📚 A-Level OCR Maths: Partial Differentiation Essentials | A-Level OCR 数学：偏微分 考点精讲
Partial differentiation extends ordinary differentiation to functions of several variables, a central topic in OCR A-Level Further Mathematics. It allows us to measure how a function changes when we vary just one of its independent variables while keeping the others fixed. Mastery of this area opens the door to multivariable optimisation, error analysis and even quantum mechanics.
偏微分将普通微分推广到多变量函数，是 OCR A-Level 进阶数学的核心主题。它使我们能够衡量仅改变一个自变量而固定其他自变量时函数如何变化。掌握这一领域将为进一步学习多元优化、误差分析乃至量子力学打开大门。
1. What is Partial Differentiation? | 什么是偏微分？
ect to x is obtained by differentiating f as if y were a constant. The notation ∂f/∂x or fx is used. Geometrically, ∂f/∂x gives the slope of the tangent line to the surface in the x‑direction.
对于函数 z = f(x, y)，关于 x 的偏导数是将 y 视为常数对 f 求导得到的。符号 ∂f/∂x 或 fx 表示这一点。从几何上看，∂f/∂x 给出曲面在 x 方向的切线的斜率。
Similarly, the partial derivative with respect to y, ∂f/∂y or fy, treats x as constant. The process simply applies the usual differentiation rules while regarding all other variables as constants.
类似地，关于 y 的偏导数 ∂f/∂y 或 fy 将 x 视为常数。这一过程只是运用通常的求导法则，同时将所有其他变量当作常数处理。
2. Computing First-Order Partials | 一阶偏导数的计算
derivative of sin(xy) w.r.t. x is y cos(xy) (chain rule). Hence ∂f/∂x = 3x²y + y cos(xy).
对于 f(x,y)=x³y + sin(xy)，求 ∂f/∂x：逐项求导，x³y 关于 x 的导数为 3x²y（y 视为常数），sin(xy) 关于 x 的导数为 y cos(xy)（链式法则）。因此 ∂f/∂x = 3x²y + y cos(xy)。
+ x cos(xy). Always practise with polynomial, exponential and trigonometric combinations.
对于 ∂f/∂y，将 x 视为常数：∂/∂y(x³y)=x³，∂/∂y(sin(xy))=x cos(xy)。因此 ∂f/∂y = x³ + x cos(xy)。务必多加练习多项式、指数函数与三角函数的组合。
3. Higher-Order Derivatives & Clairaut’s Theorem | 高阶偏导数与克莱罗定理
at if the second-order mixed partials are continuous, they are equal: ∂²f/∂x∂y = ∂²f/∂y∂x.
二阶偏导数写作 ∂²f/∂x²、∂²f/∂y² 以及混合偏导数 ∂²f/∂x∂y 和 ∂²f/∂y∂x。克莱罗定理指出，若二阶混合偏导数连续，则它们相等：∂²f/∂x∂y = ∂²f/∂y∂x。
This symmetry is a powerful check in exam problems. For f(x,y)=x²ey, we have fxx=2ey, fyy=x²ey, and fxy=fyx=2x ey.
这一对称性是考试中强大的检验工具。对于 f(x,y)=x²ey，有 fxx=2ey，fyy=x²ey，且 fxy=fyx=2x ey。
4. Chain Rule for Several Variables | 多元链式法则
f z = f(u,v) where u and v are functions of t, then the total derivative dz/dt is given by dz/dt = (∂z/∂u)(du/dt) + (∂z/∂v)(dv/dt) , the partial derivatives are obtained similarly: ∂z/∂x = (∂z/∂u)(∂u/∂x) + (∂z/∂v)(∂v/∂x).
若 z = f(u,v) 且 u 与 v 是 t 的函数，则全导数 dz/dt 由 dz/dt = (∂z/∂u)(du/dt) + (∂z/∂v)(dv/dt) 给出。如果 u 和 v 是 x 与 y 的函数，偏导数类似地得到：∂z/∂x = (∂z/∂u)(∂u/∂x) + (∂z/∂v)(∂v/∂x)。
These patterns extend to any number of intermediate variables. The OCR exam often embeds the chain rule within a word problem, such as a rate of change of volume under varying dimensions.
这些模式可推广到任意数量的中间变量。OCR 考试常将链式法则嵌入文字题，例如尺寸变化时体积的变化率。
5. Implicit Partial Differentiation | 隐函数偏微分
,y,z)=0 defines z implicitly as a function of x and y. Provided ∂F/∂z ≠ 0, we have ∂z/∂x = –(∂F/∂x)/(∂F/∂z) and ∂z/∂y = –(∂F/∂y)/(∂F/∂z).
方程 F(x,y,z)=0 将 z 隐含地定义为 x 和 y 的函数。若 ∂F/∂z ≠ 0，则有 ∂z/∂x = –(∂F/∂x)/(∂F/∂z) 以及 ∂z/∂y = –(∂F/∂y)/(∂F/∂z)。
For instance, if x² + y² + z² = 1, then let F = x² + y² + z² – 1 = 0. Hence ∂z/∂x = –x/z and ∂z/∂y = –y/z. This avoids solving for z explicitly.
例如，若 x² + y² + z² = 1，令 F = x² + y² + z² – 1 = 0。于是 ∂z/∂x = –x/z，∂z/∂y = –y/z。这样就避免了显式求解 z。
6. Stationary Points of a Function of Two Variables | 二元函数的驻点
ously: ∂f/∂x = 0 and ∂f/∂y = 0. These points can be local maxima, minima or saddle points.
当两个一阶偏导数同时为零时，即 ∂f/∂x = 0 且 ∂f/∂y = 0，该点即为驻点。这些点可能是局部极大值、极小值或鞍点。
Solving the system fx=0, fy=0 often involves simultaneous equations. For f(x,y)=x³ – 3xy + y³, we find fx=3x² – 3y = 0 ⇒ y = x², and fy= –3x + 3y² = 0 ⇒ x = y², leading to (0,0) and (1,1).
求解方程组 fx=0, fy=0 常涉及联立方程。对于 f(x,y)=x³ – 3xy + y³，由 fx=3x² – 3y = 0 得 y = x²，由 fy= –3x + 3y² = 0 得 x = y²，从而得到 (0,0) 和 (1,1)。
7. Classifying Stationary Points Using the Discriminant | 使用判别式进行驻点分类
Define the second-derivative discriminant D = fxx fyy – (fxy)² evaluated at the stationary point. The classification rules are summarised below.
定义二阶导数判别式 D = fxx fyy – (fxy)²，并在驻点处求值。分类规则总结如下。
| Condition | Classification |
|---|---|
| D > 0 and fxx > 0 | Local minimum |
| D > 0 and fxx < 0 | Local maximum |
| D < 0 | Saddle point |
| D = 0 | Inconclusive (further investigation needed) |
For f(x,y)=x³ – 3xy + y³ at (1,1): fxx=6x=6, fyy=6y=6, fxy= –3, so D = 6×6 – (–3)² = 36 – 9 = 27 > 0, and fxx > 0, hence a local minimum.
对于 f(x,y)=x³ – 3xy + y³ 在 (1,1) 处：fxx=6x=6，fyy=6y=6，fxy= –3，因此 D = 6×6 – (–3)² = 36 – 9 = 27 > 0，且 fxx > 0，故为局部极小点。
8. Constrained Optimisation: Lagrange Multipliers | 约束优化：拉格朗日乘数法
of f(x,y) subject to a constraint g(x,y)=0, we introduce a Lagrange multiplier λ and solve ∇f = λ ∇g together with g=0. The auxiliary function is L(x,y,λ) = f(x,y) – λ g(x,y).
为求解 f(x,y) 在约束条件 g(x,y)=0 下的驻值，我们引入拉格朗日乘子 λ，并求解 ∇f = λ ∇g 与 g=0 联立。辅助函数为 L(x,y,λ) = f(x,y) – λ g(x,y)。
0. This gives three equations to determine the critical coordinates (x,y) and λ. λ itself often has a real-world interpretation, such as a marginal rate of change.
然后令 ∂L/∂x = 0，∂L/∂y = 0 以及 ∂L/∂λ = 0。这样就得到三个方程，可确定关键坐标 (x,y) 和 λ。λ 本身常具有实际意义，例如边际变化率。
9. Worked Application Example | 应用题示例
sum is 100. Let f = xyz subject to x+y+z=100. Using two independent variables: set z = 100 – x – y, then maximise P = xy(100 – x – y).
问题：求三个正数 x、y、z 的最大乘积，其和为 100。令 f = xyz，约束条件为 x+y+z=100。使用两个自变量：设 z = 100 – x – y，然后最大化 P = xy(100 – x – y)。
Compute Px = y(100 – 2x – y) = 0, Py = x(100 – x – n z = 100/3. The discriminant confirms a maximum. Thus equal numbers maximise the product.
计算 Px = y(100 – 2x – y) = 0，Py = x(100 – x – 2y) = 0。正数解给出 x = y = 100/3，进而 z = 100/3。判别式确认其为最大值。因此相等数值使乘积最大。
10. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧
- Always check that you have treated the other variables as constants when computing a partial derivative.
- 在计算偏导数时，务必确认已将其他变量视为常数。
- Do not forget the chain rule in mixed compositions; introduce intermediate variables clearly.
- 在混合复合函数中不要忘记链式法则；清晰引入中间变量。
- After finding stationary points, verify that you have evaluated the discriminant D correctly and identified the nature of each point.
- 找到驻点后，应验证正确计算了判别式 D 并确定了每个点的性质。
- Lagrange problems, ensure the constraint is set to zero, g(x,y)=0, before constructing L.
- 处理拉格朗日问题时，确保在构建 L 之前将约束写成 g(x,y)=0 的形式。
- Practice with a variety of functions: polynomials, exponentials, trig, and those requiring product/quotient rules.
- 练习各种函数：多项式、指数函数、三角函数，以及需要乘积法则和商法则的函数。
Exam questions frequently combine finding partial derivatives, locating stationary points and classifying them within a single long problem. Time management and systematic working are key.
考题经常将求偏导数、寻找驻点并分类组合在一个长题中。时间管理和系统的解题步骤是关键。
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