📚 Derivatives of Inverse Hyperbolic Functions | 反双曲函数的导数
Inverse hyperbolic functions arise naturally in integration, mechanics, and advanced problem-solving. Differentiating them requires careful attention to their domains and the use of standard results. This article provides a complete AQA-aligned guide to their derivatives, with derivations, worked examples, and exam tips.
反双曲函数在积分、力学以及高阶解题中自然出现。对它们求导需要仔细关注定义域并熟练运用标准结果。本文提供一份完全对齐 AQA 考纲的指南，涵盖推导、例题与考试技巧。
1. Recall: Inverse Hyperbolic Functions | 回顾：反双曲函数
The hyperbolic functions sinh, cosh and tanh are defined using exponential functions. Their inverses are denoted sinh⁻¹ x, cosh⁻¹ x and tanh⁻¹ x, or sometimes arsinh x, arcosh x and artanh x. Each inverse has a restricted domain and range.
双曲函数 sinh、cosh、tanh 由指数函数定义。它们的反函数记为 sinh⁻¹ x、cosh⁻¹ x、tanh⁻¹ x，有时也写作 arsinh x、arcosh x、artanh x。每个反函数都有受限的定义域与值域。
- sinh⁻¹ x is defined for all real x, with range all real numbers.
- cosh⁻¹ x is defined for x ≥ 1, with principal value range [0, ∞).
- tanh⁻¹ x is defined for |x| < 1, with range all real numbers.
- sinh⁻¹ x 对所有实数 x 有定义，值域为全体实数。
- cosh⁻¹ x 定义在 x ≥ 1，主值值域为 [0, ∞)。
- tanh⁻¹ x 定义在 |x| < 1，值域为全体实数。
These inverses can also be expressed in logarithmic form, which is often useful when differentiating or integrating.
这些反函数还可以用对数形式表达，在求导或积分时往往非常有用。
2. Standard Derivative Results | 标准导数结果
but you must know how to apply them. Here f(x) is a differentiable function, and u = f(x).
以下导数在 AQA 公式册中给出，但你必须知道如何应用。这里 f(x) 是可导函数，且 u = f(x)。
| Function | Derivative | Condition |
| sinh⁻¹ x | 1 / √x² + 1 | all real x |
| cosh⁻¹ x | 1 / √x² - 1 | x > 1 |
| tanh⁻¹ x | 1 / (1 - x²) | |x| < 1 |
| coth⁻¹ x | 1 / (1 - x²) | |x| > 1 |
| sech⁻¹ x | -1 / [x √1 - x²] | 0 < x < 1 |
| csch⁻¹ x | -1 / [|x| √1 + x²] | x ≠ 0 |
Notice that sinh⁻¹ and cosh⁻¹ derivatives are positive square roots, while tanh⁻¹ and coth⁻¹ share the same algebraic form but with different domains.
注意 sinh⁻¹ 与 cosh⁻¹ 的导数为正的平方根，而 tanh⁻¹ 与 coth⁻¹ 的代数形式相同但定义域不同。
3. Differentiating sinh⁻¹ x | 求 sinh⁻¹ x 的导数
erentiate with respect to y: dx/dy = cosh y. Since cosh² y - sinh² y = 1, we have cosh y = √1 + sinh² y = √1 + x². The positive root is taken because cosh y ≥ 1 for all y.
设 y = sinh⁻¹ x，则 x = sinh y。对 y 求导：dx/dy = cosh y。由于 cosh² y - sinh² y = 1，得 cosh y = √1 + sinh² y = √1 + x²。取正根，因为对所有 y 有 cosh y ≥ 1。
Therefore dy/dx = 1 / cosh y = 1 / √1 + x². This derivative is always positive and is defined for every real x.
因此 dy/dx = 1 / cosh y = 1 / √1 + x²。该导数始终为正，且对所有实数 x 有定义。
d/dx (sinh⁻¹ x) = 1 / √x² + 1
4. Differentiating cosh⁻¹ x | 求 cosh⁻¹ x 的导数
hen x = cosh y, so dx/dy = sinh y. Using the identity cosh² y - sinh² y = 1 gives sinh y = √cosh² y - 1 = √x² - 1. Since y ≥ 0, sinh y ≥ 0, so the positive root is correct.
设 y = cosh⁻¹ x，其中 x ≥ 1，y ≥ 0。则 x = cosh y，所以 dx/dy = sinh y。利用恒等式 cosh² y - sinh² y = 1 得 sinh y = √cosh² y - 1 = √x² - 1。由于 y ≥ 0，sinh y ≥ 0，因此取正根是正确的。
Thus dy/dx = 1 / sinh y = 1 / √x² - 1. The domain is x > 1; at x = 1 the derivative is infinite, so it is excluded.
于是 dy/dx = 1 / sinh y = 1 / √x² - 1。定义域为 x > 1；在 x = 1 处导数为无穷大，因此排除该点。
d/dx (cosh⁻¹ x) = 1 / √x² - 1, x > 1
5. Differentiating tanh⁻¹ x | 求 tanh⁻¹ x 的导数
Let y = tanh⁻¹ x, with |x| = sech² y. Since sech² y = 1 - tanh² y = 1 - x², we directly obtain dy/dx = 1 / (1 - x²).
设 y = tanh⁻¹ x，其中 |x| < 1。则 x = tanh y。对 y 求导得 dx/dy = sech² y。由于 sech² y = 1 - tanh² y = 1 - x²，直接得到 dy/dx = 1 / (1 - x²)。
Note that 1 - x² is positive because |x| < 1. The derivative is symmetric and tends to infinity as x approaches ±1.
注意因为 |x| < 1，所以 1 - x² 为正。该导数具有对称性，且当 x 趋向 ±1 时趋向无穷。
d/dx (tanh⁻¹ x) = 1 / (1 - x²), |x| < 1
6. coth⁻¹, sech⁻¹ and csch⁻¹ | 反双曲余切、正割与余割
For coth⁻¹ x, defined for |x| lgebraic result as tanh⁻¹, but with a different domain: d/dx (coth⁻¹ x) = 1 / (1 - x²) for |x| > 1. This value is negative because the denominator is negative.
对于 coth⁻¹ x，定义域为 |x| > 1，类似的隐函数求导得到与 tanh⁻¹ 相同的代数结果，但定义域不同：d/dx (coth⁻¹ x) = 1 / (1 - x²)，其中 |x| > 1。由于分母为负，该导数值为负。
o x = sech y. Using dx/dy = -sech y tanh y and simplifying gives d/dx (sech⁻¹ x) = -1 / [x √1 - x²].
对于 sech⁻¹ x，其中 0 < x < 1，设 y = sech⁻¹ x，则 x = sech y。利用 dx/dy = -sech y tanh y 并化简得 d/dx (sech⁻¹ x) = -1 / [x √1 - x²]。
For csch⁻¹ x, x ≠ 0, the derivative is d/dx (csch⁻¹ x) = -1 / [|x| √1 + x²]. The absolute value is essential because the slope of csch⁻¹ x changes sign across the origin.
对于 csch⁻¹ x，x ≠ 0，导数为 d/dx (csch⁻¹ x) = -1 / [|x| √1 + x²]。绝对值至关重要，因为 csch⁻¹ x 的斜率在原点两侧符号不同。
7. Logarithmic Forms and Differentiation | 对数形式与求导
Each inverse hyperbolic function can be written as a logarithm:
每个反双曲函数都可以写成对数形式：
sinh⁻¹ x = ln(x + √x² + 1)
cosh⁻¹ x = ln(x + √x² - 1), x ≥ 1
tanh⁻¹ x = ½ ln((1 + x) / (1 - x)), |x| < 1
Differentiating these logarithmic expressions should reproduce the standard derivative results. For example, d/dx ln(x + √x² + 1) = [1 + x/√x² + 1] / [x + √x² + 1] = 1/√x² + 1, using algebraic simplification.
对这些对数表达式求导应能重现标准导数结果。例如，d/dx ln(x + √x² + 1) = [1 + x/√x² + 1] / [x + √x² + 1] = 1/√x² + 1，这里运用了代数化简。
Knowing these logarithmic forms can help you verify derivatives and solve equations involving inverse hyperbolic functions.
掌握这些对数形式有助于验证导数结果，并求解涉及反双曲函数的方程。
8. The Chain Rule with Inverse Hyperbolic Functions | 反双曲函数的链式法则
argument is not simply x, use the chain rule: if y = f⁻¹(u) and u = g(x), then dy/dx = f⁻¹'(u) · du/dx. This is the most common application in exam questions.
当自变量不是简单的 x 时，需要使用链式法则：若 y = f⁻¹(u) 且 u = g(x)，则 dy/dx = f⁻¹'(u) · du/dx。这是考试中最常见的应用。
For example, differentiate y = sinh⁻¹(3x). Let u = 3x, so dy/du = 1/√u² + 1 and du/dx = 3. Hence dy/dx = 3 / √9x² + 1.
例如，求 y = sinh⁻¹(3x) 的导数。令 u = 3x，则 dy/du = 1/√u² + 1，du/dx = 3。因此 dy/dx = 3 / √9x² + 1。
h⁻¹(x²). With u = x², dy/du = 1/(1 - u²), du/dx = 2x. Thus dy/dx = 2x / (1 - x⁴), provided |x| < 1.
另一个例子：y = tanh⁻¹(x²)。令 u = x²，则 dy/du = 1/(1 - u²)，du/dx = 2x。因此 dy/dx = 2x / (1 - x⁴)，前提是 |x| < 1。
Always simplify the square root expressions carefully, especially when u itself contains a root or a powered term.
务必仔细化简根式表达式，尤其是当 u 本身含有根号或幂项时。
9. Worked Example: Full Differentiation | 综合求导例题
Find dy/dx for y = cosh⁻¹(e2x).
求 y = cosh⁻¹(e2x) 的导数。
Let u = e2x. Then y = cosh⁻¹ u, so dy/du = 1 / √u² - 1 and du/dx = 2e2x.
令 u = e2x，则 y = cosh⁻¹ u，所以 dy/du = 1 / √u² - 1，du/dx = 2e2x。
Therefore dy/dx = 2e2x / √e4x - 1. The domain condition requires e2x > 1, i.e. x > 0.
因此 dy/dx = 2e2x / √e4x - 1。定义域条件要求 e2x > 1，即 x > 0。
dy/dx = 2e2x / √e4x - 1, x > 0
Check: since cosh⁻¹ u is only defined for u ≥ 1, the derivative is real only when e2x ≥ 1. At x = 0, the denominator is zero; hence the strict inequality x > 0 is needed.
检查：因为 cosh⁻¹ u 仅在 u ≥ 1 时定义，所以导数仅在 e2x ≥ 1 时为实数。在 x = 0 处分母为零；因此需要严格不等式 x > 0。
10. Connection to Integrals | 与积分的联系
Derivatives of inverse hyperbolic functions give standard integrals that are not immediately obvious:
反双曲函数的导数给出了不直接可见的标准积分：
| Integral | Result |
| ∫ 1 / √x² + a² dx | sinh⁻¹(x/a) + C |
| ∫ 1 / √x² - a² dx | cosh⁻¹(x/a) + C, x > a |
| ∫ 1 / (a² - x²) dx | (1/a) tanh⁻¹(x/a) + C, |x| < a |
These results are frequently tested in AQA integration questions. Recognising an integrand as the derivative of an inverse hyperbolic function can save time.
这些结果在 AQA 积分题中经常出现。识别出被积函数是某个反双曲函数的导数可以节省时间。
For example, ∫ 1 / √x² + 6x + 13 dx can be solved by completing the square to obtain a form involving sinh⁻¹.
例如，∫ 1 / √x² + 6x + 13 dx 可通过配方转化为涉及 sinh⁻¹ 的形式来求解。
11. Common Mistakes and Pitfalls | 常见错误与陷阱
Students often lose marks on inverse hyperbolic derivatives for the following reasons:
学生在反双曲函数求导中常常因以下原因失分：
- Forgetting that cosh⁻¹ x requires x > 1 and its derivative is undefined at x = 1.
- Mixing up the signs: coth⁻¹ x has the same algebraic derivative as tanh⁻¹ x, but the domain is |x| > 1, making the derivative negative.
- Omitting the absolute value in d/dx (csch⁻¹ x) = -1 / [|x| √1 + x²].
- in rule incorrectly when the argument is not a simple x, e.g. writing d/dx sinh⁻¹(x²) = 1/√x² + 1 instead of 2x/√x⁴ + 1.
- Forgetting to simplify expressions like √x² = |x| when x might be negative.
- 忘记 cosh⁻¹ x 要求 x > 1，且其导数在 x = 1 处无定义。
- 混淆符号：coth⁻¹ x 与 tanh⁻¹ x 的导数代数形式相同，但定义域为 |x| > 1，故导数为负。
- 在 d/dx (csch⁻¹ x) = -1 / [|x| √1 + x²] 中省略绝对值。
- 当自变量不是简单 x 时错误使用链式法则，例如将 d/dx sinh⁻¹(x²) 写成 1/√x² + 1 而非 2x/√x⁴ + 1。
- 忘记化简形如 √x² = |x| 的表达式，尤其当 x 可能为负时。
Always write the domain condition next to your derivative to show full understanding.
在写出导数时，务必将定义域条件一并写出，以展示完整理解。
12. Exam Tips and Summary | 考试技巧与总结
On the AQA A-level Mathematics paper, questions on inverse hyperbolic derivatives usually appear in the calculus section. They often require combining the chain rule with algebraic manipulation.
在 AQA A-Level 数学试卷中，反双曲函数求导通常出现在微积分部分。它们常需要将链式法则与代数运算结合。
Key points to remember:
需要记住的关键点：
- Memorise the standard derivatives: sinh⁻¹, cosh⁻¹, tanh⁻¹, and their domains.
- Use implicit differentiation if you forget a formula, starting from x = sinh y, etc.
- Always apply the chain rule for composite arguments.
- Watch out for absolute values and domain restrictions.
- Link derivatives to integrals to solve integration problems faster.
- 记住标准导数：sinh⁻¹、cosh⁻¹、tanh⁻¹ 及其定义域。
- 若忘记公式，可用隐函数求导，从 x = sinh y 等出发。
- 对复合自变量务必应用链式法则。
- 注意绝对值和定义域限制。
- 将导数与积分联系起来，更快地解决积分问题。
Mastering these derivatives will also help you with hyperbolic identities and differential equations later in the course.
掌握这些导数还将帮助你在后续课程中处理双曲恒等式和微分方程。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图