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The Derivative of xⁿ for Positive Integer Powers | 正整数幂函数xⁿ的导数

2026-08-30 16:55 · 本文部分内容由AI辅助创作

📚 The Derivative of xⁿ for Positive Integer Powers | 正整数幂函数xⁿ的导数

In IB Mathematics, the derivative of xⁿ is one of the most fundamental results in calculus. For any positive integer n, the derivative is n xⁿ⁻¹. This article explains why this rule works, derives it from first principles, and applies it to exam-style problems.

在IB数学中,xⁿ的导数是最基础的结果之一。对于任何正整数n,其导数都是 n xⁿ⁻¹。本文将解释这条规则为何成立,从第一性原理出发进行推导,并应用于考试风格的问题。

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

Before we can differentiate xⁿ, we must recall the formal definition of the derivative. For a function f, its derivative at x is the limit of the average rate of change over an interval that shrinks to zero.

在微分xⁿ之前,我们必须回顾导数的正式定义。对于函数f,它在x处的导数是区间缩小至零时平均变化率的极限。

f′(x) = lim (h→0) [f(x+h) − f(x)] / h

This limit is often called differentiating from first principles. If the limit exists, f is said to be differentiable at x.

这个极限通常称为"从第一性原理求导"。如果极限存在,就称f在x处可导。


2. Expanding (x + h)ⁿ | 展开 (x + h)ⁿ

The key step in differentiating xⁿ from first principles is to expand (x + h)ⁿ. We use the binomial theorem, which expresses the power as a sum of terms with decreasing powers of x.

从第一性原理微分xⁿ的关键步骤是展开(x + h)ⁿ。我们使用二项式定理,将该幂表示为x的降幂项之和。

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

For example, when n = 3, the expansion is (x + h)³ = x³ + 3x²h + 3xh² + h³.

例如,当n = 3时,展开式为(x + h)³ = x³ + 3x²h + 3xh² + h³。


3. Deriving the General Formula | 推导一般公式

Substitute the expansion into the first-principles definition. The first term xⁿ cancels with −f(x), leaving n xⁿ⁻¹ h plus terms that each contain at least h².

将展开式代入第一性原理的定义。第一项xⁿ与−f(x)相消,留下 n xⁿ⁻¹ h 加上每一项至少含有h²的后续项。

After dividing by h, we obtain n xⁿ⁻¹ plus terms involving h. As h approaches zero, these terms vanish.

除以h后,我们得到 n xⁿ⁻¹ 加上含有h的项。当h趋近于零时,这些项消失。

d/dx (xⁿ) = n xⁿ⁻¹

This result is the power rule for positive integer exponents.

这个结果就是正整数指数的幂法则。


4. Applying the Power Rule | 应用幂法则

Now that we have the formula, we can differentiate any positive integer power quickly. The table below shows some common cases.

现在我们有了公式,就能快速微分任何正整数幂。下表展示了一些常见情形。

f(x) f′(x)
x¹ 1
x² 2x
x³ 3x²
x⁵ 5x⁴
x¹⁰ 10x⁹

Notice that x⁰ = 1 for x ≠ 0, so the formula works even when n = 1.

注意x⁰ = 1(x ≠ 0),所以公式在n = 1时同样成立。


5. Combining with Constant Multiples | 常数倍与幂法则结合

Often the function you need to differentiate is multiplied by a constant. The constant multiple rule states that the derivative of c·f(x) is c·f′(x).

通常需要微分的函数会乘以一个常数。常数倍法则指出,c·f(x)的导数是c·f′(x)。

d/dx (c xⁿ) = c n xⁿ⁻¹


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6. The Sum and Difference Rules | 和差法则

When a function is a sum or difference of power terms, differentiate term by term.

当函数是幂项的和或差时,逐项求导。

d/dx (u + v) = du/dx + dv/dx, d/dx (u − v) = du/dx − dv/dx

For example, f(x) = x⁴ + x² has f′(x) = 4x³ + 2x.

例如,f(x) = x⁴ + x² 的导数为 f′(x) = 4x³ + 2x。

This rule works because the derivative is a linear operator; the limit of a sum is the sum of the limits.

该规则之所以成立,是因为导数是一种线性算子;和的极限等于极限的和。


7. Tangents and Normals | 切线与法线

One of the main uses of the derivative is finding the equation of the tangent line to a curve at a given point.

导数的主要用途之一是在给定点处求曲线的切线方程。

he tangent at x = a is n aⁿ⁻¹. The tangent line then has equation y − aⁿ = n aⁿ⁻¹ (x − a).

若y = xⁿ,在x = a处切线的斜率为 n aⁿ⁻¹。因此切线方程为 y − aⁿ = n aⁿ⁻¹ (x − a)。

The normal line is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent's slope.

法线与切线垂直,因此其斜率为切线斜率的负倒数。


8. Higher-Order Derivatives | 高阶导数

Applying the power rule repeatedly gives higher-order derivatives.

重复应用幂法则可以得到高阶导数。

In general, the k-th derivative of xⁿ is n(n−1)···(n−k+1) xⁿ⁻ᵏ.

一般地,xⁿ的k阶导数为 n(n−1)···(n−k+1) xⁿ⁻ᵏ。


9. Common Pitfalls | 常见误区


10. Exam-Style Practice | 考试风格练习

Let us solve a typical IB question. Find the derivative of f(x) = 5x³ − 2x⁴ + x² − 7.

让我们解决一道典型的IB题目。求 f(x) = 5x³ − 2x⁴ + x² − 7 的导数。

Apply the sum and difference rules term by term:

逐项应用和差法则:

f′(x) = 15x² − 8x³ + 2x

The constant −7 has derivative zero. Always double-check by re-reading the question and confirming the exponents.

常数−7的导数为零。务必重读题目并确认指数来检查答案。

nother common question asks for the gradient of y = x⁸ at x = 1. The gradient is 8·1⁷ = 8.

另一个常见问题是求 y = x⁸ 在 x = 1 处的梯度。梯度为 8·1⁷ = 8。


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