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Mastering Differentiation for Edexcel A-Level Maths | Edexcel A-Level 数学:掌握微分法

📚 Mastering Differentiation for Edexcel A-Level Maths | Edexcel A-Level 数学:掌握微分法

Differentiation is one of the most heavily examined topics in the Edexcel A-Level Mathematics specification. It appears in Pure Mathematics 1 and 2, and it underpins applications such as tangents, normals, stationary points, optimisation and connected rates of change.

微分是 Edexcel A-Level 数学大纲中考查最频繁的主题之一。它出现在 Pure Mathematics 1 和 2 中,并且是切线、法线、驻点、最优化以及相关变化率等应用的基础。


1. The Gradient Function and First Principles | 梯度函数与第一原理

The derivative f'(x) measures the gradient of a curve y = f(x) at any given point. It can be defined as a limit of the gradient of a chord as the chord gets infinitely short.

导数 f'(x) 测量曲线 y = f(x) 在任意给定点的梯度。它可定义为当割线无限变短时,割线梯度的极限。

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

This formula is called differentiation from first principles. In the exam you may be asked to use it for simple polynomials such as x² or x³.

这个公式称为从第一原理求导。考试中可能要求你对简单多项式(如 x² 或 x³)使用该公式。


2. The Power Rule and Basic Derivatives | 幂法则与基本导数

For any real power n, the power rule states that if y = xⁿ then dy/dx = n xⁿ⁻¹. To apply it, multiply by the power, then subtract one from the power.

对于任意实数次幂 n,幂法则指出:若 y = xⁿ,则 dy/dx = n xⁿ⁻¹。应用时先乘以幂指数,再将指数减一。

Common results are:

常见结果如下:

  • y = x⁵ ⇒ dy/dx = 5x⁴
  • y = 3x² ⇒ dy/dx = 6x
  • y = 7 ⇒ dy/dx = 0
  • y = 4x ⇒ dy/dx = 4

Remember that constants can be taken outside the derivative, and the derivative of a constant alone is zero.

记住,常数可以提到导数符号外,单独常数的导数为零。


3. From First Principles at a Point | 在一点处从第一原理求导

You can also find the gradient at a specific point x = a by evaluating f'(a) using the limit definition. It often requires expanding brackets and cancelling the h in the denominator.

你还可以使用极限定义计算特定点 x = a 处的梯度 f'(a)。这通常需要展开括号并约去分母中的 h。

For f(x) = x², f'(x) = lim (h → 0) [(x + h)² − x²] / h = lim (h → 0) [2xh + h²] / h = lim (h → 0) (2x + h) = 2x.

对于 f(x) = x²,f'(x) = lim (h → 0) [(x + h)² − x²] / h = lim (h → 0) [2xh + h²] / h = lim (h → 0) (2x + h) = 2x。


4. The Chain Rule | 链式法则

The chain rule is used when differentiating a function of a function. If y = [u(x)]ⁿ then dy/dx = n[u(x)]ⁿ⁻¹ × u'(x).

链式法则用于复合函数求导。若 y = [u(x)]ⁿ,则 dy/dx = n[u(x)]ⁿ⁻¹ × u'(x)。

A simple way to remember it is: differentiate the outer function, then multiply by the derivative of the inner function.

一个简单的记忆方法是:先对外层函数求导,再乘以内层函数的导数。

dy/dx = (dy/du) × (du/dx)

Example: y = (3x² + 1)⁵ gives dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

示例:y = (3x² + 1)⁵,则 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。


5. The Product and Quotient Rules | 乘积法则与商法则

For two functions u(x) and v(x), the product rule states:

对于两个函数 u(x) 和 v(x),乘积法则为:

y = uv ⇒ dy/dx = u’v + uv’

The quotient rule states:

商法则为:

y = u/v ⇒ dy/dx = (u’v − uv’) / v²

When using the quotient rule, keep the denominator squared and be careful with the negative sign in the numerator.

使用商法则时,分母要变成原分母的平方,并注意分子中的负号。

Example: y = x² sin x gives dy/dx = 2x sin x + x² cos x.

示例:y = x² sin x,则 dy/dx = 2x sin x + x² cos x。


6. Tangents and Normals | 切线与法线

The gradient of the tangent to y = f(x) at x = a is m = f'(a). The equation of the tangent is y − f(a) = m(x − a).

曲线 y = f(x) 在 x = a 处切线的梯度为 m = f'(a)。切线方程为 y − f(a) = m(x − a)。

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal −1/m. Use the same point (a, f(a)) to form the normal equation.

法线垂直于切线,因此其梯度为负倒数 −1/m。使用同一点 (a, f(a)) 建立法线方程。

Example: For y = x² at x = 1, f'(1) = 2. Tangent: y − 1 = 2(x − 1). Normal: y − 1 = −1/2 (x − 1).

示例:对于 y = x² 在 x = 1 处,f'(1) = 2。切线:y − 1 = 2(x − 1)。法线:y − 1 = −1/2 (x − 1)。


7. Stationary Points and the Second Derivative Test | 驻点与二阶导数判别法

Stationary points occur where dy/dx = 0. To classify them, you can use the second derivative d²y/dx² or a sign table for dy/dx.

驻点出现在 dy/dx = 0 的位置。判断驻点类型,可以使用二阶导数 d²y/dx²,也可以使用 dy/dx 的符号表。

  • d²y/dx² > 0 ⇒ local minimum.
  • d²y/dx² < 0 ⇒ local maximum.
  • d²y/dx² = 0 ⇒ test is inconclusive; use a sign table.

If the second derivative is positive, the curve is convex and the point is a minimum; if negative, the curve is concave and the point is a maximum.

如果二阶导数为正,曲线是凸的,该点为极小值;如果为负,曲线是凹的,该点为极大值。


8. Increasing and Decreasing Functions | 增函数与减函数

A function is increasing on an interval if f'(x) > 0 for all x in that interval. It is decreasing if f'(x) < 0.

若在某个区间内 f'(x) > 0,则函数在该区间上递增;若 f'(x) < 0,则函数递减。

To find intervals of increase or decrease, solve f'(x) > 0 or f'(x) < 0. Exam questions often combine this with quadratic inequalities.

要求递增或递减区间,可分别解不等式 f'(x) > 0 或 f'(x) < 0。考试题常将这一内容与二次不等式结合。

Example: f(x) = x³ − 3x gives f'(x) = 3x² − 3 = 3(x² − 1). Increasing when x < −1 or x > 1; decreasing when −1 < x < 1.

示例:f(x) = x³ − 3x,f'(x) = 3x² − 3 = 3(x² − 1)。当 x < −1 或 x > 1 时递增;当 −1 < x < 1 时递减。


9. Optimisation Problems | 最优化问题

Optimisation problems ask you to maximise or minimise a quantity such as area, volume or cost. You first write the quantity as a function of one variable, then set its derivative equal to zero.

最优化问题要求你最大化或最小化面积、体积或成本等量。首先将该量写成单变量函数,然后令其导数等于零。

Always verify that your stationary point is the required maximum or minimum, using the second derivative test or a sign test.

始终要验证所得驻点是否为所需的最大值或最小值,可使用二阶导数判别法或符号判别法。

Typical steps: define variables; form an equation linking them; eliminate one variable; differentiate; solve dy/dx = 0; justify the nature.

典型步骤为:定义变量;建立变量间关系;消去一个变量;求导;解 dy/dx = 0;说明驻点性质。


10. Connected Rates of Change | 相关变化率

If two quantities x and y both change with time t, their rates of change are connected by the chain rule: dy/dt = (dy/dx) × (dx/dt).

如果两个量 x 和 y 都随时间 t 变化,它们的变化率由链式法则联系:dy/dt = (dy/dx) × (dx/dt)。

In a typical problem, you are given one rate and asked to find another. Identify the missing derivative dy/dx, compute it, then multiply by the known rate dx/dt.

在典型问题中,给出一个变化率,要求另一个。找出缺少的导数 dy/dx,计算它,再乘以已知变化率 dx/dt。

Example: A spherical balloon expands so dr/dt = 0.5 cm s⁻¹. Since V = 4/3 π r³, dV/dt = 4π r² × dr/dt.

示例:一个球形气球膨胀,dr/dt = 0.5 cm s⁻¹。由于 V = 4/3 π r³,dV/dt = 4π r² × dr/dt。


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