微分：定义与基本性质

Unit 1 built the limit. Unit 2 uses it to define the derivative 导数 – the exact rate of change at a point.

Over an interval, the average rate of change is a difference quotient 差商. Two equivalent forms appear:

The first uses a step of size $h$ from $a$; the second uses two points $x$ and $a$. Both compute $\dfrac{\text{change in output}}{\text{change in input}}$ over the interval.

The instantaneous 瞬时 rate of change at $x=a$ is what the difference quotient approaches as the interval shrinks to zero. This limit is the derivative at $a$, written $f'(a)$:

provided the limit exists.

Let the point $a$ vary and the derivative becomes a new function:

This is the definition of the derivative (sometimes called differentiating “by first principles” 用定义求导). Its value at each $x$ is the instantaneous rate of change there.

Worked example. Use the definition to find $f'(x)$ for $f(x)=x^2+3x$.

Write $f(x+h)$ in full, subtract $f(x)$, and simplify before letting $h\to 0$:

Every term without $h$ cancels, so you can divide by $h$; that is what removes the $\tfrac{0}{0}$.

Common notations 记号 for the derivative of $y=f(x)$ are:

The derivative can be represented graphically, numerically, analytically, and verbally – be ready to move between them.

Geometric meaning. The derivative at a point is the slope 斜率 of the tangent line 切线 to the graph there. So the tangent line at $x=a$ passes through $\big(a, f(a)\big)$ with slope $f'(a)$:

Writing this line is a routine exam task, so keep the point-slope form ready.

Worked example. Find the tangent line to $f(x)=x^3-2x$ at $x=2$.

• The point: $f(2)=2^3-2(2)=8-4=4$, so the line passes through $(2,4)$.
• The slope: $f'(x)=3x^2-2$, so $f'(2)=3(4)-2=10$.
• The line: $y-4=10(x-2)$, or $y=10x-16$.

Use the derivative only for the slope; the point comes from $f$ itself.

Common mistakes. Using $f'(2)=10$ as the $y$-coordinate of the point, and leaving $x$ in the slope: $y-4=(3x^2-2)(x-2)$ is not a line. The slope of a tangent line is a number, $f'(a)$.

You do not always have a formula. When a function is given by a table 表格 or a graph, estimate the derivative $f'(a)$ with a difference quotient over a small interval around $a$. A table with values on both sides of $a$ gives the best estimate:

Technology (a calculator) can also estimate a derivative at a point.

Exam skill (appears almost every year). Questions such as “Approximate $M'(7.5)$ using the average rate of change of $M$ over the interval $5 \le t \le 10$” ask for exactly this difference quotient. Show the setup:

Full credit needs the numbers plugged in and the correct units 单位 (output units per input unit), since these come from real-world models.

Worked example. Water is pumped into a tank. $M(t)$ is the volume in the tank, in litres, $t$ minutes after the pump starts.

$t$ (minutes)	$0$	$5$	$10$	$15$
$M(t)$ (litres)	$12$	$20$	$26$	$29$

Estimate $M'(7.5)$. The closest table inputs on each side of $7.5$ are $5$ and $10$:

Meaning: about $7.5$ minutes after the start, the volume is rising at about $1.2$ litres per minute.

Common mistakes. Writing $M(10)-M(5)=6$ and forgetting to divide by $10-5$; dividing by the wrong difference; leaving out the units (litres per minute); and using a one-sided interval such as $5\le t\le 15$ when the table has the closest values on both sides of $7.5$.

Differentiability is stronger than continuity. The key relationship:

If $f$ is differentiable 可导 at a point, then $f$ is continuous 连续 there.

So differentiability implies continuity. The reverse is false: a continuous function can fail to be differentiable. Two ways this happens:

• A corner 尖点: the left and right difference-quotient limits disagree, as with $f(x)=|x|$ at $x=0$.
• A vertical tangent 垂直切线: the slope is infinite (no real number), as with $f(x)=\sqrt[3]{x}$ at $x=0$.

Also, a point outside the domain of $f$ cannot be in the domain of $f'$. Use the contrapositive on the exam: if $f$ is not continuous at $a$, then $f$ is not differentiable at $a$.

Common mistake. Arguing “continuous, so differentiable”. The function $|x|$ is continuous at $0$ but has a corner there. Continuity is needed for differentiability, but it is not enough.

From here we use rules instead of the limit definition each time. The power rule 幂法则 handles any power of $x$:

For arbitrary real powers this formula applies on $x>0$; integer powers and some rational powers have larger real domains. Check the domain and differentiability at the point, especially at $x=0$. Rewrite roots as powers before applying the rule.

These rules let you differentiate term by term:

• Constant: $\dfrac{d}{dx}\,k = 0$ (a constant does not change).
• Constant multiple 常数倍: $\dfrac{d}{dx}\big[k\,f(x)\big] = k\,f'(x)$.
• Sum / difference: $\dfrac{d}{dx}\big[f(x)\pm g(x)\big] = f'(x)\pm g'(x)$.

Combined with the power rule, they differentiate any polynomial 多项式 term by term. Example:

Learn these four building-block derivatives by heart:

Note the minus sign on the derivative of cosine, and that $e^{x}$ is its own derivative.

A limit that is really a derivative (LIM-3.A.1). Sometimes a limit is secretly the definition of a known derivative. If you recognize

for a function $f$ whose derivative you know, just evaluate $f'(a)$. For example, $\displaystyle \lim_{h\to 0}\frac{\sin\!\big(\tfrac{\pi}{2}+h\big)-1}{h} = \left.\frac{d}{dx}\sin x\right|_{x=\pi/2} = \cos\tfrac{\pi}{2} = 0$.

A product of two functions is not differentiated by multiplying the derivatives. Use the product rule 乘积法则:

"Derivative of the first times the second, plus the first times the derivative of the second." Example: Exam questions often build a new function from given pieces, e.g. $k(x) = \big(f(x)\big)^2 g(x)$, and ask for $k'(x)$ at a point, you to combine rules while reading values from a table.

For a quotient, use the quotient rule 商法则:

"Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared." The order matters because of the minus sign, so write the numerator carefully. Example:

Worked example: rules from a table. You are given $f(3)=2$, $f'(3)=-1$, $g(3)=5$ and $g'(3)=4$.

• For $p(x)=f(x)\,g(x)$: $p'(3)=f'(3)g(3)+f(3)g'(3)=(-1)(5)+(2)(4)=3$.
• For $q(x)=\dfrac{f(x)}{g(x)}$:

Common mistakes. Multiplying the derivatives: $p'(3)\neq f'(3)\,g'(3)=-4$. Writing the quotient-rule numerator in the wrong order ($fg'-f'g$ gives $+\tfrac{13}{25}$, the wrong sign). Forgetting to square the denominator.

The remaining trigonometric derivatives are not memorized separately – you rewrite them with identities 恒等式 and apply the quotient (or product) rule. For instance, $\tan x = \dfrac{\sin x}{\cos x}$, so the quotient rule gives

The same method (writing $\cot x=\tfrac{\cos x}{\sin x}$, $\sec x=\tfrac{1}{\cos x}$, $\csc x=\tfrac{1}{\sin x}$) gives $-\csc^2 x$, $\sec x\tan x$, and $-\csc x\cot x$.

Higher-order derivatives. Differentiating $f'$ again gives the second derivative 二阶导数 $f''(x)$ (or $\tfrac{d^2y}{dx^2}$) – the rate of change of the rate of change. An exam part like “Find $k''(3)$” just means differentiate twice, then substitute. You can also estimate a second derivative from a table by applying the average-rate-of-change method to the $f'$ values.

Worked example. Differentiate $g(x)=\dfrac{\sin x}{x}$ with the quotient rule $\left(\tfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$: take $u=\sin x$, $v=x$, giving $g'(x)=\dfrac{x\cos x-\sin x}{x^2}$. Keep the order $u'v-uv'$ in the numerator — swapping the terms flips the sign and loses the mark.

• The derivative is the slope of the tangent — the limit of the secant slope $\tfrac{f(x+h)-f(x)}{h}$ as $h\to0$.
• Memorise the rules: power, product, quotient, and the derivatives of $\sin$, $\cos$, $e^x$, and $\ln x$.
• Differentiability implies continuity, but not the reverse (a corner or cusp is continuous yet not differentiable).
• Distinguish average rate of change (secant slope over an interval) from instantaneous rate (the derivative at a point).
• Give a tangent-line equation as $y-f(a)=f'(a)(x-a)$.