📚 Partial Differentiation: Key Exam Points for IB & Edexcel Maths | 偏微分考点精讲：IB与Edexcel数学
Partial differentiation is a central topic in multivariable calculus, appearing prominently in the IB Analysis & Approaches HL Calculus option and in Edexcel A-Level Further Mathematics Core Pure. Mastering partial derivatives opens the door to understanding surfaces, optimisation, and rates of change in functions of several variables. This article condenses the essential exam skills, theorems, and common pitfalls into a bilingual revision guide, helping you tackle both straightforward computations and challenging applied problems.
偏微分是多变量微积分的核心内容，在IB Analysis & Approaches HL 微积分选修模块和Edexcel A-Level Further Mathematics Core Pure中都有显著体现。掌握偏导数是理解曲面、最优化以及多元函数变化率的关键。本文将这些必考技巧、定理和常见易错点浓缩为一篇中英双语复习指南，帮助你同时应对简单计算与具有挑战性的应用题。
1. What is Partial Differentiation? | 什么是偏微分？
When a function depends on two or more independent variables, such as z = f(x, y), we cannot simply use the ordinary derivative d/dx. Instead, we compute partial derivatives by differentiating with respect to one variable while treating all other variables as constants. The notation uses the curly symbol ∂ rather than d. For example, the partial derivative of f with respect to x is written as ∂f/∂x or f_x.
当一个函数依赖于两个或更多自变量时，例如 z = f(x, y)，我们不能直接使用普通导数 d/dx。取而代之的是计算偏导数——对其中一个变量求导，同时将所有其他变量视为常数。记号使用弯曲符号 ∂ 而非 d。例如，f 关于 x 的偏导数记作 ∂f/∂x 或 f_x。
Geometrically, ∂f/∂x at a point (a, b) gives the slope of the tangent line to the curve formed by slicing the surface z = f(x, y) with the plane y = b. It measures the instantaneous rate of change of f in the positive x-direction while y is fixed.
从几何上看，∂f/∂x 在点 (a, b) 处的值等于用平面 y = b 截取曲面 z = f(x, y) 所得曲线在该点切线的斜率。它衡量的是在 y 固定时，f 沿 x 正方向的瞬时变化率。
Both IB and Edexcel exams expect you to find first-order partial derivatives quickly and accurately. Typical expressions involve polynomials, exponentials, logarithms, and trigonometric functions. Always remember: when differentiating with respect to x, treat y as a constant, and vice versa.
IB 和 Edexcel 的考试都要求你能够快速准确地求出一阶偏导数。常见式子包括多项式、指数、对数和三角函数。永远记住：对 x 求偏导时，将 y 视为常数；反之亦然。
2. First-Order Partial Derivatives | 一阶偏导数
Computing first-order partials is the foundation of all that follows. For a function f(x, y), you need both ∂f/∂x and ∂f/∂y. If the function is given explicitly, simply apply the standard differentiation rules (power, product, quotient, chain) while keeping the inactive variable frozen.
计算一阶偏导数是后续所有内容的基础。对于函数 f(x, y)，你需要同时求出 ∂f/∂x 和 ∂f/∂y。如果函数以显式形式给出，只需在冻结非求导变量的同时，运用常规的求导法则（幂函数、乘法、除法、链式法则）即可。
Consider f(x, y) = x³y² + 5xy − 3y. Then ∂f/∂x = 3x²y² + 5y, and ∂f/∂y = 2x³y + 5x − 3. Notice how terms that are independent of x vanish in ∂f/∂x, and terms purely in y are differentiated with respect to y in ∂f/∂y.
考虑 f(x, y) = x³y² + 5xy − 3y。那么 ∂f/∂x = 3x²y² + 5y，而 ∂f/∂y = 2x³y + 5x − 3。注意与 x 无关的项在 ∂f/∂x 中消失，而纯 y 的项在 ∂f/∂y 中会被求导。
A common exam pitfall is forgetting to treat y as a constant when differentiating something like sin(xy) with respect to y. For ∂/∂y [sin(xy)], the result is x cos(xy) because the derivative of xy with respect to y is x. Both IB and Edexcel often test this with trigonometric and exponential composite functions.
一个常见的考试陷阱是：在对 y 求导时忘记把 x 视为常数，例如 sin(xy) 对 y 求偏导，结果应为 x cos(xy)，因为 xy 对 y 的导数为 x。IB 和 Edexcel 都常在三角和指数复合函数中考察这一点。
Table of common first-order partial derivatives for quick revision:
快速复习常用一阶偏导数表格：
| f(x, y) | ∂f/∂x | ∂f/∂y |
|---|---|---|
| xⁿ yᵐ | n xⁿ⁻¹ yᵐ | m xⁿ yᵐ⁻¹ |
| e^(ax+by) | a e^(ax+by) | b e^(ax+by) |
| ln(xy) | 1/x | 1/y |
| sin(ax) cos(by) | a cos(ax) cos(by) | −b sin(ax) sin(by) |
3. Higher-Order Partial Derivatives and Clairaut’s Theorem | 高阶偏导数与克莱罗定理
Once you have first-order partials, you can differentiate again to obtain second-order partial derivatives. There are four possible combinations: ∂²f/∂x², ∂²f/∂y², ∂²f/∂x∂y, and ∂²f/∂y∂x. Mixed partials are the ones where you differentiate with respect to different variables in succession.
一旦求得一阶偏导数，你可以再次求导得到二阶偏导数。共有四种可能的组合：∂²f/∂x²、∂²f/∂y²、∂²f/∂x∂y 和 ∂²f/∂y∂x。混合偏导数是指依次对不同变量求导的结果。
Clairaut’s theorem states that if the second-order mixed partial derivatives are continuous, then the order of differentiation does not matter: ∂²f/∂x∂y = ∂²f/∂y∂x. This symmetry is heavily used in exam questions to simplify calculations and verify results.
克莱罗定理指出，如果二阶混合偏导数连续，那么求导次序无关紧要：∂²f/∂x∂y = ∂²f/∂y∂x。这一对称性在考题中被广泛用于简化计算和验证结果。
For example, if f(x, y) = x⁴y³, then f_xx = ∂/∂x (4x³y³) = 12x²y³, f_yy = ∂/∂y (3x⁴y²) = 6x⁴y, and the mixed partial f_xy = ∂/∂y (4x³y³) = 12x³y², which equals f_yx = ∂/∂x (3x⁴y²) = 12x³y². Many exam questions ask you to verify this equality explicitly.
例如，若 f(x, y) = x⁴y³，则 f_xx = ∂/∂x (4x³y³) = 12x²y³，f_yy = ∂/∂y (3x⁴y²) = 6x⁴y，而混合偏导 f_xy = ∂/∂y (4x³y³) = 12x³y²，与 f_yx = ∂/∂x (3x⁴y²) = 12x³y² 相等。很多考题会明确要求你验证这一等式。
IB candidates often encounter second partials when classifying stationary points using the Hessian determinant. Edexcel papers may ask for ∂²f/∂x∂y directly, so comfortable notation switching (f_xy, f_yx, ∂²f/∂x∂y) is essential.
IB 考生在利用海森行列式对驻点进行分类时经常需要二阶偏导。Edexcel 试卷可能直接要求计算 ∂²f/∂x∂y，因此熟练切换记号（f_xy、f_yx、∂²f/∂x∂y）至关重要。
4. The Chain Rule for Multivariable Functions | 多元函数链式法则
The multivariable chain rule describes how a function of several variables changes when the independent variables themselves depend on other parameters. If z = f(x, y) and x = x(t), y = y(t), then the total derivative dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt).
多元函数链式法则描述了当自变量本身依赖于其他参数时，多变量函数如何变化。若 z = f(x, y) 且 x = x(t), y = y(t)，则全导数 dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)。
In a more general setting, if x and y are functions of two variables u and v, then ∂z/∂u = (∂z/∂x)(∂x/∂u) + (∂z/∂y)(∂y/∂u), and analogously for ∂z/∂v. This tree-diagram structure is a must-know for both IB and Edexcel exams, especially when dealing with polar coordinates or parametric surfaces.
在更一般的情况下，如果 x 和 y 是 u 和 v 两个变量的函数，那么 ∂z/∂u = (∂z/∂x)(∂x/∂u) + (∂z/∂y)(∂y/∂u)，对 ∂z/∂v 也有类似公式。这种树状图结构是 IB 和 Edexcel 考试必须掌握的内容，尤其在处理极坐标或参数曲面时。
Typical exam question: given w = xy², x = r cosθ, y = r sinθ, find ∂w/∂r and ∂w/∂θ. Applying the chain rule gives ∂w/∂r = y² · cosθ + 2xy · sinθ = r² sin²θ cosθ + 2r² cosθ sin²θ, which simplifies beautifully. These questions test both differentiation accuracy and algebraic manipulation.
典型考题：已知 w = xy²，x = r cosθ，y = r sinθ，求 ∂w/∂r 和 ∂w/∂θ。运用链式法则可得 ∂w/∂r = y² · cosθ + 2xy · sinθ = r² sin²θ cosθ + 2r² cosθ sin²θ，能够化简得很简洁。这类题目同时考察求导准确性和代数处理能力。
Always draw a dependency diagram to avoid missing terms. IB examiners especially value clarity of method, while Edexcel mark schemes reward correct partial derivatives and substitution.
务必画出变量依赖关系图以避免遗漏项。IB 考官尤为看重方法的清晰性，而 Edexcel 评分标准奖励正确的偏导数和代换。
5. Implicit Partial Differentiation | 隐函数偏微分
When a relation F(x, y, z) = 0 defines z implicitly as a function of x and y, we can find ∂z/∂x and ∂z/∂y without solving for z explicitly. Assuming F is continuously differentiable, the implicit function theorem gives ∂z/∂x = − (∂F/∂x) / (∂F/∂z), provided ∂F/∂z ≠ 0.
当关系式 F(x, y, z) = 0 隐式地将 z 定义为 x 和 y 的函数时，我们可以不求 z 的显式表达式而直接求出 ∂z/∂x 和 ∂z/∂y。假设 F 连续可微，由隐函数定理可得 ∂z/∂x = − (∂F/∂x) / (∂F/∂z)，前提是 ∂F/∂z ≠ 0。
For example, if x³ + y³ + z³ + 6xyz = 1, then let F(x, y, z) = x³ + y³ + z³ + 6xyz − 1. Compute ∂F/∂x = 3x² + 6yz, ∂F/∂z = 3z² + 6xy. Hence ∂z/∂x = − (3x² + 6yz) / (3z² + 6xy). A similar expression holds for ∂z/∂y. This technique avoids messy explicit solutions.
例如，若 x³ + y³ + z³ + 6xyz = 1，令 F(x, y, z) = x³ + y³ + z³ + 6xyz − 1。计算 ∂F/∂x = 3x² + 6yz，∂F/∂z = 3z² + 6xy。于是 ∂z/∂x = − (3x² + 6yz) / (3z² + 6xy)。对 ∂z/∂y 也有类似式子。这种方法避免了繁琐的显式求解。
In IB and Edexcel exams, implicit partial differentiation frequently appears when finding tangent planes to surfaces defined implicitly. You might also be asked to evaluate a partial derivative at a specific point, requiring you to calculate the values of x, y, z and then plug them in.
在 IB 和 Edexcel 的考试中，当求隐式定义曲面的切平面时，常常要用到隐函数偏微分。题目也可能要求在指定点计算偏导数的值，这就需要先算出 x, y, z 的数值再代入。
6. Total Differential and Error Estimation | 全微分与误差估计
The total differential of a function z = f(x, y) is defined as dz = (∂f/∂x) dx + (∂f/∂y) dy. It approximates the true change Δz when small changes dx = Δx and dy = Δy occur. This linear approximation is extremely useful in error analysis and sensitivity studies.
函数 z = f(x, y) 的全微分定义为 dz = (∂f/∂x) dx + (∂f/∂y) dy。它可用以近似当发生微小变化 dx = Δx 和 dy = Δy 时真实的变化量 Δz。这种线性近似在误差分析和敏感度研究中极为有用。
If x and y are measured with small uncertainties Δx and Δy, the propagated absolute error in z can be estimated by Δz ≈ |∂f/∂x| Δx + |∂f/∂y| Δy. The maximum possible error is usually taken as this sum of absolute values; IB questions often ask for percentage error as well.
如果 x 和 y 的测量值带有微小不确定度 Δx 和 Δy，则 z 的传播绝对误差可由 Δz ≈ |∂f/∂x| Δx + |∂f/∂y| Δy 估算。通常将最大可能误差取为这些绝对值之和；IB 试题也经常要求计算百分比误差。
A typical problem: The radius r and height h of a cylinder are measured as 5 ± 0.1 cm and 10 ± 0.2 cm. Estimate the maximum error in the calculated volume V = πr²h. Here ∂V/∂r = 2πrh, ∂V/∂h = πr². Plugging in gives ΔV ≈ |2π·5·10|(0.1) + |π·25|(0.2) = 10π + 5π = 15π cm³. Such questions appear in both IB SL/HL applications and Edexcel statistics or further maths models.
一个典型问题：圆柱体的半径 r 和高 h 的测量值分别为 5 ± 0.1 cm 和 10 ± 0.2 cm，试估计计算出的体积 V = πr²h 的最大误差。这里 ∂V/∂r = 2πrh，∂V/∂h = πr²。代入得 ΔV ≈ |2π·5·10|(0.1) + |π·25|(0.2) = 10π + 5π = 15π cm³。此类问题在 IB 应用以及 Edexcel 统计或进阶数学模型中均有出现。
7. Tangent Planes and Normal Lines | 切平面与法线
For a surface given explicitly by z = f(x, y) at a point (x₀, y₀, z₀), the equation of the tangent plane is z − z₀ = f_x(x₀, y₀)(x − x₀) + f_y(x₀, y₀)(y − y₀). The normal line has direction ratios given by ( −f_x(x₀, y₀), −f_y(x₀, y₀), 1 ).
对于由 z = f(x, y) 显式给出的曲面，在点 (x₀, y₀, z₀) 处的切平面方程为 z − z₀ = f_x(x₀, y₀)(x − x₀) + f_y(x₀, y₀)(y − y₀)。法线的方向比由 ( −f_x(x₀, y₀), −f_y(x₀, y₀), 1 ) 给出。
If the surface is defined implicitly by F(x, y, z) = 0, the tangent plane at (x₀, y₀, z₀) is F_x(x₀, y₀, z₀)(x − x₀) + F_y(x₀, y₀, z₀)(y − y₀) + F_z(x₀, y₀, z₀)(z − z₀) = 0. The normal vector is ∇F evaluated at the point. This vector formulation is standard in both IB and Edexcel.
如果曲面由隐式方程 F(x, y, z) = 0 定义，在 (x₀, y₀, z₀) 处的切平面为 F_x(x₀, y₀, z₀)(x − x₀) + F_y(x₀, y₀, z₀)(y − y₀) + F_z(x₀, y₀, z₀)(z − z₀) = 0。法向量为在该点计算的梯度 ∇F。这种向量形式在 IB 和 Edexcel 中都是标准内容。
Exam questions often require you to find where the tangent plane is horizontal (i.e. f_x = f_y = 0) or to verify that a given line is normal. Edexcel Further Pure papers may embed this in a broader optimisation strategy, while IB includes it in the Calculus option topic of multivariable functions.
考题常要求你找出切平面水平的点（即 f_x = f_y = 0），或验证某给定直线为法线。Edexcel Further Pure 试卷可能将此嵌入更广泛的优化策略中，而 IB 则将其纳入多元函数微积分选修专题。
8. Stationary Points of Functions of Two Variables | 二元函数的驻点
A stationary point of f(x, y) occurs where both first-order partial derivatives vanish simultaneously: f_x = 0 and f_y = 0. These points are candidates for local maxima, minima, or saddle points. Solving the system of equations often requires solving simultaneous non-linear equations.
f(x, y) 的驻点发生在两个一阶偏导数同时为零的地方：f_x = 0 且 f_y = 0。这些点是局部极大值、极小值或鞍点的候选点。求解该方程组往往需要解联立的非线性方程。
For example, to locate stationary points of f(x, y) = x³ + 3xy² − 3x² − 3y² + 4, set f_x = 3x² + 3y² − 6x = 0 and f_y = 6xy − 6y = 6y(x − 1) = 0. From f_y = 0, either y = 0 or x = 1. Substituting back yields the stationary points (0,0), (2,0), and (1,1), (1,−1). This systematic approach is a core skill in both IB and Edexcel multivariable optimisation problems.
例如，求 f(x, y) = x³ + 3xy² − 3x² − 3y² + 4 的驻点，令 f_x = 3x² +
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply