## Calculus 8th Edition

$y=e$
The slope of the tangent can be calculated by taking first derivative of the function $y=\frac{e^{x}}{x}$ $y'=\frac{xe^{x}-e^{x}}{x^{2}}=\frac{e^{x}(x-1)}{x^{2}}$ Slope of the tangent at $(1,e)$ is given as follows: $y'|_{(1,e)}=0$ The equation of tangent at $(1,e)$ is $(y-e)=0$ Hence, $y=e$