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10. Derivatives of Inverse Hyperbolic Functions | 反双曲函数的导数

2026-08-21 21:45 · 本文部分内容由AI辅助创作

📚 10. Derivatives of Inverse Hyperbolic Functions | 反双曲函数的导数

In AQA A-Level Mathematics, especially within further calculus topics, you are expected to differentiate inverse hyperbolic functions such as arsinh x, arcosh x and artanh x. These results look similar to the derivatives of inverse trigonometric functions, but they have no sign changes inside the square roots. This article works through the derivations, the standard forms, chain-rule extensions and exam-style applications.

在 AQA A-Level 数学中,特别是在进阶微积分主题里,你需要掌握反双曲函数(如 arsinh x、arcosh x 和 artanh x)的求导。这些结果看起来与反三角函数的导数相似,但根号内没有符号变化。本文将推导这些公式、介绍标准形式、链式法则扩展以及考试型应用。


1. Inverse Hyperbolic Functions and Notation | 反双曲函数及其记号

Inverse hyperbolic functions are written as arsinh x, arcosh x and artanh x, or equivalently as sinh⁻¹ x, cosh⁻¹ x and tanh⁻¹ x. The notation sinh⁻¹ x must not be confused with (sinh x)⁻¹, which equals 1/sinh x or cosech x.

反双曲函数记作 arsinh x、arcosh x 和 artanh x,也可以等价写作 sinh⁻¹ x、cosh⁻¹ x 和 tanh⁻¹ x。注意不要把 sinh⁻¹ x 与 (sinh x)⁻¹ 混淆,后者等于 1/sinh x 或 cosech x。

The principal domains and ranges are important because they determine the sign of the square root in each derivative formula.

主定义域和主值域非常重要,因为它们决定了每个导数公式中平方根前的符号。


2. Derivative of arsinh x | arsinh x 的导数

en sinh y = x. Differentiate both sides with respect to x, using the chain rule on sinh y.

设 y = arsinh x,则 sinh y = x。两边对 x 求导,对 sinh y 使用链式法则。

cosh y × dy/dx = 1

cosh y. To express this in terms of x, use the identity cosh² y − sinh² y = 1, so cosh y = √1 + sinh² y = √1 + x². Since cosh y is always positive, there is no need for a ± sign.

整理得 dy/dx = 1/cosh y。为了用 x 表示,利用恒等式 cosh² y − sinh² y = 1,所以 cosh y = √1 + sinh² y = √1 + x²。因为 cosh y 始终为正,所以不需要加 ± 号。

d/dx (arsinh x) = 1 / √x² + 1

This derivative is defined for all real x.

这个导数对所有实数 x 都有定义。


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3. Derivative of arcosh x | arcosh x 的导数

= arcosh x, so cosh y = x with y ≥ 0. Differentiating both sides gives sinh y × dy/dx = 1.

设 y = arcosh x,则 cosh y = x,且 y ≥ 0。两边求导得 sinh y × dy/dx = 1。

sinh y × dy/dx = 1

Therefore dy/dx = 1/sinh y. For y > 0, sinh y is positive and equals √cosh² y − 1 = √x² − 1. Hence the standard result is:

因此 dy/dx = 1/sinh y。当 y > 0 时,sinh y 为正,并且等于 √cosh² y − 1 = √x² − 1。于是得到标准结果:

d/dx (arcosh x) = 1 / √x² − 1, x > 1

1, the derivative has a vertical tangent there, so the differentiable domain is x > 1.

虽然 arcosh x 在 x = 1

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