Differentiation: Rules and Calculations | 微分法：常见函数求导规则与计算

📚 Differentiation: Rules and Calculations | 微分法：常见函数求导规则与计算

Differentiation is a cornerstone of A-Level Mathematics and a powerful tool for modelling rates of change in physics, economics, and engineering. It allows us to find the gradient of a curve at any given point by calculating the instantaneous rate of change of one variable with respect to another. This article provides a comprehensive summary of the essential derivative rules, common function derivatives, and step-by-step calculation strategies tailored for A-Level students.

微分法是 A-Level 数学的核心内容，也是在物理、经济及工程领域中建立变化率模型的有力工具。它使我们能够通过计算一个变量相对于另一个变量的瞬时变化率，来找到曲线上任意一点的切线斜率。本文为 A-Level 学生系统梳理了常用求导法则、常见函数的导数公式以及分步计算策略。

1. The Derivative from First Principles | 导数的基本定义

Before memorizing rules, it is essential to understand where the derivative comes from. The gradient of a chord between two points on a curve, f(x), at x and x+h, is given by the difference quotient. As h approaches zero, this chord becomes the tangent line at the point, and the limit gives the exact gradient.

在记忆求导法则之前，理解导数的来源至关重要。曲线 f(x) 上两点 x 与 x+h 之间割线的斜率由差商给出。当 h 趋于零时，该割线趋近于该点处的切线，而该极限值给出了精确的斜率。

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

For example, if f(x) = x², then f(x+h) = x² + 2xh + h². Substituting this into the formula and simplifying yields f'(x) = 2x. This foundational concept often appears in A-Level Paper 1 or Paper 2 as a proof question.

例如，若 f(x) = x²，则 f(x+h) = x² + 2xh + h²。将其代入公式并化简，可得 f'(x) = 2x。这一基础概念经常出现在 A-Level 试卷一或试卷二的证明题中。

2. The Power Rule | 幂函数求导法则

The power rule is the most frequently used rule in differentiation. For any real constant n, the derivative of x raised to the power n is obtained by multiplying by the power and then reducing the power by one.

幂函数法则是求导中最常用的规则。对任意实数常数 n，x 的 n 次幂的导数，只需将指数乘以系数，再将指数减一即可。

If y = xⁿ, then dy/dx = n·xⁿ⁻¹

This rule applies to positive, negative, and fractional powers. For instance, the derivative of x⁵ is 5x⁴. For reciprocal functions like 1/x³, rewrite as x⁻³ first, then apply the rule to get -3x⁻⁴, which can be rewritten as -3/x⁴. Similarly, √x can be written as x^(1/2), giving a derivative of (1/2)x^(-1/2).

该法则适用于正指数、负指数和分数指数。例如，x⁵ 的导数为 5x⁴。对于象 1/x³ 这样的倒数函数，先将其改写为 x⁻³，再套用法则得到 -3x⁻⁴，可化简为 -3/x⁴。同理，√x 可以写成 x^(1/2)，其导数为 (1/2)x^(-1/2)。

3. Constant Multiple and Sum/Difference Rules | 常数倍与和差法则

Differentiation is a linear operation. This means that the derivative of a constant multiplied by a function is simply the constant multiplied by the derivative of the function. Moreover, the derivative of a sum or difference of functions is the sum or difference of their individual derivatives.

求导是一种线性运算。这意味着常数乘以函数的导数，等于该常数乘以函数的导数。此外，函数和或差的导数，等于各函数导数的和或差。

d/dx(c·u) = c·(du/dx), d/dx(u ± v) = du/dx ± dv/dx

This enables us to differentiate polynomials term by term. For example, if y = 3x² + 4x – 5, then dy/dx = 6x + 4. Remember that the derivative of any constant, such as -5, is zero because a constant does not change. This principle is especially useful before applying product or quotient rules, as simplifying first can greatly reduce complexity.

这使我们能够对多项式逐项求导。例如，若 y = 3x² + 4x – 5，则 dy/dx = 6x + 4。请牢记常数的导数为零，例如 -5 的导数是 0，因为常数项没有变化。这一原理在应用乘积或商法则前尤为重要，因为先化简可以大大降低计算复杂度。

4. The Product Rule | 乘积法则

When differentiating a function that is the product of two separate functions of x, such as y = u·v, we cannot simply differentiate each factor and multiply the results. Instead, we apply the product rule, which states that the derivative is the first function times the derivative of the second, plus the second function times the derivative of the first.

当我们对两个关于 x 的函数乘积（如 y = u·v）进行求导时，不能简单地对各自求导再相乘。相反，我们需要应用乘积法则：其导数等于第一个函数乘以第二个函数的导数，加上第二个函数乘以第一个函数的导数。

If y = u·v, then dy/dx = u·(dv/dx) + v·(du/dx)

A common mnemonic is “Left d Right plus Right d Left”. For example, let y = x²·sin x. Here, u = x² and v = sin x. Then du/dx = 2x and dv/dx = cos x. Substituting these into the product rule gives dy/dx = x²·cos x + 2x·sin x.

一个常见的记忆口诀是”左乘右导加右乘左导”。例如，设 y = x²·sin x。这里 u = x²，v = sin x。则 du/dx = 2x，dv/dx = cos x。代入乘积法则公式，得到 dy/dx = x²·cos x + 2x·sin x。

5. The Quotient Rule | 商法则

For a function expressed as the quotient of two differentiable functions, we use the quotient rule. This rule provides a formula for the derivative of y = u/v in terms of u, v, and their derivatives.

对于表示为两个可微函数之比的函数，我们使用商法则。该法则给出了 y = u/v 的导数关于 u、v 及其导数的公式。

If y = u/v, then dy/dx = [v·(du/dx) – u·(dv/dx)] / v²

The order of operations in the numerator is critical: “Low d High minus High d Low over Low squared”. For example, consider y = (x² + 1) / (x – 1). Let u = x² + 1 and v = x – 1. Then du/dx = 2x and dv/dx = 1. Substituting yields dy/dx = [(x – 1)(2x) – (x² + 1)(1)] / (x – 1)². Expanding the numerator gives (2x² – 2x – x² – 1) / (x – 1)², which simplifies to (x² – 2x – 1) / (x – 1)².

分子中运算的顺序至关重要：”分母乘分子导数减分子乘分母导数，除以分母平方”。例如，考虑 y = (x² + 1) / (x – 1)。设 u = x² + 1，v = x – 1。则 du/dx = 2x，dv/dx = 1。代入公式得 dy/dx = [(x – 1)(2x) – (x² + 1)(1)] / (x – 1)²。展开分子得 (2x² – 2x – x² – 1) / (x – 1)²，化简得 (x² – 2x – 1) / (x – 1)²。

6. The Chain Rule | 链式法则

The chain rule is used to differentiate composite functions, often referred to as “functions of a function”. If y is a function of u, and u is a function of x, then the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x.

链式法则用于对复合函数求导，通常称为”函数的函数”。若 y 是 u 的函数，u 是 x 的函数，那么 y 关于 x 的导数就等于 y 关于 u 的导数乘以 u 关于 x 的导数。

dy/dx = dy/du · du/dx

For example, to differentiate y = (3x + 1)⁵, we let u = 3x + 1, so y = u⁵. Then du/dx = 3 and dy/du = 5u⁴. Therefore, dy/dx = 5u⁴ · 3 = 15(3x + 1)⁴. The chain rule is essential for differentiating exponential, logarithmic, and trigonometric composite functions, as shown in the following sections.

例如，对 y = (3x + 1)⁵ 求导，令 u = 3x + 1，则 y = u⁵。于是 du/dx = 3，dy/du = 5u⁴。因此，dy/dx = 5u⁴ · 3 = 15(3x + 1)⁴。链式法则是求解指数函数、对数函数和三角函数复合形式的关键，具体见以下章节。

7. Derivatives of eˣ and ln x | 指数与对数函数求导

The exponential function eˣ is unique in that its derivative is itself. For a general exponential function aˣ, we use the natural logarithm to express it as e^(x·ln a). In A-Level, we commonly encounter e^(kx), whose derivative is k·e^(kx). The natural logarithm function, ln x, has a simple derivative of 1/x.

指数函数 eˣ 的一个独特性质是它的导数等于其本身。对于一般指数函数 aˣ，我们可以利用自然对数将其表示为 e^(x·ln a)。在 A-Level 中，我们最常见到 e^(kx)，其导数为 k·e^(kx)。自然对数函数 ln x 的导数很简单，为 1/x。

f(x)	f'(x)
eˣ	eˣ
e^(kx)	k·e^(kx)
ln x	1/x

Using the chain rule, the derivative of e^(2x) is 2e^(2x), and the derivative of e^(ax+b) is a·e^(ax+b). When applying the chain rule to ln, the derivative of ln(f(x)) is f'(x)/f(x). For instance, d/dx ln(x² + 1) = 2x/(x² + 1).

利用链式法则，e^(2x) 的导数为 2e^(2x)，e^(ax+b) 的导数为 a·e^(ax+b)。将链式法则应用于对数函数时，ln(f(x)) 的导数为 f'(x)/f(x)。例如，d/dx ln(x² + 1) = 2x/(x² + 1)。

8. Derivatives of Trigonometric Functions | 三角函数求导

In A-Level Mathematics, it is crucial to remember the derivatives of the three basic trigonometric functions. These results are derived from first principles and are valid when x is measured in radians. The derivative of sin x is cos x, the derivative of cos x is -sin x, and the derivative of tan x is sec² x.

在 A-Level 数学中，牢记三个基本三角函数的导数至关重要。这些结果由基本定义推导而来，并且仅在 x 以弧度为单位时成立。sin x 的导数为 cos x，cos x 的导数为 -sin x，tan x 的导数为 sec² x。

f(x)	f'(x)
sin x	cos x
cos x	-sin x
tan x	sec² x

For composite trigonometric functions, the chain rule is applied. For instance, the derivative of sin(2x) is 2cos(2x). Similarly, the derivative of cos(3x + 1) is -3sin(3x + 1). It is also worth noting that these functions have periodic derivatives, and remembering the basic angle relationships can help simplify expressions before differentiation.

对于复合三角函数，需要应用链式法则。例如，sin(2x) 的导数为 2cos(2x)。同理，cos(3x + 1) 的导数为 -3sin(3x + 1)。值得注意的是，这些函数的导数具有周期性，熟记基本角度关系有助于在求导前化简表达式。

9. Second Order Derivatives | 二阶导数

Differentiating a function once gives the first derivative, dy/dx, which represents the rate of change. Differentiating again yields the second order derivative, written as d²y/dx² or f”(x). This represents the rate of change of the gradient and is used to determine the nature of stationary points (maximum, minimum, or point of inflection).

对函数求一次导数得到一阶导数 dy/dx，它表示变化率。再次求导得到二阶导数，记作 d²y/dx² 或 f”(x)。它表示斜率的变化率，常用来判断驻点的性质（极大值、极小值或拐点）。

If y = f(x), then f”(x) = d/dx (dy/dx)

For example, if y = x³ – 2x², then the first derivative is dy/dx = 3x² – 4x. Differentiating this result gives the second derivative: d²y/dx² = 6x – 4. In exam questions, you may be asked to find f”(x) and then evaluate it at a specific point to determine whether a curve is convex or concave at that point.

例如，若 y = x³ – 2x²，则一阶导数为 dy/dx = 3x² – 4x。对该结果再次求导得到二阶导数：d²y/dx² = 6x – 4。在考试题目中，常要求先求出 f”(x)，然后代入特定点来判断曲线在该点是凸还是凹。

10. Worked Exam-Style Examples | 考试题型精讲

Let us consolidate these rules with a few classic A-Level style problems. The key is to identify which rule to apply before jumping into the calculation. Examine the structure of the function: is it a product, quotient, or composite?

我们通过几道经典的 A-Level 题型来巩固这些法则。关键是先观察函数的结构，判断应该使用哪条法则，再进行计算。它是乘积、商，还是复合形式？

Example 1 | 示例 1: Differentiate y = (x² + 1) / eˣ.

Solution | 解答: This is a quotient, so apply the quotient rule with u = x² + 1 and v = eˣ. We have du/dx = 2x and dv/dx = eˣ. Thus, dy/dx = [eˣ(2x) – (x² + 1)eˣ] / (eˣ)² = [2x – x² – 1] / eˣ.

这是商的形式，因此应用商法则，其中 u = x² + 1，v = eˣ。可得 du/dx = 2x，dv/dx = eˣ。因此，dy/dx = [eˣ(2x) – (x² + 1)eˣ] / (eˣ)² = [2x – x² – 1] / eˣ。

Example 2 | 示例 2: Differentiate y = sin(2x) · x³.

Solution | 解答: This is a product, so apply the product rule. Let u = sin(2x) and v = x³. Then du/dx = 2cos(2x) (using the chain rule) and dv/dx = 3x². Therefore, dy/dx = sin(2x)·3x² + x³·2cos(2x) = 3x²sin(2x) + 2x³cos(2x).

这是积的形式，因此应用乘积法则。设 u = sin(2x)，v = x³。则 du/dx = 2cos(2x)（利用链式法则），dv/dx = 3x²。因此，dy/dx = sin(2x)·3x² + x³·2cos(2x) = 3x²sin(2x) + 2x³cos(2x)。

Exam Tip | 考试提示: Always simplify algebraic expressions before differentiating. For example, rewriting x·√x as x^(3/2) makes applying the power rule straightforward. Also, ensure your calculator is set to radians when dealing with trigonometric differentiation. Finally, do not forget to show your working, as method marks are awarded even if the final answer is incorrect.

考试提示：在求导前务必先化简代数表达式。例如，将 x·√x 改写为 x^(3/2) 会使应用幂函数法则变得直接明了。同时，在处理三角函数求导时，请确保计算器设置为弧度模式。最后，务必写出解题过程，因为即使最终答案有误，方法分仍然会被计入。

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Differentiation: Rules and Calculations | 微分法：常见函数求导规则与计算