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我记得sinh,cosh的初等表示，但老是忘了arcsinh, arccosh的初等表示，有没有什么...

Creato il: 11 agosto 2025

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Domanda

我记得sinh,cosh的初等表示，但老是忘了arcsinh, arccosh的初等表示，有没有什么办法能帮助记忆呢？

Domanda

我记得sinh,cosh的初等表示，但老是忘了arcsinh, arccosh的初等表示，有没有什么办法能帮助记忆呢？

Pensiero

Clarifying hyperbolic functions

The user asked about memorizing elementary expressions for arcsinh and arccosh. Here's a quick breakdown:

  • sinh x = (e^x - e^{-x}) / 2
  • cosh x = (e^x + e^{-x}) / 2
  • arcsinh x = ln(x + sqrt(x^2 + 1))
  • arccosh x = ln(x + sqrt(x^2 - 1))

It's interesting that arcsinh has a +1 while arccosh has a -1 under the square root — because cosh² - sinh² = 1!

Thinking through arcsinh and arccosh

Okay, so for arcsinh, the formula emerges from solving for y via e^y. It’s like the inverse of sinh, leading to: arcsinh(x) = ln[x + sqrt(1 + x^2)].

A pattern is forming, and I wonder if I can use a mnemonic. Maybe something like “s for square plus” since "s" looks like a sideways "+"?

For arccosh, cosh is bigger than 1, so you get sqrt(x^2 - 1).

Memory trick for derivatives

A neat memory trick comes from the derivatives:

  • d/dx of arcsinh x = 1 / sqrt(x^2 + 1)
  • d/dx of arccosh x = 1 / sqrt(x^2 - 1)

The derivative signs clue us in on the sqrt forms for each — arcsinh has a + and arccosh a -.

You can integrate the derivatives to recall the logarithmic forms, with arcsinh x = ln[x + sqrt(x^2 + 1)].

Short memory trick for inverses

For arcsinh, start with t = e^y, then solve x = (t - 1/t) / 2 to get arcsinh x = ln(x + sqrt(x^2 + 1)).

For arccosh, use the same approach, t = (x + 1/t) / 2, and solve similarly, leading to arccosh x = ln(x + sqrt(x^2 - 1)).

A helpful trick: think of cosh with “+” and sinh with “-“ under the square roots.

Thought for 59s

Risposta

有！用两条小口诀 + 一步“万能代换”就够了：

结论（可做闪记卡）

  • arcsinh⁡x=ln⁡ ⁣(x+x2+1)\operatorname{arcsinh}x=\ln\!\big(x+\sqrt{x^2+1}\big)（对一切实数 xx）
  • arccosh⁡x=ln⁡ ⁣(x+x2−1)\operatorname{arccosh}x=\ln\!\big(x+\sqrt{x^2-1}\big)（实数情形需 x≥1x\ge1）

口诀 1：s 对应 +1，c 对应 −1
记住恒等式 cosh⁡2−sinh⁡2=1\cosh^2-\sinh^2=1。倒过来时，sinh\textbf{sinh} 的反函数根号里是 x2+1x^2\textbf{+}1，cosh\textbf{cosh} 的反函数根号里是 x2−1x^2\textbf{−}1。

口诀 2：零点校验
arcsinh⁡0=0⇒ln⁡(0+1)=0\operatorname{arcsinh}0=0\Rightarrow \ln(0+\sqrt{1})=0；
arccosh⁡1=0⇒ln⁡(1+0)=0\operatorname{arccosh}1=0\Rightarrow \ln(1+\sqrt{0})=0。一算就知对不对。

万能代换（记忆推导法）
忘了就从定义出发，令 t=eyt=e^y（其中 yy 是你要求的反函数值）：

  • sinh⁡y=ey−e−y2=x⇒2x=t−1t⇒t2−2xt−1=0⇒t=x+x2+1⇒y=ln⁡t\sinh y=\frac{e^y-e^{-y}}2=x\Rightarrow 2x=t-\frac1t\Rightarrow t^2-2xt-1=0\Rightarrow t=x+\sqrt{x^2+1}\Rightarrow y=\ln t。
  • cosh⁡y=ey+e−y2=x⇒2x=t+1t⇒t2−2xt+1=0⇒t=x+x2−1⇒y=ln⁡t\cosh y=\frac{e^y+e^{-y}}2=x\Rightarrow 2x=t+\frac1t\Rightarrow t^2-2xt+1=0\Rightarrow t=x+\sqrt{x^2-1}\Rightarrow y=\ln t（选正根且 x≥1x\ge1）。

再补一个“导数版提示”也很好记：
ddxarcsinh⁡x=1x2+1\frac{d}{dx}\operatorname{arcsinh}x=\frac1{\sqrt{x^2+1}}，
ddxarccosh⁡x=1x−1x+1\frac{d}{dx}\operatorname{arccosh}x=\frac1{\sqrt{x-1}\sqrt{x+1}}。
看到根号里的 ±1\pm1，就想起上面的两个公式。

把这三点写在卡片上或笔记页边，几次复习就再也忘不掉啦。

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