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Show that cosh:$[0, \infty) \to [1, \infty)$ is strictly increasing bijection and that cosh: $(-\infty, 0] \to [1, \infty)$ is strictly decreasing bijection. Show that by inverse functions differentiability cosh has branches that differentiate: arcosh: $[1, \infty) \to [0, \infty)$ and arcosh: $[1, \infty) \to (-\infty, 0]$ and that depending on the branch,

$D(arcosh y) = {1 \over \sqrt {y^2-1}}$ for all $y \in [1, \infty)$ or

$D(arcosh y) = -{1 \over \sqrt {y^2-1}}$ for all $y \in [1, \infty)$

My work: I've proven the increasing and decreasing part, but how do I use the rule, I cant really get through $1 \over f´(x)$ (= $ 1 \over sinh(x)$)

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This is because $\;\DeclareMathOperator{\argch}{Argcosh}u=\argch y\iff \cosh u =y$ and $y\ge 0$ in the first case, $y\le 0$ in the second case. Thus, in the second case: $$(\argch)'(y)=\frac1{\sinh y}=\frac 1{-\sqrt{\cosh^2y-1}}=-\frac 1{\sqrt{y^2-1}}$$ since $\sinh y\le 0$ if $y\le 0$.

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    $\text{Argcosh}$ or $\text{Arccosh}$? – 2017-01-27
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    @H. Ergül: I'm fairly old-style. `Arg` stands for (and reads as) `‘Argument’`. Actually, the truely old-style notation should be `Argch`, but since the notations are now `cosh, sinh`, not `ch, sh`, this is a compromise. – 2017-01-27
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    I'm also a little bit greenhorn :) – 2017-01-27