\(y=\cosh x \implies y=\frac {1}{2}(e^x+e^{-x})\)

\(y=\sinh x \implies y=\frac {1}{2}(e^x-e^{-x})\)

\[y=\tanh x =\frac {e^{\displaystyle {2x}}-1}{e^{\displaystyle {2x}}+1}\] \[y=\frac {e^x-e^{-x}}{e^x+e^{-x}}=\frac {1-e^{-2x}}{1+e^{-2x}}\] as \(x\longrightarrow \infty ,\quad y=1\) as \(x\longrightarrow -\infty ,\quad y=-1\)