## Calculus: Early Transcendentals 8th Edition

$y''=\frac{1}{e^2}$
Remember that $e$ is simply a constant value. To find what $y$ is when $x=0$, plug it into the equation. $(0)y+e^y=e$ $e^y=e$ $y=1$ So we now know our $(x,y)$ coordinate is $(0,1)$. Now we use implicit differentiation to find the first and second derivatives. $(xy+e^y=e)'$ (1) $xy'+y+e^y\dot\thinspace y'=0$ Solve for $y'$ $y'=\frac{-y}{x+e^y}$ $y'=\frac{-1}{0+e^1}=\frac{-1}{e}$ From equation (1): $(xy'+y+e^y\dot\thinspace y'=0)'$ (2) $xy''+y'+y'+e^y\dot\thinspace y'y'+e^y\dot\thinspace y''=0$ Solve for $y''$ and plug in $y'=\frac{-1}{e}$, $x=0$, and $y=1$. $y''=\frac{-2y'-e^y\dot\thinspace y'y'}{x+e^y}$ $y''=\frac{1}{e^2}$