📚 Derivatives of Trigonometric Functions | 三角函数的导数
In IB Mathematics, differentiating trigonometric functions is a central skill that bridges algebraic manipulation and the behaviour of periodic phenomena. Mastery of these derivatives requires a solid understanding of limits in radians, the chain rule, and the interplay between sine, cosine, and the other four circular functions. This article systematically covers the derivatives of all six trigonometric functions, explores techniques such as implicit differentiation and inverse trigonometric differentiation, and provides exam-focused strategies to avoid common errors.
在IB数学课程中，对三角函数进行微分是一项核心技能，它将代数运算与周期现象的行为联系起来。要熟练掌握这些导数，就需要深刻理解弧度制下的极限、链式法则，以及正弦、余弦和其他四个圆函数之间的互动关系。本文系统性地涵盖了六种三角函数的导数，探讨了隐函数微分和反三角函数微分等技巧，并提供了应考策略，帮助避免常见错误。
1. The Derivative of sin x | 正弦函数的导数
he first-principles definition with the limit as h → 0, we rely on two special limits: limh→0 (sin h)/h = 1 and limh→0 (cos h - 1)/h = 0, which are valid only when the angle is measured in radians.
所有三角函数微分的基础是正弦函数的导数。利用从第一原理出发的极限定义（当 h → 0），我们需要依赖两个重要极限：limh→0 (sin h)/h = 1 和 limh→0 (cos h - 1)/h = 0。这两个极限只有在角度以弧度为单位时才成立。
s h → 0, the first term vanishes and the second approaches cos x, so d/dx (sin x) = cos x.
从差商 [sin(x+h) - sin x]/h 出发，利用和角公式展开 sin(x+h)，得到 sin x cos h + cos x sin h。减去 sin x 并除以 h，得到 sin x · (cos h - 1)/h + cos x · (sin h)/h。当 h → 0 时取极限，第一项趋近于零，第二项趋近于 cos x，因此 d/dx (sin x) = cos x。
In IB exams, you may be asked to prove this derivative using first principles or to apply it directly. Remember that the argument must be in radians; if a problem provides degrees, convert to radians first or adjust the derivative with the factor π/180.
在IB考试中，你可能会被要求用第一原理证明该导数，或直接应用它。切记自变量的单位必须是弧度；如果题目给出的是角度，请先转换为弧度，或在求导时乘以系数 π/180。
2. The Derivative of cos x | 余弦函数的导数
that cos x = sin(π/2 - x) and applying the chain rule. If y = cos x, then dy/dx = -sin x.
余弦函数的导数可以通过类似的极限过程求得，也可以利用恒等式 cos x = sin(π/2 - x)，并应用链式法则得到。如果 y = cos x，那么 dy/dx = -sin x。
os(π/2 - x) = -sin x. This approach elegantly connects the derivatives of sine and cosine.
利用余函数恒等式，令 u = π/2 - x，则 y = sin u。于是 dy/du = cos u，du/dx = -1，从而 dy/dx = -cos u = -cos(π/2 - x) = -sin x。这一方法巧妙地将正弦和余弦的导数联系起来。
ssions like cos(kx) with the chain rule. Always check the sign: d/dx (cos 5x) = -5 sin 5x.
负号常常让学生犯错，尤其是在结合链式法则对 cos(kx) 之类表达式求导时。务必检查符号：d/dx (cos 5x) = -5 sin 5x。
3. Derivatives of tan x, cot x, sec x, csc x | 正切、余切、正割、余割函数的导数
(cos x · cos x - sin x · (-sin x)) / cos²x = (cos²x + sin²x) / cos²x = 1 / cos²x = sec²x.
其余四个三角函数的导数最有效的推导方法是使用商法则，或者将它们用 sin x 和 cos x 表示。对于 tan x = sin x / cos x，运用商法则得到 (cos x · cos x - sin x · (-sin x)) / cos²x = (cos²x + sin²x) / cos²x = 1 / cos²x = sec²x。
(cot x) = -csc²x via the quotient rule or by applying the chain rule to cot x = (tan x)⁻¹.
因此，d/dx (tan x) = sec²x。类似地，cot x = cos x / sin x，通过商法则可求得 d/dx (cot x) = -csc²x，也可以利用链式法则对 cot x = (tan x)⁻¹ 求导得到。
d/dx (sec x) = sec x tan x. For csc x = 1 / sin x, we obtain d/dx (csc x) = -csc x cot x.
对于 sec x = 1 / cos x，使用链式法则或将其改写为 (cos x)⁻¹，可得 d/dx (sec x) = sec x tan x。对于 csc x = 1 / sin x，可得到 d/dx (csc x) = -csc x cot x。
A compact summary table is invaluable for memorisation:
一份简明的总结表格对于记忆非常有帮助：
| Function f(x) | Derivative f'(x) |
|---|---|
| sin x | cos x |
| cos x | -sin x |
| tan x | sec² x |
| cot x | -csc² x |
| sec x | sec x tan x |
| csc x | -csc x cot x |
Notice the symmetry: co-functions (cos, cot, csc) have negative derivatives, and the derivatives of tan, cot, sec, csc all involve pairs of functions.
请注意对称性：余函数（cos、cot、csc）的导数带有负号，而 tan、cot、sec、csc 的导数都包含成对的函数。
4. Chain Rule with Trigonometric Functions | 三角函数的链式法则
rule is essential. If y = sin(u), where u is a function of x, then dy/dx = cos(u) · du/dx.
当三角函数内部含有一个线性或非线性函数时，链式法则至关重要。如果 y = sin(u)，其中 u 是 x 的函数，那么 dy/dx = cos(u) · du/dx。
) · (6x² - 5). The pattern extends to all six functions: d/dx [tan(g(x))] = sec²(g(x)) · g'(x), and so on.
例如，y = sin(2x³ - 5x)，则 dy/dx = cos(2x³ - 5x) · (6x² - 5)。这一模式可以推广到所有六种三角函数：d/dx [tan(g(x))] = sec²(g(x)) · g'(x)，依此类推。
g 2 sin(3x) · cos(3x) · 3 = 6 sin(3x) cos(3x) = 3 sin(6x) using the double-angle identity.
要留意嵌套的链式法则：对 sin²(3x) 求导时，需要将其视为 (sin(3x))²。外函数是 u²，得到 2 sin(3x) · cos(3x) · 3 = 6 sin(3x) cos(3x)，再利用二倍角公式可简化为 3 sin(6x)。
Always simplify the final answer where possible, as IB mark schemes often reward simplification using trigonometric identities.
随时尽可能地简化最终结果，因为IB评分标准通常会奖励利用三角恒等式进行简化的做法。
5. Product and Quotient Rules Involving Trigonometric Functions | 涉及三角函数的乘积和商法则
e are indispensable. For y = x² sin x, the product rule gives dy/dx = 2x sin x + x² cos x.
当函数相乘或相除时，乘积法则和商法则必不可少。对于 y = x² sin x，运用乘积法则得到 dy/dx = 2x sin x + x² cos x。
x). Recognising the Pythagorean identity sin²x+cos²x=1 dramatically simplifies the result.
对于像 y = sin x / (1 + cos x) 这样的分式，运用商法则： +cos x)² = (cos x + cos²x + sin²x) / (1+cos x)² = (cos x + 1) / (1+cos x)² = 1 / (1+cos x)。利用毕达哥拉斯恒等式 sin²x+cos²x=1 能极大地简化结果。
s combining product and chain rules, break the expression into manageable parts. For y = eˣ cos 2x, use the product rule with u = eˣ, v = cos 2x, so v' = -2 sin 2x; the answer is eˣ cos 2x - 2 eˣ sin 2x = eˣ (cos 2x - 2 sin 2x).
在同时涉及乘积和链式法则的问题中，将表达式分解为可处理的部分。对于 y = eˣ cos 2x，运用乘积法则，令 u = eˣ，v = cos 2x，则 v' = -2 sin 2x；最终结果为 eˣ cos 2x - 2 eˣ sin 2x = eˣ (cos 2x - 2 sin 2x)。
6. Higher-Order Derivatives | 高阶导数
rst derivative is cos x, the second derivative is -sin x, the third is -cos x, and the fourth returns to sin x. The pattern repeats every four differentiations.
三角函数的周期性带来了循环的高阶导数。对于 y = sin x，一阶导数是 cos x，二阶导数是 -sin x，三阶导数是 -cos x，四阶导数又回到 sin x。每求导四次，这一模式就会重复一次。
aviour is extremely useful when finding the nth derivative: express n modulo 4 and match the phase.
对于 y = cos x，其循环为 cos x → -sin x → -cos x → sin x → cos x。这种循环性质在求 n 阶导数时极为有用：将 n 对 4 取模，然后匹配相应的相位即可。
In IB questions, you might be asked to evaluate f(n) xⁿ (cos x) = cos(x + nπ/2). These compact forms are derived from the phase-shift property.
在IB试题中，你可能会被要求计算给定函数在 π/3 处的 n 阶导数值 f(n)(π/3)。掌握好这些周期：dⁿ/dxⁿ (sin x) = sin(x + nπ/2)，dⁿ/dxⁿ (cos x) = cos(x + nπ/2)。这些紧凑的形式来自相位移动的性质。
7. Using Trigonometric Identities to Simplify Differentiation | 使用三角恒等式简化求导
x, whose derivative is -sin x, instead of applying the product rule to the original form.
有时在求导之前，先用恒等式改写原函数会更简单。例如，y = sin x cot x 可简化为 y = cos x，其导数为 -sin x，而无需对原始形式运用乘积法则。
x). However, using substitution reveals a simpler path: directly y = (1/4) tan⁴x + c for integration, but for differentiation, simply treat it as product rule with careful simplification.
像 y = tan³x sec²x 这样的函数，如果你注意到 sec²x 是 tan x 的导数，处理起来就会简单很多。令 u = tan x，则 du/dx = sec²x，函数变为 u³ du/dx，这虽利于积分，但求导时可直接运用乘积法则：dy/dx = d/dx (tan³x) · sec²x + tan³x · d/dx (sec²x)。然而，利用代换可揭示更简单的方法：直接求导并运用链式法则和乘积法则，再仔细简化即可。
to get (1 - cos x)/ sin²x = csc²x - cot x csc x, which sometimes has a simpler derivative.
另一个常见技巧是，将 1/(1+ cos x) 分子分母同乘以 1 - cos x，化为 (1 - cos x)/ sin²x = csc²x - cot x csc x，这样有时求导会更简单。
8. Implicit Differentiation with Trigonometric Functions | 隐函数求导与三角函数
y, differentiate both sides with respect to x: cos(x + y) · (1 + dy/dx) = 2x y + x² dy/dx.
当关系式中同时含有 x 和 y 的正弦或余弦时，就需要使用隐函数求导。对于 sin(x + y) = x²y，对等式两边关于 x 求导：cos(x + y) · (1 + dy/dx) = 2x y + x² dy/dx。
Collect all dy/dx terms on one side: cos(x+y) + cos(x+y) dy/dx = 2xy + x² dy/dx ⇒ dx: dy/dx (cos(x+y) - x²) = 2xy - cos(x+y), so dy/dx = (2xy - cos(x+y)) / (cos(x+y) - x²).
将包含 dy/dx 的项移到一边：cos(x+y) + cos(x+y) dy/dx = 2xy + x² dy/dx ⇒ cos(x+y) dy/dx - x² dy/dx = 2xy - cos(x+y)。提取 dy/dx：dy/dx (cos(x+y) - x²) = 2xy - cos(x+y)，所以 dy/dx = (2xy - cos(x+y)) / (cos(x+y) - x²)。
IB HL papers may feature implicit relations involving trigonometric functions alongside exponentials or logarithms. The key is to remember to apply the chain rule each time you differentiate a trigonometric function of y, yielding an extra dy/dx factor.
IB高水平试卷可能会出现同时包含三角函数、指数函数或对数函数的隐式关系。关键是要记住，每次对关于 y 的三角函数求导时，都要运用链式法则，从而产生额外的 dy/dx 因子。
9. Derivatives of Inverse Trigonometric Functions (HL) | 反三角函数的导数（高水平）
in x, arccos x, and arctan x are essential. Using implicit differentiation on y = arcsin x ⇔ sin y = x, differentiate with respect to x: cos y · dy/dx = 1 ⇒ dy/dx = 1/cos y. Since cos y = √1 - sin²y = √1 - x² for y in the principal range [-π/2, π/2], we obtain d/dx (arcsin x) = 1/√1 - x².
对于高水平学生而言，arcsin x、arccos x 和 arctan x 的导数是必考内容。对 y = arcsin x ⇔ sin y = x 运用隐函数求导，关于 x 求导得：cos y · dy/dx = 1 ⇒ dy/dx = 1/cos y。由于在值域 [-π/2, π/2] 内 cos y = √1 - sin²y = √1 - x²，因此 d/dx (arcsin x) = 1/√1 - x²。
Similarly, d/dx (arccos x) = -1/√1 - x² gn appears because arccos x decreases on its domain. For arctan x, start from y = arctan x ⇒ tan y = x, so sec²y dy/dx = 1 ⇒ dy/dx = 1/sec²y = 1/(1+tan²y) = 1/(1+x²).
类似地，d/dx (arccos x) = -1/√1 - x²。出现负号是因为 arccos x 在其定义域内是递减的。对于 arctan x，由 y = arctan x ⇒ tan y = x，于是 sec²y dy/dx = 1 ⇒ dy/dx = 1/sec²y = 1/(1+tan²y) = 1/(1+x²)。
se inverse derivatives are frequently combined with the chain rule: d/dx [arcsin(3x)] = 3/√1 - 9x². The domain restrictions must be observed; the derivative of arcsin x is defined only for |x| < 1.
这些反函数的导数经常与链式法则一起出现：d/dx [arcsin(3x)] = 3/√1 - 9x²。必须注意定义域的限制；arcsin x 的导数仅在 |x| < 1 时有定义。
10. Applications to Tangent and Normal Lines | 切线与法线的应用
the slope at x = π/2 is dy/dx = 1 - sin(π/2) = 1 - 1 = 0, indicating a horizontal tangent.
导数给出了曲线在某一点处切线的斜率。给定一条包含正弦或余弦的曲线，例如 y = x + cos x，在 x = π/2 处的斜率为 dy/dx = 1 - sin(π/2) = 1 - 1 = 0，这表明有一条水平切线。
The equation of the tangent line at (x₀, y₀) is y - y₀ = m (x - x₀), where m = f' ne is perpendicular, with slope -1/m. For y = 2 sin x at x = π/6, y₀ = 1, m = 2 cos(π/6) = √3, so the tangent is y - 1 = √3 (x - π/6) and the normal has slope -1/√3.
在点 (x₀, y₀) 处的切线方程为 y - y₀ = m (x - x₀)，其中 m = f'(x₀)。法线与切线垂直，其斜率为 -1/m。对于 y = 2 sin x，在 x = π/6 处，y₀ = 1，m = 2 cos(π/6) = √3，因此切线为 y - 1 = √3 (x - π/6)，法线的斜率为 -1/√3。
inding the slopes of both tangents at the intersection point and using the formula tan θ = |(m₁ - m₂)/(1 + m₁m₂)|.
IB考试题可能会要求计算两条相交曲线之间的夹角，这就需要求出交点处两条切线的斜率，并利用公式 tan θ = |(m₁ - m₂)/(1 + m₁m₂)| 进行计算。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
s x and sin x derivatives: d/dx (cos x) = -sin x, not sin x. Always double-check the sign.
对 cos x 和 sin x 求导时忘记负号：d/dx (cos x) = -sin x，而不是 sin x。请务必仔细检查符号。
ly hold in radians. If a problem states x in degrees, convert x to radian measure or introduce the factor π/180.
使用角度而非弧度：标准的 sin x → cos x 导数关系仅在弧度制下成立。如果题目中的 x 以度为单位，请将其转换为弧度，或引入系数 π/180。
Misapplying the chain rule: For sin²x, the derivative is 2 sin x cos x (or sin 2x), not 2 cos x. Correctly identify the outer power function.
链式法则运用错误：对于 sin²x，其导数为 2 sin x cos x（或 sin 2x），而不是 2 cos x。要正确识别外层的幂函数。
= 1, double-angle formulas, and reciprocal identities to present the neatest final answer.
忽略简化：利用 sin²x + cos²x = 1、二倍角公式以及倒数恒等式来呈现最整洁的最终答案。
Not restricting domains for inverse trig derivatives: When writing the derivative of arcsin x, ensure the absolute value of x is less than 1 and avoid algebraic errors with the square root.
未限制反三角函数的导数定义域：在写出 arcsin x 的导数时，务必确保 x 的绝对值小于 1，并避免在处理平方根时出现代数错误。
Practice with past IB paper questions, paying attention to how many marks are allocated for each step. Often, an initial correct application of the chain rule or product rule earns method marks even if a small slip occurs later.
利用历届IB真题进行练习，留意每一步所分配的分值。通常，即使后续出现小失误，最初正确运用链式法则或乘积法则也能获得方法分。
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