📚 ESAT Mathematics: Calculus & Differentiation | ESAT数学：微积分与微分专题
Calculus is the mathematics of change and motion, and differentiation sits at its very core. For ESAT candidates, mastering differentiation is not merely a syllabus requirement — it is the engine that powers physics kinematics questions, optimisation problems, and countless engineering applications. This guide walks through every differentiation technique you need, from first principles to implicit differentiation, with exam-specific insights throughout.
微积分是研究变化与运动的数学，而微分正是其核心。对于ESAT考生而言，掌握微分不仅是考纲要求，更是解决物理运动学问题、优化问题以及无数工程应用的关键引擎。本指南系统梳理你所需的一切求导技巧——从第一性原理到隐函数微分，全程贯穿ESAT考试要点。
1. What Is Differentiation? | 什么是微分？
nt x. Physically, it measures the instantaneous rate of change: velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity.
微分衡量的是函数随输入变化而变化的快慢。在几何上，导数 f′(x) 表示曲线 y = f(x) 在给定点 x 处切线的斜率。在物理上，它度量瞬时变化率：速度是位移对时间的导数，加速度是速度对时间的导数。
ESAT insight: Expect questions where a graph or a physical scenario requires you to identify which quantity is the derivative of another — velocity from a displacement-time graph is a classic example.
ESAT要点：考试中常出现这样的题目——根据图形或物理情境判断哪个量是另一个量的导数，位移-时间图像中的速度就是一个经典例子。
2. Limits and First Principles | 极限与第一性原理
The formal definition of the derivative is the limit of the average rate of change as the interval approaches zero. For a function f(x), the derivative at point x is defined as:
f′(x) = lim(h→0) [f(x+h) − f(x)] / h
For f(x) = x², we apply the definition directly:
f′(x) = lim(h→0) [(x+h)² − x²] / h = lim(h→0) (2xh + h²) / h = lim(h→0) (2x + h) = 2x
First-principles questions in ESAT typically involve expanding f(x+h), cancelling common factors, and then evaluating the limit. Practice this process for quadratic, cubic, and simple rational functions to build fluency.
ESAT中的第一性原理题通常需要展开 f(x+h)、消去公因式、再求极限。请针对二次、三次及简单有理函数反复练习这一流程，以建立熟练度。
Common mistake: forgetting that h is the variable approaching zero, not x. Keep x fixed while evaluating the limit in h.
常见错误：忘记趋近于零的变量是 h 而非 x。在求关于 h 的极限时，x 应被视为固定常数。
3. The Power Rule and Basic Operations | 幂法则与基本运算
(f + g)′ = f′ + g′. For example, if f(x) = 3x⁴ − 2x³ + x − 5, then f′(x) = 12x³ − 6x² + 1.
若 f(x) = xⁿ，则 f′(x) = n·xⁿ⁻¹。幂法则可自然推广到常数倍与函数之和：(cf)′ = c·f′，(f + g)′ = f′ + g′。例如，若 f(x) = 3x⁴ − 2x³ + x − 5，则 f′(x) = 12x³ − 6x² + 1。
ractional exponents. For instance, f(x) = 1/x = x⁻¹ gives f′(x) = −x⁻² = −1/x², and f(x) = √x = x1/2 gives f′(x) = (1/2)x−1/2 = 1/(2√x).
幂法则同样适用于负指数和分数指数。例如，f(x) = 1/x = x⁻¹ 的导数为 f′(x) = −x⁻² = −1/x²；f(x) = √x = x1/2 的导数为 f′(x) = (1/2)x−1/2 = 1/(2√x)。
ESAT tip: Rewrite functions in index form before differentiating — this turns seemingly complicated expressions into straightforward power-rule applications.
ESAT技巧：求导前先将函数改写为指数形式——这样看似复杂的表达式就变成了幂法则的直接应用。
4. Product Rule and Quotient Rule | 乘积法则与商法则
When two functions are multiplied, the derivative is not simply the product of their derivatives. The product rule states:
(fg)′ = f′g + fg′
x) = x²·sin(x) gives h′(x) = 2x·sin(x) + x²·cos(x). Note that the order of differentiation — differentiating one factor at a time while keeping the other unchanged — is the essence of this rule.
当两个函数相乘时，导数并不是两者导数的简单乘积。乘积法则表明：
(fg)′ = f′g + fg′
例如，h(x) = x²·sin(x) 的导数为 h′(x) = 2x·sin(x) + x²·cos(x)。注意，该法则的本质是——每次只对一个因子求导，另一个因子保持不变，然后将它们相加。
For division, the quotient rule states that for f(x) = u/v:
(u/v)′ = (u′v − uv′) / v²
A useful mnemonic is "low d-high minus high d-low, over low squared". For example, f(x) = x/(x² + 1):
对于除法，商法则表明，若 f(x) = u/v，则：
(u/v)′ = (u′v − uv′) / v²
一个实用的口诀是"上导下不导，减去下导上不导，除以分母平方"。例如，f(x) = x/(x² + 1)：
f′(x) = [1·(x² + 1) − x·2x] / (x² + 1)² = (1 − x²) / (x² + 1)²
ESAT practice: Product and quotient rules frequently appear alongside trigonometric and exponential functions — drill these combined questions to build speed.
ESAT练习：乘积法则和商法则经常与三角函数、指数函数结合出题——多练习这类综合题以提升速度。
5. The Chain Rule | 链式法则
The chain rule handles composite functions — functions inside functions. If y = f(u) and u = g(x), then:
dy/dx = dy/du × du/dx
Consider y = (3x² + 1)⁵. Let u = 3x² + 1. Then dy/du = 5u⁴ and du/dx = 6x, so:
链式法则用于处理复合函数——即函数嵌套函数的场景。若 y = f(u) 且 u = g(x)，则：
dy/dx = dy/du × du/dx
考虑 y = (3x² + 1)⁵。令 u = 3x² + 1，则 dy/du = 5u⁴，du/dx = 6x，因此：
dy/dx = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴
The chain rule also works in reverse for differentiation of expressions like ln(sin x) or ex³. For y = ex³, let u = x³, giving dy/dx = 3x²·ex³.
链式法则同样适用于反向情形，如 ln(sin x) 或 ex³ 的求导。对于 y = ex³，令 u = x³，得到 dy/dx = 3x²·ex³。
ESAT tip: Recognise the "outer function–inner function" structure instantly. In multiple-choice questions, the correct answer often differs from distractors only by the chain-rule factor — a missing du/dx is the most common trap.
ESAT技巧：迅速识别"外层函数-内层函数"的结构。在选择题中，正确答案与干扰项的差别往往仅在于链式法则的因子——漏掉 du/dx 是最常见的陷阱。
6. Exponential, Logarithmic and Trigonometric Derivatives | 指数、对数与三角函数的导数
Beyond polynomials, ESAT requires fluency with the derivatives of standard functions. The table below summarises the essential results:
除多项式外，ESAT要求熟练掌握标准函数的导数。下表总结了核心求导公式：
| f(x) | f′(x) |
| ex | ex |
| ekx | k·ekx |
| ln x | 1/x |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec²x |
Two key patterns emerge. First, the exponential function ex is its own derivative — a unique and elegant property. Second, trigonometric derivatives involve a sign change: the derivative of cos x is −sin x, which often causes sign errors under time pressure.
这里呈现出两个关键规律。第一，指数函数 ex 的导数是其自身——这是独一无二且优雅的性质。第二，三角函数的导数涉及符号变化：cos x 的导数是 −sin x，在时间压力下常导致符号错误。
ntegrate with the chain rule: d/dx[sin(2x)] = 2cos(2x), and d/dx[ln(3x + 1)] = 3/(3x + 1).
对于更复杂的组合，这些公式需要与链式法则结合：d/dx[sin(2x)] = 2cos(2x)，d/dx[ln(3x + 1)] = 3/(3x + 1)。
ESAT practice: Many papers include a question testing d/dx[a·ebx] or d/dx[ln(kx)] — these should be completed in under 30 seconds each.
ESAT练习：许多试卷中会出现 d/dx[a·ebx] 或 d/dx[ln(kx)] 的题目——这类题应在30秒内完成。
7. Implicit and Parametric Differentiation | 隐函数与参数方程微分
, where y is defined implicitly. Differentiate both sides with respect to x, remembering that y is a function of x:
并非所有函数都以显式形式 y = f(x) 给出。隐函数微分用于处理诸如 x² + y² = 25 的方程，其中 y 以隐式定义。对方程两边同时对 x 求导，记住 y 是 x 的函数：
2x + 2y·(dy/dx) = 0 → dy/dx = −x/y
For parametric equations, where x = f(t) and y = g(t), the chain rule provides:
dy/dx = (dy/dt) / (dx/dt)
= 3, giving dy/dx = 3/(2t). Note that the final answer is expressed in terms of t, not x.
对于参数方程，其中 x = f(t) 且 y = g(t)，链式法则给出：
dy/dx = (dy/dt) / (dx/dt)
例如，若 x = t²，y = 3t，则 dx/dt = 2t，dy/dt = 3，因此 dy/dx = 3/(2t)。注意最终答案以 t 表示，而非 x。
ESAT tip: Implicit differentiation of circles and ellipses is a recurring favourite — memorise that for x² + y² = r², the slope is always −x/y. This saves valuable time.
ESAT技巧：圆和椭圆的隐函数微分是高频考点——记住对于 x² + y² = r²，切线斜率恒为 −x/y。这会为你节省宝贵时间。
8. Second Derivatives and Their Significance | 二阶导数及其意义
The second derivative, denoted f″(x) or d²y/dx², is the derivative of the derivative. It reveals crucial information about the function's curvature and is essential for classifying stationary points.
二阶导数，记作 f″(x) 或 d²y/dx²，是导数的导数。它揭示函数曲率的关键信息，对于分类驻点至关重要。
Interpretations of the second derivative include:
二阶导数的意义包括：
Concavity: If f″(x) > 0 on an interval, the curve is concave up (shaped like a cup ∪); if f″(x) < 0, the curve is concave down (shaped like a cap ∩).
凹向：若在区间上 f″(x) > 0，曲线凹向上（形如杯 ∪）；若 f″(x) < 0，曲线凹向下（形如帽 ∩）。
Acceleration: In kinematics, d²x/dt² is acceleration — the rate of change of velocity, which itself is dx/dt.
加速度：在运动学中，d²x/dt² 是加速度——即速度（dx/dt）的变化率。
Points of inflection: oncavity. These points satisfy f″(x) = 0, but this condition is necessary, not sufficient.
拐点：当 f″(x) 改变符号时，曲线改变凹向。拐点满足 f″(x) = 0，但该条件是必要非充分的。
ls where the graph is concave up or down. Compute f″(x) = 6x − 12 and solve the inequality — this is a reliable 1-minute question.
ESAT练习：典型题目为 y = x³ − 6x² + 9x，要求判断图像在哪些区间凹向上或凹向下。计算 f″(x) = 6x − 12 并求解不等式——这是一道可靠的一分钟题。
9. Tangents, Normals and Stationary Points | 切线与法线、驻点与最值
At a given point (a, f(a)), the equation of the tangent line is found using the derivative as the slope:
y − f(a) = f′(a)(x − a)
The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: −1/f′(a), provided f′(a) ≠ 0.
在给定点 (a, f(a)) 处，切线方程利用导数作为斜率来求：
y − f(a) = f′(a)(x − a)
法线与切线垂直，因此其斜率为导数的负倒数：−1/f′(a)，前提是 f′(a) ≠ 0。
s occur where f′(x) = 0. To classify them, use either the first or second derivative test:
First derivative test: Examine the sign of f′(x) just before and after the stationary point. A change from + to − indicates a local maximum; − to + indicates a local minimum; no sign change indicates a point of inflection.
一阶导数检验：考察驻点前后 f′(x) 的符号。从 + 变为 − 表示局部极大值；从 − 变为 + 表示局部极小值；符号不变则为拐点。
Second derivative test: is a local minimum; if f″(x) < 0, a local maximum; if f″(x) = 0, the test is inconclusive.
二阶导数检验：若驻点处 f″(x) > 0，则为局部极小值；若 f″(x) < 0，则为局部极大值；若 f″(x) = 0，则该检验无法判定。
you to compute the derivative, solve f′(x) = 0, and then apply the second derivative test — a three-step process worth mastering.
ESAT技巧：驻点相关问题在数学部分几乎是必考题。通常需要你求导、解方程 f′(x) = 0、再用二阶导数检验分类——这是一个三步流程，务必熟练掌握。
10. Optimisation and ESAT Exam Strategy | 最优化问题与ESAT应试策略
Optimisation applies differentiation to real-world problems: finding the maximum or minimum value of a quantity subject to constraints. The systematic approach is:
Define the variable(s) and express the quantity to be optimised as a function of one variable.
Differentiate the function and set the derivative equal to zero: f′(x) = 0.
Solve for x and verify the nature (max or min) using the second derivative test or sign analysis.
For closed intervals, also evaluate the function at the endpoints.
最优化问题将微分应用于现实场景：在约束条件下求某个量的最大值或最小值。系统化步骤如下：
定义变量，并将待优化的量表达为单变量函数。
对函数求导并令导数为零：f′(x) = 0。
解方程求 x，并用二阶导数检验或符号分析验证性质（最大值还是最小值）。
对于闭区间问题，还需计算区间端点的函数值。
Classic ESAT-style optimisation questions include: maximising the area of a rectangle with a fixed perimeter, minimising the surface area of a cylinder for a given volume, and finding the maximum height of a projectile from a quadratic position function.
典型的ESAT风格优化题包括：固定周长下使矩形面积最大化、给定体积下使圆柱表面积最小化，以及从二次位置函数求抛射体的最大高度。
ESAT exam strategy: The mathematics component is a multiple-choice test with roughly 50 minutes for 27 questions. Differentiation questions typically alternate between routine computations and conceptual applications. Allocate no more than 1.5–2 minutes per question. If a computation feels convoluted, look for a shortcut — many questions test the product rule or chain rule in isolation, and distractors often reveal the expected method.
ESAT应试策略：数学部分为选择题，约50分钟完成27题。微分题通常在常规计算与概念应用之间交替出现。每题建议控制在1.5–2分钟以内。如果计算过于复杂，寻找捷径——许多题目只单独考查乘积法则或链式法则，干扰项往往能揭示出题者的意图。
Final revision checklist: power rule with fractional and negative indices, product and quotient rules, chain rule fluency, standard derivatives of trigonometric/exponential/logarithmic functions, implicit differentiation, parametric differentiation, second derivative and concavity, tangent/normal equations, stationary point classification, and optimisation setup. Master each block individually, then combine, and differentiation will become one of the most reliable scoring areas in your ESAT paper.
最终复习清单：幂法则（含分数与负指数）、乘积法则与商法则、链式法则熟练度、三角函数/指数/对数函数的标准导数、隐函数微分、参数方程微分、二阶导数与凹凸性、切线与法线方程、驻点分类、优化建模。逐块掌握，再融会贯通，微分将成为你在ESAT考试中最稳定的得分板块之一。
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