Differentiation: Rules, Tangents, and Stationary Points | 微分：法则、切线与驻点

📚 Differentiation: Rules, Tangents, and Stationary Points | 微分：法则、切线与驻点

Differentiation is one of the core skills in the Edexcel A Level Mathematics specification. It measures the rate of change of a function and underpins problems on gradients, tangents, normals, stationary points, and optimisation. This revision article covers the main rules, the standard derivatives you must memorise, and the most common exam applications, with worked examples and a checklist of the mistakes that cost marks every year.

微分是 Edexcel A Level 数学大纲的核心技能之一。它度量函数的变化率，是梯度、切线、法线、驻点和优化问题的基础。本篇复习文章涵盖主要求导法则、必须背熟的标准导数以及最常见的考试应用，并配有例题与每年都导致失分的错误清单。

1. What Is Differentiation? | 什么是微分？

inds the gradient of a curve at a point. If y = f(x), the derivative dy/dx, also written f' nd the tangent line is therefore the best straight-line approximation to the curve nearby.

微分求曲线在某一点的梯度。若 y = f(x)，导数 dy/dx（也写作 f'(x)）表示 y 关于 x 的瞬时变化率。从几何上看，dy/dx 是曲线在该点切线的斜率，因此切线是曲线在该点附近的最佳直线近似。

is [f(x + h) − f(x)] ÷ h, and taking the limit as h → 0 gives the gradient of the tangent.

其正式定义来自"割线收缩"：x 与 x + h 之间的平均梯度为 [f(x + h) − f(x)] ÷ h，令 h → 0 取极限即得切线斜率。

f'(x) = lim (h → 0) [ f(x + h) − f(x) ] ÷ h

In exams you usually apply standard rules rather than the limit definition. However, the limit idea explains why the derivative represents a tangent slope, and it also explains why the function must be smooth at that point for the derivative to exist.

考试中通常直接应用标准法则，而不是极限定义。但极限思想解释了为什么导数表示切线斜率，也解释了为什么函数在该点必须"光滑"导数才存在。

Notation 记号	Meaning 含义
dy/dx	Gradient function of y with respect to x｜y 关于 x 的梯度函数
f'(x)	Same thing, function notation｜同一含义，函数记号
d²y/dx²	Second derivative: differentiate twice｜二阶导数：求导两次
f'(a)	Gradient at x = a｜x = a 处的梯度

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2. The Power Rule and Basic Notation | 幂法则与基本记号

For any real constant n, the derivative of xⁿ is n xⁿ⁻¹. This is the most frequently used rule in A Level differentiation and it handles positive, negative, and fractional powers once the expression has been rewritten in index form.

对任意实常数 n，xⁿ 的导数为 n xⁿ⁻¹。这是 A Level 微分中最常用的法则；只要先把式子改写成指数形式，它就能处理正指数、负指数和分数指数。

d/dx (xⁿ) = n xⁿ⁻¹

t is zero: if y = k, then dy/dx = 0. A constant multiple is preserved: d/dx [k f(x)] = k f' n be differentiated term by term, which is what makes polynomial differentiation so quick.

常数的导数为零：若 y = k，则 dy/dx = 0。常数倍可以保留：d/dx [k f(x)] = k f'(x)。和与差可以逐项求导，这正是多项式求导如此快捷的原因。

itten first. For example, y = 1/x² becomes y = x⁻², so dy/dx = −2x⁻³ = −2/x³. Likewise y = √x becomes y = x^½, so dy/dx = ½ x⁻½ = 1 ÷ (2√x).

负指数与分数指数必须先改写。例如 y = 1/x² 写成 y = x⁻²，则 dy/dx = −2x⁻³ = −2/x³。同理 y = √x 写成 y = x^½，则 dy/dx = ½ x⁻½ = 1 ÷ (2√x)。

A common exam trap is differentiating a root or a reciprocal without rewriting it as a power. If the power is not of the form xⁿ, the power rule does not apply.

常见考试陷阱：尚未改写成幂形式就对根式或倒数求导。若式子不是 xⁿ 的形式，幂法则并不适用。

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3. Differentiating Polynomials Term by Term | 多项式逐项求导

/dx = 12x² − 4x + 5. The constant term disappears because its gradient is zero everywhere.

逐项求导后相加。例如，若 y = 4x³ − 2x² + 5x − 7，则 dy/dx = 12x² − 4x + 5。常数项消失，因为它的梯度处处为零。

ds to y = 3x³ − 5x², so dy/dx = 9x² − 10x. Expanding is normally faster and less error-prone than the product rule in this situation.

若表达式以括号乘积形式给出，一般先展开，除非题目明显要求使用乘积法则。例如 y = x²(3x − 5) 展开为 y = 3x³ − 5x²，故 dy/dx = 9x² − 10x。这种情况下展开通常比乘积法则更快、更不易出错。

ituting the x-value. For y = 4x³ − 2x² + 5x − 7 at x = 1, the gradient is 12 − 4 + 5 = 13.

求出 dy/dx 后，代入具体的 x 值即可得到该点的梯度。对于 y = 4x³ − 2x² + 5x − 7，在 x = 1 处梯度为 12 − 4 + 5 = 13。

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4. Tangents and Normals | 切线与法线

A tangent to a curve at a point P(x₁, y₁) is a straight line whose gradient equals f'(x₁). Its equation is found with the standard line formula. The normal is the line perpendicular to the tangent at the same point, so its gradient is the negative reciprocal −1 ÷ f'(x₁).

曲线在点 P(x₁, y₁) 处的切线是一条直线，其斜率等于 f'(x₁)，用标准直线方程即可写出。法线是在同一点与切线垂直的直线，因此其斜率为切线的负倒数 −1 ÷ f'(x₁)。

Tangent 切线：y − y₁ = m (x − x₁), m = f'(x₁)
Normal 法线：y − y₁ = (−1 ÷ m)(x − x₁)

is (2, 0). Differentiating gives dy/dx = 3x² − 4, and at x = 2 the gradient is 12 − 4 = 8.

例题：求曲线 y = x³ − 4x 在 x = 2 处的切线与法线。x = 2 时 y = 8 − 8 = 0，故点为 (2, 0)。求导得 dy/dx = 3x² − 4，在 x = 2 处梯度为 12 − 4 = 8。

2). Always give the equation in the form requested, usually y = mx + c or ax + by + c = 0.

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切线为 y = 8(x − 2)，即 y = 8x − 16。法线斜率为 −1/8，故为 y = −(1/8)(x − 2)。最后要按题目要求的形式给出，通常是 y = mx + c 或 ax + by + c = 0。

Two marks are routinely lost here: forgetting to calculate the y-coordinate of the point, and using the derivative as the y-intercept instead of substituting the point into the line equation.

这里常丢两分：忘记计算该点的 y 坐标；以及把导数值当作截距，而没有把点的坐标代入直线方程。

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5. Finding Stationary Points | 求驻点

A stationary point is a point where the gradient is zero, so you solve f' fied as local maxima, local minima, or points of inflection. On a graph, they are where the curve is momentarily flat before turning or flattening further.

驻点是梯度为零的点，因此解方程 f'(x) = 0。驻点分为局部极大值点、局部极小值点和拐点。在图像上，它们是曲线短暂变平、随后转向或继续变平的位置。

rentiating gives dy/dx = 3x² − 6x = 3x(x − 2). Setting this to zero gives x = 0 and x = 2.

例题：求曲线 y = x³ − 3x² + 4 的驻点坐标。求导得 dy/dx = 3x² − 6x = 3x(x − 2)。令其为零，得 x = 0 与 x = 2。

s are (0, 4) and (2, 0). Always substitute back into the original equation, never into the derivative, to find the y-coordinate.

代回原式：x = 0 时 y = 4；x = 2 时 y = 8 − 12 + 4 = 0。驻点为 (0, 4) 与 (2, 0)。务必代回原方程（而不是导数式）求 y 坐标。

Some questions give a derivative that needs factorising by grouping, or a term such as eˣ or sin x that is never zero. If a factor cannot be zero, say so explicitly, since this often carries a mark.

有些题目的导数需要用分组分解，或含有 eˣ、sin x 这类永不为零的因子。若某因子不可能为零，要明确写出来，这通常占一分。

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6. Classifying Points with the Second Derivative | 用二阶导数判定驻点性质

The second derivative, written d²y/dx² or f''(x), is found by differentiating dy/dx again. It measures how the gradient itself is changing, which tells you the direction in which the curve bends at a stationary point.

二阶导数写作 d²y/dx² 或 f''(x)，通过对 dy/dx 再求导得到。它度量梯度本身的变化方式，从而说明曲线在驻点处的弯曲方向。

If d²y/dx² < 0 at a stationary point → local maximum
If d²y/dx² > 0 at a stationary point → local minimum
If d²y/dx² = 0 → test inconclusive, use a sign test

so (0, 4) is a local maximum. At x = 2, d²y/dx² = 6 > 0, so (2, 0) is a local minimum.

继续前面的例子：y = x³ − 3x² + 4，dy/dx = 3x² − 6x，d²y/dx² = 6x − 6。x = 0 时 d²y/dx² = −6 < 0，故 (0, 4) 为局部极大值点；x = 2 时 d²y/dx² = 6 > 0，故 (2, 0) 为局部极小值点。

/dx = 3x², which is positive on both sides of x = 0, so (0, 0) is a point of inflection, not a maximum or minimum.

当 d²y/dx² = 0 时，二阶导数判定失效，必须考察 dy/dx 在该点左、右两侧的符号。对于 y = x³，dy/dx = 3x² 在 x = 0 两侧均为正，故 (0, 0) 是拐点，而非极大值或极小值点。

d²y/dx²	Conclusion 结论
Negative 负	Local maximum｜局部极大值
Positive 正	Local minimum｜局部极小值
Zero 零	Inconclusive: use sign test｜不确定：用符号检验

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7. Increasing and Decreasing Functions | 函数的增减性

A function is increasing on an interval where f'(x) > 0 and decreasing where f'(x) < 0. The boundary between the two behaviours occurs exactly at stationary points, which is why these questions are solved by finding stationary points first and then testing the intervals between them.

函数在 f'(x) > 0 的区间上递增，在 f'(x) < 0 的区间上递减。两种行为的边界恰好出现在驻点处，因此这类题目先求驻点，再检验各区间。

− 6x. The function is decreasing when 3x² − 6x < 0, that is 3x(x − 2) < 0, which gives 0 < x < 2. It is increasing for x < 0 and for x > 2.

例：对于 y = x³ − 3x² + 4，dy/dx = 3x² − 6x。函数递减的条件是 3x² − 6x < 0，即 3x(x − 2) < 0，得 0 < x < 2。函数在 x < 0 与 x > 2 上递增。

Use strict inequalities when the question says "increasing" or "decreasing" and non-strict ones only if it says "non-decreasing". Write the answer as an interval, not as a list of values, because the question is asking about a whole range of x.

当题目说"递增"或"递减"时使用严格不等号；只有在说"非递减"时才使用非严格不等号。答案应写成区间，而不是若干数值，因为题目问的是一整段 x 的范围。

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8. The Chain Rule | 链式法则

x, then the rates multiply together. In function notation, if y = f(g(x)), then dy/dx = f'(g(x)) × g'(x).

链式法则用于求"函数的函数"的导数。若 y 是 u 的函数、u 是 x 的函数，则变化率相乘。用函数记号：若 y = f(g(x))，则 dy/dx = f'(g(x)) × g'(x)。

dy/dx = (dy/du) × (du/dx)

the index by one, and the inner derivative is 3, so dy/dx = 5(3x + 2)⁴ × 3 = 15(3x + 2)⁴.

例 1：y = (3x + 2)⁵。外层幂函数把指数 5 拉下来并把指数减一，内层导数为 3，故 dy/dx = 5(3x + 2)⁴ × 3 = 15(3x + 2)⁴。

Example 2: y = √x² + 1, rewritten as y = (x² + 1)^½. Then dy/dx = ½(x² + 1)⁻½ × 2x = x ÷ √x² + 1. Rewriting roots as fractional powers before differentiating avoids most algebraic slips.

例 2：y = √x² + 1，改写为 y = (x² + 1)^½。则 dy/dx = ½(x² + 1)⁻½ × 2x = x ÷ √(x² +

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