Answer
See the details below.
Work Step by Step
Let $y=\sinh x$; then $y'=\cosh x$. Since $\cosh x$ is always positive, then $y=\sinh x$ is always increasing. Now, Let $y=\cosh x$, then $y'=\sinh x$. Since $\sin x=0$ at $x=0$, then we have a critical point $x=0$. Since, $\sinh 1\gt 0$, then $y=\cosh x$ is always increasing for all positive values of $x$ and decreasing for all negative values of $x$.
