GCSE Maths: Differentiation - Key Points | GCSE 数学：微分考点精讲

📚 GCSE Maths: Differentiation – Key Points | GCSE 数学：微分考点精讲

Differentiation is a core topic in GCSE Maths that allows us to study the rate at which a quantity changes. It is widely used to calculate the gradient of curves, find equations of tangents and normals, and solve problems involving maximum and minimum values. This article breaks down the essential differentiation skills you need, with clear explanations and worked examples.

微分是 GCSE 数学中的一个核心主题，它帮助我们研究某个量变化的快慢。微分被广泛用于计算曲线的梯度、求切线和法线的方程，以及解决与最大值和最小值有关的问题。本文将对你需要掌握的关键微分技能进行分解讲解，并提供清晰的解释和示范。

1. What is Differentiation? | 什么是微分？

Differentiation is a method used to find the gradient function, or derivative, of a given function. For a curve described by y = f(x), the derivative, written as dy/dx or f'(x), tells us the instantaneous rate of change of y with respect to x at any point. In simple terms, it gives us a formula for the slope of the tangent to the curve.

微分是一种用来求给定函数的梯度函数（也称导数）的方法。对于由 y = f(x) 描述的曲线，导数记作 dy/dx 或 f'(x)，它告诉我们在任意一点上 y 随 x 变化的瞬时速率。简单来说，它给出了曲线切线的斜率公式。

The process of differentiation turns a function for position or quantity into a function for gradient or speed. While you may have previously worked with straight-line gradients using the formula (change in y)/(change in x), differentiation extends this idea to curves where the gradient is constantly changing.

微分的过程可以将表示位置或数量的函数转化为表示梯度或速度的函数。虽然你之前可能用过 (y 的变化量)/(x 的变化量) 来计算直线的斜率，但微分将这一思想延伸到了梯度不断变化的曲线情形。

At GCSE, you mainly differentiate polynomial functions. The derivative allows you to answer questions such as ‘How steep is this curve at x = 2?’ or ‘Where does this curve have a turning point?’.

在 GCSE 阶段，你主要对多项式函数进行求导。有了导数，你就可以回答诸如 “这条曲线在 x = 2 处的陡峭程度如何？” 或者 “这条曲线在哪里有转折点？” 这类问题。

2. The Power Rule | 幂函数求导法则

The most fundamental rule in differentiation is the power rule. If y = xⁿ, then the derivative dy/dx = n xⁿ⁻¹. You simply multiply by the original power and then reduce the power by one. This rule works for any real value of n, but at GCSE you will usually see positive integer powers and sometimes fractions or negative powers.

微分中最基本的法则是幂函数求导法则。如果 y = xⁿ，那么导数 dy/dx = n xⁿ⁻¹。你只需要乘上原来的指数，然后把指数减一。这个法则对任意实数 n 都成立，但在 GCSE 中你通常遇到的是正整数次幂，有时也会遇到分数或负指数。

For example, if y = x³, then dy/dx = 3x². If y = x⁵, dy/dx = 5x⁴. If y = x, remember that x is x¹, so dy/dx = 1 × x⁰ = 1. This makes sense because the line y = x has a constant gradient of 1.

例如，如果 y = x³，那么 dy/dx = 3x²。如果 y = x⁵，dy/dx = 5x⁴。如果 y = x，记住 x 就是 x¹，所以 dy/dx = 1 × x⁰ = 1。这是合理的，因为直线 y = x 的斜率恒为 1。

When constants are involved, they simply remain unchanged as multipliers. If y = 4x², dy/dx = 4 × 2x¹ = 8x. The constant coefficient is carried through the differentiation.

当式子中含有常数因数时，它们作为乘数保持不变。例如 y = 4x²，dy/dx = 4 × 2x¹ = 8x。常系数会一直保留在求导过程中。

3. Differentiating Sums and Constant Multiples | 和与常数倍的求导

Differentiation is a linear operation, which means you can differentiate a sum term by term. If a function is made up of several terms added together, you simply differentiate each term separately and add the results. The derivative of a constant term on its own is always zero because a horizontal line has zero gradient.

微分是一种线性运算，这意味着你可以逐项对和式进行求导。如果一个函数是由若干项相加组成的，你只需对每一项分别求导，然后将结果加起来。单独常数项的导数总是为零，因为水平线的梯度为零。

For example, differentiate y = 2x³ + 5x² – 3x + 7. Using the power rule on each term: derivative of 2x³ is 6x²; derivative of 5x² is 10x; derivative of -3x is -3; derivative of the constant 7 is 0. So dy/dx = 6x² + 10x – 3.

例如，对 y = 2x³ + 5x² – 3x + 7 求导。对每一项使用幂法则：2x³ 的导数是 6x²；5x² 的导数是 10x；-3x 的导数是 -3；常数 7 的导数是 0。因此 dy/dx = 6x² + 10x – 3。

This straightforward approach works for any polynomial. Always rewrite terms in the form axⁿ before differentiating. For instance, y = 1/x can be written as x⁻¹ and then differentiated to give -1 x⁻² or -1/x².

这种直接的方法适用于任何多项式。求导前，要先将各项化为 axⁿ 的形式。例如，y = 1/x 可以改写成 x⁻¹，然后求导得到 -1 x⁻² 或 -1/x²。

4. Differentiating Polynomials | 多项式求导

To differentiate a polynomial efficiently, follow these steps: first, write each term as a power of x multiplied by a coefficient. Then apply the power rule to each term. Finally, simplify the resulting expression by combining like terms if necessary.

要高效地对多项式求导，可以遵循以下步骤：首先，将每一项写成 x 的幂次乘以系数的形式；然后对每一项应用幂法则；最后，如有必要，合并同类项进行化简。

Worked example: differentiate y = 3x⁴ – 2x³ + x² – 5x + 9. The derivative is dy/dx = 3×4x³ – 2×3x² + 2x¹ – 5 + 0 = 12x³ – 6x² + 2x – 5. Notice that the constant 9 disappears, and the x term -5x becomes -5.

示范：对 y = 3x⁴ – 2x³ + x² – 5x + 9 求导。导数为 dy/dx = 3×4x³ – 2×3x² + 2x¹ – 5 + 0 = 12x³ – 6x² + 2x – 5。注意常数 9 消失了，x 项 -5x 变成了 -5。

If the polynomial contains brackets, expand them first before differentiating. For example, y = x²(3x + 1) should be expanded to 3x³ + x², then differentiated to give 9x² + 2x. Always simplify to avoid mistakes.

如果多项式中含有括号，求导前要先展开。例如，y = x²(3x + 1) 应展开为 3x³ + x²，然后再求导得到 9x² + 2x。务必先化简以避免出错。

5. Finding the Gradient at a Point | 求曲线上一点的梯度

Once you have the derivative function dy/dx, you can find the gradient of the curve at any specific point by substituting the x-coordinate into the derivative. The result is the instantaneous gradient of the curve at that point, which is the same as the gradient of the tangent line at that x-value.

一旦你求得导数函数 dy/dx，就可以通过将 x 坐标代入导数来求出曲线上任意一点的梯度。得到的结果就是曲线在该点的瞬时梯度，也就是该 x 值处切线的梯度。

Example: for the curve y = x³ – 4x² + 6, find the gradient at x = 3. First, dy/dx = 3x² – 8x. Then substitute x = 3: gradient = 3(3)² – 8(3) = 27 – 24 = 3. So the curve has a gradient of 3 when x = 3.

示例：对于曲线 y = x³ – 4x² + 6，求 x = 3 处的梯度。首先，求导得 dy/dx = 3x² – 8x。然后代入 x = 3：梯度 = 3×(3)² – 8×3 = 27 – 24 = 3。所以曲线在 x = 3 处的梯度为 3。

This skill is often tested alongside finding equations of tangents and normals, where knowing the gradient is the crucial first step. The point itself is obtained by substituting x into the original equation to find the corresponding y-coordinate.

这一技能常与求切线和法线方程的题目一同出现，其中知道梯度是关键的第一步。点的信息可以通过将 x 代入原方程求得对应的 y 坐标来获得。

6. Equation of a Tangent | 切线方程

The tangent to a curve at a given point is the straight line that just touches the curve, having the same gradient as the curve at that point. To find its equation, you need the point of contact (x₁, y₁) and the gradient m = dy/dx evaluated at x₁. Then use the straight-line equation y – y₁ = m(x – x₁).

曲线上给定点处的切线就是恰好接触曲线，并且在该点与曲线有相同梯度的直线。要找到其方程，你需要切点坐标 (x₁, y₁) 以及在该点求得的梯度 m = dy/dx。然后使用直线方程 y – y₁ = m(x – x₁)。

Example: find the equation of the tangent to the curve y = x² + 2x at x = 1. First, find y₁: y₁ = (1)² + 2(1) = 3. Derivative: dy/dx = 2x + 2. Gradient m = 2(1) + 2 = 4. Tangent equation: y – 3 = 4(x – 1), which simplifies to y = 4x – 1.

示例：求曲线 y = x² + 2x 在 x = 1 处的切线方程。首先，求 y₁：y₁ = (1)² + 2×1 = 3。求导：dy/dx = 2x + 2。梯度 m = 2×1 + 2 = 4。切线方程：y – 3 = 4(x – 1)，化简得 y = 4x – 1。

Always ensure you substitute the x-value into the original function to find y, not into the derivative. A common mistake is putting x into dy/dx to find y-coordinate, which gives entirely wrong points.

始终要确保你将 x 值代入原函数求 y，而非代入导数。一个常见错误是把 x 代入 dy/dx 来求 y 坐标，这样会得到完全错误的点。

7. Equation of a Normal | 法线方程

The normal to a curve at a point is the straight line perpendicular to the tangent at that point. If the tangent has gradient m, the normal has gradient -1/m, provided m is not zero. The point of contact is the same, so the normal equation is y – y₁ = (-1/m)(x – x₁).

曲线在某点的法线是经过该点且垂直于切线的直线。如果切线的梯度为 m，那么法线的梯度为 -1/m，前提是 m 不为零。切点相同，因此法线方程为 y – y₁ = (-1/m)(x – x₁)。

Example: using the previous example (tangent gradient m = 4 at (1, 3)), the normal gradient is -¼. The equation of the normal is y – 3 = -¼(x – 1), which simplifies to y = -¼x + 3¼ or 4y + x = 13.

示例：利用前一示例（在点 (1, 3) 处切线梯度 m = 4），法线梯度为 -1/4。法线方程为 y – 3 = -¼(x – 1)，化简得 y = -¼x + 3¼ 或 4y + x = 13。

If the tangent gradient is zero (horizontal tangent), the normal is a vertical line x = x₁. If the tangent gradient is undefined (vertical tangent), the normal is horizontal y = y₁. At GCSE, you are most likely to work with cases where m is a non-zero number.

如果切线梯度为零（水平切线），法线就是一条竖直直线 x = x₁。如果切线梯度未定义（竖直切线），法线就是一条水平直线 y = y₁。在 GCSE 中，你最常遇到的情况是 m 为一个非零数值。

8. Stationary Points | 驻点

Stationary points are points on a curve where the gradient is zero, that is, where dy/dx = 0. At these points the tangent is horizontal. There are three types: local maximum (peak), local minimum (trough), and point of inflection (where the curve flattens but continues in the same direction).

驻点是曲线上梯度为零的点，即满足 dy/dx = 0 的点。在这些点上切线是水平的。驻点分为三类：局部极大值（峰）、局部极小值（谷）以及拐点（曲线变平但继续沿原方向延伸）。

To find stationary points, first differentiate the function, set dy/dx = 0, and solve for x. Substitute each x-value back into the original equation y = f(x) to find the corresponding y-coordinate. These (x, y) pairs are your stationary points.

要找到驻点，首先对函数求导，令 dy/dx = 0，然后解出 x。将每个 x 值代回原方程 y = f(x) 中求出对应的 y 坐标。这些 (x, y) 坐标对就是驻点。

Example: y = x³ – 3x. dy/dx = 3x² – 3. Set 3x² – 3 = 0 gives x = 1 or x = -1. Corresponding y-values: when x=1, y = -2; when x=-1, y = 2. Stationary points are (1, -2) and (-1, 2).

示例：y = x³ – 3x。求导得 dy/dx = 3x² – 3。令 3x² – 3 = 0，得 x = 1 或 x = -1。对应的 y 值：当 x=1 时，y = -2；当 x=-1 时，y = 2。驻点为 (1, -2) 和 (-1, 2)。

9. Determining Nature of Stationary Points | 判断驻点性质

After finding stationary points, you must determine whether each is a maximum, minimum or point of inflection. At GCSE, the standard method is to check the sign of dy/dx just before and just after each stationary point (often called the first derivative test).

找到驻点之后，必须判断每个驻点是极大值、极小值还是拐点。在 GCSE 中，标准的方法是检查每个驻点稍前和稍后 dy/dx 的符号（通常称为一阶导数检验法）。

For a maximum, dy/dx changes from positive to negative as x increases through the point: the curve goes up, flattens, then goes down. For a minimum, dy/dx changes from negative to positive: curve goes down, flattens, then goes up. If dy/dx has the same sign on both sides, the point is an inflection with a horizontal tangent.

对于极大值，当 x 增大经过驻点时，dy/dx 由正变负：曲线先上升，变平，然后下降。对于极小值，dy/dx 由负变正：曲线先下降，变平，然后上升。如果 dy/dx 在两侧符号相同，则该点为具有水平切线的拐点。

Using the previous example: for x = -1, check dy/dx at x = -2 (positive) and x = 0 (negative). Gradient changes from + to -, so (-1, 2) is a maximum. For x = 1, check x = 0 (negative) and x = 2 (positive): changes from – to +, so (1, -2) is a minimum.

沿用上例：对于 x = -1，检查 x = -2 时的 dy/dx（正值）和 x = 0 时的 dy/dx（负值）。梯度由正变负，因此 (-1, 2) 是极大值点。对于 x = 1，检查 x = 0（负值）和 x = 2（正值）：由负变正，因此 (1, -2) 是极小值点。

10. Applications of Differentiation | 微分应用举例

Differentiation is frequently used in GCSE word problems involving optimisation, such as maximising an area or minimising a surface area given a fixed perimeter or volume. The method typically involves writing an equation for the quantity to be optimised in terms of one variable, differentiating, finding stationary points, and then checking which one gives the maximum or minimum value.

微分在 GCSE 中的应用题中经常出现，涉及诸如在给定周长或体积的情况下最大化面积或最小化表面积的优化问题。其方法通常是：用单一变量写出待优化量的方程，求导，找出驻点，然后检验哪个驻点给出了最大值或最小值。

Example: a container in the shape of a cuboid with a square base of side x and height h is to have a volume of 500 cm³. Find the dimensions that minimise the surface area. First, express h in terms of x using volume: x²h = 500 → h = 500/x². Surface area S = 2x² + 4xh = 2x² + 2000/x. Differentiate S with respect to x: dS/dx = 4x – 2000/x². Set to zero: 4x = 2000/x² → x³ = 500 → x = ∛500 ≈ 7.94 cm, then find h.

示例：一个底面为边长 x 的正方形、高为 h 的长方体容器，体积为 500 cm³。求使表面积最小的尺寸。首先，利用体积将 h 用 x 表示：x²h = 500 → h = 500/x²。表面积 S = 2x² + 4xh = 2x² + 2000/x。对 x 求导：dS/dx = 4x – 2000/x²。令导数为零：4x = 2000/x² → x³ = 500 → x = ∛500 ≈ 7.94 cm，然后求 h。

These problems reinforce the idea that stationary points can represent optimal solutions. Once the dimensions are found, a quick check with the first derivative test confirms that the stationary point gives a minimum surface area.

这类问题强化了“驻点可以表示最优解”这一观念。一旦找到尺寸，用一阶导数检验法快速检验即可确认该驻点对应的是最小表面积。

11. Summary of Key Formulas and Tips | 关键公式与技巧总结

Here is a quick recap of the essential differentiation rules and processes for GCSE:

以下是 GCSE 微分核心法则与步骤的快速回顾：

Power rule: if y = xⁿ then dy/dx = n xⁿ⁻¹. 幂法则：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。
Constant rule: if y = c then dy/dx = 0. 常数法则：若 y = c，则 dy/dx = 0。
Sum rule: differentiate term by term. 求和法则：逐项求导。
Gradient at a point: substitute x into dy/dx. 某点的梯度：将 x 代入 dy/dx。
Tangent: y – y₁ = m(x – x₁) with m = dy/dx at (x₁, y₁). 切线：y – y₁ = m(x – x₁)，其中 m 为在 (x₁, y₁) 处的 dy/dx。
Normal: gradient = -1/m. 法线：梯度 = -1/m。
Stationary points: solve dy/dx = 0, then test sign change. 驻点：解 dy/dx = 0，然后检验符号变化。

Practise by differentiating a variety of polynomial functions and applying the concepts to tangents, normals, and optimisation problems. With regular revision, you will build confidence and accuracy for your GCSE exams.

通过对各种多项式函数求导，以及将概念应用于切线、法线和优化问题来进行练习。通过定期复习，你将为 GCSE 考试树立信心并提高准确性。

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GCSE Maths: Differentiation – Key Points | GCSE 数学：微分 考点精讲