Differentiation for Edexcel A-Level Pure Mathematics | Edexcel A-Level 纯数学：微分

📚 Differentiation for Edexcel A-Level Pure Mathematics | Edexcel A-Level 纯数学：微分

Differentiation is one of the most important tools in Edexcel A-Level Pure Mathematics. It allows you to find the gradient of a curve at any point, to analyse stationary points, and to solve real-world problems involving rates of change. This article covers the key rules, common applications, and exam techniques needed for the Edexcel specification.

微分是 Edexcel A-Level 纯数学中最重要的工具之一。它可以帮助你求出曲线上任意一点的斜率，分析驻点，并解决涉及变化率的实际问题。本文涵盖 Edexcel 考试大纲要求的关键法则、常见应用和考试技巧。

1. Average and Instantaneous Rate of Change | 平均变化率与瞬时变化率

The average rate of change of a function f(x) between x = a and x = b is given by the gradient of the chord joining the two points on the curve.

函数 f(x) 在 x = a 与 x = b 之间的平均变化率等于曲线上这两点之间弦的斜率。

Average rate of change = [f(b) – f(a)] / (b – a)

平均变化率 = [f(b) – f(a)] / (b – a)

As the interval between the two points becomes smaller, this chord gradient approaches the gradient of the tangent at a single point. This limiting value is called the instantaneous rate of change, or the derivative.

当两点之间的间隔越来越小时，这条弦的斜率会趋近于某一点处切线的斜率。这个极限值称为瞬时变化率，也就是导数。

2. Differentiation from First Principles | 第一原理求导

The derivative of f(x) can be defined formally from first principles. This is a required idea in Edexcel Pure Mathematics and is often tested as a short proof-style question.

f(x) 的导数可以从第一原理正式定义。这是 Edexcel 纯数学要求掌握的内容，经常以简短的证明题形式出现。

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

For example, if f(x) = x², expanding f(x + h) gives x² + 2xh + h². Subtracting f(x) and dividing by h gives 2x + h, which tends to 2x as h tends to 0.

例如，若 f(x) = x²，展开 f(x + h) 得到 x² + 2xh + h²。减去 f(x) 并除以 h 后得到 2x + h，当 h 趋向于 0 时其极限为 2x。

3. The Power Rule and Polynomials | 幂法则与多项式

The power rule is the most frequently used differentiation rule. For any real constant n, the derivative of xⁿ is n xⁿ⁻¹.

幂法则是微分中最常用的法则。对于任意实数常数 n，xⁿ 的导数为 n xⁿ⁻¹。

d/dx (xⁿ) = n xⁿ⁻¹

This rule can be applied term by term to any polynomial. Constants are carried along, and the derivative of a constant term is zero.

该法则可以逐项应用于任何多项式。常数可以保留为系数，常数项的导数为零。

d/dx (5x⁴ – 3x² + 2x – 7) = 20x³ – 6x + 2

Negative and fractional powers follow the same rule. For example, 1/x² can be written as x⁻², so its derivative is -2x⁻³.

负指数和分数指数也遵循同样的法则。例如，1/x² 可以写成 x⁻²，因此它的导数为 -2x⁻³。

4. Sum and Constant Multiple Rules | 和与常数倍法则

Differentiation is linear, which means you can differentiate a sum term by term and pull constant multipliers outside the derivative.

微分具有线性性质，这意味着你可以逐项求导，并将常数倍数提到导数符号之外。

d/dx [f(x) ± g(x)] = f'(x) ± g'(x)

d/dx [k f(x)] = k f'(x)

These two rules allow you to differentiate functions such as 6x⁵ + 4x³ – 9x without any further techniques. Simply differentiate each term separately and keep the coefficients.

这两个法则使你可以直接对 6x⁵ + 4x³ – 9x 这样的函数求导，只需分别对每一项求导并保留系数。

5. Product Rule | 乘积法则

When differentiating the product of two functions, the product rule must be used. If y = u v, where u and v are both functions of x, then the derivative is given by:

当对两个函数的乘积求导时，必须使用乘积法则。如果 y = u v，其中 u 和 v 都是 x 的函数，则导数为：

dy/dx = u’ v + u v’

For example, to differentiate y = x² sin x, let u = x² and v = sin x. Then u’ = 2x and v’ = cos x, so dy/dx = 2x sin x + x² cos x.

例如，对 y = x² sin x 求导时，设 u = x²，v = sin x。则 u’ = 2x，v’ = cos x，所以 dy/dx = 2x sin x + x² cos x。

In Edexcel questions, the product rule often appears with trigonometric, exponential, and logarithmic functions. Always state u, u’, v, and v’ clearly before applying the formula.

在 Edexcel 考试中，乘积法则常与三角函数、指数函数和对数函数结合出现。在套用公式前，务必先清楚地写出 u、u’、v 和 v’。

6. Quotient Rule | 商法则

The quotient rule is used to differentiate one function divided by another. If y = u / v, where u and v are functions of x, then:

商法则用于求两个函数相除的导数。如果 y = u / v，其中 u 和 v 都是 x 的函数，则：

dy/dx = (v u’ – u v’) / v²

The order in the numerator is critical: the denominator function v is multiplied by the derivative of the numerator u’ first. Reversing the order gives the wrong sign.

分子中的顺序非常关键：先由分母函数 v 乘以分子的导数 u’。如果顺序颠倒，就会得到错误的符号。

For example, if y = x³ / (x + 1), let u = x³ and v = x + 1. Then u’ = 3x² and v’ = 1, so dy/dx = [(x + 1)(3x²) – (x³)(1)] / (x + 1)².

例如，若 y = x³ / (x + 1)，设 u = x³，v = x + 1。则 u’ = 3x²，v’ = 1，所以 dy/dx = [(x + 1)(3x²) – (x³)(1)] / (x + 1)²。

7. Chain Rule | 链式法则

The chain rule is used to differentiate composite functions. If y is a function of u, and u is a function of x, then:

链式法则用于求复合函数的导数。如果 y 是 u 的函数，而 u 是 x 的函数，则：

dy/dx = dy/du × du/dx

In function notation, if y = f(g(x)), then dy/dx = f'(g(x)) g'(x). This means you differentiate the outer function and multiply by the derivative of the inner function.

用函数记号表示，如果 y = f(g(x))，则 dy/dx = f'(g(x)) g'(x)。这意味着先对外层函数求导，再乘以内层函数的导数。

For example, to differentiate y = (3x² + 1)⁵, let u = 3x² + 1. Then y = u⁵, dy/du = 5u⁴, and du/dx = 6x. Therefore dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.

例如，对 y = (3x² + 1)⁵ 求导时，设 u = 3x² + 1。则 y = u⁵，dy/du = 5u⁴，du/dx = 6x。因此 dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴。

8. Second Derivatives and Concavity | 二阶导数与凹凸性

The second derivative, written f”(x) or d²y/dx², is obtained by differentiating the first derivative. It measures the rate of change of the gradient and describes the curvature of a graph.

二阶导数写作 f”(x) 或 d²y/dx²，由一阶导数再次求导得到。它衡量斜率的变化率，并描述函数图像的弯曲情况。

d²y/dx² = d/dx (dy/dx)

If f”(x) > 0 on an interval, the curve is concave up and the gradient is increasing. If f”(x) < 0, the curve is concave down and the gradient is decreasing.

如果在某个区间内 f”(x) > 0，则曲线向上凹，且斜率在增大。如果 f”(x) < 0，则曲线向下凹，且斜率在减小。

The second derivative is also used to determine the nature of stationary points. A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum.

二阶导数还用于判断驻点的性质。二阶导数为正表示局部极小值，二阶导数为负表示局部极大值。

9. Stationary Points and Optimisation | 驻点与最优化

A stationary point occurs where the first derivative is zero, that is dy/dx = 0. At such a point, the tangent to the curve is horizontal.

当一阶导数为零，即 dy/dx = 0 时，会出现驻点。在驻点处，曲线的切线是水平的。

Solve dy/dx = 0 to find stationary points

解 dy/dx = 0 来寻找驻点

To classify a stationary point, substitute the x-value into the second derivative. If f”(x) > 0, it is a minimum point; if f”(x) < 0, it is a maximum point. If f''(x) = 0, you may need to inspect the sign of dy/dx on either side.

要对驻点进行分类，将 x 值代入二阶导数。如果 f”(x) > 0，则为极小值点；如果 f”(x) < 0，则为极大值点。如果 f''(x) = 0，则需要检查两侧 dy/dx 的符号。

Optimisation problems ask you to maximise or minimise a quantity such as area, volume, profit, or length. Write the quantity as a function of one variable, differentiate it, and solve for the stationary point.

最优化问题要求你对面积、体积、利润或长度等量求最大值或最小值。将该量写成单一变量的函数，对其进行求导，并解出驻点。

10. Practical Rates of Change | 实际变化率问题

Connected rates of change problems use the chain rule to link two different rates. A typical example involves a spherical balloon being inflated, where the radius and volume both change with time.

相关变化率问题利用链式法则将两个不同的变化率联系起来。典型例子涉及充气中的球体，其半径和体积都随时间变化。

dV/dt = dV/dr × dr/dt

For a sphere, V = 4/3 π r³, so dV/dr = 4 π r². If you know dr/dt at a particular radius, you can find dV/dt immediately.

对于球体，V = 4/3 π r³，因此 dV/dr = 4 π r²。如果已知某一半径下的 dr/dt，你就可以立即求出 dV/dt。

Edexcel questions often give one rate and ask for another. Identify the known rate, the required rate, and the derivative that connects them, then apply the chain rule carefully.

Edexcel 题目通常会给出一个速率并要求另一个速率。找出已知速率、待求速率以及连接它们的导数，然后仔细应用链式法则。

11. Gradients, Tangents, and Normals | 斜率、切线与法线

Differentiation can be used to find the equation of the tangent or the normal to a curve at a given point. First find dy/dx, substitute the x-coordinate, and use the straight-line formula y – y₁ = m(x – x₁).

微分可用于求曲线在给定点处切线或法线的方程。首先求出 dy/dx，代入 x 坐标，然后使用直线方程公式 y – y₁ = m(x – x₁)。

Tangent gradient: m = dy/dx at the point

切线斜率：m = 该点处的 dy/dx

The normal is perpendicular to the tangent. If the tangent gradient is m, the normal gradient is -1/m, provided m is not zero.

法线垂直于切线。如果切线斜率为 m，则法线斜率为 -1/m，前提是 m 不为零。

For example, to find the tangent to y = x³ – 2x at x = 1, first compute dy/dx = 3x² – 2. At x = 1, m = 3(1)² – 2 = 1. The point is (1, -1), so the tangent is y + 1 = 1(x – 1), or y = x – 2.

例如，求 y = x³ – 2x 在 x = 1 处的切线，先计算 dy/dx = 3x² – 2。当 x = 1 时，m = 3(1)² – 2 = 1。点为 (1, -1)，因此切线为 y + 1 = 1(x – 1)，即 y = x – 2。

12. Common Exam Mistakes and How to Avoid Them | 常见考试错误及避免方法

One common mistake is forgetting to rewrite roots and reciprocals as powers before differentiating. Expressions such as √x and 1/x should be written as x¹⁄² and x⁻¹ first.

一个常见错误是在求导前忘记将根式和倒数改写为幂的形式。像 √x 和 1/x 这样的表达式应首先写成 x¹⁄² 和 x⁻¹。

Another frequent error is misapplying the quotient rule by putting u’ v instead of v u’ in the numerator. Always check that the denominator function comes first in the numerator.

另一个常见错误是在商法则的分子中把 u’ v 写在了 v u’ 的前面。务必检查分母函数是否位于分子的首位。

Students also forget to multiply by the inner derivative when using the chain rule. In expressions such as (2x + 3)⁶, the derivative must include the factor 2 from the inner function.

学生也经常在使用链式法则时忘记乘以内层函数的导数。在 (2x + 3)⁶ 这类表达式中，导数必须包含来自内层函数的因子 2。

Finally, always simplify your answer where possible and present it clearly. Marks are often awarded for a correct derivative even if the final simplification is incomplete, but an incorrect sign can lose accuracy marks.

最后，尽可能简化答案并清晰书写。即使最终化简不完整，只要导数正确也常能得分，但符号错误会导致失去准确性分数。

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Differentiation for Edexcel A-Level Pure Mathematics | Edexcel A-Level 纯数学：微分