📚 Mastering Differentiation for Edexcel A-Level Maths | 爱德思 A-Level 数学微分精讲
Differentiation is one of the highest-weight topics in Edexcel A-Level Mathematics. It appears in pure mathematics papers and underpins applications in kinematics, optimisation, rates of change and curve sketching. A confident command of derivative rules and their geometric meaning is essential for top grades.
微分是爱德思 A-Level 数学中权重最高的主题之一。它出现在纯数学试卷中，也是运动学、最优化、变化率和曲线草图等应用问题的基础。扎实掌握求导法则及其几何意义，是取得高分的关键。
1. The Derivative as Gradient | 导数作为斜率
The derivative dy/dx measures the instantaneous rate of change of y with respect to x. It is defined as the limit of the average gradient Δy/Δx as Δx approaches 0. On a graph, this limit gives the gradient of the tangent to the curve at a point.
导数 dy/dx 表示 y 关于 x 的瞬时变化率。它定义为当 Δx 趋向 0 时平均斜率 Δy/Δx 的极限。在图像上，这个极限给出了曲线在某一点处切线的斜率。
dy/dx = lim (Δx → 0) Δy/Δx
Geometrically, the derivative tells you how steep the curve is at an exact point, not over an interval. A positive derivative means an increasing function, while a negative derivative means a decreasing function.
从几何角度看，导数描述的是曲线在某一点处的陡峭程度，而不是一个区间上的平均变化。导数为正表示函数递增，导数为负表示函数递减。
- Average gradient between two points: Δy/Δx = [f(x + h) − f(x)] / h
- Instantaneous gradient: the limit of average gradient as h → 0
中文要点：两点间的平均斜率是 Δy/Δx，瞬时斜率是平均斜率当 h → 0 时的极限。
2. First Principles | 第一性原理
ust not simply apply the power rule unless the question asks for the derivative by first principles.
爱德思考试经常考查从第一性原理求导。正式定义使用微小增量 h，并要求先进行代数化简，再代入 h = 0。除非题目要求从第一性原理求导，否则不能直接使用幂法则。
f'(x) = lim (h → 0) [f(x + h) − f(x)] / h
². Then [f(x + h) − f(x)] / h = (2xh + h²) / h = 2x + h. Taking the limit as h → 0 gives f'(x) = 2x.
示例：对于 f(x) = x²，展开 f(x + h) = x² + 2xh + h²。然后 [f(x + h) − f(x)] / h = (2xh + h²) / h = 2x + h。令 h → 0，得到 f'(x) = 2x。
o early, producing 0/0. Always simplify the numerator first and cancel h before taking the limit.
常见错误是过早代入 h = 0，导致出现 0/0。一定要先化简分子并约去 h，再取极限。
3. Basic Rules and Power Rule | 基本法则与幂法则
xⁿ⁻¹. Constants differentiate to zero, and sums or differences differentiate term by term.
幂法则是求导的基础。如果 y = axⁿ，则 dy/dx = n axⁿ⁻¹。常数求导为 0，和或差可以逐项求导。
If y = xⁿ, then dy/dx = n xⁿ⁻¹
ates to dy/dx = 12x³ − 10x + 2. Notice the constant −7 disappears, and 2x becomes 2x⁰ = 2.
示例：y = 3x⁴ − 5x² + 2x − 7 求导得 dy/dx = 12x³ − 10x + 2。注意常数 −7 消失，2x 变成 2x⁰ = 2。
ers in index form. For example, y = 1/x² becomes y = x⁻², so dy/dx = −2x⁻³. Similarly, y = √x becomes y = x½, so dy/dx = ½ x⁻½.
求导前应先把负指数和分数指数写成指数形式。例如 y = 1/x² 写成 y = x⁻²，所以 dy/dx = −2x⁻³。类似地，y = √x 写成 y = x½，所以 dy/dx = ½ x⁻½。
4. Chain Rule | 链式法则
ntiate the outer function, keep the inner function unchanged, then multiply by the derivative of the inner function.
链式法则用于复合函数求导。如果 y = f(u) 且 u = g(x)，则 dy/dx = dy/du × du/dx。通俗地说：对外层函数求导，保持内层函数不变，然后乘以内层函数的导数。
dy/dx = dy/du × du/dx
= u⁵. Then dy/du = 5u⁴ and du/dx = 6x. Therefore dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴.
示例：y = (3x² + 2)⁵。设 u = 3x² + 2，则 y = u⁵。于是 dy/du = 5u⁴，du/dx = 6x。因此 dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴。
for trigonometric, exponential and logarithmic composites, such as y = sin(2x + 1), y = eˣ² or y = ln(5x − 3).
链式法则对三角函数、指数函数和对数函数的复合函数至关重要，例如 y = sin(2x + 1)、y = eˣ² 或 y = ln(5x − 3)。
5. Product and Quotient Rules | 乘积法则与商法则
hen dy/dx = u dv/dx + v du/dx. The key is to keep each factor and add the two cross-terms.
两个函数相乘时使用乘积法则。如果 y = uv，其中 u 和 v 都是 x 的函数，则 dy/dx = u dv/dx + v du/dx。关键是保留每个因子并加上两个交叉项。
If y = uv, then dy/dx = u dv/dx + v du/dx
t u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x, so dy/dx = x² cos x + 2x sin x.
示例：y = x² sin x。设 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x，所以 dy/dx = x² cos x + 2x sin x。
dv/dx] / v². The numerator order is important: v du/dx comes first, then subtract u dv/dx.
当一个函数除以另一个函数时使用商法则。如果 y = u/v，则 dy/dx = [v du/dx − u dv/dx] / v²。分子的顺序很重要：先写 v du/dx，再减去 u dv/dx。
If y = u/v, then dy/dx = [v du/dx − u dv/dx] / v²
e: y = (x + 1)/(x − 2). Then dy/dx = [(x − 2)(1) − (x + 1)(1)] / (x − 2)² = (−3)/(x − 2)².
示例：y = (x + 1)/(x − 2)。则 dy/dx = [(x − 2)(1) − (x + 1)(1)] / (x − 2)² = (−3)/(x − 2)²。
6. Derivatives of Exponential, Log and Trig Functions | 指数、对数和三角函数的导数
These standard results are assumed in Edexcel exams. You must memorise them accurately, as they appear constantly in chain, product and quotient rule problems.
这些标准结果在爱德思考试中默认掌握。你必须准确记忆，因为它们经常出现在链式法则、乘积法则和商法则的问题中。
| Function | Derivative |
|---|---|
| eˣ | eˣ |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec² x |
| aˣ | aˣ ln a |
| logₐ x | 1/(x ln a) |
on, apply the chain rule automatically. For example, d/dx [sin(3x)] = 3 cos(3x), d/dx [e⁻²ˣ] = −2e⁻²ˣ, and d/dx [ln(5x)] = 1/x.
当这些函数具有线性内层函数时，要自动应用链式法则。例如 d/dx [sin(3x)] = 3 cos(3x)，d/dx [e⁻²ˣ] = −2e⁻²ˣ，而 d/dx [ln(5x)] = 1/x。
Note that ln(5x) simplifies to ln 5 + ln x, so its derivative is 0 + 1/x. This shortcut is useful, but the chain rule gives the same result.
注意 ln(5x) 可以化简为 ln 5 + ln x，所以它的导数是 0 + 1/x。这个技巧很实用，但链式法则也会得到同样的结果。
7. Tangents and Normals | 切线与法线
To find a tangent at x = a, calculate the y-coordinate y₁ = f(a) and the gradient m = f' − a). The normal is perpendicular to the tangent, so its gradient is −1/m, provided m ≠ 0.
求 x = a 处的切线，先计算 y 坐标 y₁ = f(a) 和斜率 m = f'(a)。切线方程为 y − y₁ = m(x − a)。法线垂直于切线，因此它的斜率为 −1/m，前提是 m ≠ 0。
mnormal = −1 / m_tangent
Example: For y = x³ − 3x at x = 2, f' the tangent is y − 2 = 9(x − 2). The normal gradient is −1/9, giving y − 2 = −1/9 (x − 2).
示例：对于 y = x³ − 3x 在 x = 2 处，f'(x) = 3x² − 3，所以 m = 9。该点为 (2, 2)，因此切线为 y − 2 = 9(x − 2)。法线斜率为 −1/9，即 y − 2 = −1/9 (x − 2)。
- Vertical tangent: gradient is undefined, so normal is horizontal.
- Horizontal tangent: gradient is 0, so normal is vertical.
中文要点：垂直切线的斜率不存在，因此法线是水平的；水平切线的斜率为 0，因此法线是垂直的。
8. Stationary Points and Curve Sketching | 驻点与曲线草图
ionary points occur where dy/dx = 0. To classify them, you can use the second derivative f''(x). If f''(a) > 0, the curve has a local minimum at x = a. If f''(a) < 0, it has a local maximum. If f''(a) = 0, use a sign test on either side of a.
驻点出现在 dy/dx = 0 的位置。要判断驻点类型，可以使用二阶导数 f''(x)。如果 f''(a) > 0，曲线在 x = a 处有局部极小值；如果 f''(a) < 0，则有局部极大值；如果 f''(a) = 0，需要在 a 两侧做符号测试。
f''(x) > 0 → minimum, f''(x) < 0 → maximum
Example: y = x³ − 3x has f'(x) = 3x² − 3 = 0, so x = ±1. Since f''(x) = 6x, at x = 1 we have f''(1) = 6 > 0, a minimum; at x = −1, f''(−1) = −6 < 0, a maximum.
示例：y = x³ − 3x 的 f'(x) = 3x² − 3 = 0，所以 x = ±1。由于 f''(x) = 6x，在 x = 1 处 f''(1) = 6 > 0，为极小值；在 x = −1 处 f''(−1) = −6 < 0，为极大值。
When sketching curves, find intercepts, stationary points, asymptotes and end behaviour. Label all key coordinates and indicate whether stationary points are maxima, minima or points of inflection.
绘制曲线草图时，要找出截距、驻点、渐近线和端点趋势。标记所有关键坐标，并标出驻点是极大值、极小值还是拐点。
9. Parametric and Implicit Differentiation | 参数方程与隐函数求导
). This formula works because dy/dx = dy/dt × dt/dx, and dt/dx is the reciprocal of dx/dt.
对于参数方程 x = f(t) 和 y = g(t)，斜率通过 dy/dx = (dy/dt) / (dx/dt) 求得。这个公式成立是因为 dy/dx = dy/dt × dt/dx，而 dt/dx 是 dx/dt 的倒数。
dy/dx = (dy/dt) / (dx/dt)
Example: If x = t² and y = t³, then dx/dt = 2t and dy/dt = 3t², giving
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