📚 A-Level CIE Further Maths: Partial Differentiation Key Points | A-Level CIE 进阶数学：偏微分 考点精讲
Partial differentiation extends the idea of ordinary derivatives to functions of several variables, a core topic in CIE Further Mathematics (9231). Understanding how to compute and apply partial derivatives is essential for tackling optimisation problems, rates of change, and simple partial differential equations. This revision guide covers every key concept, from first-order partial derivatives to classifying stationary points of bivariate functions, all fully aligned with the CIE syllabus.
偏微分将普通导数的概念推广到多元函数，是 CIE 进阶数学 (9231) 的核心考点。掌握偏导数的计算与应用，对于解决最优化问题、变化率问题以及简单偏微分方程至关重要。本复习指南涵盖从一阶偏导数到二元函数驻点分类的所有关键概念，全部严格对标 CIE 考纲。
1. Introduction to Partial Derivatives | 偏导数简介
If z = f(x, y) is a function of two independent variables, the partial derivative of f with respect to x, denoted by ∂f/∂x or fx, is the ordinary derivative obtained by treating y as a constant. Similarly, ∂f/∂y or fy holds x fixed. Geometrically, ∂f/∂x gives the slope of the tangent line to the surface z = f(x, y) in the x-direction.
若 z = f(x, y) 为二元函数，则 f 对 x 的偏导数记作 ∂f/∂x 或 fx，它是在将 y 视为常数时对 x 求普通导数所得。类似地，∂f/∂y 或 fy 则固定 x。几何上，∂f/∂x 表示曲面 z = f(x, y) 在 x 方向上的切线斜率。
The formal definition mirrors the single-variable derivative: ∂f/∂x = limh→0 [f(x+h, y) – f(x, y)] / h, provided this limit exists. Partial derivatives are the building blocks for all further analysis of multivariable functions.
其形式定义与一元导数类似：∂f/∂x = limh→0 [f(x+h, y) – f(x, y)] / h，只要该极限存在。偏导数是进一步分析多元函数的基石。
2. Computing First-Order Partial Derivatives | 一阶偏导数计算
To find ∂f/∂x, treat all variables except x as constants and differentiate as usual. For f(x, y) = x3y2 + sin(xy), ∂f/∂x = 3x2y2 + y·cos(xy) and ∂f/∂y = 2x3y + x·cos(xy). Always remember to apply the product, chain and trigonometric rules correctly while keeping the other variable fixed.
求 ∂f/∂x 时，将 x 以外的所有变量视为常数，正常求导即可。例如对于 f(x, y) = x3y2 + sin(xy)，∂f/∂x = 3x2y2 + y·cos(xy)，∂f/∂y = 2x3y + x·cos(xy)。务必在固定另一个变量的同时，正确使用乘积法则、链式法则和三角函数的导数公式。
For functions involving exponentials or logarithms, the same principle applies. If g(x, y) = exy + y ln x, then gx = yexy + y/x and gy = xexy + ln x. Practise with a variety of forms to build fluency.
对于含指数或对数的函数，原理不变。若 g(x, y) = exy + y ln x，则 gx = yexy + y/x，gy = xexy + ln x。多做练习，熟悉各种函数形式。
3. Second-Order Partial Derivatives | 二阶偏导数
Differentiating first-order partial derivatives again yields second-order partial derivatives. For f(x, y), we have ∂2f/∂x2 = fxx, ∂2f/∂y2 = fyy, and mixed derivatives ∂2f/∂x∂y = fxy and ∂2f/∂y∂x = fyx. Compute fxx by differentiating fx with respect to x, again treating y as constant.
对一阶偏导数再次求导即得二阶偏导数。对于 f(x, y)，有 ∂2f/∂x2 = fxx，∂2f/∂y2 = fyy，以及混合偏导数 ∂2f/∂x∂y = fxy 和 ∂2f/∂y∂x = fyx。fxx 是将 fx 再次对 x 求导，同时仍将 y 视为常数。
For example, if f(x, y) = x4y + ey sin x, then fx = 4x3y + ey cos x, so fxx = 12x2y – ey sin x. Similarly, fyy = ey sin x. Second-order derivatives are used to test concavity and classify stationary points.
例如，若 f(x, y) = x4y + ey sin x，则 fx = 4x3y + ey cos x，于是 fxx = 12x2y – ey sin x。同理，fyy = ey sin x。二阶偏导数用于判断凹凸性以及驻点分类。
4. Mixed Partial Derivatives and Symmetry | 混合偏导数与对称性
A key property is Clairaut’s theorem: if the second-order mixed partial derivatives fxy and fyx are continuous on a region, then they are equal. In CIE exams, you are often asked to verify this equality for a given function by computing both fxy and fyx and confirming they match.
一个重要性质是克莱罗定理：若二阶混合偏导数 fxy 和 fyx 在某区域内连续，则二者相等。在 CIE 考试中，常会要求对给定函数计算 fxy 和 fyx，验证它们相等。
For f(x, y) = x2y3 + x ln y, we find fx = 2xy3 + ln y ⇒ fxy = 6xy2 + 1/y. From fy = 3x2y2 + x/y ⇒ fyx = 6xy2 + 1/y. The equality fxy = fyx holds as expected.
对于 f(x, y) = x2y3 + x ln y，可得 fx = 2xy3 + ln y ⇒ fxy = 6xy2 + 1/y。由 fy = 3x2y2 + x/y ⇒ fyx = 6xy2 + 1/y。fxy = fyx 成立，符合预期。
5. The Chain Rule in Partial Differentiation | 偏微分中的链式法则
When intermediate variables are functions of another parameter, we use the tree diagram and the chain rule. If z = f(x, y), with x = x(t) and y = y(t), then the total derivative with respect to t is dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt). This extends to more variables analogously.
当中间变量又是另一个参数的函数时，需使用树形图和链式法则。若 z = f(x, y)，且 x = x(t)，y = y(t)，则对 t 的全导数为 dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)。这一法则可类似推广到更多变量。
For example, let z = x2y + y3, x = sin t, y = cos t. Then ∂z/∂x = 2xy, ∂z/∂y = x2 + 3y2. So dz/dt = 2xy·cos t + (x2 + 3y2)(–sin t). Substitute x and y to express the result in terms of t.
例如，设 z = x2y + y3，x = sin t，y = cos t。则 ∂z/∂x = 2xy，∂z/∂y = x2 + 3y2。于是 dz/dt = 2xy·cos t + (x2 + 3y2)(–sin t)。代入 x, y 即可用 t 表示结果。
6. Implicit Partial Differentiation | 隐函数偏微分
If an equation 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. The implicit function theorem gives ∂z/∂x = –Fx/Fz and ∂z/∂y = –Fy/Fz, provided Fz ≠ 0.
若方程 F(x, y, z) = 0 将 z 隐含地定义为 x 和 y 的函数，则无需解出 z 的显式即可求得 ∂z/∂x 和 ∂z/∂y。隐函数定理给出 ∂z/∂x = –Fx/Fz，∂z/∂y = –Fy/Fz，前提是 Fz ≠ 0。
Given x2z + yz3 = 5, treat F(x, y, z) = x2z + yz3 – 5. Then Fx = 2xz, Fz = x2 + 3yz2. Hence ∂z/∂x = – (2xz) / (x2 + 3yz2). This technique avoids cumbersome algebra.
对于 x2z + yz3 = 5，令 F(x, y, z) = x2z + yz3 – 5。则 Fx = 2xz，Fz = x2 + 3yz2。因此 ∂z/∂x = – (2xz) / (x2 + 3yz2)。这一技巧可避免繁琐的代数运算。
7. Total Differential and Small Changes | 全微分与微小变化
The total differential of z = f(x, y) is dz = (∂f/∂x) dx + (∂f/∂y) dy. For small finite changes Δx, Δy, the corresponding approximate change in f is Δf ≈ (∂f/∂x) Δx + (∂f/∂y) Δy. This linear approximation is widely used in error estimation and sensitivity analysis.
z = f(x, y) 的全微分为 dz = (∂f/∂x) dx + (∂f/∂y) dy。对于有限微小变化 Δx、Δy，f 的近似变化量为 Δf ≈ (∂f/∂x) Δx + (∂f/∂y) Δy。这一线性近似广泛用于误差估计和灵敏度分析。
Suppose the volume of a cylinder is V = πr2h. If r increases by 0.1 cm and h decreases by 0.2 cm, the approximate change in V is ΔV ≈ (∂V/∂r) Δr + (∂V/∂h) Δh = 2πrh
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