Answer
$f'(x) = \frac{-8x}{(x^{2}+1)^{2}}$
$f'(-1) = 2$
Work Step by Step
1. Find the derivative
$f(x) = \frac{4}{(x^{2}+1)}$
$f(x) = 4(x^{2}+1)^{-1}$
$f'(x) = -4(x^{2}+1)^{-2}(2x)$
$= \frac{4}{(x^{2}+1)^{2}}(2x)$
$f'(x) = \frac{-8x}{(x^{2}+1)^{2}}$
2. Evaluate the derivative at point $(-1, 2)$
$f'(-1) = \frac{-8(-1)}{((-1)^{2}+1)^{2}}$
$= \frac{8}{(1+1)^{2}}$
$= \frac{8}{(2)^{2}}$
$= \frac{8}{4}$
$= 2$