Answer
(a) Equation of tangent: $x-2y+2=0$.
(b) Check the graph given below
Work Step by Step
(a) $y=\frac{1}{1+x^2}$. Differentiating $y$ with respect to $x$ at $(-1,1/2)$.
Using product's rule,
$$\frac{dy}{dx}=\frac{d}{dx}(1+x^2)^{-1}=(-1)(1+x^2)^{-2}(2x)$$
$$\frac{dy}{dx}\bigg|_{x=-1}=(-1)(1+(-1)^2)^{-2}(2\times-1)=-1\times\frac{1}{2^2}\times-2=\frac{1}{2}.$$
Thus, the slope of the tangent line is $1/2$. Using point-slope form,
$$y-\frac{1}{2}=\frac{1}{2}(x+1)$$
$$\frac{2y-1}{2}=\frac{1}{2}(x+1)\implies 2y-1=x+1$$
$$x-2y+2=0$$ is the equation of the required tangent.
(b) Check the graph given below
