Proving $|\tanh x-\tanh y|\leq |x-y|$ and $|\operatorname{Argsinh}x-\operatorname{Argsinh}y|\leq |x-y|$

Question

How to prove these two inequalities for all $x, y \in \mathbb{R}$:

$$\begin{align} |\tanh x-\tanh y| &\leq |x-y| \\[4pt] |\operatorname{Argsinh}x-\operatorname{Argsinh}y| &\leq |x-y| \end{align}$$

Thank you.

Answer 1

Hints:

• The range of $\tanh x$ is $(-1,1)\;$ and $\;(\tanh)'(x)=1-\tanh^2x$,
• $(\operatorname{Argsh})'(x)=\dfrac1{\sqrt{1+x^2}}$.