Differentiation: Definition and Fundamental Properties

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

The instantaneous rate of change is the gradient of the tangent 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.

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.

Secant slopes approach the tangent slope: the derivative is the limit of average rates

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.

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 $x=0$.

Two ways a continuous function is not differentiable: a corner and a vertical tangent

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 $a$.

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

It works for whole-number powers, negative powers ($\tfrac{1}{x}=x^{-1}$), and roots ($\sqrt{x}=x^{1/2}$) – rewrite as a power first, then apply 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 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:

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 $y-f(a)=f'(a)(x-a)$.