📚 Exercise 6D.1: Differentiation of Exponential Functions | 指数函数微分练习 6D.1
Exponential functions are at the heart of many IB Mathematics problems, describing everything from population growth to radioactive decay. Exercise 6D.1 provides targeted practice on differentiating eˣ and related exponential forms, building the skills needed to tackle more advanced calculus topics. This article walks you through key techniques, common pitfalls, and step‑by‑step solutions so you can master the material with confidence.
指数函数是 IB 数学许多问题的核心，涉及从人口增长到放射性衰变的各种现象。练习 6D.1 针对 eˣ 及其相关指数形式的求导提供了专项训练，帮助你为更高等的微积分内容打下基础。本文将带你梳理关键技巧、常见错误以及分步解题思路，让你自信掌握这一部分。
1. Foundation: The Derivative of eˣ | 基础：eˣ 的导数
The simplest exponential derivative is that of f(x) = eˣ. Its uniqueness lies in the fact that the function is its own derivative: if y = eˣ, then dy/dx = eˣ. This property makes eˣ an indispensable tool in calculus.
最简单的指数函数求导是 f(x) = eˣ。它的独特之处在于该函数的导数就是它本身：若 y = eˣ，则 dy/dx = eˣ。这一性质使得 eˣ 成为微积分中不可或缺的工具。
d/dx (eˣ) = eˣ
When a constant multiplies eˣ, the derivative is simply multiplied by the same constant. For example, if y = 5eˣ, then dy/dx = 5eˣ. This linearity is essential before we introduce the chain rule.
当常数乘以 eˣ 时，导数只需乘以相同的常数。例如，若 y = 5eˣ，则 dy/dx = 5eˣ。在引入链式法则之前，这种线性性质至关重要。
2. The Chain Rule with eᵘ | 基于 eᵘ 的链式法则
tiate the outer exponential (which stays the same), then multiply by the derivative of the inner function u.
练习 6D.1 中的大多数函数都是 eᵘ 的形式，其中 u 是 x 的函数。链式法则告诉我们：如果 y = eᵘ，那么 dy/dx = eᵘ · du/dx。即，对外层指数求导（保持原样），再乘以内层函数 u 的导数。
Example: Differentiate y = e⁽³ˣ⁺²⁾. Here u = 3x + 2, so du/dx = 3. Hence dy/dx = e⁽³ˣ⁺²⁾ · 3 = 3e⁽³ˣ⁺²⁾.
示例： 对 y = e⁽³ˣ⁺²⁾ 求导。这里 u = 3x + 2，因此 du/dx = 3。于是 dy/dx = e⁽³ˣ⁺²⁾ · 3 = 3e⁽³ˣ⁺²⁾。
- Always identify the inner function u first.
- 先确定内层函数 u。
- Write out du/dx clearly to avoid sign errors.
- 清晰地写出 du/dx，以避免符号错误。
3. Differentiating aˣ (Exponential with Base a) | 对 aˣ（以 a 为底的指数函数）求导
When the base is not e but a constant a > 0, the derivative includes an ln a factor. The general rule is:
当底数不是 e 而是一个常数 a > 0 时，导数中会出现 ln a 因子。通用法则是：
d/dx (aˣ) = aˣ ln a
You can derive this by expressing aˣ as ex ln a and then applying the chain rule. This same approach is often required in 6D.1 when you encounter expressions like 2ˣ or 5²ˣ.
你可以通过将 aˣ 写成 ex ln a 来推导，然后应用链式法则。当你在 6D.1 中遇到像 2ˣ 或 5²ˣ 这样的表达式时，通常也会用到这个方法。
Example: Differentiate y = 4ˣ. Rewrite as y = ex ln 4. Then dy/dx = ex ln 4 · ln 4 = 4ˣ ln 4.
示例： 对 y = 4ˣ 求导。重写为 y = ex ln 4。则 dy/dx = ex ln 4 · ln 4 = 4ˣ ln 4。
4. Product Rule with Exponential Functions | 指数函数的积法则
ne exponentials with polynomials or trigonometric functions. For y = u·v, recall dy/dx = u'v + uv'. When one factor is eˣ or eᵘ, its derivative retains the exponential form, which simplifies factoring at the end.
6D.1 的题目常常将指数函数与多项式或三角函数结合。对于 y = u·v，记住 dy/dx = u'v + uv'。当其中一个因子为 eˣ 或 eᵘ 时，它的导数保持指数形式，从而简化最后的因式分解。
Example: y = x² e³ˣ. Let u = x², v = e³ˣ. Then u' = 2x, v' = 3e³ˣ. So dy/dx = 2x·e³ˣ + x²·3e³ˣ = e³ˣ(2x + 3x²).
示例： y = x² e³ˣ。设 u = x²，v = e³ˣ。则 u' = 2x，v' = 3e³ˣ。因此 dy/dx = 2x·e³ˣ + x²·3e³ˣ = e³ˣ(2x + 3x²)。
Factoring out the common exponential factor not only looks neater but is also expected in IB mark schemes.
提取公有的指数因式不仅使表达式更整洁，也是 IB 评分标准所期望的。
5. Quotient Rule Involving eˣ | 涉及 eˣ 的商法则
For a function expressed as a fraction y = u/v, the quotient rule gives dy/dx = (u'v − uv' erator or denominator, the algebra can become messy; however, the steps remain mechanical.
对于表示为分式 y = u/v 的函数，商法则给出 dy/dx = (u'v − uv') / v²。当分子或分母中含有指数函数时，代数运算可能会变得繁琐；但步骤依然是程序化的。
Highlight: y = eˣ / (x + 1). Set u = eˣ, v = x + 1. Then u' = eˣ, v' = 1. Derivative: (eˣ(x+1) − eˣ·1) / (x+1)² = eˣx / (x+1)².
重点示例：y = eˣ / (x + 1)。 设 u = eˣ，v = x + 1。则 u' = eˣ，v' = 1。导数为：(eˣ(x+1) − eˣ·1) / (x+1)² = eˣx / (x+1)²。
Always check if simplification (like factoring eˣ) is possible before concluding your answer.
在得出答案之前，务必检查是否可以进行化简（例如提取 eˣ 的公因子）。
6. Logarithmic Differentiation | 对数求导法
When a function has a variable in both the base and the exponent (e.g., xˣ), neither the power rule nor the exponential rule applies directly. Logarithmic differentiation provides a systematic path: take the natural log of both sides, simplify using log properties, then differentiate implicitly.
当一个函数的底数和指数都含有变量（例如 xˣ）时，幂法则和指数法则均不直接适用。对数求导法提供了一条系统路径：两边取自然对数，利用对数性质化简，然后隐式求导。
Example: y = xˣ ng both sides with respect to x: (1/y)·dy/dx = ln x + 1. Therefore dy/dx = y(ln x + 1) = xˣ(ln x + 1).
示例： y = xˣ。ln y = x ln x。两边对 x 求导：(1/y)·dy/dx = ln x + 1。因此 dy/dx = y(ln x + 1) = xˣ(ln x + 1)。
This technique also helps when differentiating complicated products or quotients that involve exponentials.
该技巧在求导含指数函数的复杂乘积或商时也十分有用。
7. Tangent and Normal Lines | 切线与法线
ngent or normal line at a given point. After finding dy/dx, evaluate it at the specified x‑ get the gradient m. Then use y − y₁ = m(x − x₁). For a normal line, the gradient is −1/m.
练习 6D.1 经常要求求出给定点的切线或法线方程。在求出 dy/dx 后，代入指定的 x 坐标求出斜率 m。然后使用 y − y₁ = m(x − x₁)。对于法线，斜率为 −1/m。
Worked case: Find the tangent to y = e²ˣ at x = ln 2. Then y₍ₗₙ₂₎ = e2 ln 2 = 4. dy/dx = 2e²ˣ, so m = 2·e2 ln 2 = 8. Tangent: y − 4 = 8(x − ln 2).
解题实例： 求 y = e²ˣ 在 x = ln 2 处的切线。此时 y₍ₗₙ₂₎ = e2 ln 2 = 4。dy/dx = 2e²ˣ，故 m = 2·e2 ln 2 = 8。切线：y − 4 = 8(x − ln 2)。
Pay attention not to confuse ln 2 as a number; leave it in exact form unless a decimal approximation is requested.
注意不要将 ln 2 混淆为一个数值；保持精确形式，除非题目要求小数近似。
8. Higher‑Order Derivatives of Exponentials | 指数函数的高阶导数
Because the derivative of eˣ is eˣ, all higher‑order derivatives of eˣ remain eˣ. For eᵏˣ, the n‑th derivative is kⁿ eᵏˣ. This pattern is extremely useful when verifying differential equations or finding Maclaurin series.
由于 eˣ 的导数为 eˣ，eˣ 的所有高阶导数仍为 eˣ。对于 eᵏˣ，n 阶导数为 kⁿ eᵏˣ。这一规律在验证微分方程或求麦克劳林级数时极为有用。
For a product like y = x eˣ, second derivative involves the product rule twice. First derivative: y' = eˣ + x eˣ. Second derivative: y'' = eˣ + eˣ + x eˣ = 2eˣ + x eˣ.
对于像 y = x eˣ 这样的乘积，二阶导数需要两次应用积法则。一阶导数：y' = eˣ + x eˣ。二阶导数：y'' = eˣ + eˣ + x eˣ = 2eˣ + x eˣ。
Tracking coefficients carefully prevents algebraic slips in recursive differentiation problems.
在递推求导问题中，仔细追踪系数可避免代数失误。
9. Tackling Mixed Functions: eˣ with sin x and cos x | 处理混合函数：eˣ 与 sin x 和 cos x
Exponential‑ ar regularly. Here you must apply the product rule while keeping signs straight. For y = eˣ sin x: y' = eˣ sin x + eˣ cos x = eˣ(sin x + cos x).
指数函数与三角函数的组合经常出现。此时需要在应用积法则的同时保持符号正确。对于 y = eˣ sin x：y' = eˣ sin x + eˣ cos x = eˣ(sin x + cos x)。
When repeated differentiation is needed, such as in integration by parts or differential equations, it is often helpful to organize results in a table.
当需要反复求导时（例如在分部积分或微分方程中），用表格整理结果通常很有帮助。
| Function | Derivative |
| eˣ sin x | eˣ(sin x + cos x) |
| eˣ cos x | eˣ(cos x − sin x) |
Recognising cyclical patterns can simplify second‑order differential equations.
识别周期性的规律可以简化二阶微分方程。
10. Common Mistakes to Avoid | 需要避免的常见错误
Even strong students slip up when rushing through exponential differentiation. Watch out for these typical pitfalls:
即使是基础扎实的学生在匆忙完成指数求导时也可能犯错。留意这些典型陷阱：
- Forgetting the chain rule: d/dx (e⁵ˣ) ≠ e⁵ˣ. Always multiply by 5.
- 忘记链式法则：d/dx (e⁵ˣ) ≠ e⁵ˣ。务必乘以 5。
- Misapplying the derivative of aˣ: writing 3ˣ as derivative 3ˣ instead of 3ˣ ln 3.
- aˣ 求导公式用错：将 3ˣ 的导数写成 3ˣ，而正确为 3ˣ ln 3。
- Errors in logarithmic differentiation: forgetting to multiply by y at the end.
- 对数求导法的错误：最后忘记乘以 y。
- Not factoring out eˣ in product/quotient answers when it would simplify the expression.
- 在积法则/商法则答案中未提取 eˣ 公因子，导致未能简化的表达式。
- Confusing the derivative of ln x with that of eˣ. d/dx (ln x) = 1/x, not 1/x · something else.
- 混淆 ln x 与 eˣ 的导数。d/dx (ln x) = 1/x，而非 1/x 乘以其他。
Double‑check each step, especially when the function involves multiple operations like chain nested inside product.
每一步都要仔细检查，尤其是当函数涉及多层运算时，如积法则内嵌套链式法则。
11. Step‑by‑Step Walkthrough of a Typical 6D.1 Problem | 典型 6D.1 题目的分步详解
Problem: Differentiate y = e²ˣ (x³ − 4x).
问题： 对 y = e²ˣ (x³ − 4x) 求导。
Solution:
Identify factors: u = e²ˣ, v = x³ − 4x.
u' = 2e²ˣ (chain rule), v' = 3x² − 4.
Apply product rule: dy/dx = u'v + uv' = 2e²ˣ(x³ − 4x) + e²ˣ(3x² − 4).
Factor e²ˣ: e²ˣ[2(x³ − 4x) + (3x² − 4)] = e²ˣ(2x³ − 8x + 3x² − 4).
Final answer: e²ˣ(2x³ + 3x² − 8x − 4).
解答：
确定因子：u = e²ˣ，v = x³ − 4x。
u' = 2e²ˣ（链式法则），v' = 3x² − 4。
应用积法则：dy/dx = u'v + uv' = 2e²ˣ(x³ − 4x) + e²ˣ(3x² − 4)。
提取 e²ˣ：e²ˣ[2(x³ − 4x) + (3x² − 4)] = e²ˣ(2x³ − 8x + 3x² − 4)。
最终答案：e²ˣ(2x³ + 3x² − 8x − 4)。
Notice that the final expression inside the brackets should be ordered by descending powers of x, which is standard mathematical presentation.
注意括号内的最终表达式应按 x 的降幂排列，这是标准的数学书写规范。
12. Connecting Differentiation to Integration (a Preview) | 连接微分与积分（前瞻）
rentiation in Exercise 6D.1 directly supports the upcoming topic of integration. Since ∫ eˣ dx = eˣ ferential equations and area problems later. The ability to spot derivatives of the form f'(x) eᶠ⁽ˣ⁾ is essential for integration by substitution.
掌握练习 6D.1 中的指数函数微分将直接为后续的积分主题提供支持。由于 ∫ eˣ dx = eˣ + C，识别这一逆过程有助于后续求解微分方程和面积问题。能够识别出形如 f'(x) eᶠ⁽ˣ⁾ 的导数形式，对于换元积分法至关重要。
For example, knowing that d/dx (ex²) = 2x ex² allows you to immediately write down ∫ 2x ex² dx = ex² + C. This seamless transition between derivative and antiderivative is a cornerstone of IB calculus.
例如，知道了 d/dx (ex²) = 2x ex²，你就可以立即写出 ∫ 2x ex² dx = ex² + C。导数与反导数之间的这种无缝衔接是 IB 微积分的基石。
Regular practice with 6D.1 builds the fluency needed to perform these reversals almost instinctively by the time you reach the integration chapters.
通过 6D.1 的持续练习，你将在学到积分章节时几乎本能地进行这种逆向运算，建立所需的熟练度。
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
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