位置（单位：秒）由 $s(t) = 3t^{2} - 4t + 1$（单位：米）给出。求该粒子在时刻 $t$ 的加速度函数。

由于 $v(t) = s^{\prime}(t)$ 且 $a(t) = v^{\prime}(t) = s^{''}(t)$，我们先求 $s(t)$ 的导数：

$$\begin{array}{cl} {s^{\prime}(t)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{s\left( {t + h} \right) - s(t)}{h}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{3\left( {t + h} \right)^{2} - 4\left( {t + h} \right) + 1 - \left( {3t^{2} - 4t + 1} \right)}{h}} \\ & {= 6t - 4.} \end{array}$$

接着，

$$\begin{array}{cl} {s^{''}(t)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{s^{\prime}\left( {t + h} \right) - s^{\prime}(t)}{h}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{6\left( {t + h} \right) - 4 - (6t - 4)}{h}} \\ & {= 6.} \end{array}$$

因此，$a = 6{\ \text{m/s}}^{2}$。

对于 $s(t) = t^{3}$，求 $a(t)$。

对于下列习题，用导数的定义求 $f^{\prime}(x)$。

54\.

$f(x) = 6$

55.

$f(x) = 2 - 3x$

56\.

$f(x) = \frac{2x}{7} + 1$

57.

$f(x) = 4x^{2}$

58\.

$f(x) = 5x - x^{2}$

59.

$f(x) = \sqrt{2x}$

60\.

$f(x) = \sqrt{x - 6}$

61.

$f(x) = \frac{9}{x}$

62\.

$f(x) = x + \frac{1}{x}$

63.

$f(x) = \frac{1}{\sqrt{x}}$

对于下列习题，利用 $y = f(x)$ 的图像描绘其导数 $f^{\prime}(x)$ 的图像。

64\. 65. 66. 67.

对于下列习题，所给极限表示函数 $y = f(x)$ 在 $x = a$ 处的导数。求 $f(x)$ 与 $a$。

68\.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\left( {1 + h} \right)^{2\text{/}3} - 1}{h}$

69.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\left\lbrack {3\left( {2 + h} \right)^{2} + 2} \right\rbrack - 14}{h}$

70\.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\text{cos}\mspace{2mu}\left( {\pi + h} \right) + 1}{h}$

71.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\left( {2 + h} \right)^{4} - 16}{h}$

72\.

$\underset{h\rightarrow 0}{\text{lim}}\frac{\lbrack 2\left( {3 + h} \right)^{2} - \left( {3 + h} \right)\rbrack - 15}{h}$

73.

$\underset{h\rightarrow 0}{\text{lim}}\frac{e^{h} - 1}{h}$

对于下列函数，

1. 画出图像，并

2. 用导数的定义证明该函数在 $x = 1$ 处不可导。

74\.

$f(x) = \left\{ \begin{array}{l}

{2\sqrt{x},0 \leq x \leq 1} \\

{3x - 1,x > 1}

\end{array} \right.$

75.

$f(x) = \left\{ \begin{array}{l}

{3,x < 1} \\

{3x,x \geq 1}

\end{array} \right.$

76\.

$f(x) = \left\{ \begin{array}{l}

{- x^{2} + 2,x \leq 1} \\

{x,x > 1}

\end{array} \right.$

77.

$f(x) = \left\{ \begin{array}{l}

{2x,x \leq 1} \\

{\frac{2}{x},x > 1}

\end{array} \right.$

对于下列图像，

1. 确定在哪些 $x = a$ 处 $\underset{x\rightarrow a}{\text{lim}}f(x)$ 存在但 $f$ 在 $x = a$ 处不连续，并且

2. 确定在哪些 $x = a$ 处函数连续但在 $x = a$ 处不可导。

78\. 79. 80.

利用所给 $f(x)$ 的图像，求 a. $f^{\prime}(-0.5)$，b. $f^{\prime}(0)$，c. $f^{\prime}(1)$，d. $f^{\prime}(2)$，e. $f^{\prime}(3)$（若存在）。

对于下列函数，利用 $f^{''}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f^{\prime}\left( {x + h} \right) - f^{\prime}(x)}{h}$ 求 $f^{''}(x)$。

81.

$f(x) = 2 - 3x$

82\.

$f(x) = 4x^{2}$

83.

$f(x) = x + \frac{1}{x}$

对于下列习题，用计算器绘制 $f(x)$ 的图像。先确定函数 $f^{\prime}(x)$，再用计算器绘制 $f^{\prime}(x)$ 的图像。

84\.

\[T\] $f(x) = - \frac{5}{x}$

85.

\[T\] $f(x) = 3x^{2} + 2x + 4.$

86\.

\[T\] $f(x) = \sqrt{x} + 3x$

87.

\[T\] $f(x) = \frac{1}{\sqrt{2x}}$

88\.

\[T\] $f(x) = 1 + x + \frac{1}{x}$

89.

\[T\] $f(x) = x^{3} + 1$

对于下列习题，就所给的每个情境说明这两个表达式的含义，务必包含单位。

1. $\frac{f\left( {x + h} \right) - f(x)}{h}$

2. $f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}$

90\.

$P(x)$ 表示某城市在 $x$ 年（以年计）时的人口。

91.

$C(x)$ 表示某游乐园中 $x$ 名顾客在特许经营上花费的总金额（以千美元计）。

92\.

$R(x)$ 表示制造 $x$ 台时钟收音机的总成本（以千美元计）。

93.

$g(x)$ 表示在测验中，经过 $x$ 小时学习后所取得的成绩（以百分点计）。

94\.

$B(x)$ 表示自 $1990$ 年起 $x$ 年后，美国大学书店中一本社会学教材的价格（以美元计）。

95.

$p(x)$ 表示在 $x$ 英尺高度处的大气压（以托为单位）。

96\.

描绘一个满足下列所有性质的函数 $y = f(x)$ 的图像：

1. $f^{\prime}(x) > 0$ for $-2 \leq x < 1$

2. $f^{\prime}(2) = 0$

3. $f^{\prime}(x) > 0$ for $x > 2$

4. $f(2) = 2$ and $f(0) = 1$

5. $\underset{x\rightarrow\text{−}\infty}{\text{lim}}f(x) = 0$ and $\underset{x\rightarrow\infty}{\text{lim}}f(x) = \infty$

6. $f^{\prime}(1)$ 不存在。

97.

假设距地面 $x$ 英尺高度处的温度 $T$（华氏度）由 $y = T(x)$ 给出。

1. 给出 $T^{\prime}(x)$ 的物理意义（含单位）。

2. 若已知 $T^{\prime}(1000) = -0.1$，解释其物理含义。

98\.

假设某公司的总利润（单位：千美元）在售出 $x$ 件商品时为 $y = P(x)$。

1. 对于 $0 < a < b$，$\frac{P(b) - P(a)}{b - a}$ 度量的是什么？单位是什么？

2. $P^{\prime}(x)$ 度量的是什么？单位是什么？

3. 假设 $P^{\prime}(30) = 5$，若售出的商品数量从 $30$ 增加到 $31$，利润大约变化多少？

99.

下图中的图像模拟了在一个人口为 $50,000$ 的小镇中，自流感首次爆发 $t$ 周后感染流感的人数 $N(t)$。

1. 描述 $N^{\prime}(t)$ 表示什么，以及当 $t$ 增大时它的变化行为。

2. 导数告诉我们这个小镇受流感爆发影响的哪些方面？

对于下列习题，使用下表，表中给出了阿波罗 $11$ 号任务中土星 $\text{V}$ 火箭在发射后 $t$ 秒时的高度 $h$。

100\.

$h^{\prime}(t)$ 的物理意义是什么？单位是什么？

101.

\[T\] 构造 $h^{\prime}(t)$ 的数值表，并在同一坐标系中画出 $h(t)$ 与 $h^{\prime}(t)$ 的图像。（*提示：*对内部点，分别估算左极限与右极限并取平均。区间 I 的内部点是指属于 I 但不是 I 的端点的元素。）

102\.

\[T\] 对数据的最佳线性拟合由 $H(t) = 7.229t - 4.905$ 给出，其中 $H$ 为火箭高度（单位：米），$t$ 为自起飞以来经过的时间。由该方程确定 $H^{\prime}(t)$。将 $H(t)$ 与所给数据画在一起，并在另一个坐标系中画出 $H^{\prime}(t)$。

103.

\[T\] 对数据的最佳二次拟合由 $G(t) = 1.429t^{2} + 0.0857t - 0.1429$ 给出，其中 $G$ 为火箭高度（单位：米），$t$ 为自起飞以来经过的时间。由该方程确定 $G^{\prime}(t)$。将 $G(t)$ 与所给数据画在一起，并在另一个坐标系中画出 $G^{\prime}(t)$。

104\.

\[T\] 对数据的最佳三次拟合由 $F(t) = 0.2037t^{3} + 2.956t^{2} - 2.705t + 0.4683$ 给出，其中 $F$ 为火箭高度（单位：米），$t$ 为自起飞以来经过的时间。由该方程确定 $F^{\prime}(t)$。将 $F(t)$ 与所给数据画在一起，并在另一个坐标系中画出 $F^{\prime}(t)$。线性、二次还是三次函数对数据的拟合最好？

105.

利用对数据的最佳线性、二次与三次拟合，确定 $H^{''}(t)、G^{''}(t)\ \text{与}\ F^{''}(t)$ 分别是什么。$H^{''}(t)、G^{''}(t)\ \text{与}\ F^{''}(t)$ 的物理意义是什么？它们的单位又是什么？

3.3 Differentiation Rules 3.3 求导法则

• 3.3.1 State the constant, constant multiple, and power rules.
• 3.3.2 Apply the sum and difference rules to combine derivatives.
• 3.3.3 Use the product rule for finding the derivative of a product of functions.
• 3.3.4 Use the quotient rule for finding the derivative of a quotient of functions.
• 3.3.5 Extend the power rule to functions with negative exponents.
• 3.3.6 Combine the differentiation rules to find the derivative of a polynomial or rational function.

• 3.3.1 叙述常数法则、常数倍法则与幂法则。
• 3.3.2 应用和差法则合并导数。
• 3.3.3 使用乘积法则求两个函数乘积的导数。
• 3.3.4 使用商法则求两个函数商的导数。
• 3.3.5 将幂法则推广到带有负指数的函数。
• 3.3.6 综合运用求导法则求多项式或有理函数的导数。

Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process. For example, previously we found that $\frac{d}{dx}\left( \sqrt{x} \right) = \frac{1}{2\sqrt{x}}$ by using a process that involved multiplying an expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate $\frac{d}{dx}\left( \sqrt[3]{x} \right)$ using the definition, while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass this process. We begin with the basics.

The Basic Rules 基本法则

The functions $f(x) = c$ and $g(x) = x^{n}$ where $n$ is a positive integer are the building blocks from which all polynomials and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resorting to the limit definition of the derivative, we must first develop formulas for differentiating these basic functions.

The Constant Rule 常数法则

We first apply the limit definition of the derivative to find the derivative of the constant function, $f(x) = c.$ For this function, both $f(x) = c$ and $f\left( {x + h} \right) = c,$ so we obtain the following result:

$$\begin{array}{cl} {f^{\prime}(x)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{c - c}{h}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{0}{h}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}0 = 0.} \end{array}$$

The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is $0.$ We restate this rule in the following theorem.

The Constant Rule 常数法则

Let $c$ be a constant.

If $f(x) = c,$ then $f^{\prime}(x) = 0.$

Alternatively, we may express this rule as

$$\frac{d}{dx}(c) = 0.$$

Applying the Constant Rule 应用常数法则

Find the derivative of $f(x) = 8.$

Solution 解答

This is just a one-step application of the rule:

$$f^{\prime}(x) = 0.$$

Find the derivative of $g(x) = -3.$

The Power Rule 幂法则

We have shown that

$$\frac{d}{dx}\left( x^{2} \right) = 2x\ \text{and}\ \frac{d}{dx}\left( x^{1\text{/}2} \right) = \frac{1}{2}x^{\text{−}{1\text{/}2}}.$$

At this point, you might see a pattern beginning to develop for derivatives of the form $\frac{d}{dx}\left( x^{n} \right).$ We continue our examination of derivative formulas by differentiating power functions of the form $f(x) = x^{n}$ where $n$ is a positive integer. We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating and proving the general rule for derivatives of functions of this form, we take a look at a specific case, $\frac{d}{dx}(x^{3}).$ As we go through this derivation, note that the technique used in this case is essentially the same as the technique used to prove the general case.

Differentiating $x^{3}$ 求 $x^{3}$ 的导数

Find $\frac{d}{dx}\left( x^{3} \right).$

Solution 解答

$$\begin{array}{clccc} {\frac{d}{dx}\left( x^{3} \right)} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{{(x + h)}^{3} - x^{3}}{h}} & & & \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{x^{3} + 3x^{2}h + 3xh^{2} + h^{3} - x^{3}}{h}} & & & \begin{array}{l} \text{Notice that the first term in the expansion of} \\ {{(x + h)}^{3}\ \text{is}\ x^{3}\ \text{and the second term is}\ 3x^{2}h.\ \text{All}} \\ {\text{other terms contain powers of}\ h\ \text{that are two or}} \\ \text{greater.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{3x^{2}h + 3xh^{2} + h^{3}}{h}} & & & \begin{array}{l} {\text{In this step the}\ x^{3}\ \text{terms have been cancelled,}} \\ {\text{leaving only terms containing}\ h.} \end{array} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{h(3x^{2} + 3xh + h^{2})}{h}} & & & {\text{Factor out the common factor of}\ h.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}(3x^{2} + 3xh + h^{2})} & & & \begin{array}{l} {\text{After cancelling the common factor of}\ h,\text{the}} \\ {\text{only term not containing}\ h\ \text{is}\ 3x^{2}.} \end{array} \\ & {= 3x^{2}} & & & {\text{Let}\ h\ \text{go to 0.}} \end{array}$$

Find $\frac{d}{dx}\left( x^{4} \right).$

As we shall see, the procedure for finding the derivative of the general form $f(x) = x^{n}$ is very similar. Although it is often unwise to draw general conclusions from specific examples, we note that when we differentiate $f(x) = x^{3},$ the power on $x$ becomes the coefficient of $x^{2}$ in the derivative and the power on $x$ in the derivative decreases by 1. The following theorem states that the power rule holds for all positive integer powers of $x.$ We will eventually extend this result to negative integer powers. Later, we will see that this rule may also be extended first to rational powers of $x$ and then to arbitrary powers of $x.$ Be aware, however, that this rule does not apply to functions in which a constant is raised to a variable power, such as $f(x) = 3^{x}.$

The Power Rule 幂法则

Let $n$ be a positive integer. If $f(x) = x^{n},$ then

$$f^{\prime}(x) = nx^{n - 1}.$$

Alternatively, we may express this rule as

$$\frac{d}{dx}x^{n} = nx^{n - 1}.$$

Proof 证明

For $f(x) = x^{n}$ where $n$ is a positive integer, we have

$$f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{{(x + h)}^{n} - x^{n}}{h}.$$ $$\text{Since}\ {(x + h)}^{n} = x^{n} + nx^{n - 1}h + \begin{pmatrix} n \\ 2 \end{pmatrix}\ x^{n - 2}h^{2} + \begin{pmatrix} n \\ 3 \end{pmatrix}\ x^{n - 3}h^{3} + \text{…} + nxh^{n - 1} + h^{n},$$

we see that

$${(x + h)}^{n} - x^{n} = nx^{n - 1}h + \begin{pmatrix} n \\ 2 \end{pmatrix}\ x^{n - 2}h^{2} + \begin{pmatrix} n \\ 3 \end{pmatrix}\ x^{n - 3}h^{3} + \text{…} + nxh^{n - 1} + h^{n}.$$

Next, divide both sides by *h*:

$$\frac{\left( {x + h} \right)^{n} - x^{n}}{h} = \frac{nx^{n - 1}h + \begin{pmatrix} n \\ 2 \end{pmatrix}\ x^{n - 2}h^{2} + \begin{pmatrix} n \\ 3 \end{pmatrix}\ x^{n - 3}h^{3} + \text{…} + nxh^{n - 1} + h^{n}}{h}.$$

Thus,

$$\frac{\left( {x + h} \right)^{n} - x^{n}}{h} = nx^{n - 1} + \begin{pmatrix} n \\ 2 \end{pmatrix}\ x^{n - 2}h + \begin{pmatrix} n \\ 3 \end{pmatrix}\ x^{n - 3}h^{2} + \text{…} + nxh^{n - 2} + h^{n - 1}.$$

Finally,

$$\begin{array}{cl} {f^{\prime}(x)} & {= \underset{h\rightarrow 0}{\text{lim}}\left( {nx^{n - 1} + \begin{pmatrix} n \\ 2 \end{pmatrix}\ x^{n - 2}h + \begin{pmatrix} n \\ 3 \end{pmatrix}\ x^{n - 3}h^{2} + \text{…} + nxh^{n - 2} + h^{n - 1}} \right)} \\ & {= nx^{n - 1}.} \end{array}$$

□

Applying the Power Rule 应用幂法则

Find the derivative of the function $f(x) = x^{10}$ by applying the power rule.

Solution 解答

Using the power rule with $n = 10,$ we obtain

$$f\prime(x) = 10x^{10 - 1} = 10x^{9}.$$

Find the derivative of $f(x) = x^{7}.$

The Sum, Difference, and Constant Multiple Rules 和、差与常数倍法则

We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions. Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract, or multiply by a constant. These rules are summarized in the following theorem.

Sum, Difference, and Constant Multiple Rules 和、差与常数倍法则

Let $f(x)$ and $g(x)$ be differentiable functions and $k$ be a constant. Then each of the following equations holds.

Sum Rule. The derivative of the sum of a function $f$ and a function $g$ is the same as the sum of the derivative of $f$ and the derivative of $g.$

$$\frac{d}{dx}\left( {f(x) + g(x)} \right) = \frac{d}{dx}\left( {f(x)} \right) + \frac{d}{dx}\left( {g(x)} \right);$$

that is,

$$\text{for}\ j(x) = f(x) + g(x),j^{\prime}(x) = f^{\prime}(x) + g^{\prime}(x).$$

Difference Rule. The derivative of the difference of a function *f* and a function *g* is the same as the difference of the derivative of *f* and the derivative of $g\text{:}$

$$\frac{d}{dx}\left( {f(x) - g(x)} \right) = \frac{d}{dx}\left( {f(x)} \right) - \frac{d}{dx}\left( {g(x)} \right);$$

that is,

$$\text{for}\ j(x) = f(x) - g(x),j^{\prime}(x) = f^{\prime}(x) - g^{\prime}(x).$$

Constant Multiple Rule. The derivative of a constant *k* multiplied by a function *f* is the same as the constant multiplied by the derivative:

$$\frac{d}{dx}\left( {kf(x)} \right) = k\frac{d}{dx}\left( {f(x)} \right);$$

that is,

$$\text{for}\ j(x) = kf(x),j^{\prime}(x) = kf^{\prime}(x).$$

Proof 证明

We provide only the proof of the sum rule here. The rest follow in a similar manner.

For differentiable functions $f(x)$ and $g(x),$ we set $j(x) = f(x) + g(x).$ Using the limit definition of the derivative we have

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{j\left( {x + h} \right) - j(x)}{h}.$$

By substituting $j\left( {x + h} \right) = f\left( {x + h} \right) + g\left( {x + h} \right)$ and $j(x) = f(x) + g(x),$ we obtain

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{\left( {f\left( {x + h} \right) + g\left( {x + h} \right)} \right) - \left( {f(x) + g(x)} \right)}{h}.$$

Rearranging and regrouping the terms, we have

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\left( {\frac{f\left( {x + h} \right) - f(x)}{h} + \frac{g\left( {x + h} \right) - g(x)}{h}} \right).$$

We now apply the sum law for limits and the definition of the derivative to obtain

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\left( \frac{f\left( {x + h} \right) - f(x)}{h} \right) + \underset{h\rightarrow 0}{\text{lim}}\left( \frac{g\left( {x + h} \right) - g(x)}{h} \right) = f^{\prime}(x) + g^{\prime}(x).$$

□

Applying the Constant Multiple Rule 应用常数倍法则

Find the derivative of $g(x) = 3x^{2}$ and compare it to the derivative of $f(x) = x^{2}.$

Solution 解答

We use the power rule directly:

$$g^{\prime}(x) = \frac{d}{dx}\left( {3x^{2}} \right) = 3\frac{d}{dx}\left( x^{2} \right) = 3\left( {2x} \right) = 6x.$$

Since $f(x) = x^{2}$ has derivative $f^{\prime}(x) = 2x,$ we see that the derivative of $g(x)$ is 3 times the derivative of $f(x).$ This relationship is illustrated in Figure 3.18.

Applying Basic Derivative Rules 应用基本求导法则

Find the derivative of $f(x) = 2x^{5} + 7.$

Solution 解答

We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:

$$\begin{array}{clccc} {f^{\prime}(x)} & {= \frac{d}{dx}\left( {2x^{5} + 7} \right)} & & & \\ & {= \frac{d}{dx}\left( {2x^{5}} \right) + \frac{d}{dx}(7)} & & & \text{Apply the sum rule.} \\ & {= 2\frac{d}{dx}\left( x^{5} \right) + \frac{d}{dx}(7)} & & & \text{Apply the constant multiple rule.} \\ & {= 2\left( {5x^{4}} \right) + 0} & & & \text{Apply the power rule and the constant rule.} \\ & {= 10x^{4}.} & & & \text{Simplify.} \end{array}$$

Find the derivative of $f(x) = 2x^{3} - 6x^{2} + 3.$

Finding the Equation of a Tangent Line 求切线方程

Find an equation of the line tangent to the graph of $f(x) = x^{2} - 4x + 6$ at $x = 1.$

Solution 解答

To find an equation of the tangent line, we need a point and a slope. To find the point, compute

$$f(1) = 1^{2} - 4(1) + 6 = 3.$$

This gives us the point $\left( {1,3} \right).$ Since the slope of the tangent line at 1 is $f^{\prime}(1),$ we must first find $f^{\prime}(x).$ Using the definition of a derivative, we have

$$f^{\prime}(x) = 2x - 4$$

so the slope of the tangent line is $f^{\prime}(1) = -2.$ Using the point-slope formula, we see that the equation of the tangent line is

$$y - 3 = -2\left( {x - 1} \right).$$

Putting the equation of the line in slope-intercept form, we obtain

$$y = -2x + 5.$$

Find an equation of the line tangent to the graph of $f(x) = 3x^{2} - 11$ at $x = 2.$ Use the point-slope form.

The Product Rule 乘积法则

Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function $f(x) = x^{2},$ whose derivative is $f^{\prime}(x) = 2x$ and not $\frac{d}{dx}(x) \cdot \frac{d}{dx}(x) = 1 \cdot 1 = 1.$

Product Rule 乘积法则

Let $f(x)$ and $g(x)$ be differentiable functions. Then

$$\frac{d}{dx}\left( {f(x)g(x)} \right) = \frac{d}{dx}\left( {f(x)} \right) \cdot g(x) + \frac{d}{dx}\left( {g(x)} \right) \cdot f(x).$$

That is,

$$\text{if}\ j(x) = f(x)g(x),\text{then}\ j^{\prime}(x) = f^{\prime}(x)g(x) + g^{\prime}(x)f(x).$$

This means that the derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.

Proof 证明

We begin by assuming that $f(x)$ and $g(x)$ are differentiable functions. At a key point in this proof we need to use the fact that, since $g(x)$ is differentiable, it is also continuous. In particular, we use the fact that since $g(x)$ is continuous, $\underset{h\rightarrow 0}{\text{lim}}g\left( {x + h} \right) = g(x).$

By applying the limit definition of the derivative to $j(x) = f(x)g(x),$ we obtain

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right)g\left( {x + h} \right) - f(x)g(x)}{h}.$$

By adding and subtracting $f(x)g(x + h)$ in the numerator, we have

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right)g\left( {x + h} \right) - f(x)g\left( {x + h} \right) + f(x)g\left( {x + h} \right) - f(x)g(x)}{h}.$$

After breaking apart this quotient and applying the sum law for limits, the derivative becomes

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\left( \frac{f\left( {x + h} \right)g\left( {x + h} \right) - f(x)g\left( {x + h} \right)}{h} \right) + \underset{h\rightarrow 0}{\text{lim}}\left( \frac{f(x)g\left( {x + h} \right) - f(x)g(x)}{h} \right).$$

Rearranging, we obtain

$$j^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\left( {\frac{f\left( {x + h} \right) - f(x)}{h} \cdot g(x + h)} \right) + \underset{h\rightarrow 0}{\text{lim}}\left( {\frac{g\left( {x + h} \right) - g(x)}{h} \cdot f(x)} \right).$$

By using the continuity of $g(x),$ the definition of the derivatives of $f(x)$ and $g(x),$ and applying the limit laws, we arrive at the product rule,

$$j^{\prime}(x) = f^{\prime}(x)g(x) + g^{\prime}(x)f(x).$$

□

Applying the Product Rule to Functions at a Point 在点处应用乘积法则

For $j(x) = f(x)g(x),$ use the product rule to find $j^{\prime}(2)$ if $f(2) = 3,f^{\prime}(2) = -4,g(2) = 1,$ and $g^{\prime}(2) = 6.$

Solution 解答

Since $j(x) = f(x)g(x),j^{\prime}(x) = f^{\prime}(x)g(x) + g^{\prime}(x)f(x),$ and hence

$$j^{\prime}(2) = f^{\prime}(2)g(2) + g^{\prime}(2)f(2) = (-4)(1) + (6)(3) = 14.$$

Applying the Product Rule to Binomials 对二项式应用乘积法则

For $j(x) = (x^{2} + 2)(3x^{3} - 5x),$ find $j^{\prime}(x)$ by applying the product rule. Check the result by first finding the product and then differentiating.

Solution 解答

If we set $f(x) = x^{2} + 2$ and $g(x) = 3x^{3} - 5x,$ then $f^{\prime}(x) = 2x$ and $g^{\prime}(x) = 9x^{2} - 5.$ Thus,

$$j^{\prime}(x) = f^{\prime}(x)g(x) + g^{\prime}(x)f(x) = \left( {2x} \right)\left( {3x^{3} - 5x} \right) + (9x^{2} - 5)(x^{2} + 2).$$

Simplifying, we have

$$j^{\prime}(x) = 15x^{4} + 3x^{2} - 10.$$

To check, we see that $j(x) = 3x^{5} + x^{3} - 10x$ and, consequently, $j^{\prime}(x) = 15x^{4} + 3x^{2} - 10.$

Use the product rule to obtain the derivative of $j(x) = 2x^{5}\left( {4x^{2} + x} \right).$

The Quotient Rule 商法则

Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that

$$\frac{d}{dx}\left( x^{2} \right) = 2x,\text{not}\ \frac{\frac{d}{dx}\left( x^{3} \right)}{\frac{d}{dx}(x)} = \frac{3x^{2}}{1} = 3x^{2}.$$

The Quotient Rule 商法则

Let $f(x)$ and $g(x)$ be differentiable functions. Then

$$\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{\frac{d}{dx}(f(x)) \cdot g(x) - \frac{d}{dx}(g(x)) \cdot f(x)}{{(g(x))}^{2}}.$$

That is,

$$\text{if}\ j(x) = \frac{f(x)}{g(x)},\text{then}\ j^{\prime}(x) = \frac{f^{\prime}(x)g(x) - g^{\prime}(x)f(x)}{{(g(x))}^{2}}.$$

The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

Applying the Quotient Rule 应用商法则

Use the quotient rule to find the derivative of $k(x) = \frac{5x^{2}}{4x + 3}.$

Solution 解答

Let $f(x) = 5x^{2}$ and $g(x) = 4x + 3.$ Thus, $f^{\prime}(x) = 10x$ and $g^{\prime}(x) = 4.$ Substituting into the quotient rule, we have

$$k^{\prime}(x) = \frac{f^{\prime}(x)g(x) - g^{\prime}(x)f(x)}{{(g(x))}^{2}} = \frac{10x\left( {4x + 3} \right) - 4(5x^{2})}{{(4x + 3)}^{2}}.$$

Simplifying, we obtain

$$k^{\prime}(x) = \frac{20x^{2} + 30x}{{(4x + 3)}^{2}}.$$

Find the derivative of $h(x) = \frac{3x + 1}{4x - 3}.$

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form $x^{k}$ where $k$ is a negative integer.

Extended Power Rule 推广的幂法则

If $k$ is a negative integer, then

$$\frac{d}{dx}\left( x^{k} \right) = kx^{k - 1}.$$

Proof 证明

If $k$ is a negative integer, we may set $n = \text{−}k,$ so that *n* is a positive integer with $k = \text{−}n.$ Since for each positive integer $n,x^{\text{−}n} = \frac{1}{x^{n}},$ we may now apply the quotient rule by setting $f(x) = 1$ and $g(x) = x^{n}.$ In this case, $f^{\prime}(x) = 0$ and $g^{\prime}(x) = nx^{n - 1}.$ Thus,

$$\frac{d}{dx}\left( x^{\text{−}n} \right) = \frac{0\left( x^{n} \right) - 1\left( {nx^{n - 1}} \right)}{\left( x^{n} \right)^{2}}.$$

Simplifying, we see that

$$\frac{d}{dx}\left( x^{\text{−}n} \right) = \frac{\text{−}nx^{n - 1}}{x^{2n}} = \text{−}nx^{{({n - 1})} - 2n} = \text{−}nx^{\text{−}n - 1}.$$

Finally, observe that since $k = \text{−}n,$ by substituting we have

$$\frac{d}{dx}\left( x^{k} \right) = kx^{k - 1}.$$

□

Using the Extended Power Rule 应用推广的幂法则

Find $\frac{d}{dx}\left( x^{-4} \right).$

Solution 解答

By applying the extended power rule with $k = -4,$ we obtain

$$\frac{d}{dx}\left( x^{-4} \right) = -4x^{-4 - 1} = -4x^{-5}.$$

Using the Extended Power Rule and the Constant Multiple Rule 综合应用推广的幂法则与常数倍法则

Use the extended power rule and the constant multiple rule to find the derivative of $f(x) = \frac{6}{x^{2}}.$

Solution 解答

It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as $f(x) = 6x^{-2}.$

$$\begin{array}{clccl} {f^{\prime}(x)} & {= \frac{d}{dx}\left( \frac{6}{x^{2}} \right) = \frac{d}{dx}\left( {6x^{-2}} \right)} & & & {\text{Rewrite}\ \frac{6}{x^{2}}\ \text{as}\ 6x^{-2}.} \\ & {= 6\frac{d}{dx}(x^{-2})} & & & \text{Apply the constant multiple rule.} \\ & {= 6(-2x^{-3})} & & & {\text{Use the extended power rule to differentiate}\ x^{-2}.} \\ & {= -12x^{-3}} & & & \text{Simplify.} \end{array}$$

Find the derivative of $g(x) = \frac{1}{x^{7}}$ using the extended power rule.

Combining Differentiation Rules 综合求导法则

As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.

Combining Differentiation Rules 综合求导法则

For $k(x) = 3h(x) + x^{2}g(x),$ find $k^{\prime}(x).$

Solution 解答

Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.

$$\begin{array}{clccc} {k^{\prime}(x)} & {= \frac{d}{dx}\left( {3h(x) + x^{2}g(x)} \right) = \frac{d}{dx}\left( {3h(x)} \right) + \frac{d}{dx}\left( {x^{2}g(x)} \right)} & & & \text{Apply the sum rule.} \\ & {= 3\frac{d}{dx}\left( {h(x)} \right) + \left( {\frac{d}{dx}\left( x^{2} \right)g(x) + \frac{d}{dx}\left( {g(x)} \right)x^{2}} \right)} & & & \begin{array}{l} \text{Apply the constant multiple rule to} \\ {\text{differentiate}\ 3h(x)\ \text{and the product}} \\ {\text{rule to differentiate}\ x^{2}g(x).} \end{array} \\ & {= 3h^{'}(x) + 2xg(x) + x^{2}g^{'}(x)} & & & \end{array}$$

Extending the Product Rule 推广乘积法则

For $k(x) = f(x)g(x)h(x),$ express $k^{\prime}(x)$ in terms of $f(x),g(x),h(x),$ and their derivatives.

Solution 解答

We can think of the function $k(x)$ as the product of the function $f(x)g(x)$ and the function $h(x).$ That is, $k(x) = \left( {f(x)g(x)} \right) \cdot h(x).$ Thus,

$$\begin{array}{clccl} {k^{\prime}(x)} & {= \frac{d}{dx}\left( {f(x)g(x)} \right) \cdot h(x) + \frac{d}{dx}\left( {h(x)} \right) \cdot \left( {f(x)g(x)} \right)} & & & \begin{array}{l} \text{Apply the product rule to the product} \\ {\text{of}\ f(x)g(x)\ \text{and}\ h(x).} \end{array} \\ & {= \left( {f^{\prime}(x)g(x) + g^{\prime}(x)f{(x))}h} \right.(x) + h^{\prime}(x)f(x)g(x)} & & & {\text{Apply the product rule to}\ f(x)g(x).} \\ & {= f^{\prime}(x)g(x)h(x) + f(x)g^{\prime}(x)h(x) + f(x)g(x)h^{\prime}\left( x\operatorname{).} \right.} & & & \text{Simplify.} \end{array}$$

Combining the Quotient Rule and the Product Rule 综合应用商法则与乘积法则

For $h(x) = \frac{2x^{3}k(x)}{3x + 2},$ find $h^{\prime}(x).$

Solution 解答

This procedure is typical for finding the derivative of a rational function.

$$\begin{matrix} {h^{'}(x)} & {= \frac{\frac{d}{dx}\left( 2x^{3}k(x) \right) \cdot (3x + 2) - \frac{d}{dx}(3x + 2) \cdot \left( 2x^{3}k(x) \right)}{(3x + 2)^{2}}} & & & \text{Apply the quotient rule.} & \\ & {= \frac{\left( 6x^{2}k(x) + k^{'}(x) \cdot 2x^{3} \right)(3x + 2) - 3\left( 2x^{3}k(x) \right)}{(3x + 2)^{2}}} & & & \begin{array}{l} \text{Apply the product rule to find} \\ {\frac{d}{dx}\left( 2x^{3}k(x) \right).\ \text{Use}\ \frac{d}{dx}(3x + 2) = 3.} \end{array} & \\ & {= \frac{-6x^{3}k(x) + 18x^{3}k(x) + 12x^{2}k(x) + 6x^{4}k^{'}(x) + 4x^{3}k^{'}(x)}{(3x + 2)^{2}}} & & & \text{Simplify.} & \\ & {= \frac{12k(x)\left( {x^{3} + x^{2}} \right) + 2k'(x)\left( {3x^{4} + 2x^{3}} \right)}{\left( {3x + 2} \right)^{2}}} & & & & \end{matrix}$$

Find $\frac{d}{dx}\left( {3f(x) - 2g(x)} \right).$

Determining Where a Function Has a Horizontal Tangent 确定函数有水平切线的位置

Determine the values of $x$ for which $f(x) = x^{3} - 7x^{2} + 8x + 1$ has a horizontal tangent line.

Solution 解答

To find the values of $x$ for which $f(x)$ has a horizontal tangent line, we must solve $f^{\prime}(x) = 0.$ Since

$$f^{\prime}(x) = 3x^{2} - 14x + 8 = \left( {3x - 2} \right)\left( {x - 4} \right),$$

we must solve $\left( {3x - 2} \right)\left( {x - 4} \right) = 0.$ Thus we see that the function has horizontal tangent lines at $x = \frac{2}{3}$ and $x = 4$ as shown in the following graph.

Finding a Velocity 求速度

The position of an object on a coordinate axis at time $t$ is given by $s(t) = \frac{t}{t^{2} + 1}.$ What is the initial velocity of the object?

Solution 解答

Since the initial velocity is $v(0) = s^{\prime}(0),$ begin by finding $s^{\prime}(t)$ by applying the quotient rule:

$$s^{\prime}(t) = \frac{1\left( {t^{2} + 1} \right) - 2t(t)}{\left( {t^{2} + 1} \right)^{2}} = \frac{1 - t^{2}}{\left( {t^{2} + 1} \right)^{2}}.$$

After evaluating, we see that $v(0) = 1.$

Find the values of $x$ for which the graph of $f(x) = 4x^{2} - 3x + 2$ has a tangent line parallel to the line $y = 2x + 3.$

Formula One Grandstands 一级方程式看台

Formula One car races can be very exciting to watch and attract a lot of spectators. Formula One track designers have to ensure sufficient grandstand space is available around the track to accommodate these viewers. However, car racing can be dangerous, and safety considerations are paramount. The grandstands must be placed where spectators will not be in danger should a driver lose control of a car (Figure 3.20).

\*\*\*\*\*\*\*\*\*\*

Safety is especially a concern on turns. If a driver does not slow down enough before entering the turn, the car may slide off the racetrack. Normally, this just results in a wider turn, which slows the driver down. But if the driver loses control completely, the car may fly off the track entirely, on a path tangent to the curve of the racetrack.

Suppose you are designing a new Formula One track. One section of the track can be modeled by the function $f(x) = x^{3} + 3x^{2} + x$ (Figure 3.21). The current plan calls for grandstands to be built along the first straightaway and around a portion of the first curve. The plans call for the front corner of the grandstand to be located at the point $\left( {-1.9,2.8} \right).$ We want to determine whether this location puts the spectators in danger if a driver loses control of the car.

1. Physicists have determined that drivers are most likely to lose control of their cars as they are coming into a turn, at the point where the slope of the tangent line is 1. Find the $\left( {x,y} \right)$ coordinates of this point near the turn.

2. Find an equation of the tangent line to the curve at this point.

3. To determine whether the spectators are in danger in this scenario, find the *x*-coordinate of the point where the tangent line crosses the line $y = 2.8.$ Is this point safely to the right of the grandstand? Or are the spectators in danger?

4. What if a driver loses control earlier than the physicists project? Suppose a driver loses control at the point $\left( {-2.5,0.625} \right).$ What is the slope of the tangent line at this point?

5. If a driver loses control as described in part 4, are the spectators safe?

6. Should you proceed with the current design for the grandstand, or should the grandstands be moved?

Section 3.3 Exercises 3.3 节习题

For the following exercises, find $f^{\prime}(x)$ for each function.

106.

$f(x) = x^{7} + 10$

107.

$f(x) = 5x^{3} - x + 1$

108.

$f(x) = 4x^{2} - 7x$

109.

$f(x) = 8x^{4} + 9x^{2} - 1$

110.

$f(x) = x^{4} + \frac{2}{x}$

111.

$f(x) = 3x\left( {18x^{4} + \frac{13}{x + 1}} \right)$

112.

$f(x) = \left( {x + 2} \right)\left( {2x^{2} - 3} \right)$

113.

$f(x) = x^{2}\left( {\frac{2}{x^{2}} + \frac{5}{x^{3}}} \right)$

114.

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

115.

$f(x) = \frac{4x^{3} - 2x + 1}{x^{2}}$

116.

$f(x) = \frac{x^{2} + 4}{x^{2} - 4}$

117.

$f(x) = \frac{x + 9}{x^{2} - 7x + 1}$

For the following exercises, find an equation of the tangent line $T(x)$ to the graph of the given function at the indicated point. Use a graphing calculator to graph the function and the tangent line.

118.

\[T\] $y = 3x^{2} + 4x + 1$ at $\left( {0,1} \right)$

119.

\[T\] $y = \frac{2}{x^{2}} + 1$ at $\left( {1,3} \right)$

120.

\[T\] $y = \frac{2x}{x - 1}$ at $\left( {-1,1} \right)$

121.

\[T\] $y = \frac{2}{x} - \frac{3}{x^{2}}$ at $\left( {1,-1} \right)$

For the following exercises, assume that $f(x)$ and $g(x)$ are both differentiable functions for all $x.$ Find the derivative of each of the functions $h(x).$

122.

$h(x) = 4f(x) + \frac{g(x)}{7}$

123.

$h(x) = x^{3}f(x)$

124.

$h(x) = \frac{f(x)g(x)}{2}$

125.

$h(x) = \frac{3f(x)}{g(x) + 2}$

For the following exercises, assume that $f(x)$ and $g(x)$ are both differentiable functions with values as given in the following table. Use the following table to calculate the following derivatives.

	$1$	$2$	$3$	$4$
$f(x)$	$3$	$5$	$−2$	$0$
$g(x)$	$2$	$3$	$−4$	$6$
$f^{\prime}(x)$	$−1$	$7$	$8$	$−3$
$g^{\prime}(x)$	$4$	$1$	$2$	$9$

	$1$	$2$	$3$	$4$
$f(x)$	$3$	$5$	$−2$	$0$
$g(x)$	$2$	$3$	$−4$	$6$
$f^{\prime}(x)$	$−1$	$7$	$8$	$−3$
$g^{\prime}(x)$	$4$	$1$	$2$	$9$

126.

Find $h^{\prime}(1)$ if $h(x) = xf(x) + 4g(x).$

127.

Find $h^{\prime}(2)$ if $h(x) = \frac{f(x)}{g(x)}.$

128.

Find $h^{\prime}(3)$ if $h(x) = 2x + f(x)g(x).$

129.

Find $h^{\prime}(4)$ if $h(x) = \frac{1}{x} + \frac{g(x)}{f(x)}.$

For the following exercises, use the following figure to find the indicated derivatives, if they exist.

130.

Let $h(x) = f(x) + g(x).$ Find

1. $h^{\prime}(1),$

2. $h^{\prime}(3),$ and

3. $h^{\prime}(4).$

131.

Let $h(x) = f(x)g(x).$ Find

1. $h^{\prime}(1),$

2. $h^{\prime}(3),$ and

3. $h^{\prime}(4).$

132.

Let $h(x) = \frac{f(x)}{g(x)}.$ Find

1. $h^{\prime}(1),$

2. $h^{\prime}(3),$ and

3. $h^{\prime}(4).$

For the following exercises,

1. evaluate $f^{\prime}(a),$ and

2. graph the function $f(x)$ and the tangent line at $x = a.$

133.

\[T\] $f(x) = 2x^{3} + 3x - x^{2},a = 2$

134.

\[T\] $f(x) = \frac{1}{x} - x^{2},a = 1$

135.

\[T\] $f(x) = x^{2} - x^{12} + 3x + 2,a = 0$

136.

\[T\] $f(x) = \frac{1}{x} - x^{2},a = -1$

137.

Find an equation of the tangent line to the graph of $f(x) = 2x^{3} + 4x^{2} - 5x - 3$ at $x = -1.$

138.

Find an equation of the tangent line to the graph of $f(x) = x^{2} + \frac{4}{x} - 10$ at $x = 8.$

139.

Find an equation of the tangent line to the graph of $f(x) = (3x - x^{2})(3 - x - x^{2})$ at $x = 1.$

140.

Find the point on the graph of $f(x) = x^{3}$ such that the tangent line at that point has an $x$ intercept of 6.

141.

Find an equation of the line passing through the point $P(3,3)$ and tangent to the graph of $f(x) = \frac{6}{x - 1}.$

142.

Determine all points on the graph of $f(x) = x^{3} + x^{2} - x - 1$ for which

1. the tangent line is horizontal

2. the tangent line has a slope of $-1.$

143.

Find a quadratic polynomial such that $f(1) = 5,f^{\prime}(1) = 3$ and $f^{''}(1) = -6.$

144.

A car driving along a freeway with traffic has traveled $s(t) = t^{3} - 6t^{2} + 9t$ meters in $t$ seconds.

1. Determine the time in seconds when the velocity of the car is 0.

2. Determine the acceleration of the car when the velocity is 0.

145.

\[T\] A herring swimming along a straight line has traveled $s(t) = \frac{t^{2}}{t^{2} + 2}$ feet in $t$ seconds.

Determine the velocity of the herring when it has traveled 3 seconds.

146.

The population in millions of arctic flounder in the Atlantic Ocean is modeled by the function $P(t) = \frac{8t + 3}{0.2t^{2} + 1},$ where $t$ is measured in years.

1. Determine the initial flounder population.

2. Determine $P^{\prime}(10)$ and briefly interpret the result.

147.

\[T\] The concentration of antibiotic in the bloodstream $t$ hours after being injected is given by the function $C(t) = \frac{2t^{2} + t}{t^{3} + 50},$ where $C$ is measured in milligrams per liter of blood.

1. Find the rate of change of $C(t).$

2. Determine the rate of change for $t = 8,12,24,$ and $36.$

3. Briefly describe what seems to be occurring as the number of hours increases.

148.

A book publisher has a cost function given by $C(x) = \frac{x^{3} + 2x + 3}{x^{2}},$ where *x* is the number of copies of a book in thousands and *C* is the cost, per book, measured in dollars. Evaluate $C^{\prime}(2)$ and explain its meaning.

149.

\[T\] According to Newton’s law of universal gravitation, the force $F$ between two bodies of constant mass $m_{1}$ and $m_{2}$ is given by the formula $F = \frac{Gm_{1}m_{2}}{d^{2}},$ where $G$ is the gravitational constant and $d$ is the distance between the bodies.

1. Suppose that $G,m_{1},\text{and}\ m_{2}$ are constants. Find the rate of change of force $F$ with respect to distance $d.$

2. Find the rate of change of force $F$ with gravitational constant $G = 6.67\ \times \ 10^{-11}$ $\text{Nm}^{2}\text{/}\text{kg}^{2},$ on two bodies 10 meters apart, each with a mass of 1000 kilograms.

3.4 Derivatives as Rates of Change 3.4 作为变化率的导数

• 3.4.1 Determine a new value of a quantity from the old value and the amount of change.
• 3.4.2 Calculate the average rate of change and explain how it differs from the instantaneous rate of change.
• 3.4.3 Apply rates of change to displacement, velocity, and acceleration of an object moving along a straight line.
• 3.4.4 Predict the future population from the present value and the population growth rate.
• 3.4.5 Use derivatives to calculate marginal cost and revenue in a business situation.

• 3.4.1 由旧值及变化量确定某个量的新值。
• 3.4.2 计算平均变化率，并说明它与瞬时变化率的区别。
• 3.4.3 将变化率应用于沿直线运动的物体的位移、速度与加速度。
• 3.4.4 由当前值与人口增长率预测未来人口。
• 3.4.5 利用导数在商业情境中计算边际成本与边际收益。

In this section we look at some applications of the derivative by focusing on the interpretation of the derivative as the rate of change of a function. These applications include acceleration and velocity in physics, population growth rates in biology, and marginal functions in economics.

Amount of Change Formula 变化量公式

One application for derivatives is to estimate an unknown value of a function at a point by using a known value of a function at some given point together with its rate of change at the given point. If $f(x)$ is a function defined on an interval $\left\lbrack {a,a + h} \right\rbrack,$ then the amount of change of $f(x)$ over the interval is the change in the $y$ values of the function over that interval and is given by

$$f\left( {a + h} \right) - f(a).$$

The average rate of change of the function $f$ over that same interval is the ratio of the amount of change over that interval to the corresponding change in the $x$ values. It is given by

$$\frac{f\left( {a + h} \right) - f(a)}{h}.$$

As we already know, the instantaneous rate of change of $f(x)$ at $a$ is its derivative

$$f^{\prime}(a) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {a + h} \right) - f(a)}{h}.$$

For small enough values of $h,f^{\prime}(a) \approx \frac{f\left( {a + h} \right) - f(a)}{h}.$ We can then solve for $f\left( {a + h} \right)$ to get the amount of change formula:

$$f(a + h) \approx f(a) + f^{\prime}(a)h.$$ (3.10)

We can use this formula if we know only $f(a)$ and $f^{\prime}(a)$ and wish to estimate the value of $f\left( {a + h} \right).$ For example, we may use the current population of a city and the rate at which it is growing to estimate its population in the near future. As we can see in Figure 3.22, we are approximating $f(a + h)$ by the $y$ coordinate at $a + h$ on the line tangent to $f(x)$ at $x = a.$ Observe that the accuracy of this estimate depends on the value of $h$ as well as the value of $f^{\prime}(a).$

Here is an interesting demonstration of rate of change.

Estimating the Value of a Function 估计函数值

If $f(3) = 2$ and $f^{\prime}(3) = 5,$ estimate $f(3.2).$

Solution 解答

Begin by finding $h.$ We have $h = 3.2 - 3 = 0.2.$ Thus,

$$f(3.2) = f\left( {3 + 0.2} \right) \approx f(3) + (0.2)f^{\prime}(3) = 2 + 0.2(5) = 3.$$

Given $f(10) = -5$ and $f^{\prime}(10) = 6,$ estimate $f(10.1).$

Motion along a Line 直线运动

Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.

Let $s(t)$ be a function giving the position of an object at time $t.$

The velocity of the object at time $t$ is given by $v(t) = s^{\prime}(t).$

The speed of the object at time $t$ is given by $\left| {v(t)} \right|.$

The acceleration of the object at $t$ is given by $a(t) = v^{\prime}(t) = s^{''}(t).$

Comparing Instantaneous Velocity and Average Velocity 比较瞬时速度与平均速度

A ball is dropped from a height of 64 feet. Its height above ground (in feet) $t$ seconds later is given by $s(t) = -16t^{2} + 64.$

1. What is the instantaneous velocity of the ball when it hits the ground?

2. What is the average velocity during its fall?

Solution 解答

The first thing to do is determine how long it takes the ball to reach the ground. To do this, set $s(t) = 0.$ Solving $-16t^{2} + 64 = 0,$ we get $t = 2,$ so it take 2 seconds for the ball to reach the ground.

1. The instantaneous velocity of the ball as it strikes the ground is $v(2).$ Since $v(t) = s^{\prime}(t) = -32t,$ we obtain $v(t) = -64\ \text{ft/s}.$

2. The average velocity of the ball during its fall is

$$v_{ave} = \frac{s(2) - s(0)}{2 - 0} = \frac{0 - 64}{2} = -32\ \text{ft/s}.$$

Interpreting the Relationship between $v(t)$ and $a(t)$ 理解 $v(t)$ 与 $a(t)$ 的关系

A particle moves along a coordinate axis in the positive direction to the right. Its position at time $t$ is given by $s(t) = t^{3} - 4t + 2.$ Find $v(1)$ and $a(1)$ and use these values to answer the following questions.

1. Is the particle moving from left to right or from right to left at time $t = 1?$

2. Is the particle speeding up or slowing down at time $t = 1?$

Solution 解答

Begin by finding $v(t)$ and $a(t).$

$v(t) = s^{\prime}(t) = 3t^{2} - 4$ and $a(t) = v^{\prime}(t) = s^{''}(t) = 6t.$

Evaluating these functions at $t = 1,$ we obtain $v(1) = -1$ and $a(1) = 6.$

1. Because $v(1) < 0,$ the particle is moving from right to left.

2. Because $v(1) < 0$ and $a(1) > 0,$ velocity and acceleration are acting in opposite directions. In other words, the particle is being accelerated in the direction opposite the direction in which it is traveling, causing $\left| {v(t)} \right|$ to decrease. The particle is slowing down.

Position and Velocity 位置与速度

The position of a particle moving along a coordinate axis is given by $s(t) = t^{3} - 9t^{2} + 24t + 4,t \geq 0.$

1. Find $v(t).$

2. At what time(s) is the particle at rest?

3. On what time intervals is the particle moving from left to right? From right to left?

4. Use the information obtained to sketch the path of the particle along a coordinate axis.

Solution 解答

1. The velocity is the derivative of the position function:

$$v(t) = s^{\prime}(t) = 3t^{2} - 18t + 24.$$

2. The particle is at rest when $v(t) = 0,$ so set $3t^{2} - 18t + 24 = 0.$ Factoring the left-hand side of the equation produces $3\left( {t - 2} \right)\left( {t - 4} \right) = 0.$ Solving, we find that the particle is at rest at $t = 2$ and $t = 4.$

3. The particle is moving from left to right when $v(t) > 0$ and from right to left when $v(t) < 0.$ Figure 3.23 gives the analysis of the sign of $v(t)$ for $t \geq 0,$ but it does not represent the axis along which the particle is moving.

Since $3t^{2} - 18t + 24 > 0$ on $\lbrack 0,2) \cup (4,\text{+}\infty),$ the particle is moving from left to right on these intervals.

Since $3t^{2} - 18t + 24 < 0$ on $\left( {2,4} \right),$ the particle is moving from right to left on this interval.

4. Before we can sketch the graph of the particle, we need to know its position at the time it starts moving $\left( t = 0) \right.$ and at the times that it changes direction $\left( {t = 2,4} \right).$ We have $s(0) = 4,s(2) = 24,$ and $s(4) = 20.$ This means that the particle begins on the coordinate axis at 4 and changes direction at 24 and 20 on the coordinate axis. The path of the particle is shown on a coordinate axis in Figure 3.24.

A particle moves along a coordinate axis. Its position at time $t$ is given by $s(t) = t^{2} - 5t + 1.$ Is the particle moving from right to left or from left to right at time $t = 3?$

Population Change 人口变化

In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and consequently can be represented by the derivative of the size of the population.

If $P(t)$ is the number of entities present in a population, then the population growth rate of $P(t)$ is defined to be $P^{\prime}(t).$

Estimating a Population 估计人口

The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now?

Solution 解答

Let $P(t)$ be the population (in thousands) $t$ years from now. Thus, we know that $P(0) = 10$ and based on the information, we anticipate $P(5) = 30.$ Now estimate $P^{\prime}(0),$ the current growth rate, using

$$P^{\prime}(0) \approx \frac{P(5) - P(0)}{5 - 0} = \frac{30 - 10}{5} = 4.$$

By applying Equation 3.10 to $P(t),$ we can estimate the population 2 years from now by writing

$$P(2) \approx P(0) + (2)P^{\prime}(0) \approx 10 + 2(4) = 18;$$

thus, in 2 years the population will be 18,000.

The current population of a mosquito colony is known to be 3,000; that is, $P(0) = 3,000.$ If $P^{\prime}(0) = 100,$ estimate the size of the population in 3 days, where $t$ is measured in days.

Changes in Cost and Revenue 成本与收益的变化

In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of the revenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and the revenue function.

If $C(x)$ is the cost of producing x items, then the marginal cost $MC(x)$ is $MC(x) = C^{\prime}(x).$

If $R(x)$ is the revenue obtained from selling $x$ items, then the marginal revenue $MR(x)$ is $MR(x) = R^{\prime}(x).$

If $P(x) = R(x) - C(x)$ is the profit obtained from selling x items, then the marginal profit $MP(x)$ is defined to be $MP(x) = P^{\prime}(x) = MR(x) - MC(x) = R^{\prime}(x) - C^{\prime}(x).$

We can roughly approximate

$$MC(x) = C^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{C\left( {x + h} \right) - C(x)}{h}$$

by choosing an appropriate value for $h.$ Since x represents objects, a reasonable and small value for $h$ is 1. Thus, by substituting $h = 1,$ we get the approximation $MC(x) = C^{\prime}(x) \approx C\left( {x + 1} \right) - C(x).$ Consequently, $C^{\prime}(x)$ for a given value of $x$ can be thought of as the change in cost associated with producing one additional item. In a similar way, $MR(x) = R^{\prime}(x)$ approximates the revenue obtained by selling one additional item, and $MP(x) = P^{\prime}(x)$ approximates the profit obtained by producing and selling one additional item.

Applying Marginal Revenue 应用边际收益

Assume that the number of barbeque dinners that can be sold, $x,$ can be related to the price charged, $p,$ by the equation $p(x) = 9 - 0.03x,0 \leq x \leq 300.$

In this case, the revenue in dollars obtained by selling $x$ barbeque dinners is given by

$$R(x) = xp(x) = x\left( {9 - 0.03x} \right) = -0.03x^{2} + 9x\ \text{for}\ 0 \leq x \leq 300.$$

Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner. Compare this to the actual revenue obtained from the sale of this dinner.

Solution 解答

First, find the marginal revenue function: $MR(x) = R^{\prime}(x) = -0.06x + 9.$

Next, use $R^{\prime}(100)$ to approximate $R(101) - R(100),$ the revenue obtained from the sale of the 101st dinner. Since $R^{\prime}(100) = 3,$ the revenue obtained from the sale of the 101st dinner is approximately \$3.

The actual revenue obtained from the sale of the 101st dinner is

$$R(101) - R(100) = 602.97 - 600 = 2.97,\text{or}\ \$ 2.97.$$

The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.

Suppose that the profit obtained from the sale of $x$ fish-fry dinners is given by $P(x) = -0.03x^{2} + 8x - 50.$ Use the marginal profit function to estimate the profit from the sale of the 101st fish-fry dinner.

Section 3.4 Exercises 第 3.4 节习题

For the following exercises, the given functions represent the position of a particle traveling along a horizontal line; $t \geq 0$.

1. Find the velocity and acceleration functions.

2. Determine the time intervals when the object is slowing down or speeding up.

150.

$s(t) = 2t^{3} - 3t^{2} - 12t + 8$

151.

$s(t) = 2t^{3} - 15t^{2} + 36t - 10$

152.

$s(t) = \frac{t}{1 + t^{2}}$

153.

A model rocket is fired vertically upward from the ground. The distance $s$ in feet that the rocket travels from the ground after $t$ seconds is given by $s(t) = -16t^{2} + 560t.$

1. Find the velocity of the rocket 3 seconds after being fired.

2. Find the acceleration of the rocket 3 seconds after being fired.

154.

A ball is thrown downward with a speed of 8 ft/s from the top of a 64-foot-tall building. After t seconds, its height above the ground is given by $s(t) = -16t^{2} - 8t + 64.$

1. Determine how long it takes for the ball to hit the ground.

2. Determine the velocity of the ball when it hits the ground.

155.

The position function $s(t) = t^{2} - 3t - 4$ represents the position of the back of a car backing out of a driveway and then driving in a straight line, where $s$ is in feet and $t$ is in seconds. In this case, $s(t) = 0$ represents the time at which the back of the car is at the garage door, so $s(0) = -4$ is the starting position of the car, 4 feet inside the garage.

1. Determine the velocity of the car when $s(t) = 0.$

2. Determine the velocity of the car when $s(t) = 14.$

156.

The position of a hummingbird flying along a straight line in $t$ seconds is given by $s(t) = 3t^{3} - 7t$ meters.

1. Determine the velocity of the bird at $t = 1$ sec.

2. Determine the acceleration of the bird at $t = 1$ sec.

3. Determine the acceleration of the bird when the velocity equals 0.

157.

A potato is launched vertically upward with an initial velocity of 100 ft/s from a potato gun at the top of an 85-foot-tall building. The position of the potato relative to the ground after $t$ seconds is given by $s(t) = -16t^{2} + 100t + 85.$

1. Find the velocity of the potato after $0.5\ \text{s}$ and $5.75\ \text{s}.$

2. Find the speed of the potato at 0.5 s and 5.75 s.

3. Determine when the potato reaches its maximum height.

4. Find the acceleration of the potato at 0.5 s and 1.5 s.

5. Determine how long the potato is in the air.

6. Determine the velocity of the potato upon hitting the ground.

158.

The position function $s(t) = t^{3} - 8t$ gives the position in miles of a freight train where east is the positive direction and $t$ is measured in hours.

1. Determine the direction the train is traveling when $s(t) = 0.$

2. Determine the direction the train is traveling when $a(t) = 0.$

3. Determine the time intervals when the train is slowing down or speeding up.

159.

The following graph shows the position $y = s(t)$ of an object moving along a straight line.

1. Use the graph of the position function to determine the time intervals when the velocity is positive, negative, or zero.

2. Sketch the graph of the velocity function.

3. Use the graph of the velocity function to determine the time intervals when the acceleration is positive, negative, or zero.

4. Determine the time intervals when the object is speeding up or slowing down.

160.

The cost function, in dollars, of a company that manufactures food processors is given by $C(x) = 200 + \frac{7}{x} + \frac{x^{2}}{7},$ where $x$ is the number of food processors manufactured.

1. Find the marginal cost function.

2. Use the marginal cost function to estimate the cost of manufacturing the thirteenth food processor.

3. Find the actual cost of manufacturing the thirteenth food processor.

161.

The price $p$ (in dollars) and the demand $x$ for a certain digital clock radio is given by the price–demand function $p = 10 - 0.001x.$

1. Find the revenue function $R(x).$

2. Find the marginal revenue function.

3. Find the marginal revenue at $x = 2000$ and $5000.$

162.

\[T\] A profit is earned when revenue exceeds cost. Suppose the profit function for a skateboard manufacturer is given by $P(x) = 30x - 0.3x^{2} - 250,$ where $x$ is the number of skateboards sold.

1. Find the exact profit from the sale of the thirtieth skateboard.

2. Find the marginal profit function and use it to estimate the profit from the sale of the thirtieth skateboard.

163.

\[T\] In general, the profit function is the difference between the revenue and cost functions: $P(x) = R(x) - C(x).$

Suppose the price-demand and cost functions for the production of cordless drills is given respectively by $p = 143 - 0.03x$ and $C(x) = 75,000 + 65x,$ where $x$ is the number of cordless drills that are sold at a price of $p$ dollars per drill and $C(x)$ is the cost of producing $x$ cordless drills.

1. Find the marginal cost function.

2. Find the revenue and marginal revenue functions.

3. Find $R^{\prime}(1000)$ and $R^{\prime}(4000).$ Interpret the results.

4. Find the profit and marginal profit functions.

5. Find $P^{\prime}(1000)$ and $P^{\prime}(4000).$ Interpret the results.

164.

A small town in Ohio commissioned an actuarial firm to conduct a study that modeled the rate of change of the town’s population. The study found that the town’s population (measured in thousands of people) can be modeled by the function $P(t) = - \frac{1}{3}t^{3} + 64t + 3000,$ where $t$ is measured in years.

1. Find the rate of change function $P^{\prime}(t)$ of the population function.

2. Find $P^{\prime}(1),P^{\prime}(2),P^{\prime}(3),$ and $P^{\prime}(4).$ Interpret what the results mean for the town.

3. Find $P^{''}(1),P^{''}(2),P^{''}(3),$ and $P^{''}(4).$ Interpret what the results mean for the town’s population.

165.

\[T\] A culture of bacteria grows in number according to the function $N(t) = 3000\left( {1 + \frac{4t}{t^{2} + 100}} \right),$ where $t$ is measured in hours.

1. Find the rate of change of the number of bacteria.

2. Find $N^{\prime}(0),N^{\prime}(10),N^{\prime}(20),$ and $N^{\prime}(30).$

3. Interpret the results in (b).

4. Find $N^{''}(0),N^{''}(10),N^{''}(20),$ and $N^{''}(30).$ Interpret what the answers imply about the bacteria population growth.

166.

The centripetal force of an object of mass $m$ is given by $F(r) = \frac{mv^{2}}{r},$ where $v$ is the speed of rotation and $r$ is the distance from the center of rotation.

1. Find the rate of change of centripetal force with respect to the distance from the center of rotation.

2. Find the rate of change of centripetal force of an object with mass 1000 kilograms, velocity of 13.89 m/s, and a distance from the center of rotation of 200 meters.

The following questions concern the population (in millions) of London by decade in the 19th century, which is listed in the following table.

Years since 1800	Population (millions)
1	0.8795
11	1.040
21	1.264
31	1.516
41	1.661
51	2.000
61	2.634
71	3.272
81	3.911
91	4.422

距 1800 年之年数	人口（百万）
1	0.8795
11	1.040
21	1.264
31	1.516
41	1.661
51	2.000
61	2.634
71	3.272
81	3.911
91	4.422

Table 3.4 Population of London Source: OpenStax. 167.

\[T\]

1. Using a calculator or a computer program, find the best-fit linear function to measure the population.

2. Find the derivative of the equation in a. and explain its physical meaning.

3. Find the second derivative of the equation and explain its physical meaning.

168.

\[T\]

1. Using a calculator or a computer program, find the best-fit quadratic curve through the data.

2. Find the derivative of the equation and explain its physical meaning.

3. Find the second derivative of the equation and explain its physical meaning.

For the following exercises, consider an astronaut on a large planet in another galaxy. To learn more about the composition of this planet, the astronaut drops an electronic sensor into a deep trench. The sensor transmits its vertical position every second in relation to the astronaut’s position. The summary of the falling sensor data is displayed in the following table.

Time after dropping (s)	Position (m)
0	0
1	−1
2	−2
3	−5
4	−7
5	−14

下落时刻（秒）	位置（米）
0	0
1	−1
2	−2
3	−5
4	−7
5	−14

169.

\[T\]

1. Using a calculator or computer program, find the best-fit quadratic curve to the data.

2. Find the derivative of the position function and explain its physical meaning.

3. Find the second derivative of the position function and explain its physical meaning.

170.

\[T\]

1. Using a calculator or computer program, find the best-fit cubic curve to the data.

2. Find the derivative of the position function and explain its physical meaning.

3. Find the second derivative of the position function and explain its physical meaning.

4. Using the result from c. explain why a cubic function is not a good choice for this problem.

The following problems deal with the Holling type I, II, and III equations. These equations describe the ecological event of growth of a predator population given the amount of prey available for consumption.

171.

\[T\] The Holling type I equation is described by $f(x) = ax,$ where $x$ is the amount of prey available and $a > 0$ is the rate at which the predator meets the prey for consumption.

1. Graph the Holling type I equation, given $a = 0.5.$

2. Determine the first derivative of the Holling type I equation and explain physically what the derivative implies.

3. Determine the second derivative of the Holling type I equation and explain physically what the derivative implies.

4. Using the interpretations from b. and c. explain why the Holling type I equation may not be realistic.

172.

\[T\] The Holling type II equation is described by $f(x) = \frac{ax}{n + x},$ where $x$ is the amount of prey available and $a > 0$ is the maximum consumption rate of the predator.

1. Graph the Holling type II equation given $a = 0.5$ and $n = 5.$ What are the differences between the Holling type I and II equations?

2. Take the first derivative of the Holling type II equation and interpret the physical meaning of the derivative.

3. Show that $f(n) = \frac{1}{2}a$ and interpret the meaning of the parameter $n.$

4. Find and interpret the meaning of the second derivative. What makes the Holling type II function more realistic than the Holling type I function?

173.

\[T\] The Holling type III equation is described by $f(x) = \frac{ax^{2}}{n^{2} + x^{2}},$ where $x$ is the amount of prey available and $a > 0$ is the maximum consumption rate of the predator.

1. Graph the Holling type III equation given $a = 0.5$ and $n = 5.$ What are the differences between the Holling type II and III equations?

2. Take the first derivative of the Holling type III equation and interpret the physical meaning of the derivative.

3. Find and interpret the meaning of the second derivative (it may help to graph the second derivative).

4. What additional ecological phenomena does the Holling type III function describe compared with the Holling type II function?

174.

\[T\] The populations of the snowshoe hare (in thousands) and the lynx (in hundreds) collected over 7 years from 1937 to 1943 are shown in the following table. The snowshoe hare is the primary prey of the lynx.

Population of snowshoe hare (thousands)	Population of lynx (hundreds)
20	10
55	15
65	55
95	60

雪鞋兔种群数量（千）	猞猁种群数量（百）
20	10
55	15
65	55
95	60

Table 3.5 Snowshoe Hare and Lynx Populations Source: OpenStax.

1. Graph the data points and determine which Holling-type function fits the data best.

2. Using the meanings of the parameters $a$ and $n,$ determine values for those parameters by examining a graph of the data. Recall that $n$ measures what prey value results in the half-maximum of the predator value.

3. Plot the resulting Holling-type I, II, and III functions on top of the data. Was the result from part a. correct?

3.5 Derivatives of Trigonometric Functions 3.5 三角函数的导数

• 3.5.1 Find the derivatives of the sine and cosine function.
• 3.5.2 Find the derivatives of the standard trigonometric functions.
• 3.5.3 Calculate the higher-order derivatives of the sine and cosine.

• 3.5.1 求正弦与余弦函数的导数。
• 3.5.2 求标准三角函数的导数。
• 3.5.3 计算正弦与余弦函数的高阶导数。

One of the most important types of motion in physics is simple harmonic motion, which is associated with such systems as an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosine functions. In this section we expand our knowledge of derivative formulas to include derivatives of these and other trigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas for the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and acceleration of simple harmonic motion.

Derivatives of the Sine and Cosine Functions 正弦函数与余弦函数的导数

We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function $f(x),$

$$f^{\prime}(x) = \underset{h\rightarrow 0}{\text{lim}}\frac{f\left( {x + h} \right) - f(x)}{h}.$$

Consequently, for values of $h$ very close to 0, $f^{\prime}(x) \approx \frac{f\left( {x + h} \right) - f(x)}{h}.$ We see that by using $h = 0.01,$

$$\frac{d}{dx}(\text{sin}\mspace{2mu} x) \approx \frac{\text{sin}\mspace{2mu}\left( {x + 0.01} \right) - \text{sin}\mspace{2mu} x}{0.01}$$

By setting $D(x) = \frac{\text{sin}\mspace{2mu}\left( {x + 0.01} \right) - \text{sin}\mspace{2mu} x}{0.01}$ and using a graphing utility, we can get a graph of an approximation to the derivative of $\text{sin}\mspace{2mu} x$ (Figure 3.25).

Upon inspection, the graph of $D(x)$ appears to be very close to the graph of the cosine function. Indeed, we will show that

$$\frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x.$$

If we were to follow the same steps to approximate the derivative of the cosine function, we would find that

$$\frac{d}{dx}(\text{cos}\mspace{2mu} x) = \text{−}\text{sin}\mspace{2mu}{x.}$$

The Derivatives of sin *x* and cos *x* sin x 与 cos x 的导数

The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.

$$\frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x$$ (3.11) $$\frac{d}{dx}\left( {\text{cos}\mspace{2mu} x} \right) = \text{−}\text{sin}\mspace{2mu} x$$ (3.12)

Proof 证明

Because the proofs for $\frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x$ and $\frac{d}{dx}\left( {\text{cos}\mspace{2mu} x} \right) = \text{−}\text{sin}\mspace{2mu} x$ use similar techniques, we provide only the proof for $\frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x.$ Before beginning, recall two important trigonometric limits we learned in Introduction to Limits:

$$\underset{h\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} h}{h} = 1\ \text{and}\ \underset{h\rightarrow 0}{\text{lim}}\frac{\text{cos}\mspace{2mu} h - 1}{h} = 0.$$

The graphs of $y = \frac{\left( {\text{sin}\mspace{2mu} h} \right)}{h}$ and $y = \frac{\left( {\text{cos}\mspace{2mu} h - 1} \right)}{h}$ are shown in Figure 3.26.

We also recall the following trigonometric identity for the sine of the sum of two angles:

$$\text{sin}\mspace{2mu}\left( {x + h} \right) = \text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} h + \text{cos}\mspace{2mu} x\mspace{2mu}\text{sin}\mspace{2mu} h.$$

Now that we have gathered all the necessary equations and identities, we proceed with the proof.

$$\begin{array}{clccl} {\frac{d}{dx}\mspace{2mu}\text{sin}\mspace{2mu} x} & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu}\left( {x + h} \right) - \text{sin}\mspace{2mu} x}{h}} & & & {\text{Apply the definition}\ \text{of the derivative.}} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\frac{\text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} h + \text{cos}\mspace{2mu} x\mspace{2mu}\text{sin}\mspace{2mu} h - \text{sin}\mspace{2mu} x}{h}} & & & \text{Use trig identity for the sine of the sum of two angles.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\left( {\frac{\text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} h - \text{sin}\mspace{2mu} x}{h} + \frac{\text{cos}\mspace{2mu} x\mspace{2mu}\text{sin}\mspace{2mu} h}{h}} \right)} & & & \text{Regroup.} \\ & {= \underset{h\rightarrow 0}{\text{lim}}\left( {\text{sin}\mspace{2mu} x\left( \frac{\text{cos}\mspace{2mu} h - 1}{h} \right) + \text{cos}\mspace{2mu} x\left( \frac{\text{sin}\mspace{2mu} h}{h} \right)} \right)} & & & {\text{Factor out}\ \text{sin}\mspace{2mu} x\ \text{and}\ \text{cos}\mspace{2mu} x.} \\ & {= \text{sin}\mspace{2mu} x{\cdot 0} + \text{cos}\mspace{2mu} x{\cdot 1}} & & & \text{Apply trig limit formulas.} \\ & {= \text{cos}\mspace{2mu} x} & & & \text{Simplify.} \end{array}$$

□

Figure 3.27 shows the relationship between the graph of $f(x) = \text{sin}\mspace{2mu} x$ and its derivative $f^{\prime}(x) = \text{cos}\mspace{2mu} x.$ Notice that at the points where $f(x) = \text{sin}\mspace{2mu} x$ has a horizontal tangent, its derivative $f^{\prime}(x) = \text{cos}\mspace{2mu} x$ takes on the value zero. We also see that where $f(x) = \text{sin}\mspace{2mu} x$ is increasing, $f^{\prime}(x) = \text{cos}\mspace{2mu} x > 0$ and where $f(x) = \text{sin}\mspace{2mu} x$ is decreasing, $f^{\prime}(x) = \text{cos}\mspace{2mu} x < 0.$

Differentiating a Function Containing sin *x* 求含 sin x 函数的导数

Find the derivative of $f(x) = 5x^{3}\text{sin}\mspace{2mu} x.$

Solution 解答

Using the product rule, we have

$$\begin{array}{cl} {f\prime(x)} & {= \frac{d}{dx}\left( {5x^{3}} \right) \cdot \text{sin}\mspace{2mu} x + \frac{d}{dx}\left( {\text{sin}\mspace{2mu} x} \right) \cdot 5x^{3}} \\ & {= 15x^{2} \cdot \text{sin}\mspace{2mu} x + \text{cos}\mspace{2mu} x \cdot 5x^{3}.} \end{array}$$

After simplifying, we obtain

$$f^{\prime}(x) = 15x^{2}\text{sin}\mspace{2mu} x + 5x^{3}\text{cos}\mspace{2mu} x.$$

Find the derivative of $f(x) = \text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x.$

Finding the Derivative of a Function Containing cos *x* 求含 cos x 函数的导数

Find the derivative of $g(x) = \frac{\text{cos}\mspace{2mu} x}{4x^{2}}.$

Solution 解答

By applying the quotient rule, we have

$$g^{\prime}(x) = \frac{(\text{−}\text{sin}\mspace{2mu} x)4x^{2} - 8x(\text{cos}\mspace{2mu} x)}{\left( {4x^{2}} \right)^{2}}.$$

Simplifying, we obtain

$$\begin{array}{cl} {g^{\prime}(x)} & {= \frac{-4x^{2}\text{sin}\mspace{2mu} x - 8x\mspace{2mu}\text{cos}\mspace{2mu} x}{16x^{4}}} \\ & {= \frac{\text{−}x\mspace{2mu}\text{sin}\mspace{2mu} x - 2\mspace{2mu}\text{cos}\mspace{2mu} x}{4x^{3}}.} \end{array}$$

Find the derivative of $f(x) = \frac{x}{\text{cos}\mspace{2mu} x}.$

An Application to Velocity 速度的应用

A particle moves along a coordinate axis in such a way that its position at time $t$ is given by $s(t) = 2\mspace{2mu}\text{sin}\mspace{2mu} t - t$ for $0 \leq t \leq 2\pi.$ At what times is the particle at rest?

Solution 解答

To determine when the particle is at rest, set $s^{\prime}(t) = v(t) = 0.$ Begin by finding $s^{\prime}(t).$ We obtain

$$s^{\prime}(t) = 2\mspace{2mu}\text{cos}\mspace{2mu} t - 1,$$

so we must solve

$$2\mspace{2mu}\text{cos}\mspace{2mu} t - 1 = 0\ \text{for}\ 0 \leq t \leq 2\pi.$$

The solutions to this equation are $t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}.$ Thus the particle is at rest at times $t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}.$

A particle moves along a coordinate axis. Its position at time $t$ is given by $s(t) = \sqrt{3}t + 2\mspace{2mu}\text{cos}\mspace{2mu} t$ for $0 \leq t \leq 2\pi.$ At what times is the particle at rest?

Derivatives of Other Trigonometric Functions 其他三角函数的导数

Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives.

The Derivative of the Tangent Function 正切函数的导数

Find the derivative of $f(x) = \text{tan}\mspace{2mu} x.$

Solution 解答

Start by expressing $\text{tan}\mspace{2mu} x$ as the quotient of $\text{sin}\mspace{2mu} x$ and $\text{cos}\mspace{2mu} x:$

$$f(x) = \text{tan}\mspace{2mu} x = \frac{\text{sin}\mspace{2mu} x}{\text{cos}\mspace{2mu} x}.$$

Now apply the quotient rule to obtain

$$f^{\prime}(x) = \frac{\text{cos}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x - (\text{−}\text{sin}\mspace{2mu} x)\text{sin}\mspace{2mu} x}{\left( {\text{cos}\mspace{2mu} x} \right)^{2}}.$$

Simplifying, we obtain

$$f^{\prime}(x) = \frac{\text{cos}^{2}x + {\ \text{sin}}^{2}x}{\text{cos}^{2}x}.$$

Recognizing that $\text{cos}^{2}x + \text{sin}^{2}x = 1,$ by the Pythagorean theorem, we now have

$$f^{\prime}(x) = \frac{1}{\text{cos}^{2}x}.$$

Finally, use the identity $\text{sec}\mspace{2mu} x = \frac{1}{\text{cos}\mspace{2mu} x}$ to obtain

$$f^{\prime}(x) = \text{sec}^{2}x.$$

Find the derivative of $f(x) = \text{cot}\mspace{2mu} x.$

The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem.

Derivatives of $\text{tan}\mspace{2mu} x,\text{cot}\mspace{2mu} x,\text{sec}\mspace{2mu} x,$ and $\text{csc}\mspace{2mu} x$ tan x、cot x、sec x 与 csc x 的导数

The derivatives of the remaining trigonometric functions are as follows:

$$\frac{d}{dx}\left( {\text{tan}\mspace{2mu} x} \right) = \text{sec}^{2}x$$ (3.13) $$\mspace{13mu}\frac{d}{dx}(\text{cot}\mspace{2mu} x) = \text{−}\text{csc}^{2}x$$ (3.14) $$\mspace{25mu}\frac{d}{dx}(\text{sec}\mspace{2mu} x) = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$$ (3.15) $$\mspace{40mu}\frac{d}{dx}(\text{csc}\mspace{2mu} x) = \text{−}\text{csc}\mspace{2mu} x\mspace{2mu}\text{cot}\mspace{2mu}{x.}$$ (3.16)

Finding the Equation of a Tangent Line 求切线方程

Find the equation of a line tangent to the graph of $f(x) = \text{cot}\mspace{2mu} x$ at $x = \frac{\text{π}}{4}.$

Solution 解答

To find an equation of the tangent line, we need a point and a slope at that point. To find the point, compute

$$f\left( \frac{\pi}{4} \right) = \text{cot}\ \frac{\pi}{4} = 1.$$

Thus the tangent line passes through the point $\left( {\frac{\pi}{4},1} \right).$ Next, find the slope by finding the derivative of $f(x) = \text{cot}\mspace{2mu} x$ and evaluating it at $\frac{\pi}{4}\text{:}$

$$f^{\prime}(x) = \text{−}\text{csc}^{2}x\ \text{and}\ f^{\prime}\left( \frac{\pi}{4} \right) = \text{−}\text{csc}^{2}\left( \frac{\pi}{4} \right) = -2.$$

Using the point-slope equation of the line, we obtain

$$y - 1 = -2\left( {x - \frac{\pi}{4}} \right)$$

or equivalently,

$$y = -2x + 1 + \frac{\pi}{2}.$$

Finding the Derivative of Trigonometric Functions 求三角函数的导数

Find the derivative of $f(x) = \text{csc}\mspace{2mu} x + x\mspace{2mu}\text{tan}\mspace{2mu} x.$

Solution 解答

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

$$f^{\prime}(x) = \frac{d}{dx}\left( {\text{csc}\mspace{2mu} x} \right) + \frac{d}{dx}(x\mspace{2mu}\text{tan}\mspace{2mu} x).$$

In the first term, $\frac{d}{dx}\left( {\text{csc}\mspace{2mu} x} \right) = \text{−}\text{csc}\mspace{2mu} x\mspace{2mu}\text{cot}\mspace{2mu} x,$ and by applying the product rule to the second term we obtain

$$\frac{d}{dx}(x\mspace{2mu}\text{tan}\mspace{2mu} x) = (1)(\text{tan}\mspace{2mu} x) + (\text{sec}^{2}x)(x).$$

Therefore, we have

$$f^{\prime}(x) = \text{−}\text{csc}\mspace{2mu} x\mspace{2mu}\text{cot}\mspace{2mu} x + \text{tan}\mspace{2mu} x + x\mspace{2mu}\text{sec}^{2}x.$$

Find the derivative of $f(x) = 2\mspace{2mu}\text{tan}\mspace{2mu} x - 3\mspace{2mu}\text{cot}\mspace{2mu} x.$

Find the slope of the line tangent to the graph of $f(x) = \text{tan}\mspace{2mu} x$ at $x = \frac{\pi}{6}.$

Higher-Order Derivatives 高阶导数

The higher-order derivatives of $\text{sin}\mspace{2mu} x$ and $\text{cos}\mspace{2mu} x$ follow a repeating pattern. By following the pattern, we can find any higher-order derivative of $\text{sin}\mspace{2mu} x$ and $\text{cos}\mspace{2mu} x.$

Finding Higher-Order Derivatives of $y = \text{sin}\mspace{2mu} x$ 求 $y = \text{sin}\mspace{2mu} x$ 的高阶导数

Find the first four derivatives of $y = \text{sin}\mspace{2mu} x.$

Solution 解答

Each step in the chain is straightforward:

$$\begin{array}{rll} y & = & {\text{sin}\mspace{2mu} x} \\ \frac{dy}{dx} & = & {\text{cos}\mspace{2mu} x} \\ \frac{d^{2}y}{dx^{2}} & = & {\text{−}\text{sin}\mspace{2mu} x} \\ \frac{d^{3}y}{dx^{3}} & = & {\text{−}\text{cos}\mspace{2mu} x} \\ \frac{d^{4}y}{dx^{4}} & = & {\text{sin}\mspace{2mu} x.} \end{array}$$

Analysis 分析

Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of sin *x* equals sin *x*, so

$$\begin{array}{l} {\frac{d^{4}}{dx^{4}}\left( {\text{sin}\mspace{2mu} x} \right) = \frac{d^{8}}{dx^{8}}\left( {\text{sin}\mspace{2mu} x} \right) = \frac{d^{12}}{dx^{12}}\left( {\text{sin}\mspace{2mu} x} \right) = \text{…} = \frac{d^{4n}}{dx^{4n}}\left( {\text{sin}\mspace{2mu} x} \right) = \text{sin}\mspace{2mu} x} \\ {\frac{d^{5}}{dx^{5}}\left( {\text{sin}\mspace{2mu} x} \right) = \frac{d^{9}}{dx^{9}}\left( {\text{sin}\mspace{2mu} x} \right) = \frac{d^{13}}{dx^{13}}\left( {\text{sin}\mspace{2mu} x} \right) = \text{…} = \frac{d^{4n + 1}}{dx^{4n + 1}}\left( {\text{sin}\mspace{2mu} x} \right) = \text{cos}\mspace{2mu} x.} \end{array}$$

For $y = \text{cos}\mspace{2mu} x,$ find $\frac{d^{4}y}{dx^{4}}.$

Using the Pattern for Higher-Order Derivatives of $y = \text{sin}\mspace{2mu} x$ 利用 $y = \text{sin}\mspace{2mu} x$ 高阶导数的规律

Find $\frac{d^{74}}{dx^{74}}\left( {\text{sin}\mspace{2mu} x} \right).$

Solution 解答

We can see right away that for the 74th derivative of $\text{sin}\mspace{2mu} x,74 = 4(18) + 2,$ so

$$\frac{d^{74}}{dx^{74}}\left( {\text{sin}\mspace{2mu} x} \right) = \frac{d^{72 + 2}}{dx^{72 + 2}}\left( {\text{sin}\mspace{2mu} x} \right) = \frac{d^{2}}{dx^{2}}\left( {\text{sin}\mspace{2mu} x} \right) = \text{−}\text{sin}\mspace{2mu} x.$$

For $y = \text{sin}\mspace{2mu} x,$ find $\frac{d^{59}}{dx^{59}}\left( {\text{sin}\mspace{2mu} x} \right).$

An Application to Acceleration 加速度的应用

A particle moves along a coordinate axis in such a way that its position at time $t$ is given by $s(t) = 2 - \text{sin}\mspace{2mu} t.$ Find $v\left( {\pi\text{/}4} \right)$ and $a\left( {\pi\text{/}4} \right).$ Compare these values and decide whether the particle is speeding up or slowing down.

Solution 解答

First find $v(t) = s^{\prime}(t)\text{:}$

$$v(t) = s^{\prime}(t) = \text{−}\text{cos}\mspace{2mu} t.$$

Thus,

$$v\left( \frac{\pi}{4} \right) = - \frac{1}{\sqrt{2}}.$$

Next, find $a(t) = v^{\prime}(t).$ Thus, $a(t) = v^{\prime}(t) = \text{sin}\mspace{2mu} t$ and we have

$$a\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}.$$

Since $v\left( \frac{\pi}{4} \right) = - \frac{1}{\sqrt{2}} < 0$ and $a\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} > 0,$ we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down.

A block attached to a spring is moving vertically. Its position at time $t$ is given by $s(t) = 2\mspace{2mu}\text{sin}\mspace{2mu} t.$ Find $v\left( \frac{5\pi}{6} \right)$ and $a\left( \frac{5\pi}{6} \right).$ Compare these values and decide whether the block is speeding up or slowing down.

Section 3.5 Exercises 3.5 节习题

For the following exercises, find $\frac{dy}{dx}$ for the given functions.

175.

$y = x^{2} - \text{sec}\mspace{2mu} x + 1$

176\.

$y = 3\mspace{2mu}\text{csc}\mspace{2mu} x + \frac{5}{x}$

177.

$y = x^{2}\text{cot}\mspace{2mu} x$

178\.

$y = x - x^{3}\text{sin}\mspace{2mu} x$

179.

$y = \frac{\text{sec}\mspace{2mu} x}{x}$

180\.

$y = \text{sin}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$

181.

$y = \left( {x + \text{cos}\mspace{2mu} x} \right)\left( {1 - \text{sin}\mspace{2mu} x} \right)$

182\.

$y = \frac{\text{tan}\mspace{2mu} x}{1 - \text{sec}\mspace{2mu} x}$

183.

$y = \frac{1 - \text{cot}\mspace{2mu} x}{1 + \text{cot}\mspace{2mu} x}$

184\.

$y = \text{cos}\mspace{2mu} x\left( {1 + \text{csc}\mspace{2mu} x} \right)$

For the following exercises, find an equation of the tangent line to each of the given functions at the indicated values of $x.$ Then use a calculator to graph both the function and the tangent line to ensure the equation for the tangent line is correct.

185.

\[T\] $f(x) = \text{−}\text{sin}\mspace{2mu} x,x = 0$

186\.

\[T\] $f(x) = \text{csc}\mspace{2mu} x,x = \frac{\pi}{2}$

187.

\[T\] $f(x) = 1 + \text{cos}\mspace{2mu} x,x = \frac{3\pi}{2}$

188\.

\[T\] $f(x) = \text{sec}\mspace{2mu} x,x = \frac{\pi}{4}$

189.

\[T\] $f(x) = x^{2} - \text{tan}\mspace{2mu} x,\ x = 0$

190\.

\[T\] $f(x) = 5\mspace{2mu}\text{cot}\mspace{2mu} x,\ x = \frac{\pi}{4}$

For the following exercises, find $\frac{d^{2}y}{dx^{2}}$ for the given functions.

191.

$y = x\mspace{2mu}\text{sin}\mspace{2mu} x - \text{cos}\mspace{2mu} x$

192\.

$y = \text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x$

193.

$y = x - \frac{1}{2}\mspace{2mu}\text{sin}\mspace{2mu} x$

194\.

$y = \frac{1}{x} + \text{tan}\mspace{2mu} x$

195.

$y = 2\mspace{2mu}\text{csc}\mspace{2mu} x$

196\.

$y = \text{sec}^{2}x$

197.

Find all $x$ values on the graph of $f(x) = -3\mspace{2mu}\text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x$ where the tangent line is horizontal.

198\.

Find all $x$ values on the graph of $f(x) = x - 2\mspace{2mu}\text{cos}\mspace{2mu} x$ for $0 < x < 2\pi$ where the tangent line has slope 2.

199.

Let $f(x) = \text{cot}\mspace{2mu} x.$ Determine the points on the graph of $f$ for $0 < x < 2\pi$ where the tangent line(s) is (are) parallel to the line $y = -2x.$

200\.

\[T\] A mass on a spring bounces up and down in simple harmonic motion, modeled by the function $s(t) = -6\mspace{2mu}\text{cos}\mspace{2mu} t$ where $s$ is measured in inches and $t$ is measured in seconds. Find the rate at which the spring is oscillating at $t = 5$ s.

201.

Let the position of a swinging pendulum in simple harmonic motion be given by $s(t) = a\mspace{2mu}\text{cos}\mspace{2mu} t + b\mspace{2mu}\text{sin}\mspace{2mu} t$ where $a$ and $b$ are constants, $t$ measures time in seconds, and $s$ measures position in centimeters. If the position is 0 cm and the velocity is 3 cm/s when $t = 0$, find the values of $a$ and $b$.

202\.

After a diver jumps off a diving board, the edge of the board oscillates with position given by $s(t) = -5\mspace{2mu}\text{cos}\mspace{2mu} t$ cm at $t$ seconds after the jump.

1. Sketch one period of the position function for $t \geq 0.$

2. Find the velocity function.

3. Sketch one period of the velocity function for $t \geq 0.$

4. Determine the times when the velocity is 0 over one period.

5. Find the acceleration function.

6. Sketch one period of the acceleration function for $t \geq 0.$

203.

The number of hamburgers sold at a fast-food restaurant in Pasadena, California, is given by $y = 10 + 5\mspace{2mu}\text{sin}\mspace{2mu} x$ where $y$ is the number of hamburgers sold and $x$ represents the number of hours after the restaurant opened at 11 a.m. until 11 p.m., when the store closes. Find $y\prime$ and determine the intervals where the number of burgers being sold is increasing.

204\.

\[T\] The amount of rainfall per month in Phoenix, Arizona, can be approximated by $y(t) = 0.5 + 0.3\mspace{2mu}\text{cos}\mspace{2mu} t,$ where $t$ is months since January. Find $y^{\prime}$ and use a calculator to determine the intervals where the amount of rain falling is decreasing.

For the following exercises, use the quotient rule to derive the given equations.

205\.

$\frac{d}{dx}(\text{cot}\mspace{2mu} x) = \text{−}\text{csc}^{2}x$

206\.

$\frac{d}{dx}(\text{sec}\mspace{2mu} x) = \text{sec}\mspace{2mu} x\mspace{2mu}\text{tan}\mspace{2mu} x$

207.

$\frac{d}{dx}(\text{csc}\mspace{2mu} x) = \text{−}\text{csc}\mspace{2mu} x\mspace{2mu}\text{cot}\mspace{2mu} x$

208.

Use the definition of derivative and the identity

$\text{cos}\mspace{2mu}\left( {x + h} \right) = \text{cos}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} h - \text{sin}\mspace{2mu} x\mspace{2mu}\text{sin}\mspace{2mu} h$ to prove that $\frac{d\left( {\text{cos}\mspace{2mu} x} \right)}{dx} = \text{−}\text{sin}\mspace{2mu} x.$

For the following exercises, find the requested higher-order derivative for the given functions.

209.

$\frac{d^{3}y}{dx^{3}}$ of $y = 3\mspace{2mu}\text{cos}\mspace{2mu} x$

210\.

$\frac{d^{2}y}{dx^{2}}$ of $y = 3\mspace{2mu}\text{sin}\mspace{2mu} x + x^{2}\text{cos}\mspace{2mu} x$

211.

$\frac{d^{4}y}{dx^{4}}$ of $y = 5\mspace{2mu}\text{cos}\mspace{2mu} x$

212\.

$\frac{d^{2}y}{dx^{2}}$ of $y = \text{sec}\mspace{2mu} x + \text{cot}\mspace{2mu} x$

213.

$\frac{d^{3}y}{dx^{3}}$ of $y = x^{10} - \text{sec}\mspace{2mu} x$

3.6 The Chain Rule 3.6 链式法则

• 3.6.1 State the chain rule for the composition of two functions.
• 3.6.2 Apply the chain rule together with the power rule.
• 3.6.3 Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
• 3.6.4 Recognize the chain rule for a composition of three or more functions.
• 3.6.5 Describe the proof of the chain rule.

• 3.6.1 陈述两个函数复合的链式法则。
• 3.6.2 将链式法则与幂法则结合使用。
• 3.6.3 在需要时正确地将链式法则与乘积法则／商法则结合使用。
• 3.6.4 识别三个或更多函数复合时的链式法则。
• 3.6.5 描述链式法则的证明。

We have seen the techniques for differentiating basic functions $(x^{n},\text{sin}\mspace{2mu} x,\text{cos}\mspace{2mu} x,\text{etc}.)$ as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as $h(x) = \text{sin}\left( x^{3} \right)$ or $k(x) = \sqrt{3x^{2} + 1}.$ In this section, we study the rule for finding the derivative of the composition of two or more functions.

Deriving the Chain Rule 推导链式法则

When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

To put this rule into context, let’s take a look at an example: $h(x) = \text{sin}\mspace{2mu}\left( x^{3} \right).$ We can think of the derivative of this function with respect to x as the rate of change of $\text{sin}\left( x^{3} \right)$ relative to the change in $x.$ Consequently, we want to know how $\text{sin}\left( x^{3} \right)$ changes as $x$ changes. We can think of this event as a chain reaction: As $x$ changes, $x^{3}$ changes, which leads to a change in $\text{sin}\mspace{2mu}\left( x^{3} \right).$ This chain reaction gives us hints as to what is involved in computing the derivative of $\ \text{sin}\left( x^{3} \right).$ First of all, a change in $x$ forcing a change in $x^{3}$ suggests that somehow the derivative of $x^{3}$ is involved. In addition, the change in $x^{3}$ forcing a change in $\text{sin}\left( x^{3} \right)$ suggests that the derivative of $\text{sin}(u)$ with respect to $u,$ where $u = x^{3},$ is also part of the final derivative.

We can take a more formal look at the derivative of $h(x) = \text{sin}\mspace{2mu}\left( x^{3} \right)$ by setting up the limit that would give us the derivative at a specific value $a$ in the domain of $h(x) = \text{sin}\mspace{2mu}\left( x^{3} \right).$

$$h^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{\text{sin}\mspace{2mu}\left( x^{3} \right) - \text{sin}\left( a^{3} \right)}{x - a}.$$

This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression $x^{3} - a^{3}$ to obtain

$$h^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{\text{sin}\mspace{2mu}\left( x^{3} \right) - \text{sin}\left( a^{3} \right)}{x^{3} - a^{3}} \cdot \frac{x^{3} - a^{3}}{x - a}.$$

From the definition of the derivative, we can see that the second factor is the derivative of $x^{3}$ at $x = a.$ That is,

$$\underset{x\rightarrow a}{\text{lim}}\frac{x^{3} - a^{3}}{x - a} = \frac{d}{dx}\left( x^{3} \right)_{x = a} = 3a^{2}.$$

However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting $u = x^{3}$ and observing that as $x\rightarrow a,u\rightarrow a^{3}\text{:}$

$$\begin{array}{cl} {\underset{x\rightarrow a}{\text{lim}}\frac{\text{sin}\mspace{2mu}\left( x^{3} \right) - \text{sin}\mspace{2mu}\left( a^{3} \right)}{x^{3} - a^{3}}} & {= \underset{u\rightarrow a^{3}}{\text{lim}}\frac{\text{sin}\mspace{2mu} u - \text{sin}\mspace{2mu}\left( a^{3} \right)}{u - a^{3}}} \\ & {= \frac{d}{du}\left( {\text{sin}\mspace{2mu} u} \right)_{u = a^{3}}} \\ & {= \text{cos}\mspace{2mu}{\left( a^{3} \right).}} \end{array}$$

Thus, $h^{\prime}(a) = \text{cos}\left( a^{3} \right) \cdot 3a^{2}.$

In other words, if $h(x) = \text{sin}\mspace{2mu}\left( x^{3} \right),$ then $h^{\prime}(x) = \text{cos}\mspace{2mu}\left( x^{3} \right) \cdot 3x^{2}.$ Thus, if we think of $h(x) = \text{sin}\left( x^{3} \right)$ as the composition $\left( {f \circ g} \right)(x) = f\left( {g(x)} \right)$ where $f(x) =$ sin $x$ and $g(x) = x^{3},$ then the derivative of $h(x) = \text{sin}\left( x^{3} \right)$ is the product of the derivative of $g(x) = x^{3}$ and the derivative of the function $f(x) = \text{sin}\mspace{2mu} x$ evaluated at the function $g(x) = x^{3}.$ At this point, we anticipate that for $h(x) = \text{sin}\mspace{2mu}\left( {g(x)} \right),$ it is quite likely that $h^{\prime}(x) = \text{cos}(g(x))g^{\prime}(x).$ As we determined above, this is the case for $h(x) = \text{sin}\mspace{2mu}\left( x^{3} \right).$

Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form to other composite functions. An informal proof is provided at the end of the section.

Let $f$ and $g$ be functions. For all x in the domain of $g$ for which $g$ is differentiable at x and $f$ is differentiable at $g(x),$ the derivative of the composite function

$$h(x) = \left( {f \circ g} \right)(x) = f\left( {g(x)} \right)$$

is given by

$$h^{\prime}(x) = f^{\prime}\left( {g(x)} \right)g^{\prime}(x).$$ (3.17)

Alternatively, if $y$ is a function of $u,$ and $u$ is a function of $x,$ then

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$$

Watch an animation of the chain rule.

Applying the Chain Rule 应用链式法则

1. To differentiate $h(x) = f\left( {g(x)} \right),$ begin by identifying $f(x)$ and $g(x).$

2. Find $f\prime(x)$ and evaluate it at $g(x)$ to obtain $f^{\prime}\left( {g(x)} \right).$

3. Find $g^{\prime}(x).$

4. Write $h^{\prime}(x) = f^{\prime}\left( {g(x)} \right) \cdot g^{\prime}(x).$

Note: When applying the chain rule to the composition of two or more functions, keep in mind that we work our way from the outside function in. It is also useful to remember that the derivative of the composition of two functions can be thought of as having two parts; the derivative of the composition of three functions has three parts; and so on. Also, remember that we never evaluate a derivative at a derivative.

The Chain and Power Rules Combined 链式法则与幂法则结合

We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form $h(x) = {(g(x))}^{n},$ we need to use the chain rule combined with the power rule. To do so, we can think of $h(x) = \left( {g(x)} \right)^{n}$ as $f\left( {g(x)} \right)$ where $f(x) = x^{n}.$ Then $f^{\prime}(x) = nx^{n - 1}.$ Thus, $f^{\prime}\left( {g(x)} \right) = n\left( {g(x)} \right)^{n - 1}.$ This leads us to the derivative of a power function using the chain rule,

$$h^{\prime}(x) = n\left( {g(x)} \right)^{n - 1}g^{\prime}(x)$$

For all values of x for which the derivative is defined, if

$$h(x) = \left( {g(x)} \right)^{n}.$$

Then

$$h^{\prime}(x) = n\left( {g(x)} \right)^{n - 1}g^{\prime}\left( x\operatorname{).} \right.$$ (3.18)

Using the Chain and Power Rules 链式法则与幂法则的应用

Find the derivative of $h(x) = \frac{1}{\left( {3x^{2} + 1} \right)^{2}}.$

Solution 解答

First, rewrite $h(x) = \frac{1}{\left( {3x^{2} + 1} \right)^{2}} = \left( {3x^{2} + 1} \right)^{-2}.$

Applying the power rule with $g(x) = 3x^{2} + 1,$ we have

$$h^{\prime}(x) = -2\left( {3x^{2} + 1} \right)^{-3}\left( {6x} \right).$$

Rewriting back to the original form gives us

$$h^{\prime}(x) = \frac{-12x}{{(3x^{2} + 1)}^{3}}.$$

Find the derivative of $h(x) = \left( {2x^{3} + 2x - 1} \right)^{4}.$

Using the Chain and Power Rules with a Trigonometric Function 对三角函数应用链式法则与幂法则

Find the derivative of $h(x) = \text{sin}^{3}x.$

Solution 解答

First recall that $\text{sin}^{3}x = \left( {\text{sin}\mspace{2mu} x} \right)^{3},$ so we can rewrite $h(x) = \text{sin}^{3}x$ as $h(x) = \left( {\text{sin}\mspace{2mu} x} \right)^{3}.$

Applying the power rule with $g(x) = \text{sin}\mspace{2mu} x,$ we obtain

$$h^{\prime}(x) = 3\left( {\text{sin}\mspace{2mu} x} \right)^{2}\text{cos}\mspace{2mu} x = 3\mspace{2mu}\text{sin}^{2}x\mspace{2mu}\text{cos}\mspace{2mu} x.$$

Finding the Equation of a Tangent Line 求切线方程

Find the equation of a line tangent to the graph of $h(x) = \frac{1}{{(3x - 5)}^{2}}$ at $x = 2.$

Solution 解答

Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the y-coordinate, substitute 2 into $h(x).$ Since $h(2) = \frac{1}{\left( {3(2) - 5} \right)^{2}} = 1,$ the point is $\left( {2,1} \right).$

For the slope, we need $h^{\prime}(2).$ To find $h^{\prime}(x),$ first we rewrite $h(x) = \left( {3x - 5} \right)^{-2}$ and apply the power rule to obtain

$$h^{\prime}(x) = -2\left( {3x - 5} \right)^{-3}(3) = -6\left( {3x - 5} \right)^{-3}.$$

By substituting, we have $h^{\prime}(2) = -6\left( {3(2) - 5} \right)^{-3} = -6.$ Therefore, the line has equation $y - 1 = -6\left( {x - 2} \right).$ Rewriting, the equation of the line is $y = -6x + 13.$

Find an equation of the line tangent to the graph of $f(x) = \left( {x^{2} - 2} \right)^{3}$ at $x = -2.$

Combining the Chain Rule with Other Rules 将链式法则与其他法则结合

Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.

Using the Chain Rule on a General Cosine Function 对一般余弦函数应用链式法则

Find the derivative of $h(x) = \text{cos}\mspace{2mu}\left( {g(x)} \right).$

Solution 解答

Think of $h(x) = \text{cos}(g(x))$ as $f\left( {g(x)} \right)$ where $f(x) = \text{cos}\mspace{2mu} x.$ Since $f^{\prime}(x) = \text{−}\text{sin}\mspace{2mu} x.$ we have $f^{\prime}\left( {g(x)} \right) = \text{−}\text{sin}\mspace{2mu}\left( {g(x)} \right).$ Then we do the following calculation.

$$\begin{array}{clccl} {h^{\prime}(x)} & {= f^{\prime}\left( {g(x)} \right)g^{\prime}(x)} & & & \text{Apply the chain rule.} \\ & {= \text{−}\text{sin}\mspace{2mu}\left( {g(x)} \right)g^{\prime}(x)} & & & {\text{Substitute}\ f^{\prime}\left( {g(x)} \right) = \text{−}\text{sin}\mspace{2mu}\left( {g(x)} \right).} \end{array}$$

Thus, the derivative of $h(x) = \text{cos}\mspace{2mu}\left( {g(x)} \right)$ is given by $h^{\prime}(x) = \text{−}\text{sin}\mspace{2mu}\left( {g(x)} \right)g^{\prime}(x).$

In the following example we apply the rule that we have just derived.

Using the Chain Rule on a Cosine Function 对余弦函数应用链式法则

Find the derivative of $h(x) = \text{cos}\mspace{2mu}\left( {5x^{2}} \right).$

Solution 解答

Let $g(x) = 5x^{2}.$ Then $g^{\prime}(x) = 10x.$ Using the result from the previous example,

$$\begin{array}{cl} {h^{\prime}(x)} & {= \text{−}\text{sin}\mspace{2mu}\left( {5x^{2}} \right) \cdot 10x} \\ & {= -10x\mspace{2mu}\text{sin}\mspace{2mu}\left( {5x^{2}} \right).} \end{array}$$

Using the Chain Rule on Another Trigonometric Function 对另一三角函数应用链式法则

Find the derivative of $h(x) = \text{sec}\mspace{2mu}\left( {4x^{5} + 2x} \right).$

Solution 解答

Apply the chain rule to $h(x) = \text{sec}\mspace{2mu}\left( {g(x)} \right)$ to obtain

$$h^{\prime}(x) = \text{sec}(g{(x))}\mspace{2mu}\text{tan}\mspace{2mu}\left( {g(x)} \right)g^{\prime}(x).$$

In this problem, $g(x) = 4x^{5} + 2x,$ so we have $g^{\prime}(x) = 20x^{4} + 2.$ Therefore, we obtain

$$\begin{array}{cl} {h^{\prime}(x)} & {= \text{sec}\mspace{2mu}\left( {4x^{5} + 2x} \right)\mspace{2mu}\text{tan}\mspace{2mu}\left( {4x^{5} + 2x} \right)\left( {20x^{4} + 2} \right)} \\ & {= (20x^{4} + 2)\text{sec}\mspace{2mu}\left( {4x^{5} + 2x} \right)\mspace{2mu}\text{tan}\mspace{2mu}\left( {4x^{5} + 2x} \right).} \end{array}$$

Find the derivative of $h(x) = \text{sin}(7x + 2).$

At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in Example 3.51 and Example 3.53. For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.

Using the Chain Rule with Trigonometric Functions 对三角函数应用链式法则

For all values of x for which the derivative is defined,

$$\begin{array}{lccl} {\frac{d}{dx}\left( {\text{sin}(g(x))}) \right.\ = \text{cos}\mspace{2mu}\left( {g\left( x) \right.}) \right.g\prime(x)} & & & {\frac{d}{dx}\mspace{2mu}\text{sin}\mspace{2mu} u\ = \text{cos}\mspace{2mu} u\frac{du}{dx}} \\ {\frac{d}{dx}\left( {\text{cos}(g(x))}) \right.\mspace{2mu} = \text{−}\text{sin}\mspace{2mu}\left( {g\left( x) \right.}) \right.g\prime(x)} & & & {\frac{d}{dx}\mspace{2mu}\text{cos}\mspace{2mu} u\mspace{2mu} = \text{−}\text{sin}\mspace{2mu} u\frac{du}{dx}} \\ {\frac{d}{dx}\left( {\text{tan}(g(x))}) \right.\ = \text{sec}^{2}\left( {g\left( x) \right.}) \right.g\prime(x)} & & & {\frac{d}{dx}\mspace{2mu}\text{tan}\mspace{2mu} u\ = \text{sec}^{2}u\frac{du}{dx}} \\ {\frac{d}{dx}\left( {\text{cot}(g(x))}) \right.\ = \text{−}\text{csc}^{2}\left( {g\left( x) \right.}) \right.g\prime(x)} & & & {\frac{d}{dx}\mspace{2mu}\text{cot}\mspace{2mu} u\ = \text{−}\text{csc}^{2}u\frac{du}{dx}} \\ {\frac{d}{dx}\left( {\text{sec}(g(x))}) \right.\ = \text{sec}(g\left( x) \right.\mspace{2mu}\text{tan}\mspace{2mu}\left( {g\left( x) \right.}) \right.g\prime(x)} & & & {\frac{d}{dx}\mspace{2mu}\text{sec}\mspace{2mu} u\ = \text{sec}\mspace{2mu} u\mspace{2mu}\text{tan}\mspace{2mu} u\frac{du}{dx}} \\ {\frac{d}{dx}\left( {\text{csc}(g(x))}) \right.\ = \text{−}\text{csc}(g\left( x) \right.)\text{cot}\mspace{2mu}\left( {g\left( x) \right.}) \right.g\prime(x)} & & & {\frac{d}{dx}\mspace{2mu}\text{csc}\mspace{2mu} u\ = \text{−}\text{csc}\mspace{2mu} u\mspace{2mu}\text{cot}\mspace{2mu} u\frac{du}{dx}.} \end{array}$$

Combining the Chain Rule with the Product Rule 将链式法则与乘积法则结合

Find the derivative of $h(x) = \left( {2x + 1} \right)^{5}\left( {3x - 2} \right)^{7}.$

Solution 解答

First apply the product rule, then apply the chain rule to each term of the product.

$$\begin{array}{clccl} {h^{\prime}(x)} & {= \frac{d}{dx}\left( {(2x + 1)}^{5} \right) \cdot \left( {3x - 2} \right)^{7} + \frac{d}{dx}\left( {(3x - 2)}^{7} \right) \cdot \left( {2x + 1} \right)^{5}} & & & \text{Apply the product rule.} \\ & {= 5{(2x + 1)}^{4} \cdot 2 \cdot \left( {3x - 2} \right)^{7} + 7{(3x - 2)}^{6} \cdot 3 \cdot {(2x + 1)}^{5}} & & & \text{Apply the chain rule.} \\ & {= 10\left( {2x + 1} \right)^{4}\left( {3x - 2} \right)^{7} + 21{(3x - 2)}^{6}{(2x + 1)}^{5}} & & & \text{Simplify.} \\ & {= \left( {2x + 1} \right)^{4}\left( {3x - 2} \right)^{6}\left( {10\left( {3x - 2} \right) + 21\left( {2x + 1} \right)} \right)} & & & {\text{Factor out}\ \left( {2x + 1} \right)^{4}\left( {3x - 2} \right)^{6}.} \\ & {= \left( {2x + 1} \right)^{4}\left( {3x - 2} \right)^{6}(72x + 1)} & & & \text{Simplify.} \end{array}$$

Find the derivative of $h(x) = \frac{x}{\left( {2x + 3} \right)^{3}}.$

Composites of Three or More Functions 三个或更多函数的复合

We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.

In general terms, first we let

$$k(x) = h\left( {f\left( {g(x)} \right)} \right).$$

Then, applying the chain rule once we obtain

$$k^{\prime}(x) = \frac{d}{dx}\left( {h(f\left( {g(x)} \right)} \right) = h\prime\left( {f\left( {g(x)} \right)} \right) \cdot \frac{d}{dx}f\left( \left( {g(x)} \right) \right).$$

Applying the chain rule again, we obtain

$$k^{\prime}(x) = h^{\prime}\left( {f\left( {g(x)} \right)f^{\prime}\left( {g(x)} \right)g^{\prime}(x)} \right).$$

For all values of x for which the function is differentiable, if

$$k(x) = h\left( {f\left( {g(x)} \right)} \right),$$

then

$$k^{\prime}(x) = h^{\prime}\left( {f\left( {g(x)} \right)} \right)f^{\prime}\left( {g(x)} \right)g^{\prime}(x).$$

In other words, we are applying the chain rule twice.

Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative at a time.

Differentiating a Composite of Three Functions 求三个函数复合的导数

Find the derivative of $k(x) = \text{cos}^{4}\left( {{7x}^{2} + 1} \right).$

Solution 解答

First, rewrite $k(x)$ as

$$k(x) = \left( {\text{cos}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)^{4}.$$

Then apply the power rule several times.

$$\begin{array}{clccl} {k^{\prime}(x)} & {= 4\left( {\text{cos}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)^{3}\left( {\frac{d}{dx}\text{cos}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)} & & & \text{Apply the chain rule.} \\ & {= 4\left( {\text{cos}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)^{3}\left( {\text{−}\text{sin}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)\left( {\frac{d}{dx}\left( {7x^{2} + 1} \right)} \right)} & & & \text{Apply the chain rule.} \\ & {= 4\left( {\text{cos}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)^{3}\left( {\text{−}\text{sin}\mspace{2mu}\left( {7x^{2} + 1} \right)} \right)\left( {14x} \right)} & & & \text{Apply the chain rule.} \\ & {= -56x\mspace{2mu}\text{sin}\mspace{2mu}(7x^{2} + 1)\text{cos}^{3}(7x^{2} + 1)} & & & \text{Simplify.} \end{array}$$

Find the derivative of $h(x) = \text{sin}^{6}\left( x^{3} \right).$

Using the Chain Rule in a Velocity Problem 在速度问题中应用链式法则

A particle moves along a coordinate axis. Its position at time t is given by $s(t) = \text{sin}\mspace{2mu}\left( {2t} \right) + \text{cos}\mspace{2mu}\left( {3t} \right).$ What is the velocity of the particle at time $t = \frac{\pi}{6}?$

Solution 解答

To find $v(t),$ the velocity of the particle at time t, we must differentiate $s(t).$ Thus,

$$v(t) = s^{\prime}(t) = 2\mspace{2mu}\text{cos}\mspace{2mu}\left( {2t} \right) - 3\mspace{2mu}\text{sin}\mspace{2mu}\left( {3t} \right).$$

Substituting $t = \frac{\pi}{6}$ into $v(t),$ we obtain $v\left( \frac{\pi}{6} \right) = -2.$

A particle moves along a coordinate axis. Its position at time $t$ is given by $s(t) = \text{sin}(4t).$ Find its acceleration at time $t.$

Proof 证明

At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example, we assume that $g(x) \neq g(a)$ for $x \neq a$ in some open interval containing $a.$ We begin by applying the limit definition of the derivative to the function $h(x)$ to obtain $h^{\prime}(a)\text{:}$

$$h^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{f\left( {g(x)} \right) - f\left( {g(a)} \right)}{x - a}.$$

Rewriting, we obtain

$$h^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{f\left( {g(x)} \right) - f\left( {g(a)} \right)}{g(x) - g(a)} \cdot \frac{g(x) - g(a)}{x - a}.$$

Although it is clear that

$$\underset{x\rightarrow a}{\text{lim}}\frac{g(x) - g(a)}{x - a} = g^{\prime}(a),$$

it is not obvious that

$$\underset{x\rightarrow a}{\text{lim}}\frac{f\left( {g(x)} \right) - f\left( {g(a)} \right)}{g(x) - g(a)} = f^{\prime}\left( {g(a)} \right).$$

To see that this is true, first recall that since g is differentiable at $a,g$ is also continuous at $a.$ Thus,

$$\underset{x\rightarrow a}{\text{lim}}g(x) = g(a).$$

Next, make the substitution $y = g(x)$ and $b = g(a)$ and use change of variables in the limit to obtain

$$\underset{x\rightarrow a}{\text{lim}}\frac{f\left( {g(x)} \right) - f\left( {g(a)} \right)}{g(x) - g(a)} = \underset{y\rightarrow b}{\text{lim}}\frac{f(y) - f(b)}{y - b} = f^{\prime}(b) = f^{\prime}\left( {g(a)} \right).$$

Finally,

$$h^{\prime}(a) = \underset{x\rightarrow a}{\text{lim}}\frac{f\left( {g(x)} \right) - f\left( {g(a)} \right)}{g(x) - g(a)} \cdot \frac{g(x) - g(a)}{x - a} = f^{\prime}\left( {g(a)} \right)g^{\prime}(a).$$

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Using the Chain Rule with Functional Values 对函数值应用链式法则

Let $h(x) = f\left( {g(x)} \right).$ If $g(1) = 4,g^{\prime}(1) = 3,$ and $f^{\prime}(4) = 7,$ find $h^{\prime}(1).$

Solution 解答

Use the chain rule, then substitute.

$$\begin{array}{clccl} {h^{\prime}(1)} & {= f^{\prime}\left( {g(1)} \right)g^{\prime}(1)} & & & \text{Apply the chain rule.} \\ & {= f^{\prime}(4) \cdot 3} & & & {\text{Substitute}\ g(1) = 4\ \text{and}\ g^{\prime}(1) = 3.} \\ & {= 7 \cdot 3} & & & {\text{Substitute}\ f\prime(4) = 7.} \\ & {= 21} & & & \text{Simplify.} \end{array}$$

Given $h(x) = f\left( {g(x)} \right).$ If $g(2) = -3,g^{\prime}(2) = 4,$ and $f^{\prime}(-3) = 7,$ find $h^{\prime}(2).$

The Chain Rule Using Leibniz’s Notation 用莱布尼茨记号表示的链式法则

As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the chain rule is used heavily in physics applications.

$\text{For}\ h(x) = f\left( {g(x)} \right),$ let $u = g(x)$ and $y = h(x) = f(u).$ Thus,

$$h^{\prime}(x) = \frac{dy}{dx},f^{\prime}\left( {g(x)} \right) = f^{\prime}(u) = \frac{dy}{du}\ \text{and}\ g^{\prime}(x) = \frac{du}{dx}.$$

Consequently,

$$\frac{dy}{dx} = h^{\prime}(x) = f^{\prime}\left( {g(x)} \right)g^{\prime}(x) = \frac{dy}{du} \cdot \frac{du}{dx}.$$

If $y$ is a function of $u,$ and $u$ is a function of $x,$ then

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$$

Taking a Derivative Using Leibniz’s Notation, Example 1 用莱布尼茨记号求导，示例 1

Find the derivative of $y = \left( \frac{x}{3x + 2} \right)^{5}.$

Solution 解答

First, let $u = \frac{x}{3x + 2}.$ Thus, $y = u^{5}.$ Next, find $\frac{du}{dx}$ and $\frac{dy}{du}.$ Using the quotient rule,

$$\frac{du}{dx} = \frac{2}{\left( {3x + 2} \right)^{2}}$$

and

$$\frac{dy}{du} = 5u^{4}.$$

Finally, we put it all together.

$$\begin{array}{clccl} \frac{dy}{dx} & {= \frac{dy}{du} \cdot \frac{du}{dx}} & & & \text{Apply the chain rule.} \\ & {= 5u^{4} \cdot \frac{2}{\left( {3x + 2} \right)^{2}}} & & & {\text{Substitute}\ \frac{dy}{du} = 5u^{4}\ \text{and}\ \frac{du}{dx} = \frac{2}{\left( {3x + 2} \right)^{2}}.} \\ & {= 5\left( \frac{x}{3x + 2} \right)^{4} \cdot \frac{2}{\left( {3x + 2} \right)^{2}}} & & & {\text{Substitute}\ u = \frac{x}{3x + 2}.} \\ & {= \frac{10x^{4}}{\left( {3x + 2} \right)^{6}}} & & & \text{Simplify.} \end{array}$$

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Taking a Derivative Using Leibniz’s Notation, Example 2 用莱布尼茨记号求导，示例 2

Find the derivative of $y = \text{tan}\mspace{2mu}\left( {4x^{2} - 3x + 1} \right).$

Solution 解答

First, let $u = 4x^{2} - 3x + 1.$ Then $y = \text{tan}\mspace{2mu} u.$ Next, find $\frac{du}{dx}$ and $\frac{dy}{du}\text{:}$

$$\frac{du}{dx} = 8x - 3\ \text{and}\ \frac{dy}{du} = \text{sec}^{2}u.$$

Finally, we put it all together.

$$\begin{array}{clccl} \frac{dy}{dx} & {= \frac{dy}{du} \cdot \frac{du}{dx}} & & & \text{Apply the chain rule.} \\ & {= \text{sec}^{2}u \cdot \left( {8x - 3} \right)} & & & {\text{Use}\ \frac{du}{dx} = 8x - 3\ \text{and}\ \frac{dy}{du} = \text{sec}^{2}u.} \\ & {= \text{sec}^{2}(4x^{2} - 3x + 1) \cdot (8x - 3)} & & & {\text{Substitute}\ u = 4x^{2} - 3x + 1.} \end{array}$$

Use Leibniz’s notation to find the derivative of $y = \text{cos}\mspace{2mu}\left( x^{3} \right).$ Make sure that the final answer is expressed entirely in terms of the variable $x.$

Section 3.6 Exercises 3.6 节习题

For the following exercises, given $y = f(u)$ and $u = g(x),$ find $\frac{dy}{dx}$ by using Leibniz’s notation for the chain rule: $\frac{dy}{dx} = \frac{dy}{du}\ \frac{du}{dx}.$

214.

$y = 3u - 6,u = 2x^{2}$

215.

$y = 6u^{3},u = 7x - 4$

216.

$y = \text{sin}\mspace{2mu} u,u = 5x - 1$

217.

$y = \text{cos}\mspace{2mu} u,u = \frac{\text{−}x}{8}$

218.

$y = \text{tan}\mspace{2mu} u,u = 9x + 2$

219.

$y = \sqrt{4u + 3},u = x^{2} - 6x$

For each of the following exercises,

1. decompose each function in the form $y = f(u)$ and $u = g(x),$ and

2. find $\frac{dy}{dx}$ as a function of $x.$

220.

$y = \left( {3x - 2} \right)^{6}$

221.

$y = \left( {3x^{2} + 1} \right)^{3}$

222.

$y = \text{sin}^{5}(x)$

223.

$y = \left( {\frac{x}{7} + \frac{7}{x}} \right)^{7}$

224.

$y = \text{tan}\mspace{2mu}\left( {\text{sec}\mspace{2mu} x} \right)$

225.

$y = \text{csc}\mspace{2mu}\left( {\pi x + 1} \right)$

226.

$y = \text{cot}^{2}x$

227.

$y = -6\mspace{2mu}\left( \sin~x \right)^{- 3}$

For the following exercises, find $\frac{dy}{dx}$ for each function.

228.

$y = \left( {3x^{2} + 3x - 1} \right)^{4}$

229.

$y = \left( {5 - 2x} \right)^{-2}$

230.

$y = \text{cos}^{3}\left( {\pi x} \right)$

231.

$y = \left( {2x^{3} - x^{2} + 6x + 1} \right)^{3}$

232.

$y = \frac{1}{\text{sin}^{2}(x)}$

233.

$y = \left( {\text{tan}\mspace{2mu} x + \text{sin}\mspace{2mu} x} \right)^{-3}$

234.

$y = x^{2}\text{cos}^{4}x$

235.

$y = \text{sin}\mspace{2mu}\left( {\text{cos}\mspace{2mu} 7x} \right)$

236.

$y = \sqrt{6 + \text{sec}\mspace{2mu}\pi x^{2}}$

237.

$y = \text{cot}^{3}\left( {4x + 1} \right)$

238.

Let $y = \left\lbrack {f(x)} \right\rbrack^{2}$ and suppose that $f^{\prime}(1) = 4$ and $\frac{dy}{dx} = 10$ for $x = 1.$ Find $f(1).$

239.

Let $y = \left( {f(x) + 5x^{2}} \right)^{4}$ and suppose that $f(-1) = -4$ and $\frac{dy}{dx} = 3$ when $x = -1.$ Find $f^{\prime}(-1)$

240.

Let $y = \left( {f(u) + 3x} \right)^{2}$ and $u = x^{3} - 2x.$ If $f(4) = 6$ and $\frac{dy}{dx} = 18$ when $x = 2,$ find $f^{\prime}(4).$

241.

\[T\] Find an equation of the tangent line to $y = \text{−}\text{sin}\mspace{2mu}\left( \frac{x}{2} \right)$ at the origin. Use a calculator to graph the function and the tangent line together.

242.

\[T\] Find an equation of the tangent line to $y = \left( {3x + \frac{1}{x}} \right)^{2}$ at the point $\left( {1,16} \right).$ Use a calculator to graph the function and the tangent line together.

243.

Find the $x$-coordinates at which the tangent line to $y = \left( {x - \frac{6}{x}} \right)^{8}$ is horizontal.

244.

\[T\] Find an equation of the line that is normal to $g(\theta) = \text{sin}^{2}\left( {\pi\theta} \right)$ at the point $\left( {\frac{1}{4},\frac{1}{2}} \right).$ Use a calculator to graph the function and the normal line together.

For the following exercises, use the information in the following table to find $h^{\prime}(a)$ at the given value for $a.$

$x$	$f(x)$	$f\prime(x)$	$g(x)$	$g\prime(x)$
0	2	5	0	2
1	1	−2	3	0
2	4	4	1	−1
3	3	−3	2	3

$x$	$f(x)$	$f\prime(x)$	$g(x)$	$g\prime(x)$
0	2	5	0	2
1	1	−2	3	0
2	4	4	1	−1
3	3	−3	2	3

245.

$h(x) = f\left( {g(x)} \right);a = 0$

246.

$h(x) = g\left( {f(x)} \right);a = 0$

247.

$h(x) = \left( {x^{4} + g(x)} \right)^{-2};a = 1$

248.

$h(x) = \left( \frac{f(x)}{g(x)} \right)^{2};a = 3$

249.

$h(x) = f\left( {x + f(x)} \right);a = 1$

250.

$h(x) = \left( {1 + g(x)} \right)^{3};a = 2$

251.

$h(x) = g\left( {2 + f\left( {x^{2}} \right)} \right);a = 1$

252.

$h(x) = f\left( {g\left( {\text{sin}\mspace{2mu} x} \right)} \right);a = 0$

253.

\[T\] The position function of a freight train is given by $s(t) = 100\left( {t + 1} \right)^{-2},$ with $s$ in meters and $t$ in seconds. At time $t = 6$ s, find the train's

1. velocity and

2. acceleration.

3. Using a. and b. is the train speeding up or slowing down?

254.

\[T\] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where $t$ is measured in seconds and $s$ is in inches:

$s(t) = -3\mspace{2mu}\text{cos}\mspace{2mu}\left( {\pi t + \frac{\pi}{4}} \right).$

1. Determine the position of the spring at $t = 1.5$ s.

2. Find the velocity of the spring at $t = 1.5$ s.

255.

\[T\] The total cost to produce $x$ boxes of Thin Mint Girl Scout cookies is $C$ dollars, where $C = 0.0001x^{3} - 0.02x^{2} + 3x + 300.$ In $t$ weeks production is estimated to be $x = 1600 + 100t$ boxes.

1. Find the marginal cost $C^{\prime}(x).$

2. Use Leibniz's notation for the chain rule, $\frac{dC}{dt} = \frac{dC}{dx} \cdot \frac{dx}{dt},$ to find the rate with respect to time $t$ that the cost is changing.

3. Use b. to determine how fast costs are increasing when $t = 2$ weeks. Include units with the answer.

256.

\[T\] The formula for the area of a circle is $A = \pi r^{2},$ where $r$ is the radius of the circle. Suppose a circle is expanding, meaning that both the area $A$ and the radius $r$ (in inches) are expanding.

1. Suppose $r = 2 - \frac{100}{\left( {t + 7} \right)^{2}}$ where $t$ is time in seconds. Use the chain rule $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$ to find the rate at which the area is expanding.

2. Use a. to find the rate at which the area is expanding at $t = 4$ s.

257.

\[T\] The formula for the volume of a sphere is $S = \frac{4}{3}\pi r^{3},$ where $r$ (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.

1. Suppose $r = \frac{1}{\left( {t + 1} \right)^{2}} - \frac{1}{12}$ where $t$ is time in minutes. Use the chain rule $\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}$ to find the rate at which the snowball is melting.

2. Use a. to find the rate at which the volume is changing at $t = 1$ min.

258.

\[T\] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function $T(x) = 94 - 10\mspace{2mu}\text{cos}\left\lbrack {\frac{\pi}{12}\left( {x - 2} \right)} \right\rbrack,$ where $x$ is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.

259.

\[T\] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function $D(t) = 5\mspace{2mu}\text{sin}\mspace{2mu}\left( {\frac{\pi}{6}t - \frac{7\pi}{6}} \right) + 8,$ where $t$ is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.

3.7 Derivatives of Inverse Functions 3.7 反函数的导数

• 3.7.1 Calculate the derivative of an inverse function.
• 3.7.2 Recognize the derivatives of the standard inverse trigonometric functions.

• 3.7.1 求反函数的导数。
• 3.7.2 识别标准反三角函数的导数。

In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric functions. This formula may also be used to extend the power rule to rational exponents.

The Derivative of an Inverse Function 反函数的导数

We begin by considering a function and its inverse. If $f(x)$ is both invertible and differentiable, it seems reasonable that the inverse of $f(x)$ is also differentiable. Figure 3.28 shows the relationship between a function $f(x)$ and its inverse $f^{-1}(x).$ Look at the point $\left( {a,f^{-1}(a)} \right)$ on the graph of $f^{-1}(x)$ having a tangent line with a slope of $\left( f^{-1} \right)^{\prime}(a) = \frac{p}{q}.$ This point corresponds to a point $\left( {f^{-1}(a),a} \right)$ on the graph of $f(x)$ having a tangent line with a slope of $f^{\prime}\left( {f^{-1}(a)} \right) = \frac{q}{p}.$ Thus, if $f^{-1}(x)$ is differentiable at $a,$ then it must be the case that

$$\left( f^{-1} \right)^{\prime}(a) = \frac{1}{f^{\prime}\left( {f^{-1}(a)} \right)}.$$

We may also derive the formula for the derivative of the inverse by first recalling that $x = f\left( {f^{-1}(x)} \right).$ Then by differentiating both sides of this equation (using the chain rule on the right), we obtain

$$1 = f^{\prime}\left( {f^{-1}(x)} \right)\left( {f^{-1})^{\prime}(x)} \right..$$

Solving for $(f^{-1})^{\prime}(x),$ we obtain

$$\left( f^{-1} \right)^{\prime}(x) = \frac{1}{f^{\prime}\left( {f^{-1}(x)} \right)}.$$ (3.19)

We summarize this result in the following theorem.

Inverse Function Theorem 反函数定理

Let $f(x)$ be a function that is both invertible and differentiable. Let $y = f^{-1}(x)$ be the inverse of $f(x).$ For all $x$ satisfying $f^{\prime}\left( {f^{-1}(x)} \right) \neq 0,$

$$\frac{dy}{dx} = \frac{d}{dx}\left( {f^{-1}(x)} \right) = \left( f^{-1} \right)^{\prime}(x) = \frac{1}{f^{\prime}\left( {f^{-1}(x)} \right)}.$$

Alternatively, if $y = g(x)$ is the inverse of $f(x),$ then

$$g'(x) = \frac{1}{f^{\prime}\left( {g(x)} \right)}.$$

Applying the Inverse Function Theorem 应用反函数定理

Use the inverse function theorem to find the derivative of $g(x) = \frac{x + 2}{x}.$ Compare the resulting derivative to that obtained by differentiating the function directly.

Solution 解答

The inverse of $g(x) = \frac{x + 2}{x}$ is $f(x) = \frac{2}{x - 1}.$ Since $g^{\prime}(x) = \frac{1}{f^{\prime}\left( {g(x)} \right)},$ begin by finding $f^{\prime}(x).$ Thus,

$$f^{\prime}(x) = \frac{-2}{\left( {x - 1} \right)^{2}}\ \text{and}\ f^{\prime}\left( {g(x)} \right) = \frac{-2}{\left( {g(x) - 1} \right)^{2}} = \frac{-2}{\left( {\frac{x + 2}{x} - 1} \right)^{2}} = - \frac{x^{2}}{2}.$$

Finally,

$$g^{\prime}(x) = \frac{1}{f^{\prime}\left( {g(x)} \right)} = - \frac{2}{x^{2}}.$$

We can verify that this is the correct derivative by applying the quotient rule to $g(x)$ to obtain