Differentiation from First Principles | 从第一原理求导

📚 Differentiation from First Principles | 从第一原理求导

Differentiation from first principles is the fundamental method of finding the derivative of a function using the limit definition. It formalises the intuitive idea of a gradient of a curve at a point, and underpins all the rules of differentiation you will use in A‑Level Mathematics. Mastering this technique not only helps you understand what a derivative truly means, but also prepares you for rigorous proof‑based questions in the Edexcel examination.

从第一原理求导是使用极限定义求函数导数的基本方法。它把曲线上某一点切线斜率的直观概念严格地形式化，并构成你将在 A‑Level 数学中使用的所有求导法则的基础。掌握这一技巧不仅能帮助你真正理解导数的含义，还能为你应对 Edexcel 考试中严谨的证明题做好准备。

1. The Idea of a Limit | 极限的概念

The derivative is defined as the limit of the average rate of change of a function as the interval shrinks to zero. We start with two points on the curve y = f(x): a fixed point A with coordinates (x, f(x)), and a nearby point B with coordinates (x+h, f(x+h)). The gradient of the chord AB is (f(x+h) – f(x)) / h. As h approaches 0, the chord becomes the tangent, and its gradient becomes the derivative f'(x).

导数被定义为当区间趋近于零时函数平均变化率的极限。我们从曲线 y = f(x) 上的两个点开始：一个固定点 A 坐标为 (x, f(x))，以及一个邻近点 B 坐标为 (x+h, f(x+h))。弦 AB 的斜率为 (f(x+h) – f(x)) / h。当 h 趋近于 0 时，弦变为切线，其斜率就成为导数 f'(x)。

This limiting process is written as:

此极限过程写作：

f'(x) = limₕ→₀ [f(x+h) – f(x)] / h

You must be comfortable using this definition to derive the derivatives of basic functions, and to prove that standard differentiation rules are valid.

你必须熟练运用这一定义来推导基本函数的导数，并证明标准求导规则是有效的。

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2. Differentiating xⁿ from First Principles | 从第一原理求 xⁿ 的导数

One of the simplest and most important examples is differentiating f(x) = xⁿ, where n is a positive integer. We begin by expanding (x+h)ⁿ using the binomial theorem. The first two terms are xⁿ + n xⁿ⁻¹ h, and all the remaining terms contain h² or higher powers.

最简单也是最重要的例子之一是对 f(x) = xⁿ（其中 n 为正整数）求导。我们首先用二项式定理展开 (x+h)ⁿ。前两项为 xⁿ + n xⁿ⁻¹ h，其余所有项都含有 h² 或更高次幂。

Subtract f(x) = xⁿ and divide by h:

减去 f(x) = xⁿ 并除以 h：

[ (x+h)ⁿ – xⁿ ] / h = [n xⁿ⁻¹ h + terms in h², h³, …] / h

Cancelling h gives n xⁿ⁻¹ + terms containing h. As h → 0, all terms with h vanish, leaving only n xⁿ⁻¹. Hence, if f(x) = xⁿ then f'(x) = n xⁿ⁻¹. This proof works for any real n when using more advanced expansions, but for A‑Level you will usually see it for positive integer powers.

約去 h 后得到 n xⁿ⁻¹ 加上含有 h 的项。当 h → 0 时，所有含 h 的项消失，只留下 n xⁿ⁻¹。因此，若 f(x) = xⁿ，则 f'(x) = n xⁿ⁻¹。这一证明对任意实数 n 也成立（当使用更高级的展开时），但在 A‑Level 中通常只对正整数幂出现。

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3. Differentiating sin x using Limits | 利用极限求 sin x 的导数

To differentiate f(x) = sin x from first principles, we need the compound‑angle formula and two special limits. Write f(x+h) – f(x) = sin(x+h) – sin x. Using the identity sin A – sin B = 2 cos((A+B)/2) sin((A–B)/2), we obtain:

要由第一原理求 f(x) = sin x 的导数，我们需要用到和角公式和两个重要极限。写出 f(x+h) – f(x) = sin(x+h) – sin x。利用恒等式 sin A – sin B = 2 cos((A+B)/2) sin((A–B)/2)，我们得到：

sin(x+h) – sin x = 2 cos(x + h/2) sin(h/2)

Divide by h:

除以 h：

[sin(x+h) – sin x] / h = cos(x + h/2) × [sin(h/2) / (h/2)]

Now take the limit as h → 0. The limit of sin θ / θ as θ → 0 is 1, and cos(x + h/2) → cos x. Hence f'(x) = cos x × 1 = cos x. This derivation is a classic exam question and requires careful justification of the small‑angle limit.

现在取 h → 0 的极限。当 θ → 0 时，sin θ / θ 的极限为 1，且 cos(x + h/2) → cos x。因此 f'(x) = cos x × 1 = cos x。这一推导是经典的考题，需要小心地说明小角极限的合理性。

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4. Differentiating cos x from First Principles | 从第一原理求 cos x 的导数

Similarly, for f(x) = cos x, we use cos(x+h) – cos x = –2 sin(x + h/2) sin(h/2). After dividing by h and rearranging:

类似地，对于 f(x) = cos x，我们使用 cos(x+h) – cos x = –2 sin(x + h/2) sin(h/2)。除以 h 并整理后：

[cos(x+h) – cos x] / h = –sin(x + h/2) × [sin(h/2) / (h/2)]

Taking the limit h → 0 gives –sin x × 1 = –sin x. The negative sign is important and often tested. Knowing both sine and cosine first‑principle proofs solidifies your understanding of trigonometric limits.

取 h → 0 的极限得到 –sin x × 1 = –sin x。负号非常重要，常被考查。掌握正弦与余弦的第一原理证明能巩固你对三角极限的理解。

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5. Proving the Sum Rule | 证明和的求导法则

If we have two functions u(x) and v(x) and define f(x) = u(x) + v(x), we can prove f'(x) = u'(x) + v'(x) directly from the definition. Write:

如果我们有两个函数 u(x) 和 v(x)，并定义 f(x) = u(x) + v(x)，我们可以直接从定义证明 f'(x) = u'(x) + v'(x)。写出：

f(x+h) – f(x) = [u(x+h)+v(x+h)] – [u(x)+v(x)] = [u(x+h)-u(x)] + [v(x+h)-v(x)]

Divide by h and take limits separately. Because the limit of a sum is the sum of the limits (provided both exist), we obtain u'(x) + v'(x). This simple proof illustrates the power of the first‑principle approach in establishing differentiation rules.

除以 h 并分别取极限。由于和的极限等于极限之和（只要两者都存在），我们得到 u'(x) + v'(x)。这一简单的证明展示了第一原理方法在建立求导法则中的威力。

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6. Proving the Product Rule | 证明乘法法则

For f(x) = u(x) v(x), the first‑principle proof is a little more involved. Start with:

对于 f(x) = u(x) v(x)，第一原理证明稍微复杂一些。从下式开始：

f(x+h) – f(x) = u(x+h)v(x+h) – u(x)v(x)

Add and subtract u(x+h)v(x) in the numerator:

在分子中加上并减去 u(x+h)v(x)：

= u(x+h)[v(x+h)–v(x)] + v(x)[u(x+h)–u(x)]

Now divide by h and take the limit h → 0. Using continuity of u (since differentiable functions are continuous), u(x+h) → u(x). The result is u(x) v'(x) + v(x) u'(x). This proof is often examined in the context of proving the derivative of x² sin x or similar products.

现在除以 h 并取 h → 0 的极限。利用 u 的连续性（因为可导函数必连续），u(x+h) → u(x)。结果是 u(x) v'(x) + v(x) u'(x)。这一证明常结合对 x² sin x 等乘积求导的题目进行考查。

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7. The Chain Rule via First Principles (Outline) | 从第一原理看链式法则（概述）

The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) g'(x). A rigorous first‑principle proof uses the definition of the derivative and considers the limit of [f(g(x+h)) – f(g(x))]/h. It requires careful handling of cases where g(x+h) = g(x) near the point, but the standard approach assumes g'(x) ≠ 0 and uses the fact that g is continuous.

链式法则指出，若 y = f(g(x))，则 dy/dx = f'(g(x)) g'(x)。严格的第一原理证明利用导数的定义，并考虑 [f(g(x+h)) – f(g(x))]/h 的极限。它需要小心处理在点附近 g(x+h) = g(x) 的情况，但标准方法假设 g'(x) ≠ 0 并利用 g 的连续性。

For A‑Level, you do not normally need to replicate the full epsilon‑delta chain rule proof, but you should understand how the definition can be adapted: multiply and divide by g(x+h) – g(x). This highlights why the derivative of the outer function is evaluated at g(x).

在 A‑Level 中，你通常不需要完全复现 ε-δ 的链式法则证明，但应该理解如何改写定义：乘以并除以 g(x+h) – g(x)。这突显出为什么外层函数的导数在 g(x) 处取值。

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8. Differentiating eˣ from First Principles | 从第一原理求 eˣ 的导数

The exponential function f(x) = eˣ has the remarkable property that its derivative is itself. Using first principles:

指数函数 f(x) = eˣ 有一个卓越的性质：它的导数就是它本身。利用第一原理：

f'(x) = limₕ→₀ [eˣ⁺ʰ – eˣ] / h = eˣ limₕ→₀ (eʰ – 1) / h

The limit limₕ→₀ (eʰ – 1)/h = 1 is a defining property of the number e. Thus f'(x) = eˣ. This proof may be requested in an exam where you are given that limit. It is a beautiful example of how the natural base simplifies calculus.

极限 limₕ→₀ (eʰ – 1)/h = 1 是数字 e 的一个定义性质。因此 f'(x) = eˣ。这个证明可能在考试中要求，并给你这个极限。这是一个优美的例子，说明自然底数如何简化微积分。

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9. Differentiating aˣ (Extension) | 求 aˣ 的导数（扩展）

For a general exponential aˣ, first principles gives f'(x) = aˣ limₕ→₀ (aʰ – 1)/h. This limit is ln a. Hence d/dx (aˣ) = aˣ ln a. You can prove this by writing a = eˡⁿ ᵃ and applying the chain rule, but the limit definition unifies the approach.

对于一般的指数函数 aˣ，第一原理给出 f'(x) = aˣ limₕ→₀ (aʰ – 1)/h。这个极限是 ln a。因此 d/dx (aˣ) = aˣ ln a。你可以通过将 a 写成 eˡⁿ ᵃ 并应用链式法则来证明，但极限定义将方法统一了起来。

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10. Common Mistakes and Exam Tips | 常见错误与应试技巧

Many students lose marks by not showing the limit explicitly, or by prematurely setting h = 0 before cancelling. Always write the full difference quotient, simplify algebraically, and only then apply the limit. When using trigonometric identities, be meticulous with the factor of 1/2 inside the sine and cosine arguments.

许多学生因为没有明确写出极限，或在尚未約分时就过早地令 h = 0 而失分。一定要写出完整的差商，先进行代数化简，然后再取极限。使用三角恒等式时，要仔细处理好正弦和余弦参数中的 1/2 因子。

Additionally, Edexcel often asks for a specific function’s derivative “using first principles”. If the question says “using differentiation from first principles”, you must start from the limit definition; quoting standard rules without proof will earn no credit. Practice writing the argument step‑by‑step, and be prepared to comment on the significance of terms that tend to zero.

另外，Edexcel 经常要求“使用第一原理”求某个特定函数的导数。如果题目说“从第一原理求导”，你必须从极限定义出发；直接引用标准公式而未经证明将不会得分。练习逐步写出论证过程，并准备好解释哪些项趋近于零的意义。

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11. Practice Question Walkthrough | 练习题解析

Question: Using differentiation from first principles, find the derivative of f(x) = x² + 3x.

题目：利用从第一原理求导，求 f(x) = x² + 3x 的导数。

Solution: f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h.
f(x+h) – f(x) = (x² + 2xh + h² + 3x + 3h) – (x² + 3x) = 2xh + h² + 3h.
Divide by h: (2x + h + 3).
Limit as h → 0: f'(x) = 2x + 3.

解答：f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h。
f(x+h) – f(x) = (x² + 2xh + h² + 3x + 3h) – (x² + 3x) = 2xh + h² + 3h。
除以 h： (2x + h + 3)。
当 h → 0 时的极限： f'(x) = 2x + 3。

Notice how each algebraic step is clearly shown, and only after simplification is the limit taken. This is the standard expected in Edexcel mark schemes.

注意每一个代数步骤都清晰地展示了出来，而且只有在化简之后才取极限。这是 Edexcel 评分标准中期望的格式。

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12. Connecting First Principles to Graphs | 将第一原理与图像联系起来

Understanding first principles also deepens your graphical intuition. The gradient of the chord is the average rate of change, while the derivative is the instantaneous rate of change. As h gets smaller, the chord better approximates the tangent. Visualising this can help you spot errors: if you forget an h term, the limit may be incorrect.

理解第一原理还能加深你对图像的直觉。弦的斜率是平均变化率，而导数是瞬时变化率。随着 h 变小，弦越来越逼近切线。想象这个过程可以帮助你发现错误：如果你遗漏了某个含 h 的项，极限很可能是错误的。

Sketch the curve, plot the two points, and imagine them merging. This geometric interpretation is at the heart of calculus and is often assessed when explaining why a derivative does not exist at a corner or cusp.

画出曲线，标出两个点，并想象它们逐渐重合。这种几何解释是微积分的核心，在解释为什么某些角点或尖点处导数不存在时也经常被考查。

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