Exercise E.2: Mastering Differentiation of Exponential & Logarithmic Functions | 习题E.2：掌握指数与对数函数的微分

📚 Exercise E.2: Mastering Differentiation of Exponential & Logarithmic Functions | 习题E.2：掌握指数与对数函数的微分

In IB Mathematics, whether you are following the Analysis & Approaches (AA) or Applications & Interpretation (AI) route, Exercise E.2 typically marks a critical turning point. It consolidates your understanding of differentiation by applying the rules to exponential and logarithmic functions. Mastering this exercise ensures you can handle everything from simple derivatives like d/dx(eˣ) to complex combinations involving the chain, product, and quotient rules with ln(x) and aˣ. This article unpacks the key concepts, formulas, worked examples, and common pitfalls associated with Exercise E.2, giving you a comprehensive revision resource.

在IB数学中，无论你选择分析与方法(AA)还是应用与解释(AI)路径，习题E.2通常标志着一个关键的转折点。它通过将求导法则应用于指数函数和对数函数来巩固你的微分知识。掌握这个练习能确保你从容应对从简单的导数如 d/dx(eˣ) 到涉及链式法则、乘积法则和商法则与 ln(x) 和 aˣ 的复杂组合。本文将拆解与习题E.2相关的核心概念、公式、例题和常见陷阱，为你提供一份全面的复习资料。

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1. What Is Exercise E.2 All About? | 习题E.2是关于什么的？

Exercise E.2 in most IB textbooks is the second set of problems in the chapter devoted to exponential and logarithmic functions. After learning the theory behind eˣ, ln x, and their properties, you now apply differential calculus. The problems start with straightforward derivative computation and gradually introduce applications such as finding tangents, normals, and linked rates of change. Many questions will also require you to simplify expressions using exponent and log laws before differentiating.

在大多数IB教材中，习题E.2是指数和对数函数章节中的第二组练习。在学习eˣ、ln x及其性质背后的理论后，你现在开始应用微分学。题目从直接求导开始，逐步引入应用，如求切线、法线以及相关变化率。许多题目还要求你在求导前先利用指数和对数定律化简表达式。

The key skills assessed include: applying the derivatives of eˣ and ln x; combining these with the chain rule for more complex arguments; using product and quotient rules when these functions multiply or divide other functions; and interpreting derivatives in context. A solid performance on Exercise E.2 is a strong predictor of success on the IB exam's calculus section.

评估的关键技能包括：应用eˣ和ln x的导数公式；将这些公式与链式法则结合以处理更复杂的复合函数；当这些函数与其他函数相乘或相除时使用乘积法则和商法则；以及在具体情境中解释导数的意义。在习题E.2上扎实的表现是IB考试微积分部分成功的有力预测。

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2. Recap of Fundamental Derivative Rules | 基础求导规则回顾

Before tackling Exercise E.2, you must have the core derivative rules at your fingertips. These rules are built upon throughout the exercise and appear in almost every solution. Make sure you can recall them instantly.

在攻克习题E.2之前，你必须熟练掌握核心求导规则。这些规则贯穿整个练习，几乎出现在每一个解答中。确保你能立刻回忆起它们。

Rule (English)	规则（中文）	Formula
Constant Rule	常数规则	d/dx (c) = 0
Power Rule	幂规则	d/dx (xⁿ) = n xⁿ⁻¹
Constant Multiple	常数倍法则	d/dx [c f(x)] = c f'(x)
Sum/Difference Rule	和差规则	d/dx [f(x) ± g(x)] = f'(x) ± g'(x)
Product Rule	乘积法则	d/dx [u v] = u' v + u v'
Quotient Rule	商法则	d/dx (u/v) = (u' v - u v') / v²
Chain Rule	链式法则	dy/dx = dy/du × du/dx

The chain rule is especially important for Exercise E.2 because exponentials and logarithms rarely appear with just x as the argument; you will see eˣ⁺² or ln(3x+1) very often. Understanding how to break these down into inner and outer functions is essential.

链式法则对习题E.2尤为重要，因为指数和对数函数的自变量很少只是x；你经常见到 eˣ⁺² 或 ln(3x+1) 这类复合形式。理解如何将其拆分为内层和外层函数至关重要。

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3. Derivative of the Natural Exponential Function | 自然指数函数的导数

The single most important result in Exercise E.2 is that the derivative of eˣ is eˣ itself. This unique property makes eˣ the building block for modelling continuous growth and decay.

习题E.2中最重要的结论是 eˣ 的导数就是其自身 eˣ。这一独特性质使得 eˣ 成为模拟连续增长与衰减的基本构成模块。

d/dx (eˣ) = eˣ

When the exponent is a function of x, say u(x), the chain rule gives:

当指数是x的函数，例如 u(x) 时，链式法则给出：

d/dx (eᵤ) = eᵤ × du/dx

As an example, differentiate y = e²ˣ. Here u = 2x, so du/dx = 2. Hence dy/dx = e²ˣ × 2 = 2e²&# rn is repeated throughout Exercise E.2: always multiply by the derivative of the exponent.

例如，对 y = e²ˣ 求导。这里 u = 2x，du/dx = 2。因此 dy/dx = e²ˣ × 2 = 2e²ˣ。这种模式在习题E.2中反复出现：始终记住乘以指数的导数。

You may also meet exponential functions with bases other than e, such as 2ˣ or 10ˣ. The formula is d/dx (aˣ) = aˣ ln a. In the IB syllabus, you can derive this by writing aˣ as e^{x ln a} and then applying the chain rule.

你可能还会遇到以非e为底的指数函数，如 2ˣ 或 10ˣ。其公式为 d/dx (aˣ) = aˣ ln a。在IB大纲中，你可以通过将 aˣ 写成 e^{x ln a} 然后应用链式法则来推导。

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4. Derivative of the Natural Logarithmic Function | 自然对数函数的导数

The derivative of ln x is 1/x for x > 0. Exercise E.2 builds heavily on this result, often combining it with the chain rule to handle arguments like ln(2x-5) or even ln(cos x).

ln x 的导数是 1/x (x > 0)。习题E.2大量运用这一结论，并经常将其与链式法则结合来处理如 ln(2x-5) 甚至 ln(cos x) 这样的复合函数。

d/dx (ln x) = 1/x

For a general differentiable function u(x) > 0, the chain rule yields:

对于一般的可微函数 u(x) > 0，链式法则给出：

d/dx (ln u) = (1/u) × du/dx = u' / u

Example: Find the derivative of f(x) = ln(5x² + 1). Let u = 5x² + 1, then u' = 10x. Therefore f'(x) = 10x / (5x&#0B2; + 1). Many IB questions ask you to leave the answer in this simplified rational form.

例如：求 f(x) = ln(5x² + 1) 的导数。设 u = 5x² + 1，则 u' = 10x。因此 f'(x) = 10x / (5x² + 1)。许多IB题目要求你将答案保持在这种简化的有理函数形式。

ln x, whose derivative is simply 0 + 1/x = 1/x. This trick saves time and reduces errors.

有时你需要在求导前利用对数定律进行化简。例如，ln(2x) = ln 2 + ln x，其导数仅为 0 + 1/x = 1/x。这一技巧可以节省时间并减少错误。

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5. Mastering the Chain Rule with e and ln | 掌握含e和ln的链式法则

function. Always identify the outer function (e^ or ln) and the inner function explicitly.

习题E.2本质上是对指数和对数函数应用链式法则的专项训练。一个典型的错误是忘记乘以内层函数的导数。务必明确识别外层函数 (e^ 或 ln) 和内层函数。

Consider y = e^{sin x}. Outer: e^u, inner: u = sin x. Then dy/dx = e^{sin x} × cos x. Similarly, for y = ln(x³ - 4x), outer: ln u, inner: u = x³ - 4x. This gives dy/dx = (3x² - 4) / (x³ - 4x).

考虑 y = e^{sin x}。外层：e^u，内层：u = sin x。那么 dy/dx = e^{sin x} × cos x。类似地，对于 y = ln(x³ - 4x)，外层：ln u，内层：u = x³ - 4x。从而 dy/dx = (3x² - 4) / (x³ - 4x)。

Harder problems in Exercise E.2 will layer multiple rules. For example, y = x² e^5x requires product rule first, then chain rule for the derivative of e^5x. Writing out the steps systematically will help you secure full marks.

习题E.2中较难的题目会叠加多个法则。例如，y = x² e^5x 需要先使用乘积法则，再对 e^5x 的导数使用链式法则。把步骤系统地写出来将帮助你拿到满分。

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6. Product and Quotient Rule Combinations | 乘积法则与商法则的组合

As Exercise E.2 progresses, you will differentiate functions where exponentials or logarithms are multiplied or divided by polynomials, trigonometric functions, or even other exponentials. Setting up u and v clearly is half the battle.

随着习题E.2的推进，你将开始求导这样一类函数：指数或对数函数与多项式、三角函数甚至其他指数函数相乘或相除。清晰地设定 u 和 v 是成功的一半。

Example: f(x) = x ln x. Let u = x, v = ln x. Then u' = 1, v' = 1/x. By product rule, f'(x) = (1)(ln x) + (x)(1/x) = ln x + 1. This result is elegant and frequently appears.

例如：f(x) = x ln x。设 u = x, v = ln x。则 u' = 1, v' = 1/x。根据乘积法则，f'(x) = (1)(ln x) + (x)(1/x) = ln x + 1。这个结果非常简洁且经常出现。

Quotient rule: g(x) = eˣ / x. u = eˣ, v = x, u' = eˣ, v' = 1. Then g'(x) = (eˣ × x - eˣ × 1) / x² = eˣ(x - 1) / x&#. Always leave the numerator factorised if possible, as IB mark schemes often reward this.

商法则：g(x) = eˣ / x。u = eˣ, v = x, u' = eˣ, v' = 1。那么 g'(x) = (eˣ × x - eˣ × 1) / x² = eˣ(x - 1) / x²。如果可能，始终保持分子因式分解的形式，因为IB的评分方案常常会奖励这一点。

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7. Implicit Differentiation with e and ln | 含e和ln的隐函数微分

oblems, especially in HL, will present equations where y is defined implicitly, such as e&# tiate both sides with respect to x, treating y as a function of x and using the chain rule for eʸ and ln x.

有些习题E.2中的题目（尤其是HL部分）会给出y被隐式定义的方程，例如 eʸ + xy = ln x。你需要对等式两边关于x求导，将y视为x的函数，并对 eʸ 和 ln x 使用链式法则。

Differentiate eʸ + xy = ln x. Left: d/dx(eʸ) = e&# y/dx; d/dx(xy) = x dy/dx + y (product rule). Right: d/dx(ln x) = 1/x. Collect dy/dx terms:

对 eʸ + xy = ln x 进行求导。左：d/dx(eʸ) = eʸ dy/dx；d/dx(xy) = x dy/dx + y（乘积法则）。右：d/dx(ln x) = 1/x。合并 dy/dx 项：

eʸ dy/dx + x dy/dx + y = 1/x ⇒ dy/dx = (1/x - y) / (eʸ + x)

Implicit differentiation questions like this test your ability to manage several rules at once. Writing the derivative of each term step by step prevents algebraic slips.

像这样的隐函数微分题目考查你同时运用多个法则的能力。对每一项的导数进行分步书写可以避免代数失误。

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8. Higher-Order Derivatives | 高阶导数

Exercise E.2 occasionally asks for the second or third derivative to study concavity or for differential equations. Since the derivative of eˣ is eˣ, higher-order derivatives of functions like e𝒌ˣ follow a predictable pattern.

习题E.2偶尔会要求求二阶或三阶导数以研究凹凸性或为微分方程服务。由于 eˣ 的导数就是 eˣ，像 e𝒌ˣ 这样的函数的高阶导数遵循可预测的模式。

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