Question

Please give steps and explanation!

Answer 1

`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`

now, let's simply plug all those guys in the point-slope form for that tangent line at x = 0.

`(\stackrel{x_1}{0}~,~\stackrel{y_1}{1})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{1}{2} \\\\\\ \begin{array}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`