#### Answer

$y=\frac{1}{2}x+1$

#### Work Step by Step

We have our original function below:
$f(x)=\frac{1}{1+x^2}$
We see that we have a quotient, so let's remember the quotient rule:
$
\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}
$
Using this, let's find the derivative of both the numerator and denominator of $f(x)$:
Numerator = $1$
Derivative of Numerator = $0$
Denominator = $1+x^2$
Derivative of Denominator = $2x$
Now, let's plug that into the quotient rule:
$f'(x)=\frac{0(1+x^2)-2x(1)}{(1+x^2)^2}$
And simplify:
$f'(x)=\frac{-2x}{(1+x^2)^2}$
Now we have the derivative. Let's now use the point $(-1,\frac{1}{2})$ to find the slope. To do that, we need to plug in $x=1$ for all $x$ in $f'(x)$:
$f'(-1)=\frac{-2(-1)}{(1+(-1)^2)^2} = \frac{2}{4} = \frac{1}{2}$
So now we know that the slope ($m$) equals $\frac{1}{2}$
So we can use the point-slope form of a line to now get an equation:
$y-y_1=m(x-x_1)$
Now we plug in our points. Remember from the point $(-1,\frac{1}{2})$, $x=-1$ and $y=\frac{1}{2}$
$y-\frac{1}{2}=\frac{1}{2}(x+1)$
Now we just have to solve for $y$:
$y-\frac{1}{2}=\frac{1}{2}x+\frac{1}{2}$
$y=\frac{1}{2}x+1$
And that's our equation for the tangent line of $f(x)=\frac{1}{1+x^2}$ through the point $(-1,\frac{1}{2})$.
To get a graph, just use a graphing utility and graph the original function and the tangent line.