Investigation 6 - The derivative of ln(x) | 探究6 - ln(x)的导数

📚 Investigation 6 - The derivative of ln(x) | 探究6 - ln(x)的导数

icle consolidates that investigation, providing a rigorous yet accessible derivation alongside several alternative proofs and applications.

在IB数学中，探究6通常聚焦于发现自然对数函数 ln(x) 的导数。通过数值探索、图形分析和正式的极限论证，你最终会得到简洁的结果 d/dx ln(x) = 1/x。本文整合这一探究，提供严谨且易于理解的推导，同时给出几种替代证明及应用。

1. The Discovery: Numerical Approximation | 探究：数值逼近

2, 3. The table below shows the results, which strongly suggest that the slope equals 1/x.

探究6的一个典型起点是使用差商来估计 ln(x) 在几个点处的导数。例如，取 h = 0.001，计算 x = 1, 2, 3 时的 [ln(x+h) − ln(x)]/h。下表的结果强烈暗示斜率等于 1/x。

These numerical estimates are close to 1, 1/2, and 1/3 respectively. This pattern is the first clue that the derivative of ln(x) might simply be the reciprocal 1/x.

这些数值估计分别接近 1、1/2 和 1/3。这一规律首次提示 ln(x) 的导数可能就是倒数 1/x。

2. From First Principles: Setting Up the Difference Quotient | 第一原理：建立差商

To derive the derivative analytically, we start from the limit definition: f'(x) = lim_h→0 +h) − f(x)]/h, provided the limit exists. For f(x) = ln(x) and for x > 0, this becomes lim_h→0 [ln(x+h) − ln(x)]/h. The restriction x > 0 is essential because the natural logarithm is only defined for positive arguments.

为了进行解析推导，我们从极限定义出发：f'(x) = lim_h→0 [f(x+h) − f(x)]/h，假设极限存在。对于 f(x) = ln(x) 且 x > 0，这变为 lim_h→0 [ln(x+h) − ln(x)]/h。x > 0 的限制非常关键，因为自然对数只对正的参数有定义。

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3. Simplifying the Logarithmic Expression | 简化对数表达式

(1 + h/x). Therefore, the difference quotient can be written as (1/h) · ln(1 + h/x). Moving the factor 1/h inside the logarithm as an exponent yields ln[(1 + h/x)^1/h]. This manipulation sets the stage for a crucial limit.

利用对数商的属性，ln(x+h) − ln(x) = ln((x+h)/x) = ln(1 + h/x)。因此，差商可以写作 (1/h) · ln(1 + h/x)。将因子 1/h 作为指数移到对数内，得到 ln[(1 + h/x)^1/h]。这一变形为关键极限铺平了道路。

4. The Pivotal Limit and Euler's Number e | 关键极限与欧拉数 e

We want to evaluate lim_h→0 ln[(1 + h/x)^1/h 0 as well, and 1/h = 1/(ux). The expression becomes (1/x) · (1/u) ln(1+u) = (1/x) ln[(1+u)^1/u]. Because the natural logarithm is continuous, we can interchange limit and logarithm: the limit equals (1/x) · ln( lim_u→0 (1+u)^1/u ). The limit inside is the classic definition of Euler's number e.

我们希望计算 lim_h→0 ln[(1 + h/x)^1/h]。引入代换 u = h/x，当 h → 0 时，u → 0，同时 1/h = 1/(ux)。表达式变为 (1/x) · (1/u) ln(1+u) = (1/x) ln[(1+u)^1/u]。因为自然对数连续，我们可以交换极限与对数的顺序：该极限等于 (1/x) · ln( lim_u→0 (1+u)^1/u )。里面的极限正是欧拉数 e 的经典定义。

lim_u→0 (1 + u)^1/u = e

Recalling that ln(e) = 1, the derivative simplifies dramatically.

回想 ln(e) = 1，导数被大幅简化。

5. Obtaining d/dx ln(x) = 1/x | 得到 d/dx ln(x) = 1/x

Substituting the value of the limit back, we obtain f' ce, the derivative of ln(x) exists for every x > 0 and is precisely the reciprocal function.

将极限值代回，我们得到 f'(x) = (1/x) · ln(e) = (1/x) · 1 = 1/x。因此，ln(x) 的导数对每个 x > 0 都存在，且恰好是倒数函数。

d/dx ln(x) = 1/x, x > 0

This result is one of the most celebrated in calculus, linking logarithms and algebraic power functions through differentiation.

这一结果是微积分中最著名的结论之一，通过求导将对数与代数幂函数联系起来。

6. Alternative Proof Using Inverse Functions | 使用反函数的替代证明

Because ln(x) is defined as the inverse of e^x, we can start with y = ln(x) ⇔ x = e^y. Differentiating x = e^y with respect to y gives dx/dy = e^y. By the inverse function theorem, dy/dx = 1 / (dx/dy) = 1 / e^y. Replacing e^y with x instantly yields dy/dx = 1/x.

因为 ln(x) 被定义为 e^x 的反函数，我们可以从 y = ln(x) ⇔ x = e^y 出发。对 x = e^y 关于 y 求导得 dx/dy = e^y。由反函数定理，dy/dx = 1 / (dx/dy) = 1 / e^y。将 e^y 替换

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