## Calculus (3rd Edition)

$$y=\cosh x\Longrightarrow y^{(100)}=\cosh x, \quad y^{(101)}=\sinh x.$$
Since $(\cosh x)'=\sinh x$ and $(\sinh x)'=\cosh x$, then the even differentiaition of $\cosh x$ will give $\cosh x$ and the odd differentiation will give $\sinh x$. That is $$y=\cosh x\Longrightarrow y^{(100)}=\cosh x, \quad y^{(101)}=\sinh x.$$