D$.\underset{x+-}{lim}\frac{2x^2}{x^2-1}=\underset{x+-}{lim}\frac{2}{1-\frac{1}{x^2}}=$

Therefore, the line $y=$

is a horizontal asymptote.

Since the denominator is $0$ when $x=+-1,$ we compute the following limits$.$

$\underset{x{1}^{+}}{lim}\frac{2x^2}{x^2-1}=$

$\underset{x{1}^{-}}{lim}\frac{2x^2}{x^2-1}=-$

$\underset{x-1^{+}}{lim}\frac{2x^2}{x^2-1}=$

$\underset{x-1^{-}}{lim}\frac{2x^2}{x^2-1}=$

Therefore, the lines $x=1$ and $x=$

$x$ are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the following figure.

E$.f^{\text{'}}(x)=\frac{4x(x^2-1)-2x^2*2x}{(x^2-1{)}^2}=$