IB Math: Differentiation of Inverse Trigonometric Functions | IB数学：反三角函数导数计算方法

📚 IB Math: Differentiation of Inverse Trigonometric Functions | IB数学：反三角函数导数计算方法

Inverse trigonometric functions appear frequently in IB Mathematics AA HL (Analysis and Approaches, Higher Level) as part of the calculus syllabus. Students are expected to differentiate functions such as arcsin x, arccos x, and arctan x, often combined with the chain rule. This article provides a systematic method for differentiating these functions, including derivations, common patterns, and worked examples aligned with IB exam style.

反三角函数在 IB 数学 AA HL（分析与方法，高级水平）的微积分部分中频繁出现。学生需要掌握对 arcsin x、arccos x 和 arctan x 等函数的求导，并常与链式法则结合使用。本文提供一套系统化的求导方法，包括公式推导、常见题型模式以及贴合 IB 考试风格的例题解析。

1. Why Inverse Trigonometric Derivatives Matter | 为什么反三角函数导数很重要

In IB Mathematics AA HL, differentiation of inverse trigonometric functions appears in Paper 1 (no calculator) and Paper 2 (calculator allowed). These derivatives are often used in more complex expressions involving chains, products, and implicit differentiation. A solid understanding saves time and prevents sign errors, which are common pitfalls.

在 IB 数学 AA HL 中，反三角函数的求导出现在 Paper 1（不允许使用计算器）和 Paper 2（允许使用计算器）中。这些导数经常出现在涉及链式法则、乘法和隐函数求导的复杂表达式中。扎实掌握这些公式不仅能节省时间，还能避免常见的符号错误。

2. The Six Basic Derivatives | 六个基本导数公式

The first step is memorising the six standard results. The three principal inverse trigonometric functions in IB are arcsin x, arccos x, and arctan x. The reciprocal functions arccsc x, arcsec x, and arccot x are less common but may appear in extension questions.

第一步是熟记六个标准结果。IB 中主要的三个反三角函数是 arcsin x、arccos x 和 arctan x。它们的倒数形式 arccsc x、arcsec x 和 arccot x 不太常见，但可能出现在拓展题中。

d/dx [arcsin x] = 1 / √(1 − x²)

d/dx [arccos x] = −1 / √(1 − x²)

d/dx [arctan x] = 1 / (1 + x²)

d/dx [arccsc x] = −1 / (|x| √(x² − 1))

d/dx [arcsec x] = 1 / (|x| √(x² − 1))

d/dx [arccot x] = −1 / (1 + x²)

Notice the relationship: d/dx[arcsin x] and d/dx[arccos x] differ only by sign. Similarly, d/dx[arctan x] and d/dx[arccot x] differ only by sign. This symmetry reduces memorisation burden significantly.

注意其中的关系：arcsin x 与 arccos x 的导数仅相差一个正负号。同样地，arctan x 与 arccot x 的导数也只相差一个正负号。这种对称性可以大大减轻记忆负担。

3. Deriving d/dx[arcsin x] | 推导 arcsin x 的导数

Although the IB formula booklet provides these derivatives, understanding their derivation is valuable for conceptual clarity and for solving implicit differentiation problems. Let y = arcsin x. Then sin y = x. Differentiate both sides with respect to x:

虽然 IB 公式手册中提供了这些导数公式，但理解其推导过程对概念理解和解决隐函数求导问题非常有帮助。设 y = arcsin x，则 sin y = x。对等号两边关于 x 求导：

cos y · (dy/dx) = 1

Since cos y = √(1 − sin² y) = √(1 − x²) for y in the principal range [−π/2, π/2], we obtain:

由于在 y 的主值区间 [−π/2, π/2] 内，cos y = √(1 − sin² y) = √(1 − x²)（取正值），因此得到：

dy/dx = 1 / √(1 − x²)

The same method works for arccos x, with the principal range [0, π] giving sin y ≥ 0, but the derivative of cos y introduces a negative sign.

同样的方法适用于 arccos x，其主值区间为 [0, π]，此时 sin y ≥ 0，但由于 cos y 的导数为负，结果会引入一个负号。

4. Deriving d/dx[arctan x] | 推导 arctan x 的导数

Let y = arctan x, so tan y = x. Differentiate implicitly:

设 y = arctan x，则 tan y = x。隐函数求导：

sec² y · (dy/dx) = 1

Since sec² y = 1 + tan² y = 1 + x², we get:

因为 sec² y = 1 + tan² y = 1 + x²，所以：

dy/dx = 1 / (1 + x²)

This derivative has no square root, which makes it algebraically easier to work with in integrals as well. In fact, ∫ dx/(1+x²) = arctan x + C is a direct consequence.

这个导数不包含根号，因此在代数和积分中更容易处理。事实上，∫ dx/(1+x²) = arctan x + C 正是其直接推论。

5. The Chain Rule Extension | 链式法则的推广

In IB exams, you will almost never differentiate a plain arcsin x. Instead, the argument is usually a function of x, such as arcsin(2x) or arctan(x²). The chain rule gives:

在 IB 考试中，你几乎不会直接对单纯的 arcsin x 求导。相反，自变量通常是 x 的函数，例如 arcsin(2x) 或 arctan(x²)。链式法则给出：

d/dx [arcsin u] = u′ / √(1 − u²)

d/dx [arccos u] = −u′ / √(1 − u²)

d/dx [arctan u] = u′ / (1 + u²)

where u = f(x) and u′ = f′(x). This is the single most important formula pattern for this topic.

其中 u = f(x)，u′ = f′(x)。这是本专题最重要的公式模式。

Example 1: Differentiate y = arcsin(3x).

例 1：求 y = arcsin(3x) 的导数。

Here u = 3x, u′ = 3. Therefore:

这里 u = 3x，u′ = 3。因此：

dy/dx = 3 / √(1 − 9x²)

No simplification is usually required beyond this form. IB marking schemes accept this answer directly.

通常不需要进一步化简。IB 评分标准直接接受这种形式的答案。

6. Worked Example: arctan with Quadratic Argument | 例题：含二次型自变量的 arctan

Example 2: Differentiate y = arctan(x² + 1).

例 2：求 y = arctan(x² + 1) 的导数。

Let u = x² + 1, so u′ = 2x. Applying the formula:

设 u = x² + 1，则 u′ = 2x。代入公式：

dy/dx = 2x / (1 + (x² + 1)²)

This can be expanded, but not necessarily. The denominator becomes 1 + (x⁴ + 2x² + 1) = x⁴ + 2x² + 2. Both forms are acceptable in IB exams.

可以展开但并非必须。分母变为 1 + (x⁴ + 2x² + 1) = x⁴ + 2x² + 2。两种形式在 IB 考试中均可接受。

Example 3: Differentiate y = arcsin(√x).

例 3：求 y = arcsin(√x) 的导数。

Here u = x^(1/2), u′ = 1/(2√x). Thus:

这里 u = x^(1/2)，u′ = 1/(2√x)。因此：

dy/dx = [1/(2√x)] / √(1 − x) = 1 / (2√x √(1 − x))

Notice that the domain is 0 ≤ x ≤ 1, consistent with the square root and arcsin arguments.

注意定义域为 0 ≤ x ≤ 1，这与根号内和 arcsin 自变量的要求一致。

7. Product and Quotient Rule Combinations | 乘积与商法则结合

IB AA HL questions often combine inverse trigonometric derivatives with the product rule or quotient rule. For example, differentiate y = x² arctan x.

IB AA HL 的题目经常将反三角函数导数与乘积法则或商法则结合。例如，求 y = x² arctan x 的导数。

Using the product rule:

使用乘积法则：

dy/dx = 2x · arctan x + x² · [1/(1 + x²)]

Final answer:

最终答案：

dy/dx = 2x arctan x + x² / (1 + x²)

The derivative of arctan x introduces a rational term, so be careful to keep the product structure clear.

arctan x 的导数引入了有理项，因此务必保持乘积结构清晰。

Example 4: Differentiate y = arcsin x / x.

例 4：求 y = arcsin x / x 的导数。

Using the quotient rule:

使用商法则：

dy/dx = [x · (1/√(1 − x²)) − arcsin x · 1] / x²

dy/dx = [x / √(1 − x²) − arcsin x] / x²

This form is fully simplified and accepted by IB examiners.

这种形式已完全化简，IB 考官认可。

8. Implicit Differentiation Involving Inverse Trig Functions | 涉及反三角函数的隐函数求导

Sometimes an equation involves a mix of y and inverse trigonometric functions, requiring implicit differentiation. Consider xy + arcsin y = x³. Differentiate both sides with respect to x:

有时方程同时包含 y 和反三角函数，需要使用隐函数求导。例如 xy + arcsin y = x³。对等号两边关于 x 求导：

y + x(dy/dx) + [1 / √(1 − y²)] · (dy/dx) = 3x²

Factor out dy/dx:

提取 dy/dx：

dy/dx [x + 1/√(1 − y²)] = 3x² − y

dy/dx = (3x² − y) / [x + 1/√(1 − y²)]

Optional simplification: multiply numerator and denominator by √(1 − y²).

可选化简：分子分母同乘 √(1 − y²)。

dy/dx = (3x² − y)√(1 − y²) / [x√(1 − y²) + 1]

IB examiners prefer the second form as it is rationalised, but the first is also correct.

IB 考官更喜欢第二种有理化后的形式，但第一种也是正确的。

9. Common Mistakes and Domain Restrictions | 常见错误与定义域限制

Students frequently make three types of errors when differentiating inverse trigonometric functions. Recognising them early helps avoid losing marks.

学生在求反三角函数导数时常犯三类错误。及早识别这些错误有助于避免失分。

Missing the minus sign: d/dx[arccos x] = −1/√(1 − x²), not +1/√(1 − x²).
Forgetting the chain rule: d/dx[arcsin(2x)] = 2/√(1 − 4x²), not 1/√(1 − 4x²).
Ignoring the domain: arcsin x is only defined for |x| ≤ 1, and arctan x is defined for all real x.

漏掉负号：d/dx[arccos x] = −1/√(1 − x²)，而不是 +1/√(1 − x²)。
忘记链式法则：d/dx[arcsin(2x)] = 2/√(1 − 4x²)，而不是 1/√(1 − 4x²)。
忽略定义域：arcsin x 仅在 |x| ≤ 1 时有定义，而 arctan x 对所有实数 x 均有定义。

For arcsin u, the domain requires |u| ≤ 1. For arctan u, there is no restriction on u, but the derivative always produces a positive denominator.

对于 arcsin u，定义域要求 |u| ≤ 1。对于 arctan u，u 没有限制，但导数总是产生正分母。

10. Summary Table for Quick Revision | 快速复习总结表

Function \| 函数	Derivative \| 导数
arcsin u	u′ / √(1 − u²)
arccos u	−u′ / √(1 − u²)
arctan u	u′ / (1 + u²)
arccsc u	−u′ / (\|u\| √(u² − 1))
arcsec u	u′ / (\|u\| √(u² − 1))
arccot u	−u′ / (1 + u²)

This table summarises all six cases with the chain rule included. Keep it handy during revision.

此表总结了包含链式法则的六种情况。复习时请随时查阅。

11. Practice Questions | 练习题

Try these four questions before checking the answers below.

请先尝试以下四道题目，再对照下方答案。

Q1: Differentiate y = arccos(5x).
Q2: Differentiate y = arctan(x³).
Q3: Differentiate y = arcsin(x) · arctan(x).
Q4: Find dy/dx if e^x · arcsin y = x.

题 1：求 y = arccos(5x) 的导数。
题 2：求 y = arctan(x³) 的导数。
题 3：求 y = arcsin(x) · arctan(x) 的导数。
题 4：若 e^x · arcsin y = x，求 dy/dx。

Solutions:

答案：

Q1: u = 5x, u′ = 5, so dy/dx = −5 / √(1 − 25x²).

题 1：u = 5x，u′ = 5，所以 dy/dx = −5 / √(1 − 25x²)。

Q2: u = x³, u′ = 3x², so dy/dx = 3x² / (1 + x⁶).

题 2：u = x³，u′ = 3x²，所以 dy/dx = 3x² / (1 + x⁶)。

Q3: Using the product rule:

题 3：使用乘积法则：

dy/dx = [1/√(1 − x²)] · arctan x + arcsin x · [1/(1 + x²)]

Q4: Differentiate implicitly:

题 4：隐函数求导：

e^x arcsin y + e^x · [1/√(1 − y²)] · (dy/dx) = 1

Solve for dy/dx:

解出 dy/dx：

dy/dx = √(1 − y²) [e^(−x) − arcsin y]

12. Final Exam Tips | 考前终极提示

For IB AA HL Paper 1, do not rely on a calculator for these derivatives. Practise recognising which formula applies the moment you see the notation. When differentiating, first identify the outer inverse trigonometric function, then the inner function u, then compute u′ before substituting into the formula.

对于 IB AA HL Paper 1，请不要依赖计算器来完成这些求导。练习看到表达式的瞬间就能识别应使用哪个公式。求导时，先确定外在的反三角函数，再确定内层函数 u，然后计算 u′，最后代入公式。

Also remember that the derivative of arccos x is the negative of arcsin x, which is a common trick in multiple-choice style questions. For arctan, the derivative is always positive, and the denominator is always a sum of squares — never a difference.

还要记住，arccos x 的导数是 arcsin x 导数的相反数，这是选择题中常见的考点。对于 arctan，导数始终为正，且分母始终是平方和——永远不会是平方差。

Finally, always check the domain of the original function. If the argument of arcsin lies outside [−1, 1], the function is undefined, and so is its derivative. Practising with past IB papers will build speed and confidence.

最后，始终检查原函数的定义域。如果 arcsin 的自变量超出 [−1, 1]，则函数无定义，导数也不存在。通过练习历年 IB 真题可以提升速度和信心。

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IB Math: Differentiation of Inverse Trigonometric Functions | IB数学：反三角函数导数计算方法