Example 4.2.2: Differentiating a Quadratic from First Principles | 示例4.2.2：用第一原理求二次函数的导数

📚 Example 4.2.2: Differentiating a Quadratic from First Principles | 示例4.2.2：用第一原理求二次函数的导数

In this structured lesson, we will examine Example 4.2.2 from the AQA A-Level Mathematics syllabus. The example asks us to differentiate the quadratic function f(x) = x² from first principles. This exercise is central to understanding how derivatives are constructed and is often a starting point in the study of calculus.

在本节结构化课程中，我们将研究AQA A-Level数学大纲中的示例4.2.2。该例题要求我们用第一原理求二次函数 f(x) = x² 的导数。这个练习对于理解导数是如何构建的至关重要，也是学习微积分的常见起点。

1. What Are First Principles? | 什么是第一原理？

The derivative of a function f at a point x is defined as the limit of the average rate of change of f over an interval as the interval length tends to zero. This is called differentiation from first principles because it relies only on the limit definition, not on rules such as the power rule.

函数 f 在某点 x 的导数被定义为：当区间长度趋于零时，f 在该区间上平均变化率的极限。这被称为“从第一原理求导”，因为它只依赖于极限定义，而不依赖幂法则等规则。

f'(x) = lim_h→0 [f(x+h) − f(x)] / h

Here, h represents a small change in x. The quotient inside the limit is called the difference quotient.

这里的 h 表示 x 的一个微小变化。极限内的商称为差商。

2. The Example: Differentiating f(x) = x² | 例题：求 f(x) = x² 的导数

Example 4.2.2 states: “Use first principles to differentiate f(x) = x².” We will apply the limit definition step by step. This will produce an expression for the derivative f'(x) as a function of x.

示例4.2.2写道：“利用第一原理求 f(x) = x² 的导数。”我们将逐步应用极限定义，从而得到导函数 f'(x) 关于 x 的表达式。

3. Setting Up the Difference Quotient | 建立差商

We begin by writing f(x+h). Since f(x) = x², we substitute x+h wherever x appears:

我们首先写出 f(x+h)。因为 f(x) = x²，所以用 x+h 替换出现 x 的位置：

f(x+h) = (x+h)²

The difference quotient is then:

那么差商为：

[f(x+h) − f(x)] / h = [(x+h)² − x²] / h

4. Expanding (x+h)² | 展开 (x+h)²

To simplify the quotient, we must expand the square. Using the binomial pattern (a+b)² = a² + 2ab + b², we get:

为了简化这个商，我们必须展开平方。利用完全平方公式 (a+b)² = a² + 2ab + b²，我们得到：

(x+h)² = x² + 2xh + h²

Substitute this back into the difference quotient:

将这个结果代回差商：

[x² + 2xh + h² − x²] / h

5. Simplifying the Quotient | 化简差商

We now cancel the x² and −x² terms in the numerator. This is a crucial step because it removes the singularity that would otherwise appear as h → 0.

现在我们消去分子中的 x² 和 −x² 项。这是关键一步，因为它消除了在 h→0 时可能出现的奇异性。

(2xh + h²) / h

Factor h out of the numerator:

从分子中提出公因子 h：

h(2x + h) / h

Since h ≠ 0 (we are considering values approaching zero but never equal to zero), we cancel h:

由于 h ≠ 0（我们考虑的是趋近于零但永远不等于零的值），因此我们约去 h：

2x + h

6. Taking the Limit as h → 0 | 令 h 趋近于零并取极限

Now we allow h to approach zero. The expression 2x + h has a simple limit: as h gets arbitrarily small, 2x + h gets arbitrarily close to 2x.

现在我们令 h 趋近于零。表达式 2x + h 的极限很简单：当 h 任意变小时，2x + h 会任意接近 2x。

f'(x) = lim_h→0 (2x + h) = 2x

Therefore, the derivative of x² is 2x. This result forms the foundation of the power rule, which states that d/dx(xⁿ) = nxⁿ⁻¹.

因此，x² 的导数是 2x。这个结果为幂法则奠定了基础，幂法则指出 d/dx(xⁿ) = nxⁿ⁻¹。

7. The Derivative as a Function | 导函数

Notice that the result f'(x) = 2x is itself a function. It gives the slope of the tangent line at any point on the curve y = x². For example, at x = 1, the slope is 2; at x = −3, the slope is −6.

注意，结果 f'(x) = 2x 本身是一个函数。它给出了曲线 y = x² 上任意一点处切线的斜率。例如，在 x = 1 处，斜率为 2；在 x = −3 处，斜率是 −6。

This illustrates that the derivative is not just a number calculated at one point; it is a function that describes how the original function changes everywhere.

这说明导数不仅仅是在某一点计算的数值，而是一个描述原函数在所有位置如何变化的函数。

8. Geometric Meaning | 几何意义

The difference quotient [f(x+h) − f(x)] / h represents the slope of a secant line through the points (x, f(x)) and (x+h, f(x+h)). As h shrinks, this secant line rotates and approaches the tangent line at x. The limiting slope is the derivative.

差商 [f(x+h) − f(x)] / h 表示过点 (x, f(x)) 和 (x+h, f(x+h)) 的割线的斜率。随着 h 缩小，割线旋转并逐步逼近点 x 处的切线。极限斜率就是导数。

For f(x) = x², the tangent line at x = 2 has slope 4; its equation is y = 4x − 4. We can verify this by checking that the line touches the curve only at that point.

对于 f(x) = x²，在 x = 2 处的切线斜率为 4，其方程为 y = 4x − 4。我们可以验证该直线只在这一点与曲线相切。

9. Common Mistakes to Avoid | 需要避免的常见错误

When working through first-principles differentiation, students often make the following errors:

在处理第一原理求导时，学生常犯以下错误：

Incorrectly expanding (x+h)² as x² + h², omitting the cross term 2xh. This leads to a wrong quotient.

错误地将 (x+h)² 展开为 x² + h²，遗漏了交叉项 2xh。这会导致差商错误。
Setting h = 0 before cancelling h. The limit process only lets h approach zero, not equal zero.

在约去 h 之前就令 h = 0。极限过程只让 h 趋近于零，而不是等于零。
Forgetting to write “lim_h→0” until after the cancellation. Omitting the limit notation in intermediate steps is formally incorrect.

在约分之后才写“lim h→0”。在中间步骤省略极限记号在形式上是错误的。
Forgetting to subtract f(x) completely, especially when f(x) is itself a sum of terms.

忘记完全减去 f(x)，尤其是当 f(x) 本身是多项和的时候。

10. Checking the Result at a Point | 在特定点检验结果

We can check f'(x) = 2x using the gradient of y = x² at a known point. At x = 0, the derivative is 0. This matches the horizontal tangent at the vertex of the parabola. For x = 3, f'(3) = 6, meaning a tangent line at x = 3 is quite steep.

我们可以利用 y = x² 在已知点的梯度来检验 f'(x) = 2x。在 x = 0 处，导数为 0，这与抛物线顶点处的水平切线一致。在 x = 3 处，f'(3) = 6，说明 x = 3 处的切线相当陡峭。

This check helps confirm that our algebraic manipulation is correct and that the limit step has been handled properly.

这种检验有助于确认我们的代数运算正确，并且极限步骤得到了正确处理。

11. Practice and Extension | 练习与拓展

Try applying first principles to f(x) = x² + 3x. You should obtain f'(x) = 2x + 3. Also try f(x) = 3x², which gives f'(x) = 6x. These exercises build fluency with the limit definition.

尝试用第一原理求 f(x) = x² + 3x 的导数，你应该得到 f'(x) = 2x + 3。再试试 f(x) = 3x²，结果是 f'(x) = 6x。这些练习能帮助你熟练掌握极限定义。

In A-Level examinations, first-principles questions may be limited to linear and quadratic functions, so mastering these examples is essential. The same method can be extended to cubic functions, but the algebra becomes more involved.

在A-Level考试中，第一原理求导的题目通常限于一次函数和二次函数，因此掌握这些示例至关重要。同样的方法可以扩展到三次函数，但代数运算会变得更复杂。

12. Summary | 总结

Example 4.2.2 shows that differentiating f(x) = x² from first principles involves the following steps: set up the difference quotient, expand, simplify, cancel h, and take the limit as h → 0. The result is f'(x) = 2x.

示例4.2.2展示了用第一原理求 f(x) = x² 的导数涉及的步骤：建立差商、展开、化简、约去 h，并令 h → 0 取极限。结果是 f'(x) = 2x。

This procedure not only yields the derivative but also reinforces the conceptual definition of a derivative as a limit of rates of change. Mastery of this foundational example prepares you for more advanced differentiation techniques.

这一过程不仅得到了导数，还强化了导数作为变化率极限的概念定义。掌握这一基础示例，将为学习更高级的求导技巧做好准备。

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Example 4.2.2: Differentiating a Quadratic from First Principles | 示例4.2.2：用第一原理求二次函数的导数