## Calculus 8th Edition

$$y=2x-e$$
Given $$y=x\ln x,\ \ \ (e,e)$$ Since $$y'=x\frac{1}{x}+\ln x=1+\ln x$$ Then $m=y'\bigg|_{(e,e)}=2$, hence the tangent line given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-e}{x-e}&=2\\ y-e&=2x-2e\\ y&=2x-e \end{align*}