Category: edexcel-alevel-maths,edexcel-alevel-maths-cn

Introduction to Differentiation — 微分入门

Differentiation is one of the two central pillars of calculus, alongside integration. At its core, differentiation allows us to determine the rate at which one quantity changes with respect to another. For students of Edexcel A-Level Mathematics, mastering differentiation is essential, as it appears throughout the pure mathematics syllabus from basic gradient calculations to sophisticated optimisation problems and parametric equations.

微分是微积分的两大核心支柱之一，与积分并列。从本质上讲，微分使我们能够确定一个量相对于另一个量的变化率。对于学习爱德思 A-Level 数学的学生来说，掌握微分至关重要，因为它贯穿于纯数学课程大纲的始终，从基本的梯度计算到复杂的优化问题和参数方程。

The concept of a derivative originated from the need to precisely describe the slope of a curve at any given point. While a straight line has a constant gradient, a curve’s steepness varies continuously. Sir Isaac Newton and Gottfried Wilhelm Leibniz independently developed the mathematical framework for differentiation in the 17th century, providing the tools to tackle problems that had puzzled mathematicians for centuries.

导数的概念源于精确描述曲线在任意给定点处的斜率的需求。虽然直线具有恒定的梯度，但曲线的陡峭程度会不断变化。艾萨克·牛顿爵士和戈特弗里德·威廉·莱布尼茨在十七世纪各自独立地发展了微分的数学框架，为解决困扰数学家几个世纪的问题提供了工具。

First Principles — 第一原理

Every differentiation technique taught at A-Level ultimately derives from the definition of the derivative from first principles. The derivative of a function f(x) at a point x is defined as the limit of the difference quotient as h approaches zero:

A-Level 阶段教授的每一种微分技巧最终都源于从第一原理出发的导数定义。函数 f(x) 在点 x 处的导数定义为差商的极限，当 h 趋近于零时：

f'(x) = lim(h→0) [f(x+h) – f(x)] / h

This definition captures the essential idea of finding the gradient of the tangent to a curve. By taking the chord between two points on the curve and allowing the distance between them to shrink infinitely, we obtain the instantaneous rate of change. Edexcel exam papers frequently test students on proving the derivatives of simple functions such as x squared and x cubed from first principles.

这一定义捕捉了求曲线切线梯度的本质思想。通过在曲线上取两点之间的弦，并让它们之间的距离无限缩小，我们得到瞬时变化率。爱德思考卷经常测试学生从第一原理证明简单函数（如 x 的平方和 x 的立方）的导数。

The Power Rule and Basic Differentiation — 幂法则与基本微分

The most fundamental rule of differentiation is the power rule. For any function of the form f(x) = x to the power of n, where n is a real number, the derivative is f'(x) = n times x to the power of n minus 1. This elegant rule forms the foundation for differentiating polynomials and rational functions.

最基本的微分法则是幂法则。对于任何形式为 f(x) = x 的 n 次方的函数，其中 n 为实数，其导数为 f'(x) = n 乘以 x 的 n 减 1 次方。这一优雅的法则构成了多项式函数和有理函数微分的基础。

Alongside the power rule, students must master several companion rules. The constant rule states that the derivative of any constant is zero. The constant multiple rule allows us to factor out coefficients: the derivative of c times f(x) is c times f'(x). The sum and difference rules enable us to differentiate term by term: the derivative of f(x) plus or minus g(x) is f'(x) plus or minus g'(x).

除了幂法则，学生还必须掌握几条配套规则。常数法则规定任何常数的导数为零。常数倍数法则允许我们将系数提出：c 乘以 f(x) 的导数是 c 乘以 f'(x)。和差法则使我们能够逐项微分：f(x) 加减 g(x) 的导数是 f'(x) 加减 g'(x)。

These basic rules enable the differentiation of any polynomial. For example, to differentiate f(x) = 4x to the power of 5 minus 3x cubed plus 2x minus 7, we apply the rules term by term to obtain f'(x) = 20x to the power of 4 minus 9x squared plus 2. The constant term disappears, and each power of x reduces by one while being multiplied by the original exponent.

这些基本规则使得任何多项式的微分成为可能。例如，对 f(x) = 4x 的五次方减 3x 的三次方加 2x 减 7 进行微分，我们逐项应用规则得到 f'(x) = 20x 的四次方减 9x 的平方加 2。常数项消失，x 的每个幂次减一，同时乘以原有的指数。

The Chain Rule — 链式法则

When functions are composed, we cannot simply differentiate each part independently. The chain rule addresses this by telling us how to differentiate a function 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 equals the derivative of y with respect to u multiplied by the derivative of u with respect to x. In Leibniz notation: dy/dx = (dy/du) multiplied by (du/dx).

当函数是复合形式时，我们不能简单地独立微分每个部分。链式法则通过告诉我们如何对函数的函数进行微分来解决这个问题。如果 y 是 u 的函数，而 u 是 x 的函数，那么 y 对 x 的导数等于 y 对 u 的导数乘以 u 对 x 的导数。用莱布尼茨符号表示：dy/dx = (dy/du) 乘以 (du/dx)。

The chain rule is perhaps the most widely applicable differentiation technique at A-Level. It appears in problems involving brackets raised to powers, trigonometric functions of linear expressions, and exponentials with linear exponents. A typical Edexcel question might ask students to differentiate y = (2x plus 1) to the power of 6, where letting u = 2x plus 1 gives dy/dx = 6(2x plus 1) to the power of 5 multiplied by 2, which simplifies to 12(2x plus 1) to the power of 5.

链式法则或许是 A-Level 中应用最广泛的微分技巧。它出现在涉及括号的幂次、线性表达式的三角函数以及具有线性指数的指数函数等问题中。一道典型的爱德思题目可能要求学生微分 y = (2x 加 1) 的六次方，令 u = 2x 加 1 可得 dy/dx = 6(2x 加 1) 的五次方乘以 2，简化为 12(2x 加 1) 的五次方。

The Product Rule — 乘积法则

When two functions are multiplied together, the product rule governs their differentiation. For y = u times v, where u and v are both functions of x, the derivative is given by dy/dx = u times dv/dx plus v times du/dx. A memorable way to recall this is “the first function times the derivative of the second, plus the second function times the derivative of the first.”

当两个函数相乘时，乘积法则控制着它们的微分。对于 y = u 乘以 v，其中 u 和 v 都是 x 的函数，导数由 dy/dx = u 乘以 dv/dx 加 v 乘以 du/dx 给出。记住这个公式的记忆方法是”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”。

The product rule becomes particularly important when dealing with expressions such as x squared times sin x or e to the power of x times ln x. In these cases, neither the chain rule nor the power rule alone suffices. For example, to differentiate y = x squared times sin x, we set u = x squared and v = sin x. Then du/dx = 2x and dv/dx = cos x, giving dy/dx = x squared times cos x plus 2x times sin x, which factors to x(x times cos x plus 2 times sin x).

乘积法则在处理诸如 x 的平方乘以 sin x 或 e 的 x 次方乘以 ln x 等表达式时变得尤为重要。在这些情况下，仅凭链式法则或幂法则是不够的。例如，要微分 y = x 的平方乘以 sin x，我们设 u = x 的平方，v = sin x。则 du/dx = 2x，dv/dx = cos x，得到 dy/dx = x 的平方乘以 cos x 加 2x 乘以 sin x，可以因式分解为 x(x 乘以 cos x 加 2 乘以 sin x)。

The Quotient Rule — 商法则

When one function is divided by another, we employ the quotient rule. For y = u divided by v, where u and v are functions of x, the derivative is dy/dx = (v times du/dx minus u times dv/dx) divided by v squared. The order of terms in the numerator is critical: it must be “bottom times derivative of the top minus top times derivative of the bottom” to obtain the correct sign.

当一个函数除以另一个函数时，我们使用商法则。对于 y = u 除以 v，其中 u 和 v 是 x 的函数，导数为 dy/dx = (v 乘以 du/dx 减 u 乘以 dv/dx) 除以 v 的平方。分子中各项的顺序至关重要：必须是”分母乘以分子的导数减去分子乘以分母的导数”才能得到正确的符号。

A common Edexcel exam question involves differentiating rational functions such as y = (x squared plus 1) divided by (x minus 2). Setting u = x squared plus 1 and v = x minus 2, we have du/dx = 2x and dv/dx = 1. Applying the quotient rule yields dy/dx = ((x minus 2) times 2x minus (x squared plus 1) times 1) divided by (x minus 2) squared, which simplifies to (x squared minus 4x minus 1) divided by (x minus 2) squared.

一道常见的爱德思考题涉及有理函数的微分，如 y = (x 的平方加 1) 除以 (x 减 2)。设 u = x 的平方加 1，v = x 减 2，我们有 du/dx = 2x，dv/dx = 1。应用商法则得到 dy/dx = ((x 减 2) 乘以 2x 减 (x 的平方加 1) 乘以 1) 除以 (x 减 2) 的平方，简化为 (x 的平方减 4x 减 1) 除以 (x 减 2) 的平方。

Differentiating Trigonometric Functions — 三角函数的微分

Edexcel A-Level Mathematics requires students to know the derivatives of the six basic trigonometric functions. The derivatives of sine and cosine form a cyclic pattern: the derivative of sin x is cos x, and the derivative of cos x is negative sin x. The derivative of tan x is sec squared x, which can alternatively be written as 1 divided by cos squared x.

爱德思 A-Level 数学要求学生掌握六个基本三角函数的导数。正弦和余弦的导数形成一个循环模式：sin x 的导数是 cos x，cos x 的导数是负 sin x。tan x 的导数是 sec 平方 x，也可以写成 1 除以 cos 平方 x。

For the reciprocal trigonometric functions, students should memorise that the derivative of sec x is sec x times tan x, the derivative of cosec x is negative cosec x times cot x, and the derivative of cot x is negative cosec squared x. These results can all be derived using the quotient rule from the definitions of the functions, but knowing them by heart saves valuable time in examinations.

对于倒数三角函数，学生应记住 sec x 的导数是 sec x 乘以 tan x，cosec x 的导数是负 cosec x 乘以 cot x，cot x 的导数是负 cosec 平方 x。这些结果都可以使用商法则从函数的定义推导出来，但熟记它们可以在考试中节省宝贵的时间。

When trigonometric functions involve linear arguments, the chain rule must be applied. For instance, the derivative of sin(ax plus b) is a times cos(ax plus b), and the derivative of cos(ax plus b) is negative a times sin(ax plus b). This pattern extends naturally to the other trigonometric functions.

当三角函数涉及线性自变量时，必须应用链式法则。例如，sin(ax 加 b) 的导数是 a 乘以 cos(ax 加 b)，cos(ax 加 b) 的导数是负 a 乘以 sin(ax 加 b)。这一模式自然地扩展到其他三角函数。

Exponential and Logarithmic Differentiation — 指数函数与对数函数的微分

The exponential function e to the power of x occupies a special place in calculus because it is its own derivative. The derivative of e to the power of x is simply e to the power of x. More generally, the derivative of e to the power of kx is k times e to the power of kx, by the chain rule. For exponential functions with other bases, the derivative of a to the power of x is a to the power of x times ln a.

指数函数 e 的 x 次方在微积分中占有特殊地位，因为它是它自身的导数。e 的 x 次方的导数就是 e 的 x 次方。更一般地，根据链式法则，e 的 kx 次方的导数是 k 乘以 e 的 kx 次方。对于以其他数为底的指数函数，a 的 x 次方的导数是 a 的 x 次方乘以 ln a。

The natural logarithm function has a beautifully simple derivative: the derivative of ln x is 1 divided by x, defined for x greater than zero. For ln(kx), the chain rule gives 1 divided by x as well, since the factor k cancels. For logarithms with other bases, the derivative of log base a of x is 1 divided by (x times ln a).

自然对数函数有一个非常简洁的导数：ln x 的导数是 1 除以 x，定义域为 x 大于零。对于 ln(kx)，链式法则给出的结果也是 1 除以 x，因为因子 k 会抵消。对于以其他数为底的对数，以 a 为底 x 的对数的导数是 1 除以 (x 乘以 ln a)。

Applications: Tangents, Normals, and Stationary Points — 应用：切线、法线与驻点

One of the most direct applications of differentiation is finding the equation of the tangent and normal to a curve at a given point. The derivative at a point gives the gradient of the tangent. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent’s gradient. Given a point (x1, y1) on the curve, the tangent equation is y minus y1 equals m times (x minus x1), where m is the derivative evaluated at x1.

微分最直接的应用之一是求曲线在给定点处的切线和法线方程。某一点的导数给出了切线的梯度。法线垂直于切线，因此其梯度是切线梯度的负倒数。给定曲线上一点 (x1, y1)，切线方程为 y 减 y1 等于 m 乘以 (x 减 x1)，其中 m 是在 x1 处求值的导数。

Stationary points occur where the first derivative equals zero, meaning the tangent is horizontal. These points can be classified as local maxima, local minima, or points of inflection. The second derivative test provides a straightforward classification method: a positive second derivative indicates a minimum, a negative second derivative indicates a maximum, and a zero second derivative requires further investigation using the first derivative test.

驻点出现在一阶导数等于零的位置，意味着切线是水平的。这些点可以分为局部极大值、局部极小值或拐点。二阶导数检验提供了一种直接的分类方法：正的二阶导数表示极小值，负的二阶导数表示极大值，零二阶导数需要使用一阶导数检验进一步分析。

Parametric Differentiation — 参数微分

When a curve is defined parametrically with x = f(t) and y = g(t), we cannot differentiate y directly with respect to x. Instead, we use the chain rule in the form dy/dx = (dy/dt) divided by (dx/dt). This technique is particularly important for Edexcel A-Level, appearing regularly in examination questions involving curves defined by trigonometric or rational parametric equations.

当曲线以参数形式定义，x = f(t) 且 y = g(t) 时，我们不能直接对 y 关于 x 求导。相反，我们使用链式法则的形式 dy/dx = (dy/dt) 除以 (dx/dt)。这一技巧对于爱德思 A-Level 尤为重要，经常出现在涉及由三角或理性参数方程定义的曲线的考题中。

To find the equation of a tangent to a parametric curve at a specific value of the parameter t, we first calculate dx/dt and dy/dt, then form dy/dx. Evaluating at the given t gives the gradient, and substituting t into the original parametric equations gives the coordinates of the point of tangency. The standard line equation form then completes the solution.

要找到参数曲线在参数 t 的特定值处的切线方程，我们首先计算 dx/dt 和 dy/dt，然后构造 dy/dx。在给定 t 处求值得到梯度，将 t 代入原始参数方程得到切点坐标。然后使用标准直线方程形式完成解答。

Implicit Differentiation — 隐函数微分

Not all relationships between x and y can be expressed with y as an explicit function of x. When an equation defines y implicitly, we differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to terms involving y. For instance, when differentiating y squared with respect to x, we obtain 2y times dy/dx.

并非所有 x 和 y 之间的关系都可以将 y 表示为 x 的显函数。当方程隐式地定义 y 时，我们对两边关于 x 求导，将 y 视为 x 的函数，并对涉及 y 的项应用链式法则。例如，对 y 的平方关于 x 求导时，我们得到 2y 乘以 dy/dx。

Implicit differentiation is essential for finding gradients of curves such as circles, ellipses, and more complex algebraic curves. A typical Edexcel problem might ask for the gradient of the curve x squared plus y squared equals 25 at the point (3, 4). Implicit differentiation gives 2x plus 2y times dy/dx equals zero, so dy/dx equals negative x divided by y, which evaluates to negative 3 divided by 4 at the given point.

隐函数微分对于求圆、椭圆以及更复杂的代数曲线等曲线的梯度至关重要。一道典型的爱德思题目可能要求求曲线 x 的平方加 y 的平方等于 25 在点 (3, 4) 处的梯度。隐函数微分得到 2x 加 2y 乘以 dy/dx 等于零，所以 dy/dx 等于负 x 除以 y，在给定点处求值得负 3 除以 4。

Connected Rates of Change — 相关变化率

Many real-world problems involve quantities that change over time in interrelated ways. Connected rates of change problems use the chain rule to relate different rates. If we know how one quantity changes with time and can relate that quantity to another variable, we can determine the rate of change of the second quantity.

许多现实世界的问题涉及随时间以相互关联的方式变化的量。相关变化率问题使用链式法则将不同的变化率联系起来。如果我们知道一个量随时间如何变化，并能将该量与另一个变量关联起来，我们就可以确定第二个量的变化率。

A classic example involves a spherical balloon being inflated. If the radius r increases at a constant rate dr/dt, we can find the rate at which the volume V increases using dV/dt = (dV/dr) times (dr/dt). Since V = (4/3) times pi times r cubed, we have dV/dr = 4 times pi times r squared, giving dV/dt = 4 times pi times r squared times (dr/dt).

一个经典例子涉及正在充气的球形气球。如果半径 r 以恒定速率 dr/dt 增加，我们可以使用 dV/dt = (dV/dr) 乘以 (dr/dt) 来求体积 V 增加的速率。由于 V = (4/3) 乘以 pi 乘以 r 的立方，我们有 dV/dr = 4 乘以 pi 乘以 r 的平方，得到 dV/dt = 4 乘以 pi 乘以 r 的平方乘以 (dr/dt)。

Edexcel examination questions frequently present scenarios involving filling containers, expanding circles, or moving shadows. The key skill is identifying the appropriate chain of derivatives that connects the known rate to the unknown one, then substituting the given numerical values at the specific moment described in the question.

爱德思考题经常呈现涉及填充容器、扩展圆形或移动阴影的场景。关键技能是识别适当的导数链，将已知变化率与未知变化率连接起来，然后在题目描述的特定时刻代入给定的数值。

Exam Techniques and Common Pitfalls — 考试技巧与常见陷阱

Success in Edexcel A-Level differentiation questions requires more than knowing the rules; it demands careful attention to algebraic manipulation and sign conventions. One of the most common errors is mishandling the negative signs in trigonometric differentiation, particularly when the argument involves a negative coefficient. Students should always write out each step systematically rather than attempting to jump to the final answer.

在爱德思 A-Level 微分题目中取得成功需要的不仅仅是了解规则，还需要仔细关注代数运算和符号约定。最常见的错误之一是三角微分中处理不当的负号，尤其是当自变量涉及负系数时。学生应该系统地写出每一步，而不是试图直接跳到最终答案。

Another frequent pitfall is forgetting to simplify expressions after applying the product or quotient rule. Edexcel mark schemes often award marks for the final simplified form, and leaving answers unsimplified can cost valuable marks. Students should practise factorising their results where possible and presenting answers in their neatest algebraic form.

另一个常见陷阱是在应用乘积法则或商法则后忘记化简表达式。爱德思评分方案经常为最终的简化形式赋分，留下未简化的答案可能损失宝贵的分数。学生应尽可能练习对结果进行因式分解，并以最整洁的代数形式呈现答案。

Time management in the examination is crucial. Differentiation questions often appear in the latter parts of longer problems, building on earlier work. Students should allocate sufficient time to check their differentiation results, as an error early in a multi-part question cascades through all subsequent parts. A quick numerical check using a calculator’s derivative function can provide reassurance when time permits.

考试中的时间管理至关重要。微分题目通常出现在较长问题的后半部分，建立在前面工作的基础上。学生应分配足够的时间来检查他们的微分结果，因为多部分题目早期的错误会级联到所有后续部分中。在时间允许的情况下，使用计算器的导数功能进行快速数值检查可以提供信心保证。

Optimisation Problems — 优化问题

Optimisation is one of the most practical applications of differentiation and a staple of Edexcel A-Level exam papers. The general approach involves expressing the quantity to be optimised as a function of a single variable, differentiating to find stationary points, and then determining which stationary point gives the required maximum or minimum. Real-world constraints must also be checked to ensure the solution lies within the feasible domain.

优化是微分最实际的应用之一，也是爱德思 A-Level 考卷中的常客。一般方法包括将要优化的量表示为单一变量的函数，求导找出驻点，然后确定哪个驻点给出所需的最大值或最小值。还必须检查现实世界的约束条件，以确保解在可行域内。

A typical optimisation problem might ask for the dimensions of a rectangular enclosure that maximise area given a fixed perimeter. If the perimeter is P, and we let one side be x, the other side is (P/2 minus x), giving area A = x times (P/2 minus x). Differentiating and setting dA/dx = 0 yields x = P/4, confirming that a square maximises the area for a given perimeter. The second derivative test verifies this is indeed a maximum.

一个典型的优化问题可能要求找出在给定周长下使面积最大化的矩形围栏尺寸。如果周长为 P，设一边为 x，则另一边为 (P/2 减 x)，得到面积 A = x 乘以 (P/2 减 x)。求导并令 dA/dx = 0 得到 x = P/4，证实正方形在给定周长下最大化面积。二阶导数检验验证了这确实是最大值。

Second Order Derivatives and Concavity — 二阶导数与凹凸性

While the first derivative tells us about the rate of change of a function, the second derivative reveals information about the rate of change of the gradient itself. The second derivative, denoted f”(x) or d squared y over dx squared, indicates the concavity of the curve. When the second derivative is positive, the curve is concave upward, resembling a cup shape. When negative, it is concave downward, resembling an arch.

一阶导数告诉我们函数的变化率，而二阶导数揭示了梯度本身的变化率信息。二阶导数记作 f”(x) 或 d 平方 y 除以 dx 平方，表示曲线的凹凸性。当二阶导数为正时，曲线向上凹，类似杯形。为负时，曲线向下凹，类似拱形。

Points of inflection occur where the concavity changes sign. At a point of inflection, the second derivative equals zero and changes sign as x passes through that point. However, a second derivative of zero does not guarantee an inflection; the sign must genuinely change. Edexcel examiners frequently test this distinction, expecting students to check the sign on both sides of the candidate point rather than simply stating that f”(x) equals zero.

拐点出现在凹凸性改变符号的位置。在拐点处，二阶导数等于零，并且当 x 经过该点时符号发生变化。然而，二阶导数为零并不保证是拐点；符号必须真正改变。爱德思考官经常测试这一区别，期望学生检查候选点两侧的符号，而不是简单地陈述 f”(x) 等于零。

Modelling with Differentiation — 微分建模

Edexcel A-Level often presents modelling questions where differentiation is used to analyse real-world situations described by functions. These models might describe the height of a projectile over time, the concentration of a drug in the bloodstream, or the profit generated by a company as a function of production volume. The core skill is interpreting the mathematical results in the context of the original problem.

爱德思 A-Level 经常呈现建模问题，在这些问题中使用微分来分析由函数描述的现实情境。这些模型可以描述弹射物随时间的高度、药物在血液中的浓度，或公司作为产量函数的利润。核心技能是在原始问题的背景下解释数学结果。

For instance, if a model gives the height h(t) of a ball thrown upwards as h(t) = 20t minus 5t squared, differentiating gives the velocity v(t) = 20 minus 10t. The maximum height occurs when v(t) = 0, at t = 2 seconds, giving h(2) = 20 metres. The ball hits the ground when h(t) = 0, at t = 4 seconds (discarding t = 0). Each mathematical finding must be clearly linked back to the physical scenario.

例如，如果一个模型给出向上抛出的球的高度 h(t) 为 h(t) = 20t 减 5t 的平方，求导得到速度 v(t) = 20 减 10t。当 v(t) = 0 时达到最大高度，在 t = 2 秒时，得到 h(2) = 20 米。当 h(t) = 0 时球落地，在 t = 4 秒时（舍弃 t = 0）。每一个数学发现都必须清晰地与物理场景联系起来。

Revision Strategy for Differentiation — 微分的复习策略

Effective revision for Edexcel A-Level differentiation should combine foundational knowledge with progressive problem-solving. Begin by ensuring complete fluency with the basic rules: power rule, chain rule, product rule, and quotient rule. Without automatic recall of these, attempting more complex problems becomes inefficient and error-prone. Daily drill exercises for five to ten minutes can cement these fundamental skills.

爱德思 A-Level 微分的有效复习应将基础知识与递进式问题解决相结合。首先要确保对基本规则的完全流暢掌握：幂法则、链式法则、乘积法则和商法则。如果不能自动回忆这些规则，尝试更复杂的问题就会变得低效且容易出错。每天五到十分钟的练习可以巩固这些基本技能。

Next, work through past paper questions organised by topic. Start with straightforward differentiation of polynomials and trigonometric functions, then progress to applications such as tangents and normals, optimisation, and connected rates of change. The Edexcel website provides a wealth of past papers with mark schemes that reveal exactly what examiners expect at each stage of a solution. Pay particular attention to the “method marks” awarded for showing correct differentiation steps.

接下来，按主题整理历年真题进行练习。从简单的多项式和三角函数微分开始，然后进展到切线法线、优化和相关变化率等应用。爱德思网站提供了大量历年真题和评分方案，准确揭示了考官在解答的每个阶段期望看到的内容。特别注意为展示正确微分步骤而授予的”方法分”。

Finally, practise under timed conditions. Differentiation questions often form parts of larger problems, so speed and accuracy are both essential. A well-prepared student should be able to differentiate any standard function in under thirty seconds, leaving more time for the interpretive and problem-solving aspects of the question. Regular timed practice builds the confidence and fluency needed for examination success.

最后，在限时条件下进行练习。微分题目通常构成较大问题的一部分，因此速度和准确性都至关重要。准备充分的学生应能在三十秒内对任何标准函数进行微分，从而为问题的解释和问题解决方面留出更多时间。定期的限时练习可以培养考试成功所需的信心和流暢度。

Introduction: Why Integration by Parts Matters — 引言：为什么分部积分法如此重要

Integration by parts is arguably the single most important integration technique in the Edexcel A-Level Mathematics syllabus. It is the natural counterpart to the product rule of differentiation and provides a systematic method for integrating products of functions that cannot be simplified through substitution or algebraic manipulation alone. In Edexcel Pure Mathematics Paper 2, integration by parts questions appear consistently, typically worth between five and twelve marks. Beyond the pure mathematics context, the technique also features prominently in Mechanics problems involving variable forces, work done by non-constant forces, and the derivation of equations of motion from acceleration functions. A thorough command of integration by parts is therefore essential for achieving a top grade.

分部积分法可以说是 Edexcel A-Level 数学大纲中最重要的积分技巧。它是微分乘法法则的自然对应，提供了一种系统性的方法来对无法通过代换或代数化简来处理的函数乘积进行积分。在 Edexcel 纯数学试卷二中，分部积分法题目稳定出现，通常价值 5 到 12 分。在纯数学范围之外，该技巧也频繁出现在涉及变力、非常力做功以及从加速度函数推导运动方程的力学问题中。因此，彻底掌握分部积分法对于取得高分至关重要。

The Derivation from the Product Rule — 从乘法法则推导

The integration by parts formula is derived directly from the product rule of differentiation. Recall that for two differentiable functions u(x) and v(x), the product rule states: d/dx(uv) = u(dv/dx) + v(du/dx). If we integrate both sides with respect to x, we obtain: ∫ d/dx(uv) dx = ∫ u(dv/dx) dx + ∫ v(du/dx) dx. The left side simplifies to uv, giving: uv = ∫ u(dv/dx) dx + ∫ v(du/dx) dx. Rearranging yields the standard formula: ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx. This derivation is worth memorising because it reveals the underlying logic: we are trading one integral for another, and the technique only works when the new integral is simpler than the original.

分部积分公式直接由微分的乘法法则推导而来。回顾一下，对于两个可微函数 u(x) 和 v(x)，乘法法则为：d/dx(uv) = u(dv/dx) + v(du/dx)。如果对两边关于 x 积分，我们得到：∫ d/dx(uv) dx = ∫ u(dv/dx) dx + ∫ v(du/dx) dx。左边简化为 uv，得到：uv = ∫ u(dv/dx) dx + ∫ v(du/dx) dx。重新排列得到标准公式：∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx。这个推导值得记住，因为它揭示了底层逻辑：我们是在用一个积分交换另一个积分，只有当新积分比原积分更简单时，这个技巧才有效。

The LIATE Rule: A Systematic Approach to Choosing u — LIATE 法则：选择 u 的系统方法

The most critical decision in any integration by parts problem is the choice of u and dv. A poor choice leads to a more complicated integral and a dead end. The LIATE mnemonic provides a reliable priority order for selecting u. The acronym stands for Logarithmic functions (ln x, logₐ x), Inverse trigonometric functions (arcsin x, arccos x, arctan x), Algebraic functions (xⁿ, polynomial expressions), Trigonometric functions (sin x, cos x, tan x), and Exponential functions (eˣ, aˣ). The function type appearing earliest in LIATE should typically be chosen as u, because differentiating these functions generally simplifies them: the derivative of ln x is 1/x, which is algebraically simpler; the derivative of arcsin x is 1/√(1−x²), which opens up substitution possibilities; while differentiating an algebraic polynomial reduces its degree.

在任何分部积分问题中，最关键的决定是 u 和 dv 的选择。糟糕的选择会导致积分变得更加复杂，走进死胡同。LIATE 口诀提供了一个可靠的选择 u 的优先级顺序。该缩写代表对数函数、反三角函数、代数函数（多项式）、三角函数和指数函数。LIATE 中出现最早的函数类型通常应该被选为 u，因为对这些函数求导通常会简化它们：ln x 的导数是 1/x，代数上更简单；arcsin x 的导数是 1/√(1−x²)，为代换法打开了可能性；而对代数多项式求导会降低其次数。

Worked Example 1: Basic Polynomial times Exponential — 例题一：基本多项式乘以指数函数

Evaluate the indefinite integral ∫ x eˣ dx. Following LIATE, we note that x is Algebraic (third position) and eˣ is Exponential (fifth position). Algebraic appears earlier, so we set u = x and dv/dx = eˣ. Then du/dx = 1, so du = dx. To find v, we integrate dv/dx: v = ∫ eˣ dx = eˣ. Substituting into the formula: ∫ x eˣ dx = x·eˣ − ∫ eˣ·1 dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C. Factorising: ∫ x eˣ dx = eˣ(x − 1) + C. This is a foundational result; any polynomial multiplied by eˣ can be handled by repeated application of this approach.

计算不定积分 ∫ x eˣ dx。依照 LIATE，注意 x 是代数函数（第三位），eˣ 是指数函数（第五位）。代数函数出现更早，所以我们设 u = x，dv/dx = eˣ。那么 du/dx = 1，所以 du = dx。要求 v，我们对 dv/dx 积分：v = ∫ eˣ dx = eˣ。代入公式：∫ x eˣ dx = x·eˣ − ∫ eˣ·1 dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C。因式分解：∫ x eˣ dx = eˣ(x − 1) + C。这是一个基础结果；任何多项式乘以 eˣ 都可以通过重复应用这个方法来解决。

Worked Example 2: Polynomial times Trigonometric Function — 例题二：多项式乘以三角函数

Evaluate ∫ x sin x dx. Here x is Algebraic and sin x is Trigonometric. According to LIATE, Algebraic precedes Trigonometric, so we let u = x and dv/dx = sin x. Then du/dx = 1, giving du = dx, and v = ∫ sin x dx = −cos x. Applying the formula: ∫ x sin x dx = x(−cos x) − ∫ (−cos x)·1 dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C. To verify, differentiate: d/dx(−x cos x + sin x) = −cos x + x sin x + cos x = x sin x, which matches the original integrand.

计算 ∫ x sin x dx。这里 x 是代数函数，sin x 是三角函数。按 LIATE，代数函数在三角函数之前，所以我们令 u = x，dv/dx = sin x。那么 du/dx = 1，得 du = dx，而 v = ∫ sin x dx = −cos x。应用公式：∫ x sin x dx = x(−cos x) − ∫ (−cos x)·1 dx = −x cos x + ∫ cos x dx = −x cos x + sin x + C。验证：求导 d/dx(−x cos x + sin x) = −cos x + x sin x + cos x = x sin x，与原被积函数一致。

Worked Example 3: The Logarithm Trick — 例题三：对数函数的技巧

Evaluate ∫ ln x dx. At first glance, this appears to be a single function, not a product. However, we can always multiply by 1 without changing the value: ∫ ln x dx = ∫ 1·ln x dx. Now we have a product. Following LIATE, Logarithmic functions come first, so u = ln x and dv/dx = 1. Then du/dx = 1/x, giving du = (1/x)dx, and v = ∫ 1 dx = x. Substituting: ∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C. This is a classic result that every A-Level student should know by heart. The same trick works for inverse trigonometric functions: treat ∫ arctan x dx as ∫ 1·arctan x dx with u = arctan x.

计算 ∫ ln x dx。乍一看这像个单一函数，不是乘积。然而，我们总是可以乘以 1 而不改变值：∫ ln x dx = ∫ 1·ln x dx。现在我们有了一个乘积。按 LIATE，对数函数排在最前面，所以 u = ln x，dv/dx = 1。那么 du/dx = 1/x，得 du = (1/x)dx，而 v = ∫ 1 dx = x。代入：∫ ln x dx = x ln x − ∫ x·(1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C。这是每个 A-Level 学生都应该熟记于心的经典结果。同样的技巧适用于反三角函数：将 ∫ arctan x dx 视为 ∫ 1·arctan x dx，设 u = arctan x。

Worked Example 4: Repeated Integration by Parts — 例题四：重复分部积分

Evaluate ∫ x² eˣ dx. We set u = x² (Algebraic) and dv/dx = eˣ (Exponential). Then du/dx = 2x, giving du = 2x dx, and v = eˣ. First application: ∫ x² eˣ dx = x² eˣ − ∫ eˣ·2x dx = x² eˣ − 2∫ x eˣ dx. The new integral ∫ x eˣ dx still requires integration by parts. We apply the technique again with u = x, dv/dx = eˣ, giving du = dx, v = eˣ. Then: ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C₁. Substituting this back into the original expression: ∫ x² eˣ dx = x² eˣ − 2(x eˣ − eˣ) + C = x² eˣ − 2x eˣ + 2eˣ + C. Factorising: ∫ x² eˣ dx = eˣ(x² − 2x + 2) + C. Notice the emerging pattern: for ∫ xⁿ eˣ dx, the result is eˣ times a polynomial of degree n with alternating signs.

计算 ∫ x² eˣ dx。我们设 u = x²（代数函数），dv/dx = eˣ（指数函数）。那么 du/dx = 2x，得 du = 2x dx，而 v = eˣ。第一次应用：∫ x² eˣ dx = x² eˣ − ∫ eˣ·2x dx = x² eˣ − 2∫ x eˣ dx。新的积分 ∫ x eˣ dx 仍需要分部积分。我们再次应用该技巧，设 u = x，dv/dx = eˣ，得 du = dx，v = eˣ。那么：∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C₁。将其代回原表达式：∫ x² eˣ dx = x² eˣ − 2(x eˣ − eˣ) + C = x² eˣ − 2x eˣ + 2eˣ + C。因式分解：∫ x² eˣ dx = eˣ(x² − 2x + 2) + C。注意其中显现的模式：对于 ∫ xⁿ eˣ dx，结果是 eˣ 乘以一个带有交替符号的 n 次多项式。

Worked Example 5: The Circular Integral Pattern — 例题五：循环积分模式

Evaluate ∫ eˣ sin x dx. This is a famous case where integration by parts appears to lead in circles, but this circularity is exactly what gives us the answer. Let u = sin x (Trigonometric) and dv/dx = eˣ (Exponential). Although LIATE would suggest Trigonometric before Exponential, in practice both choices work, but one may be more convenient. With our choice, du/dx = cos x, giving du = cos x dx, and v = eˣ. First application: I = ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx. Now apply integration by parts to the new integral ∫ eˣ cos x dx. Let u = cos x, dv/dx = eˣ. Then du/dx = −sin x, giving du = −sin x dx, and v = eˣ. This gives: ∫ eˣ cos x dx = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I. Substituting back into the first equation: I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I. Adding I to both sides: 2I = eˣ sin x − eˣ cos x. Therefore: I = (1/2)eˣ(sin x − cos x) + C. This circular approach also works for ∫ eˣ cos x dx and for integrals involving products of trigonometric and exponential functions.

计算 ∫ eˣ sin x dx。这是一个著名的例子，分部积分法看似在原地绕圈，但正是这种循环性给出了答案。设 u = sin x（三角函数），dv/dx = eˣ（指数函数）。虽然 LIATE 会建议三角函数在指数函数之前，但实际上两种选择都可行，但其中一种可能更方便。按我们的选择，du/dx = cos x，得 du = cos x dx，v = eˣ。第一次应用：I = ∫ eˣ sin x dx = eˣ sin x − ∫ eˣ cos x dx。现在对新积分 ∫ eˣ cos x dx 应用分部积分法。设 u = cos x，dv/dx = eˣ。那么 du/dx = −sin x，得 du = −sin x dx，v = eˣ。得到：∫ eˣ cos x dx = eˣ cos x − ∫ eˣ(−sin x) dx = eˣ cos x + ∫ eˣ sin x dx = eˣ cos x + I。代回第一个方程：I = eˣ sin x − (eˣ cos x + I) = eˣ sin x − eˣ cos x − I。两边加 I：2I = eˣ sin x − eˣ cos x。因此：I = (1/2)eˣ(sin x − cos x) + C。这种循环方法也适用于 ∫ eˣ cos x dx 以及涉及三角函数和指数函数乘积的积分。

Worked Example 6: Definite Integration by Parts — 例题六：定积分的分部积分法

For definite integrals, the formula becomes: ∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx. The key difference is that the uv term is evaluated at the limits before subtracting the remaining integral. Consider ∫₀¹ x eˣ dx. From our earlier indefinite result, we know ∫ x eˣ dx = eˣ(x − 1). Evaluating at the limits: F(1) = e¹(1 − 1) = 0, F(0) = e⁰(0 − 1) = −1. Therefore ∫₀¹ x eˣ dx = 0 − (−1) = 1. Alternatively, applying the definite formula directly: ∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0·e⁰) − [eˣ]₀¹ = e − (e − 1) = 1. Both methods yield the same result.

对于定积分，公式变为：∫ₐᵇ u (dv/dx) dx = [uv]ₐᵇ − ∫ₐᵇ v (du/dx) dx。关键区别在于 uv 项在减去剩余积分之前需要在上下限处求值。考虑 ∫₀¹ x eˣ dx。从我们之前的不定积分结果可知 ∫ x eˣ dx = eˣ(x − 1)。在上下限处求值：F(1) = e¹(1 − 1) = 0，F(0) = e⁰(0 − 1) = −1。因此 ∫₀¹ x eˣ dx = 0 − (−1) = 1。另一种方法，直接应用定积分公式：∫₀¹ x eˣ dx = [x eˣ]₀¹ − ∫₀¹ eˣ dx = (1·e¹ − 0·e⁰) − [eˣ]₀¹ = e − (e − 1) = 1。两种方法得出相同的结果。

The Tabular Method: A Shortcut for Repeated Applications — 表格法：重复应用的捷径

When the integrand takes the form xⁿ eᵃˣ or xⁿ sin(ax) with a large value of n, performing integration by parts n times becomes tedious and error-prone. The tabular method, sometimes called the DI method or the rapid repeated integration by parts method, organises the computation into a simple table. Create two columns. In the left column, write u and repeatedly differentiate until you reach zero. In the right column, write dv and repeatedly integrate the same number of times. Then draw diagonal arrows from each left entry to the right entry one row below, alternating signs starting with positive. Multiply along each diagonal and sum the results. For ∫ x³ eˣ dx: differentiate x³ down the left column (x³, 3x², 6x, 6, 0); integrate eˣ down the right column (eˣ, eˣ, eˣ, eˣ, eˣ). The result is x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C. This method is not examinable as a separate technique in Edexcel A-Level, but it provides a reliable verification tool.

当被积函数的形式为 xⁿ eᵃˣ 或 xⁿ sin(ax) 且 n 较大时，执行 n 次分部积分法变得繁琐且容易出错。表格法，有时称为 DI 法或快速重复分部积分法，将计算组织成一个简单的表格。创建两列。在左列中，写下 u 并重复求导直到变为零。在右列中，写下 dv 并重复积分相同次数。然后从每个左列条目向下一行的右列条目画对角线箭头，从正号开始交替符号。沿每条对角线相乘并求和。对于 ∫ x³ eˣ dx：在左列对 x³ 向下求导（x³, 3x², 6x, 6, 0）；在右列对 eˣ 向下积分（eˣ, eˣ, eˣ, eˣ, eˣ）。结果为 x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C。这种方法在 Edexcel A-Level 中不作为独立的考试技巧，但它提供了可靠的验证工具。

Integration by Parts with Inverse Trigonometric Functions — 反三角函数的分部积分

Evaluate ∫ arctan x dx. Following LIATE, Inverse trigonometric functions are second in priority, so we let u = arctan x and dv/dx = 1. Then du/dx = 1/(1 + x²), giving du = dx/(1 + x²), and v = x. Applying the formula: ∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx. The remaining integral can be solved by substitution. Let t = 1 + x², then dt = 2x dx, so x dx = dt/2. Thus ∫ x/(1 + x²) dx = ∫ (1/t)·(dt/2) = (1/2) ln|t| + C = (1/2) ln(1 + x²) + C. Therefore ∫ arctan x dx = x arctan x − (1/2) ln(1 + x²) + C. This combination of integration by parts and substitution is a common pattern in Edexcel A-Level questions.

计算 ∫ arctan x dx。按 LIATE，反三角函数排在第二位，所以我们令 u = arctan x，dv/dx = 1。那么 du/dx = 1/(1 + x²)，得 du = dx/(1 + x²)，而 v = x。应用公式：∫ arctan x dx = x arctan x − ∫ x/(1 + x²) dx。剩余的积分可以通过代换法求解。令 t = 1 + x²，则 dt = 2x dx，所以 x dx = dt/2。因此 ∫ x/(1 + x²) dx = ∫ (1/t)·(dt/2) = (1/2) ln|t| + C = (1/2) ln(1 + x²) + C。因此 ∫ arctan x dx = x arctan x − (1/2) ln(1 + x²) + C。这种分部积分法与代换法的组合是 Edexcel A-Level 考题中的常见模式。

Applications in Mechanics: Work Done by Variable Forces — 力学中的应用：变力做功

Integration by parts is indispensable in Edexcel A-Level Mechanics. Consider a particle moving along the x-axis under the influence of a variable force F(x) = x e⁻ˣ. The work done by this force as the particle moves from x = 0 to x = a is given by W = ∫₀ᵃ F(x) dx = ∫₀ᵃ x e⁻ˣ dx. Let u = x, dv/dx = e⁻ˣ, so du = dx, v = −e⁻ˣ. Then W = [−x e⁻ˣ]₀ᵃ − ∫₀ᵃ (−e⁻ˣ) dx = −a e⁻ᵃ + 0 + ∫₀ᵃ e⁻ˣ dx = −a e⁻ᵃ + [−e⁻ˣ]₀ᵃ = −a e⁻ᵃ − e⁻ᵃ + 1 = 1 − e⁻ᵃ(a + 1). As a → ∞, W → 1, meaning the total work done over an infinite displacement is finite, which is a physically interesting result.

分部积分法在 Edexcel A-Level 力学中不可或缺。考虑一个粒子在变力 F(x) = x e⁻ˣ 作用下沿 x 轴运动。当粒子从 x = 0 移动到 x = a 时，该力所做的功为 W = ∫₀ᵃ F(x) dx = ∫₀ᵃ x e⁻ˣ dx。令 u = x，dv/dx = e⁻ˣ，所以 du = dx，v = −e⁻ˣ。那么 W = [−x e⁻ˣ]₀ᵃ − ∫₀ᵃ (−e⁻ˣ) dx = −a e⁻ᵃ + 0 + ∫₀ᵃ e⁻ˣ dx = −a e⁻ᵃ + [−e⁻ˣ]₀ᵃ = −a e⁻ᵃ − e⁻ᵃ + 1 = 1 − e⁻ᵃ(a + 1)。当 a → ∞ 时，W → 1，意味着在无限位移上做的总功是有限的，这是一个有趣的物理结果。

Edexcel Exam Technique and Mark Schemes — Edexcel 考试技巧与评分标准

Edexcel examiners award marks for specific steps in integration by parts questions. The mark scheme typically allocates one mark for correctly identifying u and dv/dx, one mark for finding du/dx and v, one mark for correctly substituting into the formula, one or two marks for evaluating the resulting integral, and a final mark for the correct simplified answer including the constant of integration where required. Always show your working explicitly. Write “Let u = …” and “dv/dx = …” on separate lines. For definite integrals, show the evaluation of [uv] at the limits as a separate step. If the question asks for an exact answer, leave your answer in terms of e or π rather than giving a decimal approximation. Common examiner comments note that students lose marks by omitting brackets around negative signs and by failing to simplify their final answer fully.

Edexcel 考官对分部积分题目中的特定步骤给分。评分标准通常为：正确识别 u 和 dv/dx 得一分，求出 du/dx 和 v 得一分，正确代入公式得一分，计算所得积分得一到两分，最后正确简化答案（包括所需的积分常数）得一分。务必明确展示你的解题过程。在单独的行上写”令 u = …”和”dv/dx = …”。对于定积分，将 [uv] 在上下限处的求值作为单独的步骤展示。如果题目要求精确答案，请以 e 或 π 的形式给出答案，而不是给出小数近似值。考官的常见评语指出，学生因省略负号周围的括号以及未能完全简化最终答案而失分。

Choosing Between Substitution and Integration by Parts — 在代换法和分部积分法之间选择

One of the key skills tested in Edexcel A-Level is recognising which integration technique to apply. As a general rule, if the integrand is a product of two different types of function (for example, algebraic and exponential, or logarithmic and trigonometric), integration by parts is likely the correct approach. If the integrand involves a composite function where the derivative of the inner function appears as a factor, substitution is more appropriate. For instance, ∫ x e^(x²) dx should be tackled by substitution (let u = x²) rather than integration by parts, because the derivative of x², namely 2x, appears as a factor. Meanwhile, ∫ x eˣ dx requires integration by parts because x and eˣ are unrelated function types with no derivative link. Developing the instinct to distinguish these cases comes from extensive practice with past paper questions.

Edexcel A-Level 考查的关键技能之一是识别应使用哪种积分技巧。作为一般规则，如果被积函数是两种不同类型函数的乘积（例如代数函数和指数函数，或对数函数和三角函数），分部积分法很可能是正确的方法。如果被积函数涉及复合函数，其中内部函数的导数作为一个因式出现，那么代换法更合适。例如，∫ x e^(x²) dx 应通过代换法（令 u = x²）来解决，而不是分部积分法，因为 x² 的导数 2x 作为因式出现。同时，∫ x eˣ dx 需要分部积分法，因为 x 和 eˣ 是不相关的函数类型，没有导数联系。培养区分这些情况的直觉来自于对历年真题的大量练习。

Common Mistakes and How to Avoid Them — 常见错误及其避免方法

Several recurring mistakes cost students marks on integration by parts questions. First, incorrectly choosing u and dv is the most fundamental error. If after one round of integration by parts the new integral looks more complicated than the original, you have almost certainly chosen u incorrectly. Second, sign errors are pervasive. When v = −cos x and you substitute into the formula, remember that the term is uv − ∫ v du, so the subtraction sign interacts with the negative sign in v. Write − ∫ (−cos x) dx = + ∫ cos x dx explicitly to avoid confusion. Third, for definite integrals, do not forget to evaluate [uv] at both limits before subtracting the integral. Fourth, when using the tabular method, ensure the alternating signs start with positive for the first diagonal. Fifth, always include +C for indefinite integrals; this mark is almost always awarded explicitly in the mark scheme. Finally, check your answer by differentiation. If differentiating your result does not recover the original integrand, there is a mistake somewhere.

几个反复出现的错误让学生们在分部积分题目上失分。首先，错误选择 u 和 dv 是最根本的错误。如果经过一轮分部积分后，新积分看起来比原积分更复杂，你几乎肯定选错了 u。其次，符号错误普遍存在。当 v = −cos x 且代入公式时，记住该项是 uv − ∫ v du，因此减号与 v 中的负号相互作用。明确写出 − ∫ (−cos x) dx = + ∫ cos x dx 以避免混淆。第三，对于定积分，在减去积分之前不要忘记计算 [uv] 在两个上下限上的值。第四，使用表格法时，确保交替符号从第一条对角线的正号开始。第五，对于不定积分，务必加上 +C；评分标准中几乎总是明确给这个分数。最后，通过求导检查你的答案。如果对你的结果求导不能还原原始被积函数，说明某处有错误。

Practice Questions and Exam Strategy — 练习题与考试策略

To build fluency with integration by parts, practice with a systematic progression. Begin with straightforward polynomial-exponential products such as ∫ x e²ˣ dx and ∫ x² e³ˣ dx. Move on to polynomial-trigonometric combinations like ∫ x cos 2x dx and ∫ x² sin x dx. Then tackle logarithmic integrals including ∫ x ln x dx and ∫ (ln x)² dx. Finally, attempt the circular integral patterns: ∫ e²ˣ sin 3x dx and ∫ eˣ cos 2x dx. In the exam, allocate roughly one minute per mark. If a question is worth 7 marks, you should plan to spend about 7 minutes on it. If you become stuck, move on and return later. Integration by parts questions are often placed in the middle to later sections of the paper, alongside other challenging pure mathematics topics such as differential equations and parametric integration.

要熟练掌握分部积分法，请按系统性进阶进行练习。从简单的多项式指数函数乘积开始，如 ∫ x e²ˣ dx 和 ∫ x² e³ˣ dx。接着练习多项式三角函数组合，如 ∫ x cos 2x dx 和 ∫ x² sin x dx。然后攻克对数积分，包括 ∫ x ln x dx 和 ∫ (ln x)² dx。最后，尝试循环积分模式：∫ e²ˣ sin 3x dx 和 ∫ eˣ cos 2x dx。考试中，大约每分钟一分。如果一道题值 7 分，你应该计划花大约 7 分钟在这道题上。如果你卡住了，继续往下做，稍后再回来。分部积分法题目通常出现在试卷的中后段，与其他具有挑战性的纯数学话题如微分方程和参数积分一起出现。

Summary and Key Takeaways — 总结与要点

Integration by parts is a versatile and indispensable technique for Edexcel A-Level Mathematics. The LIATE rule provides a reliable framework for choosing u, but always verify that your choice simplifies the integral. Master the five standard patterns: polynomial times exponential, polynomial times trigonometric, logarithmic functions disguised as products with 1, repeated integration by parts for higher-degree polynomials, and the circular integral pattern for products of exponential and trigonometric functions. Remember the definite integral variant of the formula and always check your work by differentiation. With disciplined practice and careful attention to algebraic signs, integration by parts becomes a reliable tool rather than a source of anxiety on exam day.

分部积分法是 Edexcel A-Level 数学中一个多功能且不可或缺的技巧。LIATE 法则为选择 u 提供了可靠的框架，但务必验证你的选择是否简化了积分。掌握五种标准模式：多项式乘以指数函数、多项式乘以三角函数、伪装成与 1 乘积的对数函数、针对高次多项式的重复分部积分，以及针对指数函数和三角函数乘积的循环积分模式。记住公式的定积分变体，并始终通过求导检查你的答案。通过有纪律的练习和对代数符号的仔细关注，分部积分法将成为一个可靠的工具，而非考试当天的焦虑来源。