Differentiating Exponentials and Logarithms | 指数函数与对数函数的求导

📚 Differentiating Exponentials and Logarithms | 指数函数与对数函数的求导

In A-Level Mathematics, differentiating exponentials and logarithms appears throughout Pure Mathematics, especially in calculus, modelling and differential equations. The natural exponential function eˣ and the natural logarithm ln x have special derivative results that simplify many calculations.

在 A-Level 数学中，指数函数与对数函数的求导贯穿纯数学的多个主题，尤其出现在微积分、建模和微分方程中。自然指数函数 eˣ 和自然对数函数 ln x 具有特殊的导数结果，可以简化许多计算。

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1. Core Derivative Results | 核心导数结果

The two fundamental results are d/dx (eˣ) = eˣ ithm case. These form the basis for nearly every question on this topic in Edexcel papers.

两个核心结果是：d/dx (eˣ) = eˣ，以及 d/dx (ln x) = 1/x，其中对数函数要求 x > 0。这两个结果是 Edexcel 试卷中几乎所有相关题目的基础。

d/dx (eˣ) = eˣ

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

The key property of eˣ is that its derivative is equal to itself. For ln x, the derivative is the reciprocal function 1/x, which is defined only for positive x.

eˣ 的关键性质是它的导数等于它本身。对于 ln x，其导数是倒数函数 1/x，仅在 x 为正时有定义。

Function	Derivative	Condition
eˣ	eˣ	all real x
ln x	1/x	x > 0

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2. Differentiating e^kx | e^kx 的求导

To differentiate y = e^kx use the chain rule with u = kx. Since dy/du = eᵘ and du/dx = k, the result is dy/dx = k e^kx.

要求 y = e^kx 的导数，可令 u = kx 并使用链式法则。因为 dy/du = eᵘ 且 du/dx = k，所以 dy/dx = k e^kx。

d/dx [e^kx] = k e^kx

is worth memorising because it appears so often in exponential growth and decay problems.

这是 e^(f(x)) 一般结果的特殊情形，但由于它经常出现在指数增长和衰减问题中，值得单独记住。

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3. Differentiating ln x and ln(kx) | ln x 与 ln(kx) 的求导

re k is a positive constant and x > 0, the chain rule gives dy/dx = (1/(kx)) × k = 1/x.

ln x 的导数是 1/x，其中 x > 0。对于 y = ln(kx)，其中 k 是正常数且 x > 0，链式法则给出 dy/dx = (1/(kx)) × k = 1/x。

d/dx [ln(kx)] = 1/x

log laws: ln(kx) = ln k + ln x. Since ln k is constant, its derivative is 0, leaving 1/x.

这也可以利用对数法则证明：ln(kx) = ln k + ln x。由于 ln k 是常数，其导数为 0，因此剩下 1/x。

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4. General Exponential and Logarithmic Bases | 一般指数与对数底数

For a general base a > 0, a ≠ 1, write aˣ = e^{x ln a}. Differentiating then gives d/dx (aˣ) = aˣ ln a.

对于一般底数 a > 0 且 a ≠ 1，可将 aˣ 写成 e^{x ln a}。然后求导得到 d/dx (aˣ) = aˣ ln a。

d/dx (aˣ) = aˣ ln a

Similarly, since logₐ x = ln x / ln a, its derivative is d/dx (logₐ x) = 1/(x ln a).

类似地，因为 logₐ x = ln x / ln a，所以它的导数为 d/dx (logₐ x) = 1/(x ln a)。

d/dx (logₐ x) = 1/(x ln a)

These general results are less common in A-Level questions than the natural exponential and natural logarithm, but they can appear in modelling or proof questions.

这些一般结果在 A-Level 题目中不如自然指数和自然对数常见，但可能出现于建模或证明题中。

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5. Chain Rule with Exponential Functions | 指数函数的链式法则

For y = e^(f(x)), the derivative is dy/dx = f '(x) e^(f(x)). Always multiply by the derivative of the exponent.

对于 y = e^(f(x))，导数为 dy/dx = f '(x) e^(f(x))。一定要乘以指数的导数。

d/dx [e^(f(x))] = f '(x) e^(f(x))

Example 1: y = e^2x, dy/dx = 2e^2x.

例 1：y = e^2x，dy/dx = 2e^2x。

Example 2: y = e^-x, dy/dx = -e^-x.

例 2：y = e^-x，dy/dx = -e^-x。

Example 3: y = e^x², dy/dx = 2x e^x².

例 3：y = e^x²，dy/dx = 2x e^x²。

In each case, the original exponential function remains in the answer, multiplied by the derivative of the exponent.

在每种情形中，原指数函数都保留在答案中，再乘以指数的导数。

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6. Chain Rule with Logarithmic Functions | 对数函数的链式法则

For y = ln(f(x)) with f(x) > 0, the derivative is dy/dx = f '(x)/f(x). This is one of the most frequently tested formulas in Edexcel papers.

对于 y = ln(f(x)) 且 f(x) > 0，导数为 dy/dx = f '(x)/f(x)。这是 Edexcel 试卷中最常考查的公式之一。

d/dx [ln(f(x))] = f '(x)/f(x), f(x) > 0

Example 1: y = ln(2x + 1), so f(x) = 2x + 1 and f '(x) = 2. Therefore dy/dx = 2/(2x + 1).

例 1：y = ln(2x + 1)，则 f(x) = 2x + 1，f '(x) = 2。因此 dy/dx = 2/(2x + 1)。

Example 2: y = ln(x² + 3), so f(x) = x² + 3 and f '(x) = 2x. Therefore dy/dx = 2x/(x² + 3).

例 2：y = ln(x² + 3)，则 f(x) = x² + 3，f '(x) = 2x。因此 dy/dx = 2x/(x² + 3)。

Example 3: y = ln(cos x), valid when cos x > 0. Here f '(x) = -sin x, so dy/dx = -sin x / cos x = -tan x.

例 3：y = ln(cos x)，在 cos x > 0 时成立。这里 f '(x) = -sin x，所以 dy/dx = -sin x / cos x = -tan x。

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7. Product and Quotient Rules | 乘积法则与商法则

Exponential and logarithmic functions often appear in products or quotients. Use the product rule or quotient rule, then simplify carefully.

指数函数和对

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