📚 GCSE WJEC Mathematics: Differentiation Revision | GCSE WJEC 数学：微分 考点精讲
Differentiation is a cornerstone of GCSE Higher Tier Mathematics, enabling us to analyse how functions change. In WJEC exams, you will be expected to differentiate polynomial functions, find gradients of curves, determine equations of tangents, and apply differentiation to solve real-world problems involving maximum and minimum values. This revision guide covers every key concept and technique you need, with clear worked examples, and exam-focused tips and the common errors that cost marks.
微分是 GCSE 高等级数学的基石，帮助我们分析函数的变化。在 WJEC 考试中，你需要掌握多项式函数的求导、求曲线梯度、确定切线方程，并应用微分解决涉及最大值和最小值的实际问题。本复习指南涵盖所有关键概念和技巧，配有清晰的示例、考试要点提示，以及最容易失分的常见错误。
1. What is Differentiation? | 什么是微分？
ith respect to another. Geometrically, the derivative at a particular point equals the gradient of the tangent to the curve at that point. For a function y = f(x), the derivative is denoted dy/dx or f'(x).
微分是求函数导数的一个过程，导数衡量的是瞬时变化率。从几何角度看，某一点的导数等于该点处曲线切线的斜率。对于函数 y = f(x)，导数记作 dy/dx 或 f'(x)。
Unlike straight-line graphs which have constant gradient, curves have varying steepness. Differentiation allows us to find the gradient at any specific point on a curve without drawing a tangent by hand. This is why it is far more powerful than simply reading a gradient off a graph.
直线图具有恒定斜率，而曲线的陡峭程度各处不同。利用微分，我们无需手动画切线即可求出曲线上任意指定点的梯度。这正是微分远强于"看图估斜率"的原因。
The notation matters in WJEC papers. The symbol dy/dx means "the derivative of y with respect to x", while f'(x) is used when the function is written in function notation. A third notation, y', appears occasionally. All three mean exactly the same thing.
在 WJEC 试卷中，符号的写法很重要。dy/dx 表示"y 关于 x 的导数"；当函数写成 f(x) 形式时，用 f'(x)；偶尔也会看到 y'。这三种写法含义完全相同。
At Higher Tier you should also understand the informal idea behind it: differentiation finds the gradient of a chord joining two nearby points and then asks what happens as the two points get infinitely close together. That limiting idea is the bridge from average rate of change to instantaneous rate of change.
在高等级考试中，你还应理解其背后的直观思想：微分先求连接相邻两点的弦的斜率，再考察当两点无限接近时斜率的变化趋势。这个极限思想正是从"平均变化率"通向"瞬时变化率"的桥梁。
2. The Power Rule: Differentiating xⁿ | 幂法则：对 xⁿ 求导
en the derivative is found by multiplying by the power and then reducing the power by one.
最基本也最重要的法则是幂法则：若 y = xⁿ，求导时先将指数乘下来作为系数，然后将指数减 1。
y = xⁿ → dy/dx = n xⁿ⁻¹
erentiating y = x⁵ gives dy/dx = 5x⁴. Differentiating y = x⁻² gives dy/dx = -2x⁻³. For y = √x = x¹ᐟ², differentiating gives dy/dx = (1/2)x⁻¹ᐟ², which can be written as 1/(2√x).
例如，对 y = x⁵ 求导得 dy/dx = 5x⁴；对 y = x⁻² 求导得 dy/dx = -2x⁻³；对于 y = √x = x¹ᐟ²，求导得 dy/dx = (1/2)x⁻¹ᐟ²，也可写成 1/(2√x)。
ten as 5x⁰, and differentiating gives 0. This aligns with the rule that the derivative of any constant is zero.
要记住常数可以写作指数为 0 的形式：y = 5 可写成 5x⁰，求导后得 0。这与"常数的导数恒为零"的规则一致。
n dy/dx = a n xⁿ⁻¹. So y = 7x³ differentiates to 21x², and y = -4x² differentiates to -8x.
同时要记住系数保持不变：若 y = axⁿ，则 dy/dx = a n xⁿ⁻¹。因此 y = 7x³ 求导得 21x²，y = -4x² 求导得 -8x。
| Function f(x) | Derivative f'(x) | Reason |
|---|---|---|
| x³ | 3x² | 3 × x3−1 |
| x⁵ | 5x⁴ | 5 × x5−1 |
| x⁻² | -2x⁻³ | -2 × x−2−1 |
| √x = x¹ᐟ² | (1/2)x⁻¹ᐟ² = 1/(2√x) | (1/2) × x1/2−1 |
| 7x³ | 21x² | coefficient 7 is kept |
| 5 | 0 | constant term, 5x⁰ |
3. Differentiating Polynomials Term by Term | 逐项求导多项式
difference is the sum or difference of the derivatives. If y = f(x) ± g(x), then dy/dx = f'(x) ± g'(x).
多项式可以逐项求导。和或差的导数等于各导数之和或差：若 y = f(x) ± g(x)，则 dy/dx = f'(x) ± g'(x)。
2. Differentiating each term in turn gives 12x³, −10x, 7 and 0, so dy/dx = 12x³ − 10x + 7.
例题：对 y = 3x⁴ − 5x² + 7x − 2 求导。逐项求导分别得 12x³、−10x、7 和 0，因此 dy/dx = 12x³ − 10x + 7。
Notice how a linear term such as 7x becomes the constant 7, and a constant term such as −2 disappears completely. WJEC examiners frequently see candidates incorrectly keeping a constant in the derivative or differentiating a linear term to x.
注意线性项 7x 求导后变成常数 7，而常数项 −2 则完全消失。WJEC 阅卷中常见错误包括：导数中保留了常数项，或把线性项错误地写成 x。
e derivative in descending powers of x for clarity, and simplify coefficients as far as possible.
再看一例：y = 6x³ + 2x² − x + 9，得 dy/dx = 18x² + 4x − 1。建议按 x 的降幂书写导数以便检查，并把系数化到最简。
4. Negative and Fractional Indices | 负指数与分数指数
Roots and reciprocals must both be converted into index form first: 1/x² = x⁻², 1/x = x⁻¹, √x = x¹ᐟ², ∛x = x¹ᐟ³, and 1/√x = x⁻¹ᐟ².
求导之前，必须先把每一项改写成 axⁿ 的形式。根式与倒数都要先化为指数形式：1/x² = x⁻²，1/x = x⁻¹，√x = x¹ᐟ²，∛x = x¹ᐟ³，1/√x = x⁻¹ᐟ²。
Worked example: differentiate y = 4/x² + 3√x. Rewriting gives y = 4x⁻² + 3x¹ᐟ². Then dy/dx = −8x⁻³ + (3/2)x⁻¹ᐟ², which may also be written as −8/x³ + 3/(2√x).
例题：对 y = 4/x² + 3√x 求导。改写为 y = 4x⁻² + 3x¹ᐟ²，则 dy/dx = −8x⁻³ + (3/2)x⁻¹ᐟ²，也可写成 −8/x³ + 3/(2√x)。
Worked example: differentiate y = 5x∛x. First note that x × x¹ᐟ³ = x⁴ᐟ³, so y = 5x⁴ᐟ³ and dy/dx = (20/3)x¹ᐟ³.
例题：对 y = 5x∛x 求导。先注意 x × x¹ᐟ³ = x⁴ᐟ³，故 y = 5x⁴ᐟ³，dy/dx = (20/3)x¹ᐟ³。
Some WJEC questions accept a final answer given in either index form or fraction/root form, but the intermediate step must show index form clearly, because that is where the method marks lie.
WJEC 部分题目接受指数形式或分数、根式形式的最终答案，但中间步骤必须清楚写出指数形式，因为过程分就在这里。
5. Finding the Gradient at a Given Point | 求指定点的梯度
To find the gradient of a curve at a particular point, differentiate, then substitute the x-coordinate of the point into the derivative.
求曲线在某一点的梯度，方法是先求导，再把该点的 x 坐标代入导数表达式。
4. Substituting x = 2 gives 3(4) − 4 = 8. So the gradient of the curve at that point is 8.
例题：求 y = x³ − 4x + 1 在 x = 2 处的梯度。求导得 dy/dx = 3x² − 4，代入 x = 2 得 3 × 4 − 4 = 8，因此曲线在该点的梯度为 8。
The reverse question is also common: find the coordinates of the points on a curve where the gradient takes a given value. Here you set dy/dx equal to that value and solve.
反向题型也很常见：求曲线上梯度等于某一给定值的点的坐标。此时令 dy/dx 等于该值并解方程即可。
−2. Remember that a square root usually produces both a positive and a negative solution.
例题：求 y = x³ − 4x + 1 上梯度为 8 的点的 x 坐标。解 3x² − 4 = 8，得 3x² = 12，x² = 4，因此 x = 2 或 x = −2。切记开平方通常会产生正负两个解。
infinity corresponds to a vertical tangent. WJEC questions sometimes describe a tangent as "parallel to the x-axis", which is simply another way of saying dy/dx = 0.
梯度为零表示切线水平；梯度不存在则对应竖直切线。WJEC 题目有时说切线"平行于 x 轴"，这不过是 dy/dx = 0 的另一种说法。
6. Equations of Tangents and Normals | 切线方程与法线方程
Once you know the gradient m at a point (x₁, y₁), the tangent is the straight line through that point with that gradient. Using the straight-line formula gives the equation of the tangent.
求出一条曲线在某点 (x₁, y₁) 处的梯度 m 后，切线就是过该点、斜率为 m 的直线。套用直线方程公式即可得到切线方程。
y − y₁ = m(x − x₁)
erentiating gives dy/dx = 2x + 3, so at x = 1 the gradient is m = 5. The tangent is therefore y − 4 = 5(x − 1), which simplifies to y = 5x − 1.
例题：求 y = x² + 3x 在 x = 1 处的切线方程。先求 y 坐标：y = 1 + 3 = 4，点为 (1, 4)。求导得 dy/dx = 2x + 3，x = 1 时 m = 5。故切线为 y − 4 = 5(x − 1)，化简得 y = 5x − 1。
The normal is the line perpendicular to the tangent at the same point. Its gradient is the negative reciprocal of the tangent gradient, so if the tangent has gradient m, the normal has gradient −1/m. Here that would be −1/5.
法线是在同一点与切线垂直的直线。其斜率是切线斜率的负倒数：若切线斜率为 m，则法线斜率为 −1/m，本例中为 −1/5。
equation. If (1, 4) satisfies y = 5x − 1, then 4 = 5 − 1 is true, so the line is correct.
检查答案既快又值得做：把已知点代回直线方程。若 (1, 4) 满足 y = 5x − 1，即 4 = 5 − 1 成立，则直线正确。
7. Stationary Points: Maximum and Minimum | 驻点：极大值与极小值
ent is horizontal. Stationary points may be local maxima (the curve is at a peak) or local minima (the curve is in a trough).
驻点是曲线上梯度为零的点，即 dy/dx = 0，此处切线水平。驻点可以是局部极大值（曲线的峰）或局部极小值（曲线的谷）。
zero and factorising: 3(x² − 4x + 3) = 0, so 3(x − 1)(x − 3) = 0, giving x = 1 and x = 3.
例题：求 y = x³ − 6x² + 9x + 1 的驻点。求导得 dy/dx = 3x² − 12x + 9。令其为零并因式分解：3(x² − 4x + 3) = 0，即 3(x − 1)(x − 3) = 0，得 x = 1 和 x = 3。
= 5, giving the point (1, 5). At x = 3: y = 27 − 54 + 27 + 1 = 1, giving the point (3, 1).
再把 x 值代回原函数求 y 坐标。x = 1 时：y = 1 − 6 + 9 + 1 = 5，得点 (1, 5)。x = 3 时：y = 27 − 54 + 27 + 1 = 1，得点 (3, 1)。
A very common error is to substitute the x-values only into the derivative rather than back into the original equation, which produces meaningless coordinates. Always return to the original function for the y-coordinate.
极常见错误是把 x 值代回导数而不是原函数，从而得到毫无意义的坐标。求 y 坐标时务必回到原函数。
8. The Second Derivative and the Nature of a Point | 二阶导数与驻点性质
The second derivative is found by differentiating the first derivative. It is written d²y/dx² or f''(x). It tells us how the gradient itself is changing, and it is the standard tool in WJEC papers for determining whether a stationary point is a maximum or a minimum.
二阶导数是对一阶导数再求一次导得到的，记作 d²y/dx² 或 f''(x)。它描述梯度本身如何变化，是 WJEC 试卷中判断驻点为极大值还是极小值的标准工具。
If d²y/dx² < 0 → the point is a maximum
If d²y/dx² > 0 → the point is a minimum
so (1, 5) is a maximum. At x = 3, d²y/dx² = 6, which is positive, so (3, 1) is a minimum.
继续前面的例题：y = x³ − 6x² + 9x + 1，dy/dx = 3x² − 12x + 9，因此 d²y/dx² = 6x − 12。x = 1 时 d²y/dx² = −6 为负，故 (1, 5) 是极大值；x = 3 时 d²y/dx² = 6 为正，故 (3, 1) 是极小值。
If the second derivative equals zero, the test is inconclusive; the point could be a maximum, a minimum or a point of inflection. In that case use the gradient test: examine the sign of dy/dx just to the left and just to the right of the stationary point. If the gradient changes from positive to negative, it is a maximum; from negative to positive, a minimum.
若二阶导数为零，则无法判定：该点可能是极大值、极小值或拐点。这时要改用梯度检验法：考察驻点左侧与右侧略偏处 dy/dx 的正负。若梯度由正变负，为极大值；由负变正，为极小值。
You can also use the shape of the curve as a sanity check. A positive cubic has a maximum first (on the left) and a minimum second (on the right), which matches exactly what we found above.
也可借助曲线形状自检。正系数三次曲线的左侧是极大值、右侧是极小值，与上面的结果完全吻合。
9. Optimisation: Real-World Maximum and Minimum Problems | 最优化：实际最大值与最小值问题
Optimisation problems ask you to maximise or minimise a quantity such as area, volume, cost or profit. The method is always the same: form an expression in one variable, differentiate, set the derivative to zero, solve, then justify the nature of the answer with the second derivative.
最优化问题要求你求面积、体积、成本或利润等量的最大值或最小值。方法固定：先建立单变量表达式，再求导、令导数为零、求解，最后用二阶导数说明结果的类型。
imises S. Since the volume is x²h = 32, the height is h = 32/x². The surface area is the base plus four sides: S = x² + 4xh = x² + 4x(32/x²) = x² + 128/x.
例题：一个无盖盒子底面为边长 x cm 的正方形，体积为 32 cm³。证明表面积 S = x² + 128/x，并求使 S 最小的 x 值。由体积 x²h = 32 得高 h = 32/x²。表面积为底面积加四个侧面：S = x² + 4xh = x² + 4x(32/x²) = x² + 128/x。
Rewriting as S = x² +
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