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2019年12月16日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2022年7月27日 · $\begingroup$ The reason why we take the positive square root for $\cosh$ is partially that $\cosh\ge0$ and it's probably inherent to the proof you're reading, but it should be noted that $\sinh^{-1}x$ has the explicit formula $\ln\left(x+\sqrt{x^2+1}\right)$, so you could just compute $\cosh\sinh^{-1}(x)$ directly in terms of elementary functions. $\endgroup$

2021年1月2日 · $\begingroup$ Well okay, but the 1968 edition (the only one I have immediate access to) actually has $\cosh^{-1}$ rather than $\operatorname{arcosh}$ ... so in light of what you say, perhaps Spiegel is insisting that the negative branch is …

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2024年3月21日 · With the regular trig functions, if I ever end up with something like $\operatorname{trig}_1(\operatorname{arctrig}_2(f(x))$, where $\text{trig}_1$ and $\text{trig}_2$ are two arbitrary trigonometric

2022年7月11日 · I've recently started learning hyperbolic functions and inverse hyperbolic functions, and I came across this equation involving inverse hyperbolic functions.

Evaluate $$\lim_{t\to\infty} (\cosh x)^{1/x}.$$ I tried to use L'Hopital's but I think I made a mess of the differentiation, and the differentiation doesn't seem like it'll help much. calculus limits

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