[tex]f(x)=\cfrac{e^x+1}{e^{x^2} + 1}\implies \cfrac{df}{dx}=\stackrel{\textit{\large quotient rule}}{\cfrac{e^x(e^{x^2}+1)~~ - ~~(e^x+1)\stackrel{chain~rule}{(e^{x^2}\cdot 2x)}}{(e^{x^2} + 1)^2}} \\\\\\ \left. \cfrac{df}{dx} \right|_{x=0}\implies \cfrac{e^0(e^{0^2}+1)~~ - ~~(e^0+1)(e^{0^2}\cdot 2(0))}{(e^{0^2} + 1)^2}\implies \cfrac{2~~ - ~~0}{4}\implies \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}} \\\\[-0.35em] ~\dotfill\\\\ f(0)=\cfrac{e^0 + 1}{e^{0^2} + 1}\implies f(0)=\cfrac{2}{2}\implies f(0)=1[/tex]
now, let's simply plug all those guys in the point-slope form for that tangent line at x = 0.
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{1})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{0})\implies y=\cfrac{1}{2}x+1[/tex]
