Differentiating Exponentials and Logarithms | 指数函数与对数函数的微分

📚 Differentiating Exponentials and Logarithms | 指数函数与对数函数的微分

Exponential and logarithmic functions are central to A‑Level calculus. Mastering their derivatives opens the door to modelling growth, decay, and many advanced techniques. This tutorial covers the essential differentiation rules for e^x, a^x, ln x, log_a x, and combinations using the chain, product, and quotient rules, all tailored to the Edexcel specification.

指数函数与对数函数是 A‑Level 微积分的核心。掌握它们的导数，能够为建立增长、衰减模型以及众多高级技巧打下基础。本教程涵盖 e^x、a^x、ln x、log_a x 的基本求导规则，以及运用链式法则、乘积法则和商法则处理它们的组合函数，内容完全切合 Edexcel 考试要求。

1. Review of Differentiation Rules & Expo/Log Functions | 微分法则回顾及指数/对数函数

Before tackling exponentials and logarithms, recall the basic differentiation rules: the power rule d/dx(xⁿ) = n xⁿ⁻¹, the sum rule, and the constant multiple rule. The natural exponential function is e^x e is the natural logarithm ln x, defined for x > 0. For any positive base a ≠ 1, we have a^x and log_a x.

在接触指数和对数函数的微分之前，先回顾基本的求导法则：幂函数法则 d/dx(xⁿ) = n xⁿ⁻¹，以及和法则与常数倍法则。自然指数函数为 e^x，其中 e ≈ 2.71828；其反函数为自然对数 ln x，定义域为 x > 0。对于任意正底数 a ≠ 1，同样有 a^x 和 log_a x。

2. The Derivative of e^x | e^x 的导数

The exponential function e^x has the remarkable property that its derivative is itself. Using first principles or the definition of e, we obtain the fundamental result:

指数函数 e^x 具有一个显著的特性：其导数等于它自身。利用导数第一定义或 e 的定义，可以得到这个基本结论：

d/dx (e^x) = e^x

This means the slope of the e^x curve at any point equals the function value at that point. It is the only non‑zero function that is its own derivative (up to a constant multiple).

这意味着 e^x 曲线上任意一点处的斜率等于该点的函数值。它是唯一一个导数等于自身的非零函数（允许相差一个常数因子）。

3. Derivative of ln x | ln x 的导数

Since ln x is the inverse of e^x, we can find its derivative via implicit differentiation. Let y = ln x, then e^y = x. Differentiating both sides with respect to x gives e^y dy/dx = 1, so dy/dx = 1/e^y = 1/x. Hence:

由于 ln x 是 e^x 的反函数，我们可以通过隐函数微分法求其导数。设 y = ln x，则 e^y = x。两边对 x 求导得 e^y dy/dx = 1，所以 dy/dx = 1/e^y = 1/x。因此：

d/dx (ln x) = 1/x,   x > 0

This formula is valid only for positive x. In some contexts we consider the derivative of ln|x|, which is 1/x for all non‑zero x, but the Edexcel specification typically focuses on x > 0.

该公式仅在 x 为正数时成立。某些情形下会考虑 ln|x| 的导数，对所有非零 x 均为 1/x，但 Edexcel 考纲通常只关注 x > 0 的情况。

4. Differentiating a^x | a^x 的微分

To differentiate a^x where a > 0, a ≠ 1, express it in base e: a^x = e^{x ln a}. Then use the chain rule:

为了对 a^x（a > 0, a ≠ 1）求导，可将其写成以 e 为底的形式：a^x = e^{x ln a}，然后运用链式法则：

d/dx (a^x) = a^x ln a

The proof is straightforward: d/dx (e^{x ln a}) = e^{x ln a} · ln a = a^x ln a. Note the special case when a = e gives ln e = 1, recovering the derivative of e^x.

证明过程很简单：d/dx (e^{x ln a}) = e^{x ln a} · ln a = a^x ln a。注意当 a = e 时，ln e = 1，恰好回到 e^x 的导数。

5. Derivative of log_a x | log_a x 的导数

Using the change‑of‑base formula, log_a x = (ln x)/(ln a). Since ln a is a constant, we differentiate:

利用换底公式，log_a x = (ln x)/(ln a)。由于 ln a 是常数，直接求导即可：

d/dx (log_a x) = 1/(x ln a),   x > 0

This formula can also be remembered as analogous to the derivative of a^x: the ln a moves to the denominator in the log case, whereas it appears in the numerator for the exponential.

该公式也可以类比 a^x 的导数来记忆：在对数情形中，ln a 出现在分母，而在指数情形中则出现在分子。

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6. Chain Rule with e^f(x) and ln(f(x)) | 链式法则应用于 e^f(x) 与 ln(f(x))

The chain rule is essential when the exponent or the argument of a logarithm is itself a function. For an exponent f(x):

当指数或对数函数的自变量本身是函数时，链式法则就必不可少了。对于指数 f(x)：

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

Similarly, for a logarithm of a function:

同理，对数复合函数的导数为：

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

le combined with the basic derivatives from Sections 2 and 3. Example: differentiate y = e^3x²+1. Let f(x)=3x²+1, then f'(x)=6x, so dy/dx = 6x e^3x²+1.

这些都是直接由链式法则结合第2、3节中的基本导数得到的结果。例如：对 y = e^3x²+1 求导。令 f(x)=3x²+1，则 f'(x)=6x，因此 dy/dx = 6x e^3x²+1。

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7. Product and Quotient Rules Involving e^x and ln x | 包含 e^x 和 ln x 的乘积与商法则

onentials or logarithms with polynomials or trigonometric terms, use the product rule (uv)' = u'v + uv' or the quotient rule (u/v)' = (u'v – uv')/v².

当函数将指数或对数与多项式或三角函数混合时，运用乘积法则 (uv)' = u'v + uv' 或商法则 (u/v)' = (u'v – uv')/v²。

Example product: y = x² e^x. Let u=x², v=e^x. Then u'=2x, v'=e^x. So dy/dx = 2x e^x + x² e^x = x e^x(2 + x).

乘积示例：y = x² e^x。令 u=x², v=e^x。则 u'=2x, v'=e^x。于是 dy/dx = 2x e^x + x² e^x = x e^x(2 + x)。

Example quotient: y = ln x / x. Using the quotient rule, u=ln x, v=x, u'=1/x, v'=1. Thus dy/dx = [(1/x)·x – ln x·1] / x² = (1 – ln x)/x².

商示例：y = ln x / x。使用商法则，u=ln x, v=x, u'=1/x, v'=1。因此 dy/dx = [(1/x)·x – ln x·1] / x² = (1 – ln x)/x²。

Always simplify the result where possible, as examiners often reward simplified answers.

尽可能将结果化简，因为考官通常会给化简后的答案加分。

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8. Implicit Differentiation with Logarithms | 隐函数微分与对数

instance, differentiate ln(y) = x² + sin x with respect to x. Treat y as a function of x:

当 x 与 y 之间的关系并未明确表示为 y=f(x) 时，隐函数微分法非常有效。例如，对 ln(y) = x² + sin x 两边关于 x 求导。将 y 视为 x 的函数：

(1/y) dy/dx = 2x + cos x   →   dy/dx = y(2x + cos x)

If the equation involves exponentials of y, such as e^y = x + ln y, differentiate term‑by‑term: e^y dy/dx = 1 + (1/y) dy/dx, then solve for dy/dx.

若方程包含关于 y 的指数项，如 e^y = x + ln y，可以逐项求导：e^y dy/dx = 1 + (1/y) dy/dx，然后解出 dy/dx。

This technique often appears in questions where you need to find the gradient of a curve at a given point.

这一技巧常出现在需要求曲线上指定点处梯度的题目中。

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9. Differentiating Functions of the Form x^x | 形如 x^x 的函数的微分（对数微分法）

A function like y = x^x (for x > 0) cannot be differentiated by ordinary power or exponential rules alone. Take natural logarithms of both sides:

像 y = x^x（x > 0）这样的函数，无法仅靠普通的幂法则或指数法则求导。应对等式两边取自然对数：

ln y = x ln x

Now differentiate implicitly with respect to x:

然后关于 x 进行隐函数求导：

(1/y) dy/dx = ln x + 1   →   dy/dx = x^x (ln x + 1)

This method, often called logarithmic differentiation, is also useful for complicated products and quotients, making them simpler to differentiate.

这种方法常被称为对数微分法，对于复杂的乘积与商也同样有用，能使求导变得简单。

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10. Applications: Simple Differential Equations | 应用：简单微分方程

ations that model growth and decay. The simplest is dy/dx = k y, whose solution is y = A e^kx, where A is a constant determined by initial conditions.

指数函数和对数函数的导数很自然地引出用于描述增长与衰减的微分方程。最简单的形式是 dy/dx = k y，其通解为 y = A e^kx，其中 A 为由初始条件确定的常数。

Another common form is dy/dx = k(y – variables. Recognising that d/dx (ln y) = (1/y) dy/dx helps in integrating factor methods.

另一常见形式是 dy/dx = k(y – a)，同样可以通过分离变量法得到指数形式的解。认识到 d/dx (ln y) = (1/y) dy/dx 会有助于积分因子法的运用。

Example: Solve dy/dx = 3y given y=2 at x=0. Separate variables: ∫ 1/y dy = ∫ 3 dx ⇒ ln|y| = 3x + C ⇒ y = A e^3x. Using y(0)=2 gives A=2, so y = 2e^3x.

示例：求解 dy/dx = 3y，且当 x=0 时 y=2。分离变量：∫ 1/y dy = ∫ 3 dx ⇒ ln|y| = 3x + C ⇒ y = A e^3x。代入 y(0)=2 得 A=2，因此 y = 2e^3x。

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11. Common Pitfalls and Examiner Tips | 常见陷阱与考官提示

• Forgetting the domain: ln x and log_a x are only defined for x>0. Write the condition if required.

  忘记定义域：ln x 与 log_a x 仅在 x>0 时有定义。如有需要，应注明条件。

• Confusing the derivative of a^x with that of x^a. The former involves ln a, the latter uses the power rule.

  混淆 a^x 与 x^a 的导数。前者需要引入 ln a，而后者使用幂法则。

• Omitting the chain rule: when differentiating e^f(x) or ln(f(x)), always multiply by f'(x).

  遗漏链式法则：对 e^f(x) 或 ln(f(x)) 求导时，务必乘以 f'(x)。

• Misapplying the quotient rule: keep track of the subtraction order – it is u'v – uv', not the reverse.

  误用商法则：注意减法顺序——总是 u'v – uv'，切勿颠倒。

• When using logarithmic differentiation, remember to multiply by y at the end to recover dy/dx.

  使用对数微分法时，最后记得乘以 y 才能得到 dy/dx。

Examiners often test these points, so careful algebraic manipulation and clear presentation are vital.

考官常常在这些细节上出题，因此细心的代数演算和清晰的书写至关重要。

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12. Summary and Key Formulae | 总结与关键公式

The core derivatives are collected below. Memorising them together with the chain rule extensions will cover the vast majority of A‑Level questions on this topic.

以下汇总核心导数公式。熟记这些公式及链式法则扩展形式，足以应对 A‑Level 中绝大部分的相关题目。

d/dx (e^x) = e^x
d/dx (ln x) = 1/x
d/dx (a^x) = a^x ln a
d/dx (log_a x) = 1/(x ln a)
d/dx [e^f(x)] = f'(x) e^f(x)
d/dx [ln(f(x))] = f'(x)/f(x)

Keep these results at your fingertips, practise with a variety of functions, and you will handle differentiation of exponentials and logarithms with confidence in the exam.

将这些结论熟稔于心，并对多样化的函数进行练习，你便能在考试中从容应对指数函数与对数函数的微分。

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