📚 The Derivative of Polynomial Functions | 多项式函数的导数计算
Differentiation is one of the most fundamental tools in A-Level Mathematics. For polynomial functions, the derivative represents the rate of change at any given point and is essential for curve sketching, optimisation, and kinematics. This guide covers all core rules and techniques required for the Edexcel specification.
微分是A-Level数学中最基础的工具之一。对于多项式函数而言，导数表示任意一点的变化率，是曲线绘图、最优化问题以及运动学中不可或缺的内容。本指南涵盖Edexcel考纲所要求的全部核心规则与技巧。
1. The Power Rule: The Foundation | 幂法则：基础中的基础
The power rule states that if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. This single rule underpins almost all polynomial differentiation. For example, the derivative of x⁵ is 5x⁴, and the derivative of x² is 2x. The exponent n can be any real number — positive, negative, or fractional — although in pure polynomial work it is usually a positive integer.
幂法则指出：若 f(x) = xⁿ，则 f'(x) = nxⁿ⁻¹。这一条规则几乎是所有多项式微分的基础。例如，x⁵ 的导数为 5x⁴，x² 的导数为 2x。指数 n 可以是任意实数——正数、负数或分数——虽然在纯多项式问题中它通常是正整数。
f(x) = xⁿ → f'(x) = nxⁿ⁻¹
When differentiating a term like x⁷, multiply the exponent (7) by the coefficient (1) and then reduce the exponent by one: 7x⁶. This operation works seamlessly whether the term is x, x³, or even a constant term x⁰ = 1.
在微分形如 x⁷ 的项时，将指数（7）乘以系数（1），然后将指数减一：得到 7x⁶。这一操作无论是对 x、x³ 还是常数项 x⁰ = 1 都同样适用。
2. Differentiating Constant Terms | 常数项的微分
A constant term, such as 5 or −3, has a derivative of zero. Geometrically, a horizontal line y = c has zero slope at every point. Since a constant can be written as cx⁰, applying the power rule gives c × 0 × x⁻¹ = 0, which confirms this result.
常数项（如 5 或 −3）的导数为零。从几何角度看，水平直线 y = c 在每个点的斜率都为零。由于常数可以写成 cx⁰，应用幂法则得到 c × 0 × x⁻¹ = 0，这也印证了这一结论。
For example, if f(x) = x³ + 2x² − 7, then f'(x) = 3x² + 4x. The derivative of −7 is simply 0, so it disappears from the resulting expression. This simplification is often the first step students perform — and forgetting it is a common source of error.
例如，若 f(x) = x³ + 2x² − 7，则 f'(x) = 3x² + 4x。−7 的导数就是 0，因此它从结果表达式中消失。这是学生解题的第一步——忘记处理常数项是常见的错误来源之一。
3. Multiplying by a Constant Coefficient | 常数系数与乘法
When a polynomial term has a constant coefficient, you differentiate the variable part and keep the coefficient intact. Formally, d/dx [k × f(x)] = k × f'(x). So the derivative of 4x³ is 4 × 3x² = 12x², and the derivative of −2x⁵ is −2 × 5x⁴ = −10x⁴.
当多项式项带有常数系数时，你只需要微分变量部分，并将系数保持不变。正式地写为 d/dx [k × f(x)] = k × f'(x)。因此 4x³ 的导数为 4 × 3x² = 12x²，−2x⁵ 的导数为 −2 × 5x⁴ = −10x⁴。
This rule is known as the constant multiple rule. It dramatically speeds up differentiation: instead of dealing with the coefficient in every conceptual step, you simply multiply the result of the power rule by the existing coefficient. For instance, 6x⁴ becomes 24x³; ½x² becomes x.
这条规则称为常数倍法则。它大大加快了微分速度：不需要在每一步概念中处理系数，你只需将幂法则的结果乘以原有系数。例如，6x⁴ 变为 24x³；½x² 变为 x。
4. Differentiating Sums and Differences | 和与差的微分
Polynomials are sums of individual terms, and differentiation is linear. This means you can differentiate term by term and then add or subtract the results. If f(x) = x² + 3x − 4, then f'(x) = 2x + 3. Each term is treated independently; the derivative of a sum equals the sum of the derivatives.
多项式是各项之和，而微分是线性的。这意味着你可以逐项微分，然后将结果相加或相减。若 f(x) = x² + 3x − 4，则 f'(x) = 2x + 3。每一项独立处理；和的导数等于导数的和。
Similarly, g(x) = 5x⁴ − 2x³ + x² − 8 gives g'(x) = 20x³ − 6x² + 2x. In the Edexcel exam, you will often be asked to differentiate a polynomial of this form without further context. Practising this skill until it is automatic is essential for success.
类似地，g(x) = 5x⁴ − 2x³ + x² − 8 给出 g'(x) = 20x³ − 6x² + 2x。在Edexcel考试中，你经常会被要求微分这种形式的多项式而不附带其他背景。将这一技巧练到自动化的程度是取得好成绩的关键。
5. Negative and Fractional Exponents | 负指数与分数指数
Although classic polynomials use non-negative integer exponents, the Edexcel specification expects you to differentiate expressions with negative and fractional powers. Rewrite the expression using index laws first. For example, 1/x² is rewritten as x⁻², whose derivative is −2x⁻³.
虽然经典多项式使用非负整数指数，但Edexcel考纲要求你能够微分带有负指数和分数指数的表达式。首先利用指数法则重写表达式。例如，1/x² 改写为 x⁻²，其导数为 −2x⁻³。
Similarly, √x can be rewritten as x^(1/2), and its derivative is ½x^(−1/2) = 1/(2√x). In the exam, it is often prudent to leave your final answer in the same index form as the original question, unless instructed otherwise.
类似地，√x 可以改写为 x^(1/2)，其导数为 ½x^(−1/2) = 1/(2√x)（即根号下x分之1）。在考试中，除非另有说明，通常建议将最终答案保持与题目相同的指数形式。
1/xⁿ = x⁻ⁿ, √x = x^½, ∛x = x^(1/3)
Some exam questions deliberately present the function in fraction or root form to test whether you can convert it before differentiating. A typical example: f(x) = 3/x + 2√x becomes f(x) = 3x⁻¹ + 2x^½, and then f'(x) = −3x⁻² + x^(−½).
有些考试题目故意以分数或根号形式呈现函数，以考查你是否能在微分之前完成转换。一个典型例子：f(x) = 3/x + 2√x 先化为 f(x) = 3x⁻¹ + 2x^½，然后得到 f'(x) = −3x⁻² + x^(−½)。
6. Differentiation from First Principles | 从第一性原理出发的微分
Differentiation from first principles is the formal definition of the derivative and a key Edexcel topic. The derivative f'(x) is defined as the limit of the gradient of a chord as the interval h tends to zero:
从第一性原理出发的微分是导数的正式定义，也是Edexcel的重点考点。导数 f'(x) 定义为当区间 h 趋近于零时割线斜率的极限：
f'(x) = limₕ→₀ [f(x+h) − f(x)] / h
For a polynomial such as f(x) = x², you compute f(x+h) = (x+h)² = x² + 2xh + h². Subtracting f(x) = x² leaves 2xh + h². Dividing by h gives 2x + h, and taking the limit h → 0 yields f'(x) = 2x.
对于诸如 f(x) = x² 的多项式，计算 f(x+h) = (x+h)² = x² + 2xh + h²。减去 f(x) = x² 后剩下 2xh + h²。除以 h 得到 2x + h，取极限 h → 0 后得到 f'(x) = 2x。
Full working is required in Edexcel questions that ask for differentiation from first principles. Show all steps clearly, including the expansion, the subtraction, the cancellation of h, and then the limiting process.
在Edexcel的题目中，凡是要求从第一性原理出发微分的题目都需要完整的过程。请清晰地写出所有步骤，包括展开、相减、消去 h，以及极限过程。许多学生的失分源于跳步或没有明确写出极限符号。
7. Derivative Notation: f'(x), dy/dx, and Gradient | 导数的记号：f'(x)、dy/dx 与斜率
Edexcel uses multiple notations for derivatives, and you must be fluent in all of them. If y = x³, then dy/dx = 3x². If f(x) = 2x⁴, then f'(x) = 8x³. Both notations represent the same concept: the derivative. The dy/dx notation is particularly useful in connected rates of change problems.
Edexcel使用多种导数记号，你必须对它们都足够熟练。若 y = x³，则 dy/dx = 3x²。若 f(x) = 2x⁴，则 f'(x) = 8x³。这两种记号都代表同一个概念：导数。dy/dx 记号在有联系的变化率问题中尤其有用。
The derivative evaluated at a specific point, such as f'(2), gives the gradient of the tangent to the curve at x = 2. For instance, if f(x) = x² − 3x + 1, then f'(x) = 2x − 3, so the gradient at x = 2 is f'(2) = 1. If the gradient is zero, the point is a stationary point.
导数在具体点的取值，如 f'(2)，给出了曲线在 x = 2 处切线的斜率。例如，若 f(x) = x² − 3x + 1，则 f'(x) = 2x − 3，因此在 x = 2 处的斜率为 f'(2) = 1。如果斜率为零，则该点为驻点。
8. Applying Differentiation: Tangents and Normals | 微分的应用：切线与法线
The derivative at a point gives the slope of the tangent line. The equation of the tangent at point (x₁, y₁) is found using y − y₁ = m(x − x₁), where m = f'(x₁). A normal is a line perpendicular to the tangent; its gradient is the negative reciprocal, −1/m.
某点的导数给出切线的斜率。在点 (x₁, y₁) 处的切线方程通过 y − y₁ = m(x − x₁) 求得，其中 m = f'(x₁)。法线是与切线垂直的直线；其斜率为切线斜率的负倒数，即 −1/m。
Example: find the tangent to f(x) = x² + 2x at x = 1. First, f(1) = 3 and f'(x) = 2x + 2, so f'(1) = 4. The tangent is y − 3 = 4(x − 1), which simplifies to y = 4x − 1. The normal therefore has gradient −1/4, giving y − 3 = −¼(x − 1).
示例：求 f(x) = x² + 2x 在 x = 1 处的切线。首先，f(1) = 3 且 f'(x) = 2x + 2，所以 f'(1) = 4。切线为 y − 3 = 4(x − 1)，化简为 y = 4x − 1。法线的斜率为 −1/4，因此其方程为 y − 3 = −¼(x − 1)。
9. Second Derivatives and Stationary Points | 二阶导数与驻点
The second derivative, written as f”(x) or d²y/dx², is obtained by differentiating the first derivative. For f(x) = 2x³ + 3x², we get f'(x) = 6x² + 6x and f”(x) = 12x + 6. The second derivative tells us about the rate of change of the gradient itself, i.e., the curvature.
二阶导数写作 f”(x) 或 d²y/dx²，通过对一阶导数再次微分得到。对于 f(x) = 2x³ + 3x²，我们有 f'(x) = 6x² + 6x，f”(x) = 12x + 6。二阶导数告诉我们梯度本身的变化率，也就是曲率。
Stationary points occur where f'(x) = 0. To classify them, evaluate f”(x) at the stationary point: if f”(x) > 0, it is a local minimum; if f”(x) < 0, it is a local maximum; if f''(x) = 0, the test is inconclusive and you should examine the sign of f'(x) on either side.
驻点出现在 f'(x) = 0 处。为对它们进行分类，在驻点处计算 f”(x)：若 f”(x) > 0，则为局部极小值；若 f”(x) < 0，则为局部极大值；若 f''(x) = 0，则该检验不具决定性，需检查 f'(x) 在两侧的符号。
f'(x) = 0 → stationary point; f”(x) > 0 → min; f”(x) < 0 → max
10. Differentiating Simple Composite Polynomials | 简单复合多项式的微分
While the full chain rule is covered elsewhere, Edexcel expects you to differentiate expressions of the form (ax + b)ⁿ. The derivative is found using the chain rule:
虽然完整的链式法则在别处有所涉及，但Edexcel要求你能够微分形如 (ax + b)ⁿ 的表达式。其导数通过链式法则求得：
d/dx [(ax + b)ⁿ] = n(ax + b)ⁿ⁻¹ × a
For example, if y = (3x + 2)⁴, then dy/dx = 4(3x + 2)³ × 3 = 12(3x + 2)³. The inner derivative (the 3) is often called the “inside derivative”. This technique extends naturally to any linear expression raised to any power.
例如，若 y = (3x + 2)⁴，则 dy/dx = 4(3x + 2)³ × 3 = 12(3x + 2)³。内部的导数（即3）常被称为”内层导数”。这一技巧自然地推广到任意一次表达式升到任意次幂的情形。
11. Common Exam Pitfalls | 常见考试陷阱
Students frequently make avoidable errors when differentiating polynomials. Forgetting that the derivative of a constant is zero, misapplying signs, or failing to simplify fractional exponents are common. Another frequent error is multiplying incorrectly: the derivative of 3x² is 6x, not 3x or 9x.
学生在微分多项式时常犯一些可以避免的错误。忘记常数项导数为零、符号处理错误、或者未能正确化简分数指数都是常见问题。另一个常见错误是乘法错误：3x² 的导数是 6x，而不是 3x 或 9x。
- Always reduce the exponent by exactly one: xⁿ → xⁿ⁻¹, not xⁿ or xⁿ⁺¹.
- Always reduce the exponent by exactly one: xⁿ → xⁿ⁻¹，而不是 xⁿ 或 xⁿ⁺¹。
- Remember to multiply by the original exponent: 5x³ → 15x².
- 记住乘以原指数：5x³ → 15x²。
- Keep track of negative signs when differentiating negative coefficients.
- 微分负系数时注意负号的跟踪。
- Do not forget the chain rule when differentiating composite functions.
- 微分复合函数时不要忘记链式法则。
In the exam, always check your answer by considering whether the degree of the derivative is one less than the degree of the original polynomial. This quick sanity check catches many errors before you move on to the next part of the question.
在考试中，始终通过检查导数的次数是否比原多项式的次数低一次来验证答案。这个快速的合理性检查能让你在进入下一问之前发现许多错误。
12. Worked Example: Full Edexcel-Style Question | 完整示例：Edexcel风格题目
Consider the function f(x) = x³ − 6x² + 9x + 2. (a) Find f'(x). (b) Find the coordinates of the stationary points and classify them. (c) Find the equation of the tangent at x = 0.
考虑函数 f(x) = x³ − 6x² + 9x + 2。(a) 求 f'(x)。(b) 求驻点坐标并分类。(c) 求在 x = 0 处的切线方程。
Solution (a): Applying the power rule term by term:
解答 (a)： 逐项应用幂法则：
f'(x) = 3x² − 12x + 9
Solution (b): Set f'(x) = 0: 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0, so x = 1 or x = 3. The corresponding y-values are f(1) = 1 − 6 + 9 + 2 = 6 and f(3) = 27 − 54 + 27 + 2 = 2. So the stationary points are (1, 6) and (3, 2).
解答 (b)： 令 f'(x) = 0：3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3) = 0，因此 x = 1 或 x = 3。对应的 y 值为 f(1) = 1 − 6 + 9 + 2 = 6，f(3) = 27 − 54 + 27 + 2 = 2。所以驻点为 (1, 6) 和 (3, 2)。
To classify, find f”(x) = 6x − 12. At x = 1, f”(1) = −6 < 0, so (1, 6) is a local maximum. At x = 3, f''(3) = 6 > 0, so (3, 2) is a local minimum.
为进行分类，求 f”(x) = 6x − 12。在 x = 1 处，f”(1) = −6 < 0，因此 (1, 6) 是局部极大值。在 x = 3 处，f''(3) = 6 > 0，因此 (3, 2) 是局部极小值。
Solution (c): At x = 0, f(0) = 2 and f'(0) = 9. The tangent is y − 2 = 9(x − 0), i.e., y = 9x + 2.
解答 (c)： 在 x = 0 处，f(0) = 2，f'(0) = 9。切线为 y − 2 = 9(x − 0)，即 y = 9x + 2。
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