📚 Derivatives of ex, sin x, cos x, ln x | 指数、正弦、余弦与自然对数的导数
In IB Mathematics, mastering the derivatives of exponential, trigonometric, and logarithmic functions is essential. These functions—eˣ, sin x, cos x, and ln x—form the backbone of calculus and appear constantly in both analysis and application problems. This article presents their rules, proofs, and typical exam-style uses.
在IB数学中，掌握指数函数、三角函数和对数函数的导数是至关重要的。eˣ、sin x、cos x 和 ln x 这些函数构成了微积分的基础，频繁出现在分析题和应用题中。本文介绍它们的求导法则、推导证明以及典型的考试题型应用。
1. The Derivative of eˣ | eˣ 的导数
The simplest and most elegant rule: the derivative of eˣ is eˣ itself.
最简单也最优雅的法则：eˣ 的导数就是它自身。
d/dx (eˣ) = eˣ
This holds because the natural exponential function is defined so that its rate of change at any point equals its value. The proof uses the limit definition and the fact that limh→0 (eʰ − 1)/h = 1.
这是因为自然指数函数的定义方式使得它在任意点的变化率就等于函数值。证明用到导数极限定义以及重要极限 limh→0 (eʰ − 1)/h = 1。
For eᵏˣ, we apply the chain rule: d/dx (eᵏˣ) = k eᵏˣ.
对于 eᵏˣ，使用链式法则：d/dx (eᵏˣ) = k eᵏˣ。
2. Derivative of sin x | sin x 的导数
The derivative of sin x is cos x. This is a fundamental result derived from the limit definition and trigonometric identities.
sin x 的导数是 cos x。这是根据极限定义和三角恒等式得出的基本结果。
d/dx (sin x) = cos x
) = sin x cos h + cos x sin h and the limits (sin h)/h → 1 and (cos h − 1)/h → 0 as h → 0.
证明要点：利用 sin(x+h) = sin x cos h + cos x sin h，并结合 h→0 时 (sin h)/h → 1 以及 (cos h − 1)/h → 0。
With the chain rule, d/dx (sin(ax + b)) = a cos(ax + b).
借助链式法则，d/dx (sin(ax + b)) = a cos(ax + b)。
3. Derivative of cos x | cos x 的导数
The derivative of cos x is −sin x. Notice the negative sign.
cos x 的导数是 −sin x。注意这里的负号。
d/dx (cos x) = −sin x
that of sin x, using cos(x+h) = cos x cos h − sin x sin h and the same two special limits.
其证明与 sin x 类似，利用 cos(x+h) = cos x cos h − sin x sin h 及相同的两个极限。
Often, IB questions ask for the second derivative or simple harmonic motion applications.
用链式法则：d/dx (cos(ax + b)) = −a sin(ax + b)。IB 题目常要求求二阶导数或简单谐运动的运用。
4. Derivative of ln x | ln x 的导数
The derivative of the natural logarithm function is 1/x for x > 0.
自然对数函数的导数是 1/x（x > 0）。
d/dx (ln x) = 1/x
This comes from the inverse relationship between eˣ and ln x. If y = ln x, then x = eʸ; differentiating implicitly gives 1 = eʸ dy/dx, so dy/dx = 1/eʸ = 1/x.
这源于 eˣ 与 ln x 的互逆关系。若 y = ln x，则 x = eʸ；隐函数求导得 1 = eʸ dy/dx，因此 dy/dx = 1/eʸ = 1/x。
does not affect the derivative. This reflects the logarithm property ln(kx) = ln k + ln x.
对于 ln(kx)，链式法则给出 d/dx (ln(kx)) = 1/(kx) × k = 1/x，这说明系数 k 不影响导数。这对应于对数性质 ln(kx) = ln k + ln x，ln k 是常数。
5. Combining the Rules | 综合运用法则
IB exam problems often combine exponential, trigonometric, and logarithmic derivatives. You must be fluent in product rule, quotient rule, and chain rule alongside these basic derivatives.
IB 试题常将指数、三角和对数导数组合起来。你必须在掌握这些基础导数的基础上，熟练运用乘法法则、除法法则和链式法则。
Example: Find the derivative of f(x) = eˣ sin x.
示例：求 f(x) = eˣ sin x 的导数。
Product rule: f′(x) = eˣ sin x + eˣ cos x = eˣ (sin x + cos x).
乘法法则：f′(x) = eˣ sin x + eˣ cos x = eˣ (sin x + cos x)。
Another example: g(x) = ln(cos x), where we need the chain rule.
再如一例：g(x) = ln(cos x)，这里需要用链式法则。
g′(x) = (1/cos x) × (−sin x) = −tan x.
g′(x) = (1/cos x) × (−sin x) = −tan x。
6. Higher-Order Derivatives | 高阶导数
These functions have cyclic or patterned higher derivatives, which is a common IB feature.
这些函数的高阶导数具有循环或规律性特征，这是 IB 常见的考点。
eˣ: all derivatives are eˣ.
eˣ 的所有阶导数都是 eˣ。
sin x: f′(x) = cos x, f″(x) = −sin x, f‴(x) = −cos x, f⁗(x) = sin x (back to start).
sin x：f′(x) = cos x，f″(x) = −sin x，f‴(x) = −cos x，f⁗(x) = sin x（回到起点）。
cos x: f′(x) = −sin x, f″(x) = −cos x, f‴(x) = sin x, f⁗(x) = cos x.
cos x：f′(x) = −sin x，f″(x) = −cos x，f‴(x) = sin x，f⁗(x) = cos x。
ln x: f′(x) = x⁻¹, f″(x) = −x⁻², f‴(x) = 2x⁻³, and in general f⁽ⁿ⁾(x) = (−1)ⁿ⁻¹ (n−1)! x⁻ⁿ for n ≥ 1.
ln x：f′(x) = x⁻¹，f″(x) = −x⁻²，f‴(x) = 2x⁻³，一般地 f⁽ⁿ⁾(x) = (−1)ⁿ⁻¹ (n−1)! x⁻ⁿ（n ≥ 1）。
7. Maclaurin Series Connection | 与麦克劳林级数的联系
Knowing these derivatives allows quick expansion of Maclaurin series, widely tested in IB HL.
掌握这些导数可快速推导麦克劳林级数，这在 IB HL 中广泛考查。
eˣ = 1 + x + x²/2! + x³/3! + …
eˣ = 1 + x + x²/2! + x³/3! + …
sin x = x − x³/3! + x⁵/5! − …
sin x = x − x³/3! + x⁵/5! − …
cos x = 1 − x²/2! + x⁴/4! − …
cos x = 1 − x²/2! + x⁴/4! − …
ln(1+x) = x − x²/2 + x³/3 − … for |x| < 1.
ln(1+x) = x − x²/2 + x³/3 − …（|x| < 1）。
ansions rely on the evaluated derivatives at x = 0, which all follow from the rules above.
这些展开式依赖于在 x = 0 处求得的导数值，而所有导数值都源自上述法则。
8. Graphical Interpretations | 图形解释
Derivatives give slope. For eˣ or ln x, the slope 1/x is always positive but decreases to zero, reflecting the flattening of the logarithmic curve.
导数代表斜率。对 eˣ，斜率等于 y 值，因此图形始终递增，且在 (0,1) 处切线斜率为 1。对 sin x，斜率为 cos x，因此 sin x 的极大值出现在 cos x = 0 处，等等。对 ln x，斜率 1/x 总为正但趋近于零，反映了对数曲线的趋于平缓。
9. Derivative of aˣ and logₐ x | aˣ 和 logₐ x 的导数
Although the topic focuses on eˣ and ln x, IB sometimes extends to general exponentials and logarithms. These can be converted to base e:
虽然本主题聚焦于 eˣ 和 ln x，但 IB 偶尔会拓展到一般指数和对数。可通过换底公式转化为以 e 为底：
d/dx (aˣ) = aˣ ln a, d/dx (logₐ x) = 1/(x ln a)
These follow because aˣ = eˣˡⁿᵃ and logₐ x = ln x / ln a.
这是因为 aˣ = eˣˡⁿᵃ，而 logₐ x = ln x / ln a。
10. Common IB Pitfalls and How to Avoid Them | IB 常见错误及避免方法
| Pitfall / 常见错误 | Correct approach / 正确方法 |
|---|---|
| Forgetting the negative sign in derivative of cos x | Remember: cos → −sin, not +sin. |
| Misapplying chain rule: e.g. d/dx (e²ˣ) = e²ˣ (wrong) | Multiply by derivative of exponent: d/dx (e²ˣ) = 2 e²ˣ. |
| Thinking derivative of ln x is x, or 1, or ln x itself | Derivative of ln x is exactly 1/x. |
| Missing domain restrictions: ln x only defined for x > 0 | Always state x > 0 or handle absolute values for ln|x|. |
| Confusing sin and cos derivative cycles in higher derivatives | Use the cyclic pattern: sin → cos → −sin → −cos → sin. |
11. Practice Exam-Style Questions | 考试题型练习
Here are short, targeted examples to solidify your skills:
以下精选练习巩固你的技能：
Differentiate f(x) = 3 e⁴ˣ − 2 sin(5x).
求导：f(x) = 3 e⁴ˣ − 2 sin(5x)。
Solution: f′(x) = 12 e⁴ˣ − 10 cos(5x).
Find f″(x) for f(x) = ln(x² + 1).
对 f(x) = ln(x² + 1) 求二阶导数。
f′(x) = 2x/(x²+1). Then f″(x) = (2(x²+1) − 2x·2x)/(x²+1)² = (2 − 2x²)/(x²+1)².
If y = eˣ cos x, show that d²y/dx² − 2 dy/dx + 2y = 0.
若 y = eˣ cos x，证明 d²y/dx² − 2 dy/dx + 2y = 0。
y′ = eˣ cos x − eˣ sin x, y″ = eˣ cos x − eˣ sin x − eˣ sin x − eˣ cos x = −2 eˣ sin x. Substitute: −2 eˣ sin x − 2(eˣ cos x − eˣ sin x) + 2 eˣ cos x = 0.
12. Summary and Key Takeaways | 总结与要点
d/dx(eˣ) = eˣ — unique self-derivative property.
d/dx(eˣ) = eˣ —— 独有的自导性。
d/dx(sin x) = cos x, d/dx(cos x) = −sin x — watch the sign.
d/dx(sin x) = cos x，d/dx(cos x) = −sin x —— 注意符号。
d/dx(ln x) = 1/x — valid for x > 0.
d/dx(ln x) = 1/x —— 对 x > 0 成立。
Always couple basic rules with chain, product, quotient rules in exams.
考试中始终将基本求导法则与链式、乘法、除法法则结合使用。
Cyclic higher-order derivatives of sin and cos simplify many problems.
sin 和 cos 的循环高阶导数能简化许多问题。
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