Derivatives of eˣ, sin x, cos x, and ln x | eˣ、sinx、cosx、lnx的导数

📚 Derivatives of eˣ, sin x, cos x, and ln x | eˣ、sinx、cosx、lnx的导数

The derivative is the heart of differential calculus. Among all functions, the exponential function eˣ, the trigonometric functions sin x and cos x, and the natural logarithm ln x form the core toolkit for IB Mathematics Analysis and Approaches (AA) and Applications and Interpretation (AI).

导数是微积分的核心。在所有函数中，指数函数 eˣ、三角函数 sinx 和 cosx、自然对数 lnx 构成了 IB 数学分析与方法（AA）以及应用与解释（AI）的核心工具包。

These four functions appear in nearly every real-world model, from population growth to sound waves and radioactive decay. Understanding their derivatives precisely and intuitively is not just an exam requirement; it is the foundation of all further calculus you will study.

这四类函数几乎出现在每一个现实世界模型中，从人口增长到声波再到放射性衰变。精确且直观地理解它们的导数不仅是考试要求，更是你今后学习一切高等微积分的基础。

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1. The Definition of the Derivative | 导数的定义

Before we differentiate any specific function, we need the definition. For a function f(x), the derivative f'(x) is defined as the limit of the average rate of change as the interval shrinks to zero:

在求任何具体函数的导数之前，我们需要定义。对于函数 f(x)，其导数 f'(x) 定义为当区间缩至零时平均变化率的极限：

f'(x) = limₕ→₀ [f(x + h) − f(x)] / h

This limit, when it exists, gives the slope of the tangent line to the curve at any point x. We will apply this definition directly to each of our four functions.

这个极限（若存在）给出曲线在任意点 x 处切线的斜率。我们将把这个定义直接应用于我们的四个函数。

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2. The Derivative of eˣ | eˣ 的导数

The single most remarkable fact in calculus is that the function f(x) = eˣ is its own derivative. That is:

微积分中最非凡的事实是函数 f(x) = eˣ 的导数就是其自身，即：

d/dx (eˣ) = eˣ

Let us verify this using the definition of the derivative:

让我们用导数的定义来验证这一点：

f'(x) = limₕ→₀ [e^(x+h) − eˣ] / h = eˣ · limₕ→₀ (eʰ − 1) / h

The limit limₕ→₀ (eʰ − 1)/h equals 1 precisely because the base e is defined so that this limit holds. This is why e is the “natural” base for exponential functions: no other base has this self-derivative property.

极限 limₕ→₀ (eʰ − 1)/h 恰好等于 1，因为底数 e 的定义正是为了保证该极限成立。这就是为什么 e 是指数函数的”自然”底数：没有其他底数具有这种自身导数性质。

In terms of the chain rule, if the exponent is a function u(x), then:

利用链式法则，若指数是一个函数 u(x)，则：

d/dx (e^(u(x))) = e^(u(x)) · u'(x)

For example, d/dx (e^(3x)) = 3e^(3x). The factor 3 appears because the inner function 3x has derivative 3.

例如，d/dx (e^(3x)) = 3e^(3x)。因子 3 的出现是因为内层函数 3x 的导数为 3。

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3. The Derivative of sin x | sinx 的导数

We now turn to the trigonometric functions. Starting with sine, the derivative is:

现在转向三角函数。从正弦开始，其导数为：

d/dx (sin x) = cos x

To prove this from first principles, we use the sum identity sin(a+b) = sin a cos b + cos a sin b:

要从第一性原理证明这一点，我们使用和角公式 sin(a+b) = sin a cos b + cos a sin b：

d/dx (sin x) = limₕ→₀ [sin(x+h) − sin x] / h
= sin x · limₕ→₀ (cos h − 1)/h + cos x · limₕ→₀ (sin h)/h

Two standard limits are needed here: limₕ→₀ (sin h)/h = 1 and limₕ→₀ (cos h − 1)/h = 0. Substituting these gives cos x, as expected.

这里需要两个标准极限：limₕ→₀ (sin h)/h = 1 和 limₕ→₀ (cos h − 1)/h = 0。代入后恰好得到 cos x。

With the chain rule, for sin(u(x)):

结合链式法则，对于 sin(u(x))：

d/dx (sin(u)) = cos(u) · u’

For example, d/dx (sin(2x)) = 2cos(2x), and d/dx (sin(x²)) = 2x·cos(x²).

例如，d/dx (sin(2x)) = 2cos(2x)，而 d/dx (sin(x²)) = 2x·cos(x²)。

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4. The Derivative of cos x | cosx 的导数

Similarly, the derivative of cosine is:

类似地，余弦的导数为：

d/dx (cos x) = −sin x

The negative sign is important and easy to forget. Using the co-function identity cos x = sin(π/2 − x), we can derive this from the derivative of sine:

这个负号非常重要且容易忘记。利用余函数关系 cos x = sin(π/2 − x)，我们可以从正弦的导数推导出来：

d/dx (cos x) = cos(π/2 − x) · (−1) = −sin x

Alternatively, applying the definition directly yields the same result. Geometrically, as x increases, cos x initially decreases (since its slope at 0 is 0 but negative just after), matching −sin x.

或者，直接应用定义也能得到相同结果。从几何上看，当 x 增大时，cos x 先减小（因为它在 0 处斜率为 0，但在紧随其后为负），这与 −sinx 相符。

With the chain rule: d/dx (cos(u)) = −sin(u) · u’. For example, d/dx (cos(5x)) = −5sin(5x).

结合链式法则：d/dx (cos(u)) = −sin(u) · u’。例如，d/dx (cos(5x)) = −5sin(5x)。

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5. The Derivative of ln x | lnx 的导数

The natural logarithm is the inverse of the exponential function. Its derivative is:

自然对数是指数函数的反函数。它的导数为：

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

One elegant derivation uses implicit differentiation. Let y = ln x, so that eʸ = x. Differentiating both sides with respect to x:

一个优雅的推导使用隐函数求导。令 y = lnx，则 eʸ = x。对两边关于 x 求导：

eʸ · dy/dx = 1 → dy/dx = 1/eʸ = 1/x

This method connects the derivative of an inverse function to the derivative of the original function. The domain restriction x > 0 is essential, since ln x is undefined for x ≤ 0.

这种方法将反函数的导数与原函数的导数联系起来。定义域限制 x > 0 至关重要，因为 lnx 在 x ≤ 0 时无定义。

Using the chain rule, for ln(u(x)):

利用链式法则，对于 ln(u(x))：

d/dx (ln(u)) = u’/u

For example, d/dx (ln(3x + 1)) = 3/(3x + 1). This “logarithmic differentiation” technique simplifies many complicated products and quotients.

例如，d/dx (ln(3x + 1)) = 3/(3x + 1)。这种”对数求导法”可以简化许多复杂的乘积和商。

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6. The Derivative of Other Bases | 其他底数的导数

Although the four functions above are the focus, IB exams often ask for general bases. We can express aˣ and logₐx in terms of e and ln:

虽然上述四个函数是重点，但 IB 考试常考一般底数。我们可以将 aˣ 和 logₐx 用 e 和 ln 表示：

aˣ = e^(x ln a) → d/dx (aˣ) = aˣ ln a

logₐ x = ln x / ln a → d/dx (logₐ x) = 1 / (x ln a)

These formulas reduce to our four core rules when a = e. They appear frequently in IB Paper 1 and Paper 2 questions, especially in contexts involving exponential growth and decay.

当 a = e 时，这些公式就简化为我们四个核心规则。它们在 IB 卷一和卷二试题中频繁出现，尤其在涉及指数增长与衰减的情境中。

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7. The Chain Rule in Action | 链式法则的实际运用

Virtually every IB question on these derivatives involves the chain rule. The rule states that if y = f(u) and u = g(x), then:

几乎所有涉及这些导数的 IB 考题都需要链式法则。该法则指出，若 y = f(u) 且 u = g(x)，则：

dy/dx = dy/du · du/dx

Here are four classic IB-style examples, each using a different outer function:

以下是四个经典的 IB 风格例题，每个使用不同的外层函数：

• y = e^(sin x). Let u = sin x; then dy/dx = e^(sin x) · cos x.

  y = e^(sinx)。令 u = sinx，则 dy/dx = e^(sinx) · cosx。

• y = sin(eˣ). Let u = eˣ; then dy/dx = cos(eˣ) · eˣ.

  y = sin(eˣ)。令 u = eˣ，则 dy/dx = cos(eˣ) · eˣ。

• y = ln(cos x). Let u = cos x; then dy/dx = (1/cos x) · (−sin x) = −tan x.

  y = ln(cosx)。令 u = cosx，则 dy/dx = (1/cosx) · (−sinx) = −tanx。

• y = cos(ln x). Let u = ln x; then dy/dx = −sin(ln x) · (1/x).

  y = cos(lnx)。令 u = lnx，则 dy/dx = −sin(lnx) · (1/x)。

Notice how each derivative combines the outer derivative with the inner derivative. This nesting is central to many IB exam questions.

注意每个导数如何将外层导数与内层导数相乘。这种嵌套是许多 IB 考题的核心。

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8. The Product and Quotient Rules | 积法则与商法则

These functions also appear inside products and quotients. The product rule: (uv)’ = u’v + uv’. The quotient rule: (u/v)’ = (u’v − uv’)/v².

这些函数也出现在乘积和商中。积法则：(uv)’ = u’v + uv’。商法则：(u/v)’ = (u’v − uv’)/v²。

Consider the IB-style problem: differentiate y = x²eˣ. Applying the product rule with u = x² and v = eˣ:

考虑一个 IB 风格问题：求 y = x²eˣ 的导数。使用积法则，令 u = x²，v = eˣ：

dy/dx = 2x·eˣ + x²·eˣ = eˣ(x² + 2x)

Another example: y = sin x / x. Using the quotient rule with u = sin x, v = x:

另一个例子：y = sinx / x。使用商法则，令 u = sinx，v = x：

dy/dx = (x·cos x − sin x)/x²

These rules are straightforward once the basic derivatives are internalised. Most IB mark schemes award a method mark for identifying which rule to use.

一旦基本导数内化于心，这些法则就非常直接。大多数 IB 评分方案会给识别使用哪一条法则的方法分。

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9. Table of Derivative Rules | 导数公式总结表

The following table summarises the rules for your revision. It is worth memorising completely:

下表总结了你需要复习的规则，值得完整记住：

Function | 函数	Derivative | 导数	Chain Rule Version | 链式法则版本
eˣ	eˣ	e^(u) · u’
sin x	cos x	cos(u) · u’
cos x	−sin x	−sin(u) · u’
ln x	1/x	u’/u
aˣ	aˣ ln a	a^(u) · ln a · u’
logₐ x	1/(x ln a)	u’/(u ln a)

You should also recall that d/dx (tan x) = sec²x, which follows from the quotient rule applied to sin x / cos x.

你还应记得 d/dx (tanx) = sec²x，它由商法则应用于 sinx / cosx 得到。

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10. Common Mistakes and Pitfalls | 常见错误与陷阱

Below are the most frequent errors IB students make with these derivatives. Avoiding them can save valuable marks:

以下是 IB 学生对这些导数最常犯的错误。避免它们可以节省宝贵的分数：

• Forgetting the minus sign in the derivative of cos x. Always write d/dx (cos x) = −sin x.

  忘记 cosx 导数中的负号。始终写出 d/dx (cosx) = −sinx。

• Confusing eˣ with xᵉ. The derivative of eˣ is eˣ, but the derivative of xᵉ is e·x^(e−1). These are completely different rules.

  混淆 eˣ 与 xᵉ。eˣ 的导数是 eˣ，但 xᵉ 的导数是 e·x^(e−1)。两者是完全不同的规则。

• Applying the chain rule when it is not needed. For example, d/dx (eˣ) is simply eˣ, not eˣ · something extra.

  在不必要时使用链式法则。例如 d/dx (eˣ) 就是 eˣ，不需要额外乘以什么。

• Forgetting the domain restriction for ln x. The formula d/dx (ln x) = 1/x only holds for x > 0.

  忘记 lnx 的定义域限制。公式 d/dx (lnx) = 1/x 仅在 x > 0 时成立。

• When differentiating ln(u), forgetting the factor u’. The answer is u’/u, not simply 1/u.

  对 ln(u) 求导时忘记因子 u’。答案是 u’/u，而不只是 1/u。

Being aware of these traps before the exam gives you a significant advantage over students who learn them the hard way.

考前意识到这些陷阱，会让你比那些靠犯错才学会的学生有显著优势。

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11. Worked IB-Style Exam Questions | IB 风格例题精解

Let us now work through three realistic exam questions showing how these rules combine.

现在让我们完整解答三道贴近真实考试的题目，展示这些规则如何组合运用。

Question 1: Find the derivative of f(x) = e^(2x) · sin(3x).

题目 1：求 f(x) = e^(2x) · sin(3x) 的导数。

We need the product rule with u = e^(2x) and v = sin(3x):

我们需要积法则，令 u = e^(2x)，v = sin(3x)：

u’ = 2e^(2x), v’ = 3cos(3x)
f'(x) = 2e^(2x)sin(3x) + 3e^(2x)cos(3x)
= e^(2x)[2sin(3x) + 3cos(3x)]

Question 2: Find the derivative of y = ln(x² + 1).

题目 2：求 y = ln(x² + 1) 的导数。

Let u = x² + 1, so u’ = 2x. Therefore:

令 u = x² + 1，则 u’ = 2x。因此：

dy/dx = 2x / (x² + 1)

Question 3: Differentiate g(x) = cos(eˣ) + e^(cos x).

题目 3：求 g(x) = cos(eˣ) + e^(cosx) 的导数。

Each term uses the chain rule independently:

每一项分别使用链式法则：

g'(x) = −sin(eˣ) · eˣ + e^(cos x) · (−sin x)

Note that the second term comes from differentiating e^(cos x) with u = cos x, so we multiply by u’ = −sin x.

注意第二项来自对 e^(cosx) 求导，其中 u = cosx，所以乘以 u’ = −sinx。

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12. Summary and Final Tips | 总结与最终建议

The four derivatives you must know cold are:

你必须熟记的四个导数是：

d/dx (eˣ) = eˣ, d/dx (sin x) = cos x, d/dx (cos x) = −sin x, d/dx (ln x) = 1/x

Master the chain rule, because it multiplies with every one of these rules. Practice recognising composite functions quickly: sin(2x), e^(x³), ln(5x), cos(1/x). Each requires one outer derivative and one inner derivative.

掌握链式法则，因为它与每一个规则相乘。练习快速识别复合函数：sin(2x)、e^(x³)、ln(5x)、cos(1/x)。每一个都需要一个外层导数和一个内层导数。

Finally, always check your answer by testing with a simple value. For instance, if you differentiate y = eˣ and get eˣ, test at x = 0: the slope should be 1 because the tangent line at (0,1) has slope 1. A quick sanity check catches many careless errors.

最后，始终用简单数值检验答案。例如，如果你求 y = eˣ 的导数并得到 eˣ，可以在 x = 0 处检验：斜率应为 1，因为 (0,1) 处切线斜率为 1。快速合理性检查能发现许多粗心错误。

Memorise the table, practise the chain rule, and avoid the common pitfalls. These four derivatives will serve you throughout IB Mathematics and beyond.

记住表格，练习链式法则，避免常见陷阱。这四个导数将贯穿 IB 数学及以后的学习，使你受益无穷。

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