📚 GCSE AQA Maths: Basics of Calculus Revision | GCSE AQA 数学：微积分基础 考点精讲
Calculus is one of the most powerful branches of mathematics, enabling us to analyse how quantities change and how they accumulate. In GCSE AQA Maths, the calculus topic introduces differentiation and basic integration, laying the groundwork for A-level study. You will learn to find gradients of curves, equations of tangents and normals, and reverse differentiation to find original functions.
微积分是数学中最强大的分支之一，使我们能够分析量如何变化以及如何累积。在 GCSE AQA 数学中，微积分专题介绍了求导和基础积分，为 A-level 学习打下基础。你将学习如何求曲线梯度、切线和法线方程，以及通过逆求导找出原函数。
1. What Is Calculus? | 什么是微积分？
Calculus is divided into two main parts: differentiation and integration. Differentiation deals with rates of change and slopes of curves, while integration deals with accumulation and areas under curves. In GCSE, you focus on differentiating polynomial functions and integrating to reverse the process.
微积分主要分为两部分：微分和积分。微分研究变化率和曲线斜率，积分研究累积和曲线下的面积。在 GCSE 中，你重点学习多项式函数的求导以及作为逆运算的积分。
A real-world example is velocity: if you know the distance–time function, differentiation gives the velocity (rate of change of distance). Similarly, integration of the velocity gives the distance travelled.
现实世界中的一个例子是速度：如果已知距离-时间函数，微分可得到速度（距离的变化率）。同样，对速度进行积分可得到所经过的距离。
2. The Gradient of a Curve | 曲线的梯度
For a straight line, the gradient is constant and easy to calculate. For a curve, the steepness changes at every point. To find the gradient at a specific point, we draw a tangent to the curve at that point and calculate its gradient. Differentiation gives us a formula to find this gradient without drawing.
对于直线，梯度是常数且容易计算。对于曲线，陡峭程度在每个点都不同。要找到某一点的梯度，我们在该点画出曲线的切线，然后计算其梯度。求导则给了我们一个无需画图就能求出该梯度的公式。
The gradient of a curve y = f(x) at any point is given by the derivative, written as dy/dx or f'(x). It is the limit of the gradient of a chord as the two points get infinitely close.
曲线 y = f(x) 任意点的梯度由导数给出，记作 dy/dx 或 f'(x)。它是在两点无限接近时，割线梯度的极限。
3. Differentiating xⁿ | 幂函数的求导
The most important rule you must learn is the power rule for differentiation: if y = xⁿ, then dy/dx = nxⁿ⁻¹. Here n can be any real number, though at GCSE we usually use positive integers and sometimes negative integers or fractions.
你必须掌握的最重要法则是幂函数求导法则：若 y = xⁿ，则 dy/dx = nxⁿ⁻¹。其中 n 可以是任何实数，但在 GCSE 中我们通常使用正整数，有时也用负整数或分数。
For example: if y = x³, then dy/dx = 3x²; if y = x⁵, dy/dx = 5x⁴. When n is a fraction, such as y = x½, dy/dx = ½ x⁻½. For negative powers: y = x⁻¹ → dy/dx = −x⁻².
例如：若 y = x³，则 dy/dx = 3x²；若 y = x⁵，则 dy/dx = 5x⁴。当 n 为分数时，如 y = x½，则 dy/dx = ½ x⁻½。对于负指数：y = x⁻¹ → dy/dx = −x⁻²。
4. Sum and Constant Multiple Rules | 和与常数倍法则
If you have a function made up of several terms, differentiate each term separately. For a constant c multiplied by a function, the derivative is c times the derivative of the function. Also, the derivative of a constant is 0.
如果一个函数由多项组成，则对每一项分别求导。对于常数 c 乘以函数，其导数是常数 c 乘以该函数的导数。此外，常数的导数为 0。
Example: For y = 3x⁴ + 2x² − 5, differentiate term by term: dy/dx = 3 × 4x³ + 2 × 2x − 0 = 12x³ + 4x.
示例：对于 y = 3x⁴ + 2x² − 5，逐项求导：dy/dx = 3 × 4x³ + 2 × 2x − 0 = 12x³ + 4x。
Another example: y = ½ x² + 5x − 3 → dy/dx = x + 5.
再如：y = ½ x² + 5x − 3 → dy/dx = x + 5。
5. Finding the Gradient at a Point | 求某点的梯度
To find the gradient of the curve at a specific x-value, first find dy/dx, then substitute the x-value into the derivative. This gives the numerical value of the gradient.
要找到曲线上某特定 x 值处的梯度，首先求出 dy/dx，然后将 x 值代入导数。这样就得到了梯度的数值。
Worked example: Given y = x³ − 2x, find the gradient at x = 1. dy/dx = 3x² − 2. At x = 1, gradient = 3(1)² − 2 = 1.
解题示例：给定 y = x³ − 2x，求 x = 1 处的梯度。dy/dx = 3x² − 2。在 x = 1 处，梯度 = 3(1)² − 2 = 1。
6. Equation of a Tangent | 切线方程
A tangent to a curve at a point is a straight line that touches the curve just there and has the same gradient as the curve at that point. To find its equation, you need the gradient m (found from dy/dx) and the coordinates (x₁, y₁) of the point. Then use the straight-line formula: y − y₁ = m(x − x₁).
曲线在某一点的切线是一条恰好在该处接触曲线，并与曲线在该点梯度相同的直线。要找到它的方程，需要梯度 m（从 dy/dx 得到）和该点的坐标 (x₁, y₁)。然后使用直线公式：y − y₁ = m(x − x₁)。
Example: Find the tangent to y = x² at the point (2, 4). dy/dx = 2x, so at x=2, m = 4. Equation: y − 4 = 4(x − 2) → y = 4x − 4.
示例：求 y = x² 在点 (2, 4) 处的切线。dy/dx = 2x，故在 x=2 处 m = 4。方程：y − 4 = 4(x − 2) → y = 4x − 4。
7. Equation of a Normal | 法线方程
The normal to a curve at a point is the line perpendicular to the tangent at that point. Its gradient is the negative reciprocal of the tangent’s gradient: m_normal = −1/m_tangent. Use the same point and straight-line formula.
Published by TutorHao | GCSE Mathematics Revision Series | aleveler.com
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