Question

Find the equation of a tangent line at the point on a curve f(x) when x=ef(x)=xln(x)+x

Answer 1

$\bf f(x)=xln(x)+x\implies \cfrac{dy}{dx}=\stackrel{product~rule}{\left[1\cdot ln(x)+x\cdot \cfrac{1}{x}\right]}+1 \\\\\\ \left. \cfrac{dy}{dx}=ln(x)+2 \right|_{x=e} \implies 1+2\implies 3 \\\\\\ \textit{when x=e, what is \underline{y}?}\qquad f(e)=eln(e)+e\implies f(e)=2e \\\\\\ (e~~,~~2e)\qquad\qquad \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2e=3(x-e) \\\\\\ y-2e=3x-3e\implies y=3x-e$