Key Topic 033: Differentiation from First Principles | 关键考点 033：从第一原理求导

📚 Key Topic 033: Differentiation from First Principles | 关键考点 033：从第一原理求导

Differentiation from first principles is the formal limit process that defines the derivative of a function at a point. It shows exactly how the gradient of a curve arises from the slopes of secant lines as two points become infinitely close.

从第一原理求导是定义函数在某点导数的正式极限过程。它精确地展示了当两个点无限接近时，曲线的梯度如何从割线的斜率产生。

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1. The Idea of a Gradient on a Curve | 曲线上梯度的概念

On a straight line, the gradient is constant and can be found as rise over run. On a curve, the slope changes continuously, so we need to find the slope at a single point by letting a secant line approach a tangent line.

在直线上，梯度是恒定的，可以用纵差除以横差来求得。在曲线上，斜率不断变化，因此我们需要通过让割线趋近于切线来求出单点处的斜率。

If two points on a curve are separated by a horizontal distance h, the secant line joining them becomes a better and better approximation to the tangent as h gets smaller and smaller.

如果曲线上的两个点之间的水平距离为 h，那么随着 h 越来越小，连接这两点的割线会越来越接近切线。

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2. The Limit Definition of the Derivative | 导数的极限定义

For a function f(x), the derivative f ‘(x) is defined by the following limit, provided the limit exists:

对于函数 f(x)，导数 f ‘(x) 由以下极限定义，前提是该极限存在：

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

Here, h is a small change in x. The expression f(x + h) − f(x) is the change in the y-values, so the quotient is the slope of the secant line joining x and x + h.

这里，h 是 x 的一个微小变化。表达式 f(x + h) − f(x) 是 y 值的变化量，因此该商式是连接 x 与 x + h 的割线的斜率。

The notation f ‘(x) is read as ‘f prime of x’. In Edexcel exams, you may also see dy/dx used to represent the same derivative when the function is written as y = f(x).

记号 f ‘(x) 读作 ‘f prime of x’。在 Edexcel 考试中，当函数写成 y = f(x) 时，你也可能看到用 dy/dx 表示同一个导数。

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3. Worked Example: f(x) = x² | 例题：f(x) = x²

We apply the limit definition directly. First, compute f(x + h):

我们直接应用极限定义。首先计算 f(x + h)：

f(x + h) = (x + h)² = x² + 2xh + h²

Then form the difference quotient:

然后构造差商：

[f(x + h) − f(x)] / h = (x² + 2xh + h² − x²) / h = (2xh + h²) / h

Cancelling h from the numerator and denominator gives 2x + h. Taking the limit as h approaches 0 produces:

从分子和分母中约去 h 得到 2x + h。当 h 趋向 0 时取极限得到：

f ‘(x) = lim (h → 0) (2x + h) = 2x

This confirms the familiar result that the derivative of x² is 2x.

这证实了熟悉的结果：x² 的导数是 2x。

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4. Worked Example: f(x) = x³ | 例题：f(x) = x³

Start by expanding (x + h)³ using the binomial expansion or repeated multiplication:

首先使用二项式展开或重复乘法展开 (x + h)³：

(x + h)³ = x³ + 3x²h + 3xh² + h³

The difference quotient becomes:

差商变为：

[f(x + h) − f(x)] / h = (x³ + 3x²h + 3xh² + h³ − x³) / h = (3x²h + 3xh² + h³) / h

Factor h out of every term in the numerator and simplify:

从分子的每一项中提取公因式 h 并化简：

= 3x² + 3xh + h²

As h tends to 0, the terms 3xh and h² both vanish, leaving:

当 h 趋于 0 时，3xh 和 h² 两项都消失，剩下：

f ‘(x) = 3x²

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5. Differentiating f(x) = 1/x | 求 f(x) = 1/x 的导数

For reciprocal functions, the difference quotient involves fractions. Write f(x + h) = 1/(x + h) and combine the two fractions:

对于倒数函数，差商涉及分式。写出 f(x + h) = 1/(x + h) 并通分两个分数：

1/(x + h) − 1/x = [x − (x + h)] / [x(x + h)] = −h / [x(x + h)]

Divide this by h and simplify:

将其除以 h 并化简：

[−h / (x(x + h))] / h = −1 / [x(x + h)]

Taking the limit as h approaches 0 gives:

取 h 趋向 0 的极限得到：

f ‘(x) = −1 / x²

This matches the power rule: d/dx [x⁻¹] = −x⁻².

这与幂法则一致：d/dx [x⁻¹] = −x⁻²。

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6. Differentiating f(x) = √x | 求 f(x) = √x 的导数

Write the square root as a half power: √x = x^½. Direct substitution into the definition gives a difference of square roots:

将平方根写成半次幂：√x = x^½。直接代入定义会得到平方根的差：

[√(x + h) − √x] / h

Multiply the numerator and denominator by the conjugate √(x + h) + √x:

将分子和分母同时乘以共轭式 √(x + h) + √x：

[(x + h) − x] / [h(√(x + h) + √x)] = h / [h(√(x + h) + √x)]

Cancelling h and taking the limit gives:

约去 h 并取极限得到：

f ‘(x) = 1 / (2√x)

This is equivalent

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