Answer
$e^{-x}+e^{x}$
Work Step by Step
$f''(x)=\displaystyle \frac{d}{dx}\left[\frac{df}{dx}\right]=\frac{d^{2}f}{dx^{2}}$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[e^{-x}+e^{x}]=-e^{-x}+e^{x}$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[ -e^{-x}+e^{x}]=-(-e^{-x})+e^{x}$
= $e^{-x}+e^{x}$