📚 The Power Rule of Differentiation: The Exponent Decreases by One | 幂函数求导：指数减少1的规则
The power rule is one of the most fundamental differentiation techniques in A-Level mathematics. It states that for any real number n, the derivative of x raised to the power n is n times x raised to the power (n − 1). This simple rule, often summarised as "bring the power down, then subtract one from the exponent," underpins a vast range of calculus problems.
幂函数求导法则是A-Level数学中最基础的求导技巧之一。它指出：对于任何实数 n，x 的 n 次幂的导数等于 n 乘以 x 的 (n−1) 次幂。这条简单的规则——通常概括为"把指数拉下来，然后指数减1"——支撑着大量微积分问题。
1. The Rule Statement | 法则的表述
iginal exponent n becomes the multiplicative coefficient, and the new exponent is exactly one less than the original.
如果 f(x) = xⁿ，其中 n 是任意实数常数，那么导数为 f ′(x) = n xⁿ⁻¹。核心思想是：原来的指数 n 变成了乘法系数，而新的指数恰好比原来的指数少1。
d/dx (xⁿ) = n xⁿ⁻¹
This rule applies for all real values of n: positive integers, negative integers, fractions, and even irrational numbers. Note that the domain of the derivative may differ from the domain of the original function, particularly when n is negative or fractional.
这条规则适用于 n 的所有实数值：正整数、负整数、分数，甚至无理数。需要注意的是，导数的定义域可能与原函数的定义域不同，尤其是当 n 为负数或分数时。
2. Why the Exponent Decreases by One | 为什么指数减少1
The reason behind the "decrease by one" lying the binomial expansion to (x+h)ⁿ, every term except the first contains a factor of h. When we divide by h and let h approach zero, only the term n xⁿ⁻¹ survives.
"减少1"这一现象的原因在于导数的定义。利用极限定义，f ′(x) = lim(h→0) [(x+h)ⁿ − xⁿ] / h，对 (x+h)ⁿ 应用二项式展开，除第一项外的每一项都含有因子 h。当我们除以 h 并让 h 趋近于零时，只有 n xⁿ⁻¹ 这一项留存下来。
xh + h². As h → 0, this approaches 3x². The exponent 3 has dropped to 2, exactly one less.
例如，展开 (x+h)³ 得到 x³ + 3x²h + 3xh² + h³。减去 x³ 并除以 h 后得到 3x² + 3xh + h²。当 h → 0 时，该表达式趋近于 3x²。指数从3降为2，恰好减少1。
3. Formal Proof Using the Binomial Theorem | 利用二项式定理的正式证明
For a positive integer n, we can prove the power rule rigorously. Start with the definition of the derivative:
对于正整数 n，我们可以严格证明幂函数求导法则。从导数的定义出发：
f ′(x) = lim(h→0) [(x+h)ⁿ − xⁿ] / h
Using the binomial theorem, (x+h)ⁿ = xⁿ + n xⁿ⁻¹ h + [n(n−1)/2] xⁿ⁻² h² + … + hⁿ. Subtracting xⁿ and dividing by h gives n xⁿ⁻¹ + [n(n−1)/2] xⁿ⁻² h + … + hⁿ⁻¹. Taking the limit as h → 0, all terms containing h vanish, leaving f ′(x) = n xⁿ⁻¹.
使用二项式定理，(x+h)ⁿ = xⁿ + n xⁿ⁻¹ h + [n(n−1)/2] xⁿ⁻² h² + … + hⁿ。减去 xⁿ 并除以 h 后得到 n xⁿ⁻¹ + [n(n−1)/2] xⁿ⁻² h + … + hⁿ⁻¹。取 h → 0 的极限，所有含 h 的项都消失，剩下 f ′(x) = n xⁿ⁻¹。
This proof works for positive integers. For other real values of n, a more advanced proof using logarithmic differentiation is typically employed, but the result is identical.
该证明适用于正整数。对于其他实数 n，通常使用对数求导法进行更高级的证明，但结果是完全相同的。
4. Working with Negative Exponents | 负指数的处理
power rule with n = −1, we get f ′(x) = (−1) x⁻² = −1/x². The exponent −1 decreases to −2.
幂函数求导法则对负指数同样适用，前提是 x ≠ 0。考虑 f(x) = x⁻¹ = 1/x。令 n = −1 应用幂函数法则，得到 f ′(x) = (−1) x⁻² = −1/x²。指数从−1降为−2。
d/dx (1/x) = −1/x² | d/dx (1/x²) = −2/x³
Notice that the coefficient becomes negative when n is negative. This makes sense intuitively: if a function is decreasing, its derivative should be negative. For x > 0, 1/x is indeed a decreasing function.
注意当 n 为负数时，系数变为负数。这从直观上是合理的：如果一个函数在递减，其导数应为负。对于 x > 0，1/x 的确是一个递减函数。
5. Fractional Exponents | 分数指数
ctional exponents represent roots, and the power rule handles them seamlessly. If f(x) = x1/2 = √x, then f ′(x) = (1/2) x−1/2 = 1/(2√x). Here the exponent 1/2 has decreased to −1/2.
分数指数代表根式，幂函数法则可以无缝处理它们。如果 f(x) = x1/2 = √x，那么 f ′(x) = (1/2) x−1/2 = 1/(2√x)。这里的指数从1/2降为−1/2。
d/dx (√x) = 1/(2√x) | d/dx (x3/2) = (3/2) x1/2
For f(x) = x3/2, the derivative is (3/2) x1/2. This result is particularly useful in kinematics when dealing with distances proportional to t3/2. Note that when x > 0, the derivative of √x is positive, consistent with the fact that √x is an increasing function.
对于 f(x) = x3/2，导数为 (3/2) x1/2。这一结果在运动学中特别有用，当处理与 t3/2 成正比的距离时。注意当 x > 0 时，√x 的导数为正，这与 √x 是递增函数的事实一致。
6. The Derivative of a Constant | 常数的导数
c is zero. We can view the constant c as c · x⁰, and the power rule gives c · 0 · x⁻¹ = 0.
幂函数法则的一个特殊情况发生在 n = 0 时。由于对所有 x ≠ 0 有 x⁰ = 1，常数函数 f(x) = c 的导数为零。我们可以将常数 c 视为 c · x⁰，则幂函数法则给出 c · 0 · x⁻¹ = 0。
d/dx (c) = 0 for any constant c
This is consistent with the geometric interpretation: the graph of a constant function is a horizontal line with slope 0 everywhere. This rule is essential when differentiating polynomials, as each constant term contributes nothing to the derivative.
这与几何解释一致：常数函数的图像是水平线，处处斜率为0。在求多项式导数时这个规则必不可少，因为每个常数项对导数没有贡献。
7. Linear Combinations and Polynomials | 线性组合与多项式
ve is f ′(x) = a n xⁿ⁻¹, where a is a constant. The coefficient a is simply carried along.
幂函数法则与求和法则、常数倍法则结合，可以逐项求多项式的导数。对于 f(x) = a xⁿ，导数为 f ′(x) = a n xⁿ⁻¹，其中 a 是常数。系数 a 只是被保留下来。
le, to differentiate f(x) = 4x⁵ − 3x² + 7x − 9, we apply the rule to each term separately:
例如，要求 f(x) = 4x⁵ − 3x² + 7x − 9 的导数，我们分别对每一项应用法则：
f ′(x) = 20x⁴ − 6x + 7
ponent 1 drops to 0, and x⁰ = 1), and the constant term −9 has derivative 0. The structure "decrease by one" is visible in every term: 5 → 4, 2 → 1, 1 → 0.
注意 7x 的导数是7（因为指数1降为0，x⁰ = 1），而常数项−9的导数为0。"减少1"的结构在每一项中清晰可见：5→4，2→1，1→0。
8. The Power Rule Combined with the Chain Rule | 幂函数法则与链式法则的结合
When the base itself is a function of x, we must use the chain rule. For a function of the form [g(x)]ⁿ, the derivative is n[g(x)]ⁿ⁻¹ · g ′(x). This is an extension of the power rule, often called the "generalised power rule."
当底数本身是 x 的函数时，我们必须使用链式法则。对于 [g(x)]ⁿ 形式的函数，导数为 n[g(x)]ⁿ⁻¹ · g ′(x)。这是幂函数法则的推广，通常称为"广义幂函数法则"。
For example, if y = (3x² + 1)⁴, then with g(x) = 3x² + 1 and g ′(x) = 6x, we obtain:
例如，若 y = (3x² + 1)⁴，令 g(x) = 3x² + 1 且 g ′(x) = 6x，我们得到：
dy/dx = 4(3x² + 1)³ · 6x = 24x(3x² + 1)³
Here the exponent 4 decreases to 3, but we must multiply by the derivative of the inner function. This technique is essential for differentiating composite functions and appears frequently in A-Level examination questions.
这里的指数4降为3，但我们必须乘以内部函数的导数。这一技巧对于求复合函数的导数至关重要，在A-Level考试题中频繁出现。
9. Common Mistakes and How to Avoid Them | 常见错误与规避方法
A frequent error is forgetting to multiply by the original coefficient. For instance, the derivative of 3x² is not 3x — the correct answer is 6x. Students sometimes also incorrectly differentiate x⁵ as 5x⁵ instead of 5x⁴, forgetting that the exponent must decrease.
一个常见错误是忘记乘以原来的系数。例如，3x² 的导数不是 3x——正确答案是 6x。学生有时也会错误地认为 x⁵ 的导数是 5x⁵ 而不是 5x⁴，忘记了指数必须减1。
| Function | 函数 | Common Error | 常见错误 | Correct Answer | 正确答案 |
| x⁴ | 4x⁴ | 4x³ |
| 5x³ | 5x² | 15x² |
| 1/x | 1/x² | −1/x² |
| √x | 1/(2√x) with no domain condition | 1/(2√x), x > 0 |
writing as x⁻² first. Always express radical or reciprocal forms as powers of x before applying the rule, and double-check that the exponent has indeed decreased by exactly one.
另一个细微的错误是求 d/dx(1/x²) 时忘记先将表达式改写为 x⁻²。在应用法则之前，始终将根式或倒数形式写成 x 的幂形式，并检查指数是否确实恰好减少了1。
10. Revealing the Emergence of Negative Exponents | 负指数的自然出现及其意义
s negative. These negative signs and exponents encode the rate of decrease of the function's slope.
当对倒数函数求导时（x > 0），幂函数法则有时会产生负指数，即使原指数的值为正。例如，d/dx(x²) = 2x 的指数1是正的，d/dx(x³) = 3x² 的指数2也是正的，但 d/dx(1/x) = −1/x² 却是负的。这些负号与负指数编码了函数斜率递减的信息，因而在优化问题中至关重要。
11. Applications in Real-World Problems | 在实际问题中的应用
ration a(t) = v ′(t) = 6t − 4. Each differentiation step reduces the highest power by one.
幂函数法则在物理学和经济学中经常出现。在运动学中，如果位移 s(t) = t³ − 2t² + 5t，那么速度 v(t) = s ′(t) = 3t² − 4t + 5，加速度 a(t) = v ′(t) = 6t − 4。每一步求导都将最高次数减少1。
C(x) = 50 + 8x + x² represents producing x units, then marginal cost C ′(x) = 8 + 2x. The "exponent decreases by one" near costs have constant marginal costs, while quadratic costs have linear marginal costs.
在经济学中，边际成本是总成本的导数。如果总成本 C(x) = 50 + 8x + x² 表示生产 x 个单位所需的成本，那么边际成本 C ′(x) = 8 + 2x。"指数减少1"的模式告诉我们：线性成本具有恒定边际成本，而二次成本具有线性边际成本。
12. Summary and Exam Tips | 总结与考试技巧
th the necessary domain restrictions. The exponent always decreases by exactly one, the original exponent becomes the coefficient, and any constant coefficient is preserved.
总结如下：幂函数法则指出 d/dx(xⁿ) = n xⁿ⁻¹ 对所有实数 n 成立，但需满足必要的定义域限制。指数总是恰好减少1，原指数变成系数，任何常数系数都会被保留。
In examinations, always rewrite radicals and reciprocals in index form (√x as x1/2, 1/x³ as x⁻³) before differentiating. Write the derivative with positive exponents after simplification when possible, or with negative exponents when appropriate. Check the domain: for negative or fractional exponents, ensure the x-values are valid for both the function and its derivative.
在考试中，求导之前始终将根式和倒数改写为指数形式（如 √x 改写为 x1/2，1/x³ 改写为 x⁻³）。化简后，如果可能，用正指数写出导数，或者在适当时使用负指数。检查定义域：对于负数或分数指数，确保 x 的值对原函数和导数都有意义。
d/dx(xⁿ) = n xⁿ⁻¹ — bring the power down, reduce the exponent by one.
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图