4: Differentiating Quadratics | 微分二次函数

📚 4: Differentiating Quadratics | 微分二次函数

Differentiation is one of the most powerful tools in A-Level mathematics. It allows us to find the rate at which a function changes at any given point, and for quadratics specifically, it gives us a straightforward way to determine gradients, turning points, and the behaviour of curves. This article covers everything you need to know about differentiating quadratic functions for Edexcel A-Level Maths.

微分是A-Level数学中最强大的工具之一。它使我们能够找到函数在任意给定点的变化率，而具体到二次函数，它为我们提供了一种直接的方法来确定梯度、转折点以及曲线的行为。本文将涵盖Edexcel A-Level数学中关于微分二次函数你所需要知道的一切。

1. Differentiation from First Principles | 从第一原理出发的微分

Before we apply rules mechanically, it is essential to understand where differentiation comes from. The derivative of a function f(x) at a point x is defined as the limit of the average rate of change as the interval shrinks to zero. This is known as differentiating from first principles.

在我们机械地应用规则之前，理解微分的来源至关重要。函数 f(x) 在点 x 处的导数被定义为当区间缩小到零时平均变化率的极限。这被称为从第一原理出发的微分。

The formal definition is:

正式定义如下：

f'(x) = lim (h → 0) [f(x + h) − f(x)] / h

For a quadratic function f(x) = ax² + bx + c, we substitute and expand. The key algebraic step is expanding (x + h)² = x² + 2xh + h². When we subtract f(x), the x², bx and c terms cancel, leaving 2xh + h² + bh. Dividing by h gives 2x + h + b, and as h approaches zero, the derivative becomes 2ax + b.

对于二次函数 f(x) = ax² + bx + c，我们代入并展开。关键的代数步骤是展开 (x + h)² = x² + 2xh + h²。当我们减去 f(x) 时，x²、bx 和 c 项相互抵消，剩下 2xh + h² + bh。除以 h 得到 2x + h + b，当 h 趋近于零时，导数变为 2ax + b。

Worked example: Differentiate f(x) = x² from first principles.

示例：从第一原理微分 f(x) = x²。

f'(x) = lim (h → 0) [(x + h)² − x²] / h = lim (h → 0) (2xh + h²) / h = lim (h → 0) (2x + h) = 2x

This confirms that the derivative of x² is 2x, and in general, the derivative of ax² is 2ax.

这证实了 x² 的导数是 2x，并且一般来说，ax² 的导数是 2ax。

2. The Power Rule | 幂法则

The power rule states that if f(x) = xⁿ, then f'(x) = n·xⁿ⁻¹. This rule applies to all real powers, including positive, negative and fractional exponents. For quadratics, we only need the cases n = 2, n = 1 and n = 0, but understanding the general rule is crucial for later topics.

幂法则指出，如果 f(x) = xⁿ，那么 f'(x) = n·xⁿ⁻¹。这个规则适用于所有实数次幂，包括正数、负数和分数指数。对于二次函数，我们只需要 n = 2、n = 1 和 n = 0 的情况，但理解一般规则对后续主题至关重要。

For a quadratic in the form ax² + bx + c, we differentiate term by term:

对于形式为 ax² + bx + c 的二次函数，我们逐项微分：

The derivative of ax² is 2ax (multiply by the power 2, then reduce the power by 1).
ax² 的导数是 2ax（乘以幂 2，然后将幂减 1）。
The derivative of bx is b (since x¹ becomes 1·x⁰ = 1).
bx 的导数是 b（因为 x¹ 变为 1·x⁰ = 1）。
The derivative of c (a constant) is 0.
c（常数）的导数是 0。

Worked example: Differentiate f(x) = 5x² − 3x + 7.

示例：微分 f(x) = 5x² − 3x + 7。

f'(x) = 2 × 5x²⁻¹ − 1 × 3x¹⁻¹ + 0 = 10x − 3

Notice that a constant term always differentiates to zero because it represents a horizontal line with no gradient.

注意常数项总是微分为零，因为它代表一条没有梯度的水平线。

3. Differentiating Quadratics in Various Forms | 微分各种形式的二次函数

Quadratics do not always appear in the standard form ax² + bx + c. You may encounter them as products of brackets, such as y = (x + 3)(x − 2), or in completed square form, y = a(x − h)² + k. Before differentiating, always expand or rewrite the expression so that you can apply the power rule term by term.

二次函数并不总是以标准形式 ax² + bx + c 出现。你可能会遇到括号乘积的形式，例如 y = (x + 3)(x − 2)，或者完全平方的形式，y = a(x − h)² + k。在微分之前，始终要先展开或重写表达式，以便逐项应用幂法则。

Example 1: Differentiate y = (x + 3)(x − 2).

示例1：微分 y = (x + 3)(x − 2)。

y = x² + x − 6, so dy/dx = 2x + 1

Example 2: Differentiate y = 2(x − 1)² + 4.

示例2：微分 y = 2(x − 1)² + 4。

y = 2(x² − 2x + 1) + 4 = 2x² − 4x + 6, so dy/dx = 4x − 4

Always check whether the expression is fully simplified before differentiating. Failing to expand brackets is one of the most common mistakes students make in exams.

在微分之前，始终检查表达式是否已完全化简。未能展开括号是学生在考试中最常见的错误之一。

4. Finding the Gradient at a Specific Point | 在特定点求梯度

The derivative dy/dx gives a general formula for the gradient at any point on the curve. To find the gradient at a particular x-coordinate, simply substitute that value into dy/dx. The result is the slope of the tangent line at that point.

导数 dy/dx 给出了曲线上任意一点的梯度的一般公式。要在特定的 x 坐标处找到梯度，只需将该值代入 dy/dx 即可。结果就是该点处切线的斜率。

Worked example: Find the gradient of the curve y = 2x² − 5x + 3 at x = 4.

示例：求曲线 y = 2x² − 5x + 3 在 x = 4 处的梯度。

dy/dx = 4x − 5

At x = 4: dy/dx = 4(4) − 5 = 16 − 5 = 11

The gradient of the curve at x = 4 is 11. This means that the tangent to the curve at the point (4, 15) has slope 11.

曲线在 x = 4 处的梯度是 11。这意味着曲线在点 (4, 15) 处的切线斜率为 11。

5. Equation of a Tangent | 切线方程

Once we know the gradient at a point and the coordinates of that point, we can find the equation of the tangent line using the point-slope form: y − y₁ = m(x − x₁). For a quadratic y = ax² + bx + c, the tangent at x = p has gradient 2ap + b.

一旦我们知道某点的梯度和该点的坐标，就可以使用点斜式求切线方程：y − y₁ = m(x − x₁)。对于二次函数 y = ax² + bx + c，在 x = p 处的切线梯度为 2ap + b。

Worked example: Find the equation of the tangent to y = x² + 3x − 1 at x = 2.

示例：求 y = x² + 3x − 1 在 x = 2 处的切线方程。

Step 1: Find the y-coordinate of the point.

步骤1：求该点的 y 坐标。

y = (2)² + 3(2) − 1 = 4 + 6 − 1 = 9, so the point is (2, 9)

Step 2: Find the gradient at x = 2.

步骤2：求 x = 2 处的梯度。

dy/dx = 2x + 3, so at x = 2: m = 2(2) + 3 = 7

Step 3: Use the point-slope form.

步骤3：使用点斜式。

y − 9 = 7(x − 2) ⇒ y = 7x − 5

The tangent line is y = 7x − 5.

切线方程为 y = 7x − 5。

6. 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. If the tangent has gradient m, the normal has gradient −1/m (provided m ≠ 0). This is because perpendicular lines satisfy m₁ × m₂ = −1.

曲线在某点的法线是在该点垂直于切线的直线。如果切线的梯度为 m，则法线的梯度为 −1/m（前提是 m ≠ 0）。这是因为垂直线的斜率满足 m₁ × m₂ = −1。

Worked example: Find the equation of the normal to y = x² − 4x + 5 at x = 1.

示例：求 y = x² − 4x + 5 在 x = 1 处的法线方程。

Step 1: Find the y-coordinate.

步骤1：求 y 坐标。

y = 1 − 4 + 5 = 2, so the point is (1, 2)

Step 2: Find the gradient of the tangent.

步骤2：求切线的梯度。

dy/dx = 2x − 4, at x = 1: m = 2(1) − 4 = −2

Step 3: The normal gradient is −1/m = −1/(−2) = 1/2.

步骤3：法线的梯度为 −1/m = −1/(−2) = 1/2。

Step 4: Write the equation.

步骤4：写出方程。

y − 2 = (1/2)(x − 1) ⇒ y = (1/2)x + 3/2

The normal line is y = (1/2)x + 3/2. Note that when the tangent gradient is zero (at a turning point), the normal is a vertical line.

法线方程为 y = (1/2)x + 3/2。注意当切线梯度为零时（在转折点处），法线是一条垂直线。

7. Stationary Points (Turning Points) | 驻点（转折点）

A stationary point occurs where the gradient is zero, i.e., where dy/dx = 0. For a quadratic, there is exactly one stationary point, which is either a minimum (if the coefficient of x² is positive) or a maximum (if it is negative). This is the vertex of the parabola.

驻点出现在梯度为零的地方，即 dy/dx = 0 处。对于二次函数，恰好有一个驻点，它要么是最小值点（如果 x² 的系数为正），要么是最大值点（如果 x² 的系数为负）。这就是抛物线的顶点。

Worked example: Find the coordinates of the stationary point of y = 2x² − 8x + 3.

示例：求 y = 2x² − 8x + 3 的驻点坐标。

Step 1: Differentiate and set equal to zero.

步骤1：微分并令其等于零。

dy/dx = 4x − 8 = 0 ⇒ x = 2

Step 2: Substitute back to find y.

步骤2：代回求 y。

y = 2(2)² − 8(2) + 3 = 8 − 16 + 3 = −5

Step 3: Determine its nature. Since the coefficient of x² is +2 (> 0), this is a minimum point.

步骤3：确定其性质。由于 x² 的系数为 +2（> 0），这是一个最小值点。

The stationary point is (2, −5), and it is a minimum. You can also confirm this by taking the second derivative, d²y/dx² = 4, which is positive, indicating a minimum.

驻点为 (2, −5)，且为最小值。你也可以通过求二阶导数 d²y/dx² = 4 来确认，该值为正，表示最小值。

8. Applications: Optimisation Problems | 应用：最优化问题

Quadratic differentiation is frequently applied to real-world optimisation problems. Whenever a quantity can be expressed as a quadratic function, its maximum or minimum value occurs at the turning point. This is particularly common in geometry, economics and physics problems.

二次微分经常应用于现实世界的最优化问题。当一个量可以表示为二次函数时，其最大值或最小值出现在转折点处。这在几何、经济和物理问题中尤为常见。

Worked example: A farmer has 100 metres of fencing to enclose a rectangle. Find the maximum area.

示例：一个农民有100米围栏来围一个矩形。求最大面积。

Let the width be x and the length be 50 − x (since perimeter = 2(x + length) = 100). The area is A = x(50 − x) = 50x − x².

设宽为 x，长为 50 − x（因为周长 = 2(x + 长) = 100）。面积为 A = x(50 − x) = 50x − x²。

dA/dx = 50 − 2x = 0 ⇒ x = 25

At x = 25, the area is A = 25 × 25 = 625 m². Since d²A/dx² = −2 < 0, this is a maximum.

当 x = 25 时，面积为 A = 25 × 25 = 625 平方米。由于 d²A/dx² = −2 < 0，这是最大值。

The maximum area is 625 m². Note that the optimum rectangle is a square.

最大面积为 625 平方米。注意最优矩形是一个正方形。

9. Second Derivative and Concavity | 二阶导数与凹凸性

The second derivative, denoted d²y/dx² or f”(x), is the derivative of the derivative. For a quadratic f(x) = ax² + bx + c, the second derivative is simply 2a. It is constant for all x, which tells us that a quadratic curve has constant curvature.

二阶导数，记为 d²y/dx² 或 f”(x)，是导数的导数。对于二次函数 f(x) = ax² + bx + c，二阶导数就是 2a。它对所有 x 都是常数，这告诉我们二次曲线具有恒定的曲率。

When 2a > 0, the curve is concave up (shaped like a ‘U’), and the stationary point is a minimum. When 2a < 0, the curve is concave down (shaped like an '∩'), and the stationary point is a maximum.

当 2a > 0 时，曲线是凹向上的（形状像 ‘U’），驻点是最小值。当 2a < 0 时，曲线是凹向下的（形状像 '∩'），驻点是最大值。

Example: For y = −3x² + 6x − 2, the first derivative is dy/dx = −6x + 6, and the second derivative is d²y/dx² = −6. Since −6 < 0, the curve is concave down everywhere, and the stationary point at x = 1 is a maximum.

示例：对于 y = −3x² + 6x − 2，一阶导数为 dy/dx = −6x + 6，二阶导数为 d²y/dx² = −6。由于 −6 < 0，曲线处处凹向下，在 x = 1 处的驻点是最大值。

The second derivative test is a quick and reliable way to classify stationary points of quadratics.

二阶导数检验是对二次函数驻点进行分类的一种快速可靠的方法。

10. Common Mistakes to Avoid | 应避免的常见错误

Students often lose marks on differentiation questions due to a handful of recurring errors. Being aware of these pitfalls will help you avoid them in your exam.

学生在微分题上常常因为少数反复出现的错误而丢分。意识到这些陷阱将帮助你在考试中避免它们。

Forgetting to differentiate every term: each term in ax² + bx + c must be differentiated separately, including constants which become zero.
忘记对每一项求导：ax² + bx + c 中的每一项都必须分别求导，包括变为零的常数项。
Incorrectly applying the power rule: remember to multiply by the power and then subtract one from the power. For x², the derivative is 2x, not x.
错误地应用幂法则：记住先乘以幂，然后从幂中减一。对于 x²，导数是 2x，而不是 x。
Not expanding brackets before differentiating: expressions like (x + 1)² must be expanded to x² + 2x + 1 first.
微分前没有展开括号：像 (x + 1)² 这样的表达式必须先展开为 x² + 2x + 1。
Confusing tangent and normal: the tangent has gradient dy/dx; the normal has gradient −1/(dy/dx).
混淆切线和法线：切线的梯度是 dy/dx；法线的梯度是 −1/(dy/dx)。
Forgetting to state whether a stationary point is a maximum or minimum.
忘记说明驻点是最大值还是最小值。

Always show your working clearly and check whether your final answer is consistent with the shape of the quadratic.

始终清晰展示你的计算过程，并检查最终答案是否与二次函数的形状一致。

11. Practice Questions | 练习题

Test your understanding with the following questions. Try to complete them without looking at the solutions, then check your answers.

用以下问题测试你的理解。先尝试不看解答完成它们，然后检查你的答案。

Question \| 题目	Answer \| 答案
1. Differentiate y = 3x² + 2x − 5	dy/dx = 6x + 2
2. Find the gradient of y = x² − 7x + 10 at x = 3	dy/dx = 2x − 7; at x = 3: −1
3. Find the equation of the tangent to y = 4x² at x = 2	dy/dx = 8x; m = 16; point (2, 16); y = 16x − 16
4. Find the stationary point of y = x² − 6x + 9	dy/dx = 2x − 6 = 0 ⇒ x = 3; y = 0; minimum (3, 0)
5. Find the normal to y = x² + 2x at x = 1	dy/dx = 2x + 2; m = 4; normal m = −1/4; y − 3 = −(1/4)(x − 1)

If you answered all five correctly, you have mastered the basics of differentiating quadratics. If not, review the relevant sections above before moving on.

如果你全部答对了五道题，你已经掌握了微分二次函数的基础。如果没有，请在继续之前复习上面相关的章节。

12. Summary and Key Takeaways | 总结与关键要点

Differentiating quadratics is a fundamental skill that underpins much of A-Level calculus. The key result is that the derivative of ax² + bx + c is always 2ax + b. From this single formula, you can find gradients at any point, construct tangent and normal lines, locate stationary points, and solve optimisation problems.

微分二次函数是一项基本技能，支撑着A-Level微积分的许多内容。关键结果是 ax² + bx + c 的导数总是 2ax + b。从这个单一公式中，你可以在任意点求梯度、构造切线和法线、定位驻点以及解决最优化问题。

The power rule: d/dx(xⁿ) = n·xⁿ⁻¹
幂法则：d/dx(xⁿ) = n·xⁿ⁻¹
The derivative of ax² + bx + c is 2ax + b
ax² + bx + c 的导数是 2ax + b
Gradient at x = p: substitute p into dy/dx
在 x = p 处的梯度：将 p 代入 dy/dx
Tangent gradient = dy/dx; Normal gradient = −1/(dy/dx)
切线梯度 = dy/dx；法线梯度 = −1/(dy/dx)
Stationary points occur where dy/dx = 0
驻点出现在 dy/dx = 0 处
Second derivative d²y/dx² = 2a determines concavity and nature of turning points
二阶导数 d²y/dx² = 2a 决定凹凸性和转折点的性质

Master these techniques and practise regularly; differentiation of quadratics will become second nature, providing a solid foundation for polynomials, exponentials and trigonometric functions in later chapters.

掌握这些技巧并定期练习；二次函数的微分将成为你的本能反应，为后续章节中的多项式、指数函数和三角函数打下坚实的基础。

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4: Differentiating Quadratics | 微分二次函数