Answer
$y=\displaystyle \frac{1}{2}$
Work Step by Step
For x=$1$, the point on the graph is $(1, f(1))$.
The slope of the tangent at x$=1$ is $m=f^{\prime}(1).$
The point-slope equation of the tangent line is
$y-y_{1}=m(x-x_{1})$
$y-f(1)=f^{\prime}(1)(x-1)$
$f(1)=\displaystyle \frac{1}{1+1}=\frac{1}{2}$
For $f^{\prime}(x)$, use the quotient rule,
$\displaystyle \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$= \displaystyle \frac{(x)^{\prime}(x^{2}+1)-(x)(x^{2}+1)^{\prime}}{(x^{2}+1)^{2}}$
$f^{\prime}(x)= \displaystyle \frac{(1)(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\frac{-x^{2}+1}{(x^{2}+1)^{2}}$
... evaluate at x=$1$ ...
$f^{\prime}(1)=0$
An equation for the tangent line is
$y-\displaystyle \frac{1}{2}=0(x-1)$
$y=\displaystyle \frac{1}{2}$
