Differentiating xⁿ | 对 xⁿ 求导

📚 Differentiating xⁿ | 对 xⁿ 求导

Differentiation is a core tool in A-Level Mathematics. For Edexcel, one of the first rules you learn is how to differentiate powers of x. This rule appears in almost every calculus question, so mastering it is essential. It also unlocks everything that follows: tangents, normals, stationary points, rates of change and optimisation problems.

微积分是 A-Level 数学的核心工具。在 Edexcel 考试中，你最先学会的规则之一就是如何对 x 的幂求导。这条规则几乎出现在所有微积分题目中，因此掌握它至关重要。它也是后续内容的基础：切线、法线、驻点、变化率以及最优化问题都依赖它。

This article starts from the formal definition of a derivative, proves the power rule for positive integer powers using the binomial expansion, then extends it to negative and fractional powers, sums of terms, gradients, tangents and stationary points.

本文从导数的形式定义出发，利用二项式展开证明正整数指数的幂函数求导法则，然后把它推广到负指数、分数指数、多项式的和差、斜率、切线以及驻点问题中。

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1. The Definition of a Derivative | 导数的定义

The derivative f'(x) measures the rate at which f(x) changes with x. Geometrically, it gives the gradient of the tangent to the curve at any point.

导数 f'(x) 衡量 f(x) 随 x 变化的速率。从几何上看，它给出曲线上任意一点处切线的斜率。

By definition, the derivative is the limit of the gradient of a chord as the change in x tends to zero. Take two points on the curve, separated horizontally by a small distance h, and work out the gradient of the straight line joining them.

根据定义，导数是当 x 的改变量趋于零时，割线斜率的极限。在曲线上取两点，它们在水平方向上相隔一个很小的距离 h，然后求连接这两点的直线的斜率。

f'(x) = limₕ→₀ [f(x + h) − f(x)] / h

This definition is sometimes called 'differentiation from first principles' g point for proving the power rule. In an exam you will rarely be asked to use it for anything more complicated than a quadratic, but you must be able to quote it and to use it confidently for simple functions.

这个定义有时被称为"用基础原理求导"（differentiation from first principles）。对于 f(x) = xⁿ，它正是证明幂函数求导法则的出发点。考试中极少要求你对复杂函数使用它，但你必须能够准确写出这个定义，并能熟练用于简单函数。

ominator h is the horizontal change, and the limit h → 0 turns the chord into the tangent.

这个公式有两种有用的读法：分子 f(x + h) − f(x) 是纵向变化量，分母 h 是横向变化量，而极限 h → 0 则把割线变成了切线。

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2. The Power Rule | 幂函数求导法则

xⁿ, the derivative is found by bringing the power down and then reducing the power by one.

幂函数求导法则是求 x 的幂的导数时最重要的捷径。它指出：若 y = xⁿ，则求导时将指数"拉下来"，然后把指数减一。

If y = xⁿ, then dy/dx = n xⁿ⁻¹

The rule converts a limit problem into pure algebra. Instead of computing a limit every time, you multiply by the old power and decrease that power by one. For example, the derivative of x³ is 3x², and the derivative of x⁵ is 5x⁴. The table below shows some common cases.

这条法则把一个极限问题转化成了纯代数运算。你不需要每次都算极限，只需乘以原来的指数，再把指数减一。例如，x³ 的导数是 3x²，x⁵ 的导数是 5x⁴。下表给出了一些常见情形。

Function / 函数	Derivative / 导数
x	1
x²	2x
x³	3x²
x⁴	4x³
x⁵	5x⁴
x⁻¹	−x⁻²
√x = x1/2	1 / (2√x) = ½x−1/2

That single general statement is what makes it so powerful: one line covers every power you will meet in the A-Level specification.

注意这条法则是对一般指数 n 陈述的，而不仅限于 n = 2 或 n = 3。正是这种一般性使它如此有力：一行公式覆盖了 A-Level 考纲中你会遇到的所有幂次。

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3. Applying the Rule: Worked Examples | 应用法则：例题

h coefficient 1. If y = x⁷, then dy/dx = 7x⁶. Multiply by 7, reduce the power from 7 to 6.

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先从最简单的情形开始：系数为 1 的单个幂。若 y = x⁷，则 dy/dx = 7x⁶。乘以 7，指数由 7 减为 6。

. If y = 5x⁴, the constant 5 stays put and only the power changes: dy/dx = 5 × 4x³ = 20x³.

接着看幂前面带数字系数的情形。若 y = 5x⁴，常数 5 保持不变，只有幂次发生变化：dy/dx = 5 × 4x³ = 20x³。

y = x⁻³, then dy/dx = −3x⁻⁴. Bring the −3 down, then subtract 1 from the index to get −4.

负指数的处理方式完全相同。若 y = x⁻³，则 dy/dx = −3x⁻⁴。把 −3 拉下来，再把指数减 1 得到 −4。

2. Substituting a value of x into the derivative always gives an instantaneous rate of change, not an average one.

最后是一道求斜率的题目。对 y = x³，dy/dx = 3x²，因此在 x = 2 处曲线的斜率为 3 × 2² = 12。把 x 的取值代入导数，得到的始终是瞬时变化率，而不是平均变化率。

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4. Proof for Positive Integer Powers | 正整数指数的证明

Let n be a positive integer. We can prove the power rule using the definition of the derivative and the binomial expansion.

设 n 为正整数。我们可以利用导数的定义和二项式展开来证明幂函数求导法则。

The binomial theorem expands the bracket (x + h)ⁿ into a sum of terms with decreasing powers of x.

二项式定理把括号 (x + h)ⁿ 展开成一系列 x 的幂次递减的项。

(x + h)ⁿ = xⁿ + n xⁿ⁻¹h + [n(n − 1) / 2] xⁿ⁻²h² + ... + hⁿ

Substitute this into the first-principles formula and subtract xⁿ from the expansion. Then divide the result by h.

将这个展开式代入基础求导公式，并从展开式中减去 xⁿ，然后除以 h。

f(x + h) − f(x) = n xⁿ⁻¹h + [n(n − 1) / 2] xⁿ⁻²h² + ... + hⁿ

[f(x + h) − f(x)] / h = n xⁿ⁻¹ + [n(n − 1) / 2] xⁿ⁻²h + ... + hⁿ⁻¹

Every term after n xⁿ⁻¹h contains at least one factor h. As h tends to zero, all those terms vanish, leaving only n xⁿ⁻¹. The proof therefore gives exactly the power rule you have been using, and it explains why the index must drop by one: the term that survives is the one whose power of h was exactly 1 before dividing.

在 n xⁿ⁻¹h 之后的每一项都至少含有一个因子 h。当 h 趋于零时，这些项全部消失，只剩下 n xⁿ⁻¹。因此这个证明给出的正是你一直在使用的幂函数求导法则，也解释了为什么指数必须减一：在除以 h 之前，h 的次数恰好为 1 的那一项是唯一存活的项。

limₕ→₀ n xⁿ⁻¹ + [n(n − 1) / 2] xⁿ⁻²h + ... + hⁿ⁻¹ = n xⁿ⁻¹

, including n = 1, where the expansion stops after two terms and the derivative of x is 1.

同样的二项式论证对任意正整数 n 都成立，包括 n = 1 的情形：此时展开式只有两项，而 x 的导数是 1。

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5. Why the Rule Also Works for Negative Powers | 为什么负指数也成立

Negative integer powers are not covered by the binomial proof above, but the rule still holds. Write x⁻ⁿ as 1 / xⁿ and treat

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