#### Answer

$$y=\cosh x\Longrightarrow y^{(100)}=\cosh x, \quad y^{(101)}=\sinh x.$$

#### Work Step by Step

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
$$y=\cosh x\Longrightarrow y^{(100)}=\cosh x, \quad y^{(101)}=\sinh x.$$