Logarithmic Form of Inverse Hyperbolic Functions | 反双曲函数的对数形式

📚 Logarithmic Form of Inverse Hyperbolic Functions | 反双曲函数的对数形式

Inverse hyperbolic functions appear naturally in calculus and integration. Expressing them in logarithmic form makes it possible to evaluate them, solve equations, and differentiate or integrate with confidence. This article explains how to derive and apply the three main logarithmic forms required for AQA A-Level Mathematics.

反双曲函数在微积分与积分问题中自然出现。将它们写成对数形式，可以方便地求值、解方程，并从容地求导和积分。本文讲解 AQA 数学大纲要求的三个主要反双曲函数对数形式的推导与运用。

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1. Definitions of Hyperbolic Functions | 双曲函数的定义

Hyperbolic functions are defined in terms of exponential functions. These definitions are the starting point for inverting the functions.

双曲函数由指数函数定义，这些定义是求反函数的出发点。

Function	Definition
sinh x	(eˣ − e⁻ˣ) ÷ 2
cosh x	(eˣ + e⁻ˣ) ÷ 2
tanh x	(eˣ − e⁻ˣ) ÷ (eˣ + e⁻ˣ)

Notice that sinh x and tanh x are odd functions, while cosh x is even. Their domains are the real numbers, but tanh x has output in (−1, 1). This will affect the domains of their inverses.

注意：sinh x 与 tanh x 是奇函数，cosh x 是偶函数。它们的定义域都是全体实数，但 tanh x 的值域为 (−1, 1)，这会影响反函数的定义域。

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2. Deriving arsinh x | 推导 arsinh x

Let y = arsinh x. Then x = sinh y, so:

设 y = arsinh x，则 x = sinh y，因此：

x = (eʸ − e⁻ʸ) ÷ 2

Multiply both sides by 2eʸ to obtain a quadratic in eʸ.

两边同乘 2eʸ，得到关于 eʸ 的二次方程。

(eʸ)² − 2x eʸ − 1 = 0

Using the quadratic formula and discarding the negative root gives eʸ = x + √(x² + 1). Therefore:

用二次求根公式并舍去负根，得 eʸ = x + √(x² + 1)。因此：

arsinh x = ln(x + √(x² + 1))

This expression is valid for every real x, because x² + 1 is always positive.

该表达式对所有实数 x 均成立，因为 x² + 1 恒为正。

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3. Deriving arcosh x | 推导 arcosh x

Let y = arcosh x. Then x = cosh y, so:

设 y = arcosh x，则 x = cosh y，因此：

x = (eʸ + e⁻ʸ) ÷ 2

Multiplying by 2eʸ gives a quadratic:

同乘 2eʸ 得到二次方程：

(eʸ)² − 2x eʸ + 1 = 0

The two formal solutions are eʸ = x ± √(x² − 1). Since y ≥ 0 for the principal inverse function, we choose the root that gives eʸ ≥ 1. This is the plus sign. Hence:

两个形式解为 eʸ = x ± √(x² − 1)。对于主值反函数 y ≥ 0，需选择使 eʸ ≥ 1 的根，即取加号。因此：

arcosh x = ln(x + √(x² − 1))

The domain is x ≥ 1, because √(x² − 1) is only real when x² − 1 ≥ 0.

定义域为 x ≥ 1，因为 √(x² − 1) 仅在 x² − 1 ≥ 0 时为实数。

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4. Deriving artanh x | 推导 artanh x

Let y = artanh x. Then x = tanh y, so:

设 y = artanh x，则 x = tanh y，因此：

x = (eʸ − e⁻ʸ) ÷ (eʸ + e⁻ʸ)

Multiply through by eʸ to rewrite the equation:

两边同乘 eʸ 重写方程：

x(e²ʸ + 1) = e²ʸ − 1

Collecting terms in e²ʸ gives e²ʸ(1 − x) = 1 + x. Taking logs:

将含 e²ʸ 的项合并，得 e²ʸ(1 − x) = 1 + x。取对数：

2y = ln((1 + x) ÷ (1 − x))

Therefore the logarithmic form is:

因此对数形式为：

artanh x = ½ ln((1 + x) ÷ (1 − x))

This is valid only for |x| < 1, so that both (1 + x) and (1 − x) are positive.

该式仅在 |x| < 1 时成立，此时 1 + x 与 1 − x 均为正。

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5. Domains and Ranges | 定义域与值域

The table below summarises where each logarithmic form is valid.

下表总结了各对数形式的适用范围。

Inverse function	Domain	Range
arsinh x	All real numbers	All real numbers
arcosh x	x ≥ 1	y ≥ 0
artanh x	−1 < x < 1	All real numbers

For arcosh x, the alternative negative root x − √(x² − 1) is positive but less than 1; choosing it would give y < 0. AQA requires the principal values with y ≥ 0.

对于 arcosh x，另一根 x − √(x² − 1) 虽为正数但小于 1；若取它为 eʸ，将得到 y < 0。AQA 要求采用主值，即 y ≥ 0。

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6. Simplifying Expressions with Log Laws | 用对数法则化简

The logarithmic forms can be combined using laws of logarithms. For example:

这些对数形式可以用对数运算法则进行合并。例如：

arsinh 2 + arsinh 3 ≠ arsinh 5

You must not treat arsinh linearly. Instead, substitute the log forms and simplify using ln(ab) = ln a + ln b.

不能把 arsinh 当作线性运算。应当代入对数形式，再利用 ln(ab) = ln a + ln b 化简。

• Example: arsinh x + arsinh y = ln((x + √(x² + 1))(y + √(y² + 1)))

  例子：arsinh x + arsinh y = ln((x + √(x² + 1))(y + √(y² + 1)))

• Example: artanh x ± artanh y can be combined as ½ ln((1 + x)(1 ± y) / ((1 − x)(1 ∓ y))).

  例子：artanh x ± artanh y 可合并为 ½ ln((1 + x)(1 ± y) / ((1 − x)(1 ∓ y)))。

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7. Solving Equations | 解方程

The log forms are especially useful for solving equations involving inverse hyperbolic functions.

在解含反双曲函数的方程时，对数形式尤为有用。

arsinh x = 2

Using the definition, this means ln(x + √(x² + 1)) = 2. Exponentiate both sides to get x + √(x² + 1) = e², then solve by isolating the square root.

由定义，ln(x + √(x² + 1)) = 2。两边取指数得 x + √(x² + 1) = e²，再分离根号解出 x。

x = (e² − e⁻²) ÷ 2 = sinh 2

This confirms that arsinh and sinh are inverse functions.

这也验证了 arsinh 与 sinh 互为反函数。

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8. Differentiation from the Logarithmic Forms | 由对数形式求导

Differentiating the logarithmic forms yields the standard derivatives. For arsinh x:

对对数形式求导可以得到标准导数。先看 arsinh x：

d/dx ln(x + √(x² + 1)) = 1 / √(x² + 1)

Similarly:

类似地：

• d/dx arcosh x = 1 / √(x² − 1), for x > 1

  d/dx arcosh x = 1 / √(x² − 1)，x > 1

• d/dx artanh x = 1 / (1 − x²), for |x| < 1

  d/dx artanh x = 1 / (1 − x²)，|x| < 1

These derivatives match the inverse function rule, providing a useful consistency check.

这些导数与反函数求导法则一致，可用于互相验证。

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9. Integration: Recognising the Forms | 积分：识别形式

Reversing the derivatives gives standard integrals. These are often tested directly in AQA papers.

将导数反向使用即得标准积分。这些积分在 AQA 考卷中经常直接考查。

∫ 1/√(x² + 1) dx = arsinh x + C

∫ 1/√(x² − 1) dx = arcosh x + C

∫ 1/(1 − x²) dx = artanh x + C

Because these integrals involve logarithms, they also lead to forms such as ½ ln|x + √(x² + 1)|, which is especially relevant when the coefficient of x² is not 1.

由于这些积分与对数相关，还可写成 ½ ln|x + √(x² + 1)| 等形式，尤其当 x² 的系数不为 1 时非常有用。

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10. Important Identities | 重要恒等式

Several identities follow directly from the log forms. For example, for x ≥ 1:

一些重要恒等式可直接由对数形式推出。例如，当 x ≥ 1 时：

arcosh x = ln(x + √(x² − 1)) = ln(1 / (x − √(x² − 1)))

This gives alternative forms useful in algebra.

这个结果提供了代数化简时的替代形式。

Also, arsinh is an odd function, so arsinh(−x) = −arsinh x. This is evident from the log form because x + √(x² + 1) is always positive and its reciprocal is −x + √(x² + 1).

arsinh 是奇函数，所以 arsinh(−x) = −arsinh x。从对数形式也可看出，因为 x + √(x² + 1) 恒正，且其倒数为 −x + √(x² + 1)。

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11. Common Pitfalls | 常见误区

• Using arcosh x for x < 1. Remember that √(x² − 1) is not real in that region.

  在 x < 1 时使用 arcosh x。注意此时 √(x² − 1) 不是实数。

• Choosing the wrong root when deriving arcosh. The expression x − √(x² − 1) is positive but ≤ 1, so it would give y ≤ 0.

  推导 arcosh 时选错根。x − √(x² − 1) 为正但不大于 1，选取它会得到 y ≤ 0。

• Forgetting that artanh has domain |x| < 1. For |x| > 1, the logarithmic argument becomes invalid.

  忘记 artanh 的定义域 |x| < 1。当 |x| > 1 时，对数自变量失去意义。

• Treating arsinh(x + y) as arsinh x + arsinh y. There is no such additive property.

  把 arsinh(x + y) 当作 arsinh x + arsinh y。反双曲函数没有这种加法性质。

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12. Worked Examples | 例题演练

Example 1: Solve arcosh x = 3.

例 1：解方程 arcosh x = 3。

Using the log form, ln(x + √(x² − 1)) = 3. Hence x + √(x² − 1) = e³. Solving for x gives x = (e³ + e⁻³) ÷ 2 = cosh 3.

由对数形式，ln(x + √(x² − 1)) = 3，故 x + √(x² − 1) = e³。解出 x 得 x = (e³ + e⁻³) ÷ 2 = cosh 3。

Example 2: Differentiate y = arsinh(2x).

例 2：对 y = arsinh(2x) 求导。

Using the chain rule with the log form or standard derivative:

使用链式法则与标准导数：

dy/dx = (1 / √((2x)² + 1)) × 2 = 2 / √(4x² + 1)

Example 3: Evaluate ∫ from 0 to 1 of 1/√(x² + 1) dx.

例 3：计算定积分 ∫₀¹ 1/√(x² + 1) dx。

Using the antiderivative arsinh x:

使用原函数 arsinh x：

[arsinh x]₀¹ = arsinh 1 − arsinh 0 = ln(1 + √2)

This illustrates how the logarithmic form gives an exact value.

该例说明了对数形式可给出精确值。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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