← 返回资讯屏轩教育 · 行业资讯

A-Level Maths: Differentiation of Quadratic Functions | A-Level数学:二次函数的求导方法

2026-08-30 11:47 · 本文部分内容由AI辅助创作

📚 A-Level Maths: Differentiation of Quadratic Functions | A-Level数学:二次函数的求导方法

s the gradient at any point on the curve. This article explains the method clearly, with worked examples and exam-style tips.

微分是A-Level数学中最强大的工具之一。对于形如 f(x) = ax² + bx + c 的二次函数,导数给出曲线上任意一点的切线斜率。本文将清晰讲解求导方法,并附有例题与应试技巧。


1. What Is a Quadratic Function? | 什么是二次函数?

x² + bx + c, where a, b and c are constants and a ≠ 0. The graph of a quadratic function is a parabola, which may open upwards (a > 0) or downwards (a < 0).

二次函数是指可以写成 f(x) = ax² + bx + c 形式的函数,其中 a、b、c 为常数,且 a ≠ 0。二次函数的图像是抛物线,开口方向由 a 决定:a > 0 时开口向上,a < 0 时开口向下。

² + 4x + 5. In this article, we focus on differentiating such functions with respect to x.

例如 f(x) = 3x² + 2x − 1 和 g(x) = −x² + 4x + 5 都是二次函数。本文重点讨论如何对这类函数关于 x 求导。


2. Understanding the Derivative | 理解导数

The derivative f'(x) measures the rate at which f(x) changes as x changes. Geometrically, f'(a) is the gradient of the tangent line to the curve y = f(x) at the point x = a.

导数 f'(x) 衡量 f(x) 随 x 变化的速率。从几何上看,f'(a) 是曲线 y = f(x) 在 x = a 处的切线斜率。

gradient is constant and equal to m. For a quadratic function, however, the gradient changes continuously along the curve, so the derivative is itself a linear function.

对于线性函数 f(x) = mx + c,斜率为常数 m。但是对于二次函数,斜率沿曲线不断变化,因此导数本身是一个线性函数。


3. The Power Rule of Differentiation | 求导的幂法则

The power rule states that if f(x) = xⁿ, then f' le applies to all real powers of x, including positive, negative and fractional exponents.

幂法则指出:如果 f(x) = xⁿ,则 f'(x) = n·xⁿ⁻¹。该法则适用于 x 的所有实数幂次,包括正指数、负指数和分数指数。

If f(x) = xⁿ, then f'(x) = n xⁿ⁻¹

mains as a multiplier: d/dx (axⁿ) = a·n·xⁿ⁻¹. This is known as the constant multiple rule.

对 axⁿ 这类项求导时,常数 a 保留为倍数:d/dx (axⁿ) = a·n·xⁿ⁻¹。这称为常数倍法则。


4. Differentiating a Quadratic Function | 二次函数的求导

To differentiate f(x) = ax² + bx + c, apply the power rule to each term separately:

要对 f(x) = ax² + bx + c 求导,只需对每一项分别使用幂法则:

Therefore, the derivative of a quadratic function is f'(x) = 2ax + b. Notice that the derivative of a quadratic is a linear function.

因此,二次函数的导数为 f'(x) = 2ax + b。注意,二次函数的导数是一个线性函数。

If f(x) = ax² + bx + c, then f'(x) = 2ax + b


5. Worked Example 1: Basic Differentiation | 例题1:基本求导

Differentiate f(x) = 5x² − 3x + 7.

求 f(x) = 5x² − 3x + 7 的导数。

For 5x², we get 5 × 2x = 10x. For −3x, we get −3. For the constant 7, the derivative is 0.

对每一项使用幂法则。5x² 的导数为 5 × 2x = 10x;−3x 的导数为 −3;常数 7 的导数为 0。

f'(x) = 10x − 3

at any point x is given by 10x − 3. For example, at x = 2, the gradient is 10(2) − 3 = 17.

这意味着曲线在任意点 x 处的斜率为 10x − 3。例如,在 x = 2 处,斜率为 10(2) − 3 = 17。


6. Worked Example 2: With Negative Coefficients | 例题2:含负系数

Differentiate f(x) = −4x² + 6x − 1.

求 f(x) = −4x² + 6x − 1 的导数。

Using the power rule for each term: the derivative of −4x² is −8x, the derivative of 6x is 6, and the derivative of −1 is 0.

对每一项使用幂法则:−4x² 的导数为 −8x,6x 的导数为 6,−1 的导数为 0。

f'(x) = −8x + 6

You can also write the derivative as f'(x) = 6 − 8x. Both forms are correct.

也可以写成 f'(x) = 6 − 8x。两种写法都是正确的。


7. Geometric Meaning of the Derivative | 导数的几何意义

For a quadratic curve y = ax² + bx + c, the derivative f' represents the slope of the tangent line at x. As x increases, the slope changes linearly.

对于二次曲线 y = ax² + bx + c,导数 f'(x) = 2ax + b 表示 x 处的切线斜率。随着 x 增大,斜率线性变化。

vertex of the parabola, the tangent is horizontal, meaning the gradient is zero. Setting f'(x) = 0 and solving for x gives the x-coordinate of the vertex: x = −b/(2a).

在抛物线的顶点处,切线是水平的,即斜率为零。令 f'(x) = 0 并求解 x,即可得到顶点的 x 坐标:x = −b/(2a)。

This is a key connection between differentiation and curve sketching. The derivative tells you where the curve is increasing (f'(x) > 0) or decreasing (f'(x) < 0).

这是微分与作图之间的关键联系。导数告诉你曲线在哪里递增(f'(x) > 0)或递减(f'(x) < 0)。


广告 / Sponsored

8. Stationary Points and the Vertex | 驻点与顶点

A stationary point occurs where f' atic function, there is exactly one stationary point, which is the vertex of the parabola.

驻点出现在 f'(x) = 0 处。对于二次函数,恰好有一个驻点,即抛物线的顶点。

Given f(x) = ax² + bx + c, set f' yields x = −b/(2a). Substituting this back into f(x) gives the y-coordinate of the vertex.

对于 f(x) = ax² + bx + c,令 f'(x) = 2ax + b = 0,解得 x = −b/(2a)。将其代入 f(x) 即可得到顶点的 y 坐标。

Vertex: x = −b/(2a), y = f(−b/(2a))

For example, for f(x) = x² − 4x + 5, the derivative is f' 4. Setting 2x − 4 = 0 gives x = 2. Then f(2) = 4 − 8 + 5 = 1, so the vertex is at (2, 1).

例如,对于 f(x) = x² − 4x + 5,导数为 f'(x) = 2x − 4。令 2x − 4 = 0 得 x = 2。于是 f(2) = 4 − 8 + 5 = 1,所以顶点为 (2, 1)。


9. Tangents and Normals to a Quadratic | 二次函数的切线与法线

n of a tangent line at any point on a quadratic curve. The tangent at x = a has gradient f'(a) and passes through the point (a, f(a)).

导数可以帮助你求二次曲线上任意一点处的切线方程。在 x = a 处的切线斜率为 f'(a),且过点 (a, f(a))。

Using the equation of a straight line, y − f(a) = f' he normal is perpendicular to the tangent, so its gradient is the negative reciprocal of f'(a).

利用直线方程 y − f(a) = f'(a)(x − a) 即可得到切线。法线与切线垂直,因此其斜率为 f'(a) 的负倒数。

For example, for f(x) = x² + 1 at x = 1, we have f(1) = 2 and f'(x) = 2x, so f' 1), which simplifies to y = 2x. The normal has gradient −1/2, giving y − 2 = −1/2(x − 1).

例如,对于 f(x) = x² + 1,在 x = 1 处,f(1) = 2,f'(x) = 2x,所以 f'(1) = 2。切线为 y − 2 = 2(x − 1),化简得 y = 2x。法线斜率为 −1/2,方程为 y − 2 = −1/2(x − 1)。


10. Common Mistakes to Avoid | 常见错误

The following mistakes frequently appear in exam work. Learn to spot and correct them:

以下错误在考试中经常出现。学会识别并纠正它们:

Always double-check each term individually after differentiating, especially when coefficients are negative or fractions.

求导后务必逐项检查,特别是当系数为负数或分数时。


11. Practice Questions | 练习

Try these questions on your own before checking the answers:

请先独立完成以下练习,再对照答案:

Question 1: Differentiate f(x) = 7x² − 2x + 4.

练习1:求 f(x) = 7x² − 2x + 4 的导数。

Question 2: Find the gradient of the curve y = x² + 3x − 2 at x = −1.

练习2:求曲线 y = x² + 3x − 2 在 x = −1 处的斜率。

estion 3: Find the coordinates of the vertex of f(x) = 2x² + 8x + 3 using differentiation.

练习3:利用微分求 f(x) = 2x² + 8x + 3 的顶点坐标。

Answers: 1) f'(x) = 14x − 2. 2) f'(−1) = 2(−1) + 3 = 1. 3) f'(x) = 4x + 8, so x = −2; f(−2) = 8 − 16 + 3 = −5, so the vertex is (−2, −5).

答案:1) f'(x) = 14x − 2。2) f'(−1) = 2(−1) + 3 = 1。3) f'(x) = 4x + 8,故 x = −2;f(−2) = 8 − 16 + 3 = −5,所以顶点为 (−2, −5)。


12. Exam Tips | 考试技巧

In exam questions, you may need to differentiate a quadratic as part of a larger problem involving tangents, rates of change or optimisation. Always write f'(x) clearly and simplify where possible.

在考试中,求二次函数导数可能只是更大题目的一个环节,例如求切线、变化率或最优化问题。务必清晰地写出 f'(x) 并尽可能化简。

Remember that the derivative of ax² is always 2ax, regardless of the value of b and c. This simple fact, combined with careful algebra, will help you avoid unnecessary errors.

记住 ax² 的导数始终是 2ax,与 b 和 c 的值无关。这一简单事实配合仔细的代数运算,能帮助你避免不必要的错误。

Finally, always check whether the original function is written in standard form ax² + bx + c. If not, expand brackets or collect like terms first, then differentiate term by term.

最后,务必检查原函数是否写成标准形式 ax² + bx + c。如果不是,需要先展开括号或合并同类项,然后逐项求导。

Published by TutorHao | Mathematics Revision Series | aleveler.com

广告 / Sponsored
【现货】剑桥国际高中教材 Cambridge IGCSE & O Level Additional Mathemati…
数学 备考教辅 · 正版现货 · 与本文内容相关
查看详情 / View on Taobao参考价 ¥198.0

更多咨询请联系16621398022(同微信)

备考资讯每日更新,关注不错过
考纲变化 · 科目分析 · 学习方法
扫码关注公众号 tutorhao
在微信里长按二维码识别关注
或截图后用「扫一扫 → 相册」选择本图
公众号:tutorhao