Differentiation for Edexcel A-Level Maths: Rules and Applications | Edexcel A-Level 数学：微分法则与应用

📚 Differentiation for Edexcel A-Level Maths: Rules and Applications | Edexcel A-Level 数学：微分法则与应用

Differentiation is one of the most important topics in Edexcel A-Level Mathematics, appearing in Pure Mathematics, Mechanics and many applied problem-solving contexts. This article covers the key rules, techniques and exam-style applications you need to master, from first principles to connected rates of change.

微分是 Edexcel A-Level 数学中最重要的主题之一，在纯数学、力学以及许多应用题中都会出现。本文涵盖你需要掌握的核心法则、技巧与考试型应用，从第一性原理到相关变化率。

1. The Gradient of a Curve | 曲线的梯度

A straight line has a constant gradient, but the gradient of a curve changes from point to point. The gradient at a point is the gradient of the tangent to the curve at that point, and it can be approximated by the gradient of a chord joining two nearby points.

直线的梯度是恒定的，但曲线的梯度随点而变化。曲线上某一点处的梯度就是该点处切线的梯度，它可以用连接两个邻近点的弦的梯度来近似。

If a curve has equation y = f(x), the average gradient between x = a and x = b is given by Δy / Δx = [f(b) – f(a)] / (b – a). As the two points get closer together, this average gradient approaches the instantaneous gradient.

如果曲线方程为 y = f(x)，在 x = a 与 x = b 之间的平均梯度为 Δy / Δx = [f(b) – f(a)] / (b – a)。当两点越来越接近时，这个平均梯度就会趋近于瞬时梯度。

2. Differentiation from First Principles | 第一性原理求导

Differentiation from first principles uses a limit to define the derivative exactly. For a function f(x), the derivative f'(x) is defined as:

第一性原理求导使用极限来精确定义导数。对于函数 f(x)，导数 f'(x) 定义为：

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

This expression represents the limit of the chord gradient as the interval h tends to zero. Edexcel often asks you to prove the derivative of x² or x³ from first principles.

该表达式表示当区间 h 趋向于 0 时弦梯度的极限。Edexcel 经常要求你从第一性原理证明 x² 或 x³ 的导数。

For f(x) = x², expand f(x + h) = x² + 2xh + h². The difference is 2xh + h², so dividing by h gives 2x + h. Letting h → 0 gives f'(x) = 2x.

对于 f(x) = x²，展开得 f(x + h) = x² + 2xh + h²。差为 2xh + h²，除以 h 得到 2x + h。令 h → 0 得到 f'(x) = 2x。

3. The Power Rule | 幂函数求导法则

The power rule is the fastest way to differentiate expressions of the form xⁿ. If y = xⁿ, then the derivative is dy/dx = n xⁿ⁻¹. This works for any real constant n, including negative and fractional powers.

幂函数求导法则是求形如 xⁿ 的表达式导数的最快方法。如果 y = xⁿ，那么导数为 dy/dx = n xⁿ⁻¹。这适用于任何实数 n，包括负指数和分数指数。

For example, if y = x⁵ then dy/dx = 5x⁴. If y = √x = x^(1/2), then dy/dx = ½ x^(−1/2). If y = 1/x² = x^(−2), then dy/dx = −2x^(−3).

例如，若 y = x⁵，则 dy/dx = 5x⁴。若 y = √x = x^(1/2)，则 dy/dx = ½ x^(−1/2)。若 y = 1/x² = x^(−2)，则 dy/dx = −2x^(−3)。

Always rewrite roots and reciprocal terms as powers of x before differentiating. This reduces careless mistakes in exams.

在求导之前，始终将根式和倒数项改写为 x 的幂。这可以减少考试中的粗心错误。

4. Constant Multiple, Sum and Difference Rules | 常数倍数、和与差法则

Differentiation is linear, which means you can differentiate term by term. If k is a constant, then d/dx [k f(x)] = k f'(x). For two functions f(x) and g(x), the derivative of their sum or difference is d/dx [f(x) ± g(x)] = f'(x) ± g'(x).

求导是线性的，这意味着你可以逐项求导。如果 k 是常数，则 d/dx [k f(x)] = k f'(x)。对于两个函数 f(x) 和 g(x)，它们的和或差的导数为 d/dx [f(x) ± g(x)] = f'(x) ± g'(x)。

For example, y = 3x⁴ – 5x² + 2x – 7 differentiates term by term to give dy/dx = 12x³ – 10x + 2. The constant term -7 differentiates to 0 because it has zero gradient.

例如，y = 3x⁴ – 5x² + 2x – 7 逐项求导得到 dy/dx = 12x³ – 10x + 2。常数项 -7 求导为 0，因为它的梯度为零。

Always remember that the derivative of a constant is zero. Many Edexcel questions hide this simple rule inside longer polynomial expressions.

始终记住常数的导数为 0。许多 Edexcel 题目会在较长的多项式表达式中隐藏这条简单法则。

5. Tangents and Normals | 切线与法线

To find the equation of a tangent to a curve at a point, first substitute the x-coordinate into dy/dx to find the gradient m. Then use y – y₁ = m(x – x₁) with the point (x₁, y₁).

要求曲线上某一点处的切线方程，首先将 x 坐标代入 dy/dx 求出梯度 m。然后在该点 (x₁, y₁) 处使用 y – y₁ = m(x – x₁)。

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of m: m_normal = -1/m. If the tangent gradient is 2, the normal gradient is -1/2.

法线垂直于切线，因此它的梯度是 m 的负倒数：m_normal = -1/m。如果切线梯度为 2，法线梯度就是 -1/2。

Example: for y = x³ – 3x at x = 1, the gradient is dy/dx = 3x² – 3 = 0, so the tangent is horizontal. The tangent equation is y = -2, and the normal is the vertical line x = 1.

示例：对于 y = x³ – 3x 在 x = 1 处，梯度为 dy/dx = 3x² – 3 = 0，所以切线是水平的。切线方程为 y = -2，法线是竖直线 x = 1。

6. Stationary Points and Local Extrema | 驻点与局部极值

Stationary points occur where dy/dx = 0. At these points the tangent is horizontal, and the curve may have a local maximum, a local minimum or a point of inflection.

驻点出现在 dy/dx = 0 的位置。在这些点处切线是水平的，曲线可能出现局部最大值、局部最小值或拐点。

To classify a stationary point, you can use the second derivative d²y/dx² or examine the sign change of dy/dx on either side of the point. A sign change from positive to negative indicates a maximum, while negative to positive indicates a minimum.

要对驻点进行分类，可以使用二阶导数 d²y/dx²，也可以考察该点两侧 dy/dx 的符号变化。符号由正变负表示极大值，由负变正表示极小值。

dy/dx sign change	Type of stationary point
+ → 0 → −	Local maximum
− → 0 → +	Local minimum
+ → 0 → + or − → 0 → −	Point of inflection

In Edexcel exams, you must show a clear method for classifying stationary points, not just state the answer.

在 Edexcel 考试中，你必须展示分类驻点的清晰方法，而不能只写出答案。

7. The Second Derivative | 二阶导数

The second derivative d²y/dx² tells you about the concavity of a curve. If d²y/dx² > 0 at a stationary point, the curve is convex and the point is a local minimum. If d²y/dx² < 0, the curve is concave and the point is a local maximum.

二阶导数 d²y/dx² 告诉你曲线的凹凸性。如果在驻点处 d²y/dx² > 0，则曲线是凸的，该点处为局部最小值。如果 d²y/dx² < 0，则曲线是凹的，该点处为局部最大值。

If d²y/dx² = 0, the second derivative test is inconclusive. In that case you should use the sign change method for dy/dx to classify the stationary point correctly.

如果 d²y/dx² = 0，二阶导数检验无法得出结论。在这种情况下，你应该使用 dy/dx 的符号变化法来正确分类驻点。

For y = x³ – 3x, dy/dx = 3x² – 3 and d²y/dx² = 6x. At x = 1, d²y/dx² = 6 > 0, so the point is a local minimum; at x = -1, d²y/dx² = -6 < 0, so the point is a local maximum.

对于 y = x³ – 3x，dy/dx = 3x² – 3，d²y/dx² = 6x。在 x = 1 处，d²y/dx² = 6 > 0，所以该点是局部最小值；在 x = -1 处，d²y/dx² = -6 < 0，所以该点是局部最大值。

8. Increasing and Decreasing Functions | 递增与递减函数

A function f(x) is increasing on an interval if f'(x) ≥ 0 for all x in that interval, and decreasing if f'(x) ≤ 0. To find intervals of increase or decrease, solve the inequality f'(x) > 0 or f'(x) < 0.

如果在一个区间内 f'(x) ≥ 0，则函数 f(x) 在该区间内递增；如果 f'(x) ≤ 0，则递减。要求递增或递减区间，需要解不等式 f'(x) > 0 或 f'(x) < 0。

Example: for f(x) = x³ – 3x, f'(x) = 3x² – 3. Solving f'(x) > 0 gives x < -1 or x > 1, so f is increasing on (-∞, -1) and (1, ∞). It is decreasing on (-1, 1).

示例：对于 f(x) = x³ – 3x，f'(x) = 3x² – 3。解 f'(x) > 0 得 x < -1 或 x > 1，因此 f 在 (-∞, -1) 和 (1, ∞) 上递增，在 (-1, 1) 上递减。

Be careful to use strict inequalities when describing intervals unless the question accepts endpoints. Edexcel mark schemes often accept either closed or open intervals for monotonicity, but consistency matters.

描述区间时要小心使用严格不等式，除非题目接受端点。Edexcel 评分标准通常允许单调区间的开闭写法，但保持一致很重要。

9. Sketching Gradient Functions | 绘制导函数图像

If you are given the graph of y = f(x), you can sketch the graph of y = f'(x) by reading off the gradient behaviour. Where f(x) has a stationary point, f'(x) crosses the x-axis.

如果给出 y = f(x) 的图像，你可以通过读取梯度行为来绘制 y = f'(x) 的图像。在 f(x) 有驻点的地方，f'(x) 穿过 x 轴。

Where f(x) is increasing, f'(x) is positive, so its graph lies above the x-axis. Where f(x) is decreasing, f'(x) is negative, so its graph lies below the x-axis.

在 f(x) 递增的地方，f'(x) 为正，因此其图像位于 x 轴上方。在 f(x) 递减的地方，f'(x) 为负，因此其图像位于 x 轴下方。

The steepness of f(x) corresponds to the y-value of f'(x). A steeper positive gradient gives a higher positive value for f'(x), and a steeper negative gradient gives a lower negative value.

f(x) 的陡峭程度对应 f'(x) 的 y 值。更陡的正梯度给出更高的正 f'(x) 值，更陡的负梯度给出更低的负 f'(x) 值。

10. Connected Rates of Change | 相关变化率

Connected rates of change use the chain rule to link two or more rates. If y depends on x and x depends on t, then dy/dt = dy/dx × dx/dt.

相关变化率使用链式法则将两个或多个变化率联系起来。如果 y 依赖于 x，x 依赖于 t，则 dy/dt = dy/dx × dx/dt。

Example: the area A of a circle is A = πr². If the radius r increases at a rate dr/dt = 0.5 cm s⁻¹, find dA/dt when r = 4 cm. First, dA/dr = 2πr. Then dA/dt = 2πr × 0.5 = πr. At r = 4, dA/dt = 4π cm² s⁻¹.

示例：圆的面积 A = πr²。如果半径 r 以 dr/dt = 0.5 cm s⁻¹ 的速度增大，求 r = 4 cm 时的 dA/dt。首先 dA/dr = 2πr。然后 dA/dt = 2πr × 0.5 = πr。当 r = 4 时，dA/dt = 4π cm² s⁻¹。

Always write down the chain rule explicitly and substitute the known values only after differentiating. This is a common source of method marks in Edexcel exams.

始终先明确写出链式法则，并在求导后再代入已知数值。这是 Edexcel 考试中常见的方法分来源。

11. Exam Technique and Common Errors | 考试技巧与常见错误

Before differentiating, simplify expressions fully. Rewrite rational terms like (x² + 1)/x as x + 1/x, and rewrite roots as fractional powers. This makes the power rule easier to apply.

求导之前，先完全化简表达式。将有理项如 (x² + 1)/x 改写为 x + 1/x，将根式改写为分数指数。这会让幂函数法则更容易应用。

Common errors include forgetting that the derivative of a constant is zero, incorrectly differentiating products as if they were sums, and misapplying the normal gradient formula. Do not multiply gradients; take the negative reciprocal for a normal.

常见错误包括忘记常数的导数为零、把乘积当作和来错误求导，以及误用法线梯度公式。不要直接相乘梯度；法线应取负倒数。

In mechanics problems, differentiation links displacement s, velocity v and acceleration a: v = ds/dt and a = dv/dt = d²s/dt². Ensure you understand these notations and their applications.

在力学问题中，求导将位移 s、速度 v 和加速度 a 联系起来：v = ds/dt，a = dv/dt = d²s/dt²。确保你理解这些记号及其应用。

12. Summary and Practice Checklist | 总结与练习清单

To master Edexcel differentiation, make sure you can complete all the following tasks confidently:

为了掌握 Edexcel 微分，确保你能自信地完成以下所有任务：

Differentiate from first principles for simple powers of x.
使用第一性原理对 x 的简单幂函数求导。
Apply the power rule to positive, negative and fractional exponents.
将幂函数法则应用于正指数、负指数和分数指数。
Find equations of tangents and normals at a point.
求一点处的切线与法线方程。
Locate and classify stationary points using dy/dx and d²y/dx².
使用 dy/dx 和 d²y/dx² 确定并分类驻点。
Determine intervals where a function is increasing or decreasing.
判断函数递增或递减的区间。
Sketch the gradient function from a given curve graph.
根据给定的曲线图像绘制导函数图像。
Solve connected rates of change using the chain rule.
使用链式法则解决相关变化率问题。

Regular exam-question practice is essential, especially with Edexcel past papers, because differentiation questions often combine several of these skills in one problem.

定期练习真题非常重要，尤其是 Edexcel 历年试卷，因为求导题经常在一个问题中综合考查上述多种技能。

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Differentiation for Edexcel A-Level Maths: Rules and Applications | Edexcel A-Level 数学：微分法则与应用