Unit A2: The
Derivative单元 A2：
导数

Find $f'(2)$ for $f(x) = x^2$ using the limit definition.

The tangent line at $(2,4)$ has slope $4$.

用极限定义求 $f(x) = x^2$ 的 $f'(2)$。

点 $(2,4)$ 处的切线斜率为 $4$。

Find $g'(3)$ for $g(x) = \dfrac{1}{x}$ from the definition. The difference quotient is a compound fraction, so clear it before taking the limit:

Now the offending factor of $h$ has cancelled, so the limit is routine:

The tangent to $y = 1/x$ at $(3, \tfrac13)$ has slope $-\tfrac19$, which is negative because the curve is decreasing there. This matches the general result $g'(x) = -1/x^2$ that the power rule will give once we extend it to negative exponents.

用定义求 $g(x) = \dfrac{1}{x}$ 的 $g'(3)$。差商是个繁分数，先把它化简再取极限：

现在那个碍事的因子 $h$ 已经约掉，极限就变得平常了：

$y = 1/x$ 在 $(3, \tfrac13)$ 处的切线斜率为 $-\tfrac19$，为负是因为曲线在那里递减。这与幂法则（power rule）推广到负指数后给出的一般结果 $g'(x) = -1/x^2$ 相符。

Find $p'(a)$ for $p(x) = x^2 - 5x$ using the second definition, factoring rather than expanding. Since $p(x) - p(a) = (x^2 - a^2) - 5(x - a)$, factor out $(x-a)$:

Taking $x \to a$ gives $p'(a) = 2a - 5$. The factoring route avoids expanding $(a+h)^2$ and is often cleaner when the point $a$ is left symbolic. Notice $p'(a) = 0$ at $a = \tfrac52$, the vertex of the parabola, where the tangent is horizontal.

用第二个定义求 $p(x) = x^2 - 5x$ 的 $p'(a)$，采用因式分解而非展开。由 $p(x) - p(a) = (x^2 - a^2) - 5(x - a)$，提出公因式 $(x-a)$：

令 $x \to a$ 得 $p'(a) = 2a - 5$。因式分解这条路避开了展开 $(a+h)^2$，在让点 $a$ 保持符号形式时通常更干净。注意当 $a = \tfrac52$ 时 $p'(a) = 0$，那是抛物线的顶点，切线在该处水平。

Differentiate $f(x) = \sqrt{x}$ from the definition. Rationalize the difference quotient:

Taking $h \to 0$ gives $f'(x) = \dfrac{1}{2\sqrt{x}}$ for $x > 0$.

用定义对 $f(x) = \sqrt{x}$ 求导。把差商有理化：

令 $h \to 0$，当 $x > 0$ 时得 $f'(x) = \dfrac{1}{2\sqrt{x}}$。

For a line $f(x) = mx + b$, the difference quotient is constant before the limit even starts:

So $f'(x) = m$ everywhere: a line has constant slope, which is geometrically obvious and a useful sanity check. The special case $m = 0$ gives a constant function $f(x) = b$ with $f'(x) = 0$, since the difference quotient is $0/h = 0$ for every $h \ne 0$. A constant has no rate of change, so its derivative is zero, not undefined.

对一条直线 $f(x) = mx + b$，差商在取极限之前就已是常数：

于是处处 $f'(x) = m$：直线有恒定斜率，这在几何上显而易见，也是个有用的检验。特例 $m = 0$ 给出常数函数 $f(x) = b$，其 $f'(x) = 0$，因为对每个 $h \ne 0$ 差商都是 $0/h = 0$。常数没有变化率，所以它的导数是零，而非无定义。

Differentiate $f(x) = x|x|$. Writing it piecewise, $f(x) = x^2$ for $x \ge 0$ and $f(x) = -x^2$ for $x < 0$. Away from the origin the power rule (derived in Section 5) gives $f'(x) = 2x$ for $x > 0$ and $f'(x) = -2x$ for $x < 0$, that is $f'(x) = 2|x|$. At $x = 0$ check the definition directly:

Both one-sided limits agree, so $f'(0) = 0$ and the derivative function is $f'(x) = 2|x|$, continuous everywhere. This is a clean example where $f$ is built from $|x|$ yet is still differentiable: the corner of $|x|$ is smoothed out by the extra factor of $x$.

对 $f(x) = x|x|$ 求导。写成分段形式：当 $x \ge 0$ 时 $f(x) = x^2$，当 $x < 0$ 时 $f(x) = -x^2$。在原点之外，幂法则（power rule，第 5 节推导）给出 $x > 0$ 时 $f'(x) = 2x$、$x < 0$ 时 $f'(x) = -2x$，即 $f'(x) = 2|x|$。在 $x = 0$ 处直接用定义检验：

两个单侧极限一致，所以 $f'(0) = 0$，导函数 $f'(x) = 2|x|$ 处处连续。这是个干净的例子：$f$ 由 $|x|$ 构成却仍可微，$|x|$ 的尖角被额外的因子 $x$ 抹平了。

Differentiate $f(x) = x^2 + 3x$ as a function, not at one point. Form the difference quotient and simplify before taking the limit:

Now $\lim_{h \to 0}(2x + h + 3) = 2x + 3$, valid for every real $x$, so $f'(x) = 2x + 3$. Carrying $x$ symbolically through the whole computation is the difference between the derivative at a point (Section 1) and the derivative as a function (this section): we do the algebra once and read off the slope at any $x$ we like, for instance $f'(0) = 3$ and $f'(-2) = -1$.

把 $f(x) = x^2 + 3x$ 当作一个函数来求导，而非在某一点求。先写出差商并在取极限前化简：

于是 $\lim_{h \to 0}(2x + h + 3) = 2x + 3$，对每个实数 $x$ 都成立，所以 $f'(x) = 2x + 3$。让 $x$ 以符号形式贯穿整个计算，正是 一点处 的导数（第 1 节）与 作为函数 的导数（本节）之别：我们只做一次代数，就能读出任意 $x$ 处的斜率，例如 $f'(0) = 3$、$f'(-2) = -1$。

Find the tangent to $f(x) = x^2$ at $x = 3$. We have $f(3) = 9$ and $f'(3) = 6$, so

求 $f(x) = x^2$ 在 $x = 3$ 处的切线。有 $f(3) = 9$、$f'(3) = 6$，于是

A ball is dropped so its height is $s(t) = 100 - 4.9 t^2$ metres after $t$ seconds. The average velocity over $[1, 1+h]$ is

Letting $h \to 0$ gives the instantaneous velocity $s'(1) = -9.8$ m/s. The sign is negative because the ball is falling, and the magnitude is exactly the average velocity with the $-4.9h$ correction term removed. In general $s'(t) = -9.8t$, so the speed grows linearly in time, the signature of constant gravitational acceleration $s''(t) = -9.8$ m/s$^2$.

一个小球自由下落，$t$ 秒后高度为 $s(t) = 100 - 4.9 t^2$ 米。区间 $[1, 1+h]$ 上的平均速度为

令 $h \to 0$ 得到瞬时速度 $s'(1) = -9.8$ m/s。符号为负是因为小球在下落，其大小恰是去掉修正项 $-4.9h$ 后的平均速度。一般地 $s'(t) = -9.8t$，所以速率随时间线性增长，这正是恒定重力加速度 $s''(t) = -9.8$ m/s$^2$ 的标志。

Suppose $C(q)$ is the cost in dollars of producing $q$ kilograms of a chemical, and at $q = 200$ the cost is $C(200) = 4500$ dollars with $C'(200) = 12$. The derivative is the marginal cost: producing the next kilogram costs about 12 dollars, since $C'$ carries units of dollars per kilogram. To estimate the cost of $203$ kg, use the tangent line as a local model:

This is the first appearance of linear approximation: the tangent line is the function's best linear stand-in near the point, and a derivative with units lets you turn a rate into a concrete prediction.

设 $C(q)$ 是生产 $q$ 千克某化学品的成本（以美元计），在 $q = 200$ 处成本 $C(200) = 4500$ 美元、$C'(200) = 12$。该导数是边际成本：再生产一千克约花 12 美元，因为 $C'$ 的单位是美元每千克。要估计 $203$ kg 的成本，用切线作为局部模型：

这是线性近似（linear approximation）的首次登场：切线是函数在该点附近最佳的线性替身，而带单位的导数让你把一个变化率转化为具体的预测。

Find both the tangent and the normal line to $f(x) = \sqrt{x}$ at $x = 4$. From Section 2, $f'(x) = \frac{1}{2\sqrt{x}}$, so $f(4) = 2$ and $f'(4) = \frac{1}{4}$. The tangent has slope $\frac14$:

The normal line is perpendicular to the tangent, so its slope is the negative reciprocal $-4$:

Perpendicular slopes multiply to $-1$ here, $\frac14 \cdot (-4) = -1$, which is the geometric check. The normal line is the direction a marble would roll if released at rest on the curve at that point.

求 $f(x) = \sqrt{x}$ 在 $x = 4$ 处的切线（tangent）与法线（normal line）。由第 2 节 $f'(x) = \frac{1}{2\sqrt{x}}$，所以 $f(4) = 2$、$f'(4) = \frac{1}{4}$。切线斜率为 $\frac14$：

法线与切线垂直，所以其斜率是负倒数 $-4$：

这里垂直斜率之积为 $-1$，即 $\frac14 \cdot (-4) = -1$，这是几何检验。法线指向一颗弹珠若在该点静止释放后会滚动的方向。

Suppose $f'(a)$ exists. For $x \neq a$ write

Taking $x \to a$, the first factor tends to $f'(a)$ and the second to $0$, so the product tends to $0$. Hence $\lim_{x\to a} f(x) = f(a)$, which is continuity at $a$.

The converse fails: $f(x) = |x|$ is continuous at $0$ but the left and right difference quotients give slopes $-1$ and $+1$, so no derivative exists there.

Contrapositive, the useful form. Reading the theorem backwards: if $f$ is discontinuous at $a$, then $f'(a)$ cannot exist. So a jump, a removable hole, or an infinite discontinuity automatically rules out differentiability there. This is often the fastest way to disqualify a point: check continuity first, and if it fails you are done.

设 $f'(a)$ 存在。对 $x \neq a$ 写出

令 $x \to a$，第一个因子趋于 $f'(a)$，第二个趋于 $0$，所以乘积趋于 $0$。于是 $\lim_{x\to a} f(x) = f(a)$，这就是在 $a$ 处的连续性。

逆命题不成立：$f(x) = |x|$ 在 $0$ 处连续，但左右差商给出斜率 $-1$ 与 $+1$，所以那里没有导数。

逆否命题，最有用的形式。把定理倒过来读：若 $f$ 在 $a$ 处不连续，则 $f'(a)$ 不可能存在。所以跳跃、可去间断或无穷间断都自动排除了那里的可微性。这往往是判定一点不可微的最快办法：先查连续性，一旦不成立就完事了。

Let $f(x) = |x|$. Compute the one-sided derivatives at $0$ from the definition. For $h > 0$, $|0+h| = h$, so

For $h < 0$, $|0+h| = -h$, so

Since $1 \ne -1$, the two-sided limit does not exist and $f'(0)$ is undefined, even though $f$ is continuous at $0$. The graph has a corner: a sudden change of slope with no single tangent.

设 $f(x) = |x|$。用定义计算 $0$ 处的单侧导数。当 $h > 0$ 时 $|0+h| = h$，所以

当 $h < 0$ 时 $|0+h| = -h$，所以

由于 $1 \ne -1$，双侧极限不存在，$f'(0)$ 无定义，尽管 $f$ 在 $0$ 处连续。图像有一个尖角：斜率骤变，没有唯一的切线。

Let $f(x) = x^{1/3}$. At $x = 0$ the difference quotient is

As $h \to 0$ from either side, $h^{2/3} \to 0^+$, so the quotient tends to $+\infty$. The limit is not a finite number, so $f'(0)$ does not exist; geometrically the tangent is vertical. Notice $f$ is continuous at $0$ (indeed $\lim_{x\to 0} x^{1/3} = 0 = f(0)$), so this is a second way continuity can hold while differentiability fails, distinct from the corner case.

设 $f(x) = x^{1/3}$。在 $x = 0$ 处差商为

当 $h$ 从两侧趋于 $0$ 时，$h^{2/3} \to 0^+$，所以该商趋于 $+\infty$。极限不是有限数，所以 $f'(0)$ 不存在；几何上切线是竖直的。注意 $f$ 在 $0$ 处连续（确实 $\lim_{x\to 0} x^{1/3} = 0 = f(0)$），所以这是连续成立而可微失效的第二种方式，与尖角情形不同。

For $f(x) = x^n$, the binomial theorem gives

Subtract $x^n$, divide by $h$, and every surviving term except the first still carries a factor of $h$:

As $h \to 0$ all the trailing terms vanish, leaving $f'(x) = n x^{n-1}$.

Why this is rigorous. The expansion has finitely many terms (exactly $n+1$ of them), so the limit of the sum is the sum of the limits with no convergence worries. Every term past the first contains a positive power of $h$, hence tends to $0$, and the first term $n x^{n-1}$ has no $h$ in it. This is a genuine proof, not a heuristic, for every positive integer $n$.

对 $f(x) = x^n$，二项式定理给出

减去 $x^n$、除以 $h$，除第一项外每个保留下来的项都仍带有因子 $h$：

当 $h \to 0$ 时所有后续项都消失，剩下 $f'(x) = n x^{n-1}$。

为何这是严格的。这个展开只有有限多项（恰好 $n+1$ 项），所以和的极限等于极限的和，无需担心收敛（convergence）问题。第一项之后的每一项都含 $h$ 的正幂，因而趋于 $0$，而第一项 $n x^{n-1}$ 中不含 $h$。对每个正整数 $n$，这都是真正的证明，而非启发式论证。

The $x \to a$ form gives a second, arguably cleaner, proof of the power rule. Use the algebraic identity

where the second factor is a sum of exactly $n$ terms. Dividing by $x - a$ removes the singularity:

Now let $x \to a$. Each of the $n$ terms tends to $a^{n-1}$, so the sum tends to $n\,a^{n-1}$. Hence $f'(a) = n a^{n-1}$. This route avoids binomial coefficients entirely and generalizes to the proof that $\frac{d}{dx}x^{m/n}$ obeys the same rule once roots are in play.

$x \to a$ 形式给出幂法则的第二个、可以说更干净的证明。使用代数恒等式

其中第二个因子恰是 $n$ 项之和。除以 $x - a$ 即消去奇异性：

现在令 $x \to a$。这 $n$ 项每项都趋于 $a^{n-1}$，所以和趋于 $n\,a^{n-1}$。于是 $f'(a) = n a^{n-1}$。这条路径完全避开了二项式系数，并可推广到证明带根式时 $\frac{d}{dx}x^{m/n}$ 也遵循同一法则。

Differentiate $f(x) = 4x^3 - 7x^2 + 2x - 9$ term by term. The sum and constant-multiple rules let you handle each monomial separately:

The constant $-9$ contributes $0$, and the linear term $2x$ contributes its slope $2$. Every polynomial is differentiable everywhere, and its derivative is again a polynomial of one lower degree.

对 $f(x) = 4x^3 - 7x^2 + 2x - 9$ 逐项求导。求和法则与常数倍法则让你能分别处理每个单项式：

常数 $-9$ 贡献 $0$，一次项 $2x$ 贡献其斜率 $2$。每个多项式都处处可微，其导数又是一个次数低一阶的多项式。

Rewrite roots and reciprocals as powers before differentiating. For $f(x) = \dfrac{1}{x^2} + \sqrt[3]{x}$, write $f(x) = x^{-2} + x^{1/3}$ and apply the power rule with $n = -2$ and $n = \tfrac13$:

The power rule $\frac{d}{dx}x^n = nx^{n-1}$ in fact holds for every real exponent $n$, a fact proved in full generality in Unit A4 using logarithmic differentiation. Here we simply note that the negative-power case agrees with Worked Example 1.2, where $\frac{d}{dx}\frac1x = -x^{-2}$.

求导前先把根式和倒数改写成幂。对 $f(x) = \dfrac{1}{x^2} + \sqrt[3]{x}$，写成 $f(x) = x^{-2} + x^{1/3}$，分别取 $n = -2$ 和 $n = \tfrac13$ 应用幂法则：

幂法则 $\frac{d}{dx}x^n = nx^{n-1}$ 实际上对每个实指数 $n$ 都成立，这一事实在 Unit A4 用对数求导法以完全一般的形式证明。这里我们只指出负指数情形与例题 1.2 一致，那里 $\frac{d}{dx}\frac1x = -x^{-2}$。

To see the general proof in miniature, compute $\frac{d}{dx}x^3$ from the definition. Expand $(x+h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$, subtract $x^3$, and divide by $h$:

Every term except $3x^2$ still carries a factor of $h$, so as $h \to 0$ they vanish and $\frac{d}{dx}x^3 = 3x^2$. Compare this with the general binomial argument: the surviving term is always the one whose binomial coefficient is $\binom{n}{1} = n$, here $\binom{3}{1} = 3$, giving the $nx^{n-1}$ pattern.

为了看到一般证明的缩影，用定义计算 $\frac{d}{dx}x^3$。展开 $(x+h)^3 = x^3 + 3x^2 h + 3x h^2 + h^3$，减去 $x^3$，再除以 $h$：

除 $3x^2$ 外每项都仍带因子 $h$，所以当 $h \to 0$ 时它们消失，得 $\frac{d}{dx}x^3 = 3x^2$。把它与一般的二项式论证对比：留存下来的总是那个二项式系数为 $\binom{n}{1} = n$ 的项，这里是 $\binom{3}{1} = 3$，给出 $nx^{n-1}$ 的模式。

For $f(x) = x^3$: $f'(x) = 3x^2$, $f''(x) = 6x$, $f'''(x) = 6$, and $f^{(4)}(x) = 0$.

对 $f(x) = x^3$：$f'(x) = 3x^2$，$f''(x) = 6x$，$f'''(x) = 6$，$f^{(4)}(x) = 0$。

Take $f(x) = 2x^4 - x^2$. Differentiating repeatedly:

Each derivative lowers the degree by one, so a degree-$n$ polynomial has $f^{(n)}$ a nonzero constant and $f^{(n+1)} = 0$ identically. Here $n = 4$, and indeed $f^{(4)} = 48 = 2\cdot 4!$ (the leading coefficient $2$ times $4!$), after which everything is zero.

取 $f(x) = 2x^4 - x^2$。反复求导：

每求一次导次数降一，所以 $n$ 次多项式的 $f^{(n)}$ 是非零常数，而 $f^{(n+1)} = 0$ 恒成立。这里 $n = 4$，确实 $f^{(4)} = 48 = 2\cdot 4!$（首项系数 $2$ 乘 $4!$），此后全为零。

For $f(x) = \dfrac{1}{x} = x^{-1}$, the power rule gives a sign-alternating, factorial-growing pattern:

The pattern is $f^{(n)}(x) = (-1)^n \, n! \, x^{-(n+1)}$. Unlike a polynomial, this never becomes zero, because $1/x$ is not a polynomial. Spotting such a closed form for $f^{(n)}$ is exactly what makes Taylor series (Unit B7) computable.

对 $f(x) = \dfrac{1}{x} = x^{-1}$，幂法则给出一个符号交替、按阶乘增长的模式：

其模式为 $f^{(n)}(x) = (-1)^n \, n! \, x^{-(n+1)}$。与多项式不同，它永不归零，因为 $1/x$ 不是多项式。能为 $f^{(n)}$ 看出这样一个闭式，正是泰勒级数（Taylor series，Unit B7）可计算的关键。

Both $x^{2/3}$ and $x^{1/3}$ have a vertical feature at the origin, but they differ. For $f(x) = x^{2/3}$ the difference quotient is

which tends to $+\infty$ as $h \to 0^+$ and to $-\infty$ as $h \to 0^-$. The two sides run to opposite infinities, so the graph comes to a sharp point: a cusp. For $g(x) = x^{1/3}$ the difference quotient $h^{-2/3} = 1/h^{2/3}$ tends to $+\infty$ from both sides, so the graph rises smoothly through a vertical tangent without a point. A cusp has one-sided slopes of opposite sign; a vertical tangent has the same sign on both sides.

$x^{2/3}$ 和 $x^{1/3}$ 在原点处都有竖直特征，但二者不同。对 $f(x) = x^{2/3}$，差商为

当 $h \to 0^+$ 时趋于 $+\infty$，当 $h \to 0^-$ 时趋于 $-\infty$。两侧奔向相反的无穷，所以图像收成一个尖点：尖点（cusp）。对 $g(x) = x^{1/3}$，差商 $h^{-2/3} = 1/h^{2/3}$ 从两侧都趋于 $+\infty$，所以图像经过一条竖直切线平滑上升、不形成尖点。尖点的单侧斜率符号相反；竖直切线两侧符号相同。

Consider $f$ defined by $f(x) = x^2 \sin(1/x)$ for $x \ne 0$ and $f(0) = 0$. This $f$ is differentiable everywhere, including at $0$, where

by the squeeze theorem, since $|h\sin(1/h)| \le |h|$. Yet for $x \ne 0$ the derivative is $f'(x) = 2x\sin(1/x) - \cos(1/x)$, which has no limit as $x \to 0$ because the $\cos(1/x)$ term oscillates. So $f'$ exists everywhere but is itself discontinuous at $0$. The lesson: a derivative need not be continuous, which is why the class of continuously differentiable functions ($C^1$) is a genuine restriction beyond merely differentiable.

考虑这样定义的 $f$：当 $x \ne 0$ 时 $f(x) = x^2 \sin(1/x)$，且 $f(0) = 0$。这个 $f$ 处处可微，在 $0$ 处也是，那里

由夹逼定理（squeeze theorem）得到，因为 $|h\sin(1/h)| \le |h|$。然而当 $x \ne 0$ 时导数为 $f'(x) = 2x\sin(1/x) - \cos(1/x)$，当 $x \to 0$ 时它没有极限，因为 $\cos(1/x)$ 项不停振荡。所以 $f'$ 处处存在，但它自身在 $0$ 处不连续。教训是：导数未必连续，这正是连续可微函数类（$C^1$）相比仅仅可微是个实实在在的更强限制的原因。

Computing every derivative from the limit definition is correct but slow. Unit A3 packages the recurring limits into the product, quotient, and chain rules, and Unit A4 extends them to the transcendental functions. Everything there is justified by the single limit you practiced in this unit.

每个导数都用极限定义来算是正确的，但很慢。Unit A3 把反复出现的极限打包成乘积法则、商的法则与链式法则，Unit A4 又把它们推广到超越函数。那里的一切都由你在本单元练习过的这一个极限所证成。