GCSE WJEC Mathematics: Basics of Calculus | GCSE WJEC 数学：微积分基础考点精讲

📚 GCSE WJEC Mathematics: Basics of Calculus | GCSE WJEC 数学：微积分基础考点精讲

Calculus is a branch of mathematics that studies how things change. It is divided into two main areas: differentiation, which finds rates of change and slopes of curves, and integration, which finds areas under curves and the accumulation of quantities. In GCSE WJEC Mathematics, you are introduced to the fundamental ideas of differentiation and basic integration, providing a foundation for further study.

微积分是研究变化的一个数学分支。它主要分为两部分：微分用于求解变化率和曲线斜率；积分用于计算曲线下的面积和数量的累积。在 GCSE WJEC 数学中，你将学习微分和基本积分的基础概念，为进阶学习打下基础。

1. What is Calculus? | 什么是微积分？

Calculus allows us to model and analyse real-world situations involving motion, growth and optimisation. The two central operations – differentiation and integration – are inverse processes: differentiation gives the gradient of a function, while integration recovers the original function from its gradient. At GCSE level, we focus on simple algebraic functions and polynomial expressions.

微积分让我们能够对涉及运动、增长和最优化的现实问题进行建模与分析。两个核心运算——微分和积分——是互逆的过程：微分给出函数的梯度，而积分则根据梯度恢复原函数。在 GCSE 阶段，我们主要处理简单的代数函数和多项式表达式。

2. Gradient of a Curve | 曲线的斜率

For a straight line, the gradient is constant. For a curve, the gradient varies from point to point. The gradient at a particular point on the curve y = f(x) is given by the derivative, denoted f'(x) or dy/dx. Geometrically, this is the slope of the tangent touching the curve at that point. Understanding this concept is the first step towards differentiation.

对于直线，斜率是恒定的。对于曲线，斜率随点的位置而变化。曲线 y = f(x) 在某点的斜率由导数给出，记作 f'(x) 或 dy/dx。从几何上看，这就是在该点接触曲线的切线的斜率。理解这一概念是学习微分的第一步。

3. The Power Rule for Differentiation | 幂函数的微分法则

The power rule is the most important differentiation rule at GCSE. It states that if y = xⁿ, then the derivative is dy/dx = n xⁿ⁻¹. You multiply by the original power and then reduce the power by one.

幂法则是 GCSE 阶段最重要的微分法则。该法则指出，若 y = xⁿ，则其导数为 dy/dx = n xⁿ⁻¹。你先乘以原来的指数，然后将指数减一。

If y = xⁿ, then dy/dx = n xⁿ⁻¹

The table below shows some common functions and their derivatives using the power rule.

下表列出了一些常见函数及其通过幂法则得到的导数。

Function f(x)	Derivative f'(x)
x²	2x
x³	3x²
x⁵	5x⁴
x⁻¹	−x⁻²
x^½	½ x^−½

Note that the power rule also works for negative and fractional powers, once you rewrite roots and reciprocals using indices.

注意，幂法则同样适用于负指数和分数指数，只要你将根式和倒数用指数形式重新表示。

4. Differentiating Polynomials | 多项式的微分

To differentiate a polynomial, you apply the power rule to each term separately. If f(x) = a xⁿ + b x^m + c, then f'(x) = a·n xⁿ⁻¹ + b·m x^m−1. Constant terms become zero because the derivative of a constant vanishes.

要对多项式进行微分，你须对每一项分别应用幂法则。若 f(x) = a xⁿ + b x^m + c，那么 f'(x) = a·n xⁿ⁻¹ + b·m x^m−1。常数项的导数为零，因为常数的导数总为零。

Example: Differentiate y = 4x³ − 2x² + 5x − 7.

例：对 y = 4x³ − 2x² + 5x − 7 进行微分。

dy/dx = 4×3 x² − 2×2 x + 5 − 0 = 12x² − 4x + 5. Always show the step-by-step working to avoid mistakes with coefficients.

dy/dx = 4×3 x² − 2×2 x + 5 − 0 = 12x² − 4x + 5。务必给出逐步计算过程，以避免系数错误。

5. Tangent to a Curve | 曲线的切线

Once you know the derivative, you can find the equation of the tangent at any point on the curve. The derivative evaluated at x = a gives the gradient m of the tangent. Then you use the point (a, f(a)) and the straight-line formula y − y₁ = m(x − x₁).

一旦你知道了导数，就可以求出曲线上任意一点处的切线方程。将 x = a 代入导数，即得切线的斜率 m。然后利用点 (a, f(a)) 和直线公式 y − y₁ = m(x − x₁) 求得方程。

Example: For the curve y = x² + 3x, find the tangent at x = 2.

例：对于曲线 y = x² + 3x，求 x = 2 处的切线方程。

First, differentiate: dy/dx = 2x + 3. At x = 2, m = 2(2) + 3 = 7. The point on the curve is (2, 2²+6) = (2,10). The tangent equation is y − 10 = 7(x − 2), which simplifies to y = 7x − 4.

首先求导：dy/dx = 2x + 3。当 x = 2 时，m = 2(2) + 3 = 7。曲线上的点为 (2, 10)。切线方程为 y − 10 = 7(x − 2)，化简得 y = 7x − 4。

6. Stationary Points | 驻点

A stationary point occurs where the gradient of the curve is zero, i.e. dy/dx = 0. These points can be maximums, minimums or points of inflection. To find stationary points, differentiate the function, set the derivative equal to zero and solve for x. Substitute back to get the y-coordinates.

驻点出现在曲线斜率为零处，即 dy/dx = 0。这些点可能是极大值点、极小值点或拐点。要找出驻点，先对函数求导，令导数为零并解出 x 值，再将 x 代回原函数得到 y 坐标。

Example: Find the stationary points of y = x³ − 3x.

例：求 y = x³ − 3x 的驻点。

dy/dx = 3x² − 3. Set 3x² − 3 = 0 → x² = 1 → x = 1 or x = −1. The corresponding y values are y(1) = −2, y(−1) = 2. So the stationary points are (1, −2) and (−1, 2).

dy/dx = 3x² − 3。令 3x² − 3 = 0 → x² = 1 → x = 1 或 x = −1。相应的 y 值为 y(1) = −2，y(−1) = 2。因此驻点为 (1, −2) 和 (−1, 2)。

7. Nature of Stationary Points (Second Derivative) | 驻点的性质（二阶导数检验）

To determine whether a stationary point is a maximum or minimum, use the second derivative f”(x). If f”(x) > 0 at the point, the curve is concave up and it is a minimum. If f”(x) < 0, the curve is concave down and it is a maximum. If f''(x) = 0, further investigation is needed.

要判断驻点是极大还是极小，需使用二阶导数 f”(x)。若在该点 f”(x) > 0，曲线呈凹向上，该点为极小值点。若 f”(x) < 0，曲线呈凹向下，该点为极大值点。若 f''(x) = 0，则需进一步分析。

Continuing the example y = x³ − 3x, we have the second derivative d²y/dx² = 6x. At x = 1, f”(1) = 6 > 0, hence a minimum at (1, −2). At x = −1, f”(−1) = −6 < 0, hence a maximum at (−1, 2). This method confirms the nature efficiently.

继续上面的例子 y = x³ − 3x，其二阶导数为 d²y/dx² = 6x。在 x = 1 处，f”(1) = 6 > 0，因此 (1, −2) 为极小值点。在 x = −1 处，f”(−1) = −6 < 0，因此 (−1, 2) 为极大值点。该方法能高效确认驻点性质。

8. Introduction to Integration (Reverse Differentiation) | 积分入门（微分的逆运算）

Integration is the reverse process of differentiation. If we know the derivative f'(x), finding the original function f(x) is called integration. The result is an antiderivative, and because the derivative of a constant is zero, we always add a constant of integration C.

积分是微分的逆过程。若已知导数 f'(x)，确定原函数 f(x) 的过程就称为积分。所得结果是一个原函数，并且由于常数的导数为零，我们总是加上一个积分常数 C。

For example, if f'(x) = 2x, then F(x) = x² + C. The rule for integrating powers is essentially the reverse of the power rule for derivatives.

例如，若 f'(x) = 2x，则 F(x) = x² + C。幂函数积分的法则本质上是导数幂法则的逆操作。

∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, for n ≠ −1

To integrate a polynomial, integrate term by term and include the constant at the end.

对于多项式的积分，逐项积分并在最后加上常数 C。

9. Indefinite Integrals & Constant of Integration | 不定积分与积分常数

An indefinite integral gives a family of functions differing by a constant. When we write ∫ f(x) dx, we mean all antiderivatives of f(x). In practical problems, an extra condition (like a point on the curve) allows you to find the specific value of C.

不定积分给出的是一族仅相差一个常数的函数。当我们写 ∫ f(x) dx 时，指的是 f(x) 的所有原函数。在实际问题中，附加条件（例如曲线上的一个点）可让你求出具体的 C 值。

Example: Given dy/dx = 6x² and the curve passes through (1, 5), find y.

例：已知 dy/dx = 6x² 且曲线经过点 (1, 5)，求 y。

Integrate: y = ∫ 6x² dx = 6×(x³/3) + C = 2x³ + C. Using y(1)=5: 2(1)³ + C = 5 → C = 3. So y = 2x³ + 3.

积分得 y = ∫ 6x² dx = 6×(x³/3) + C = 2x³ + C。利用 y(1)=5：2(1)³ + C = 5 → C = 3。因此 y = 2x³ + 3。

10. Definite Integrals and Area Under a Curve | 定积分与曲线下的面积

A definite integral calculates the exact area between a curve, the x-axis and two vertical lines x = a and x = b. It is written as ∫ₐ^b f(x) dx. The fundamental theorem of calculus tells us to find an antiderivative F(x), then evaluate F(b) − F(a).

定积分计算的是曲线、x 轴以及两条竖直线 x = a 和 x = b 之间所围成的准确面积，记作 ∫ₐ^b f(x) dx。微积分基本定理告诉我们，先找出一个原函数 F(x)，然后计算 F(b) − F(a)。

∫ₐ^b f(x) dx = [F(x)]ₐ^b = F(b) − F(a)

Example: Find the area under y = 2x + 1 from x = 1 to x = 3.

例：求 y = 2x + 1 在 x = 1 到 x = 3 区间下的面积。

Integrate: F(x) = x² + x. Then area = F(3) − F(1) = (9+3) − (1+1) = 12 − 2 = 10 square units. Always check whether the curve goes below the x-axis, as that would need careful handling of signs.

积分得 F(x) = x² + x。面积 = F(3) − F(1) = (9+3) − (1+1) = 12 − 2 = 10 平方单位。务必检查曲线是否落到 x 轴下方，如有则需要小心处理符号。

11. Applications and Problem-Solving | 应用与解题技巧

In WJEC GCSE, calculus is often applied to motion: if displacement s is a function of time t, then velocity v = ds/dt and acceleration a = dv/dt. You can also use integration to go from acceleration to velocity to displacement, given initial conditions. Optimisation problems (maximising area or minimising cost) also rely on finding stationary points via differentiation.

在 WJEC GCSE 中，微积分常应用于运动学：若位移 s 是时间 t 的函数，则速度 v = ds/dt，加速度 a = dv/dt。你也可以利用积分，在给定初始条件下，从加速度求出速度再求位移。最优化问题（最大化面积或最小化成本）也依赖于通过微分寻找驻点。

Tip: Always label what you are differentiating with respect to. For motion, clearly write ds/dt and dv/dt so the examiner knows you understand the rates of change. For optimisation, ensure you check the second derivative to confirm you have found the desired maximum or minimum.

提示：务必标明求导是相对于哪个变量。对于运动问题，清楚地写出 ds/dt 和 dv/dt，让考官知道你对变化率的理解。对于最优化，确保检验二阶导数来证实你找到了所需的极大值或极小值。

12. Summary & Key Formulas | 总结与关键公式

Master these core results for GCSE WJEC calculus success:

掌握以下核心结果，在 GCSE WJEC 微积分中取得成功：

Power Rule: d/dx (xⁿ) = n xⁿ⁻¹
Polynomial derivative: differentiate term by term.
Tangent line: y − y₁ = m(x − x₁) where m = dy/dx at the point.
Stationary points: solve dy/dx = 0.
Second derivative test: f”(x) > 0 → minimum; f”(x) < 0 → maximum.
Indefinite integral: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1).
Definite integral: ∫ₐ^b f(x) dx = F(b) − F(a).

中文关键公式清单：

幂函数求导：d/dx (xⁿ) = n xⁿ⁻¹
多项式导数：逐项求导。
切线方程：y − y₁ = m(x − x₁)，其中 m 为该点导数。
驻点：解 dy/dx = 0。
二阶导数检验：f”(x) > 0 → 极小；f”(x) < 0 → 极大。
不定积分：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)。
定积分：∫ₐ^b f(x) dx = F(b) − F(a)。

Practice with past paper questions and always show your working clearly – the method marks are as important as the final answer.

用历年真题反复练习，并始终清晰地展示解题步骤——过程分与最终答案同样重要。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

GCSE WJEC Mathematics: Basics of Calculus | GCSE WJEC 数学：微积分基础 考点精讲