#### Answer

$k'(x) = \frac{-1}{(2+x)^{2}}$
$slope = -\frac{1}{16}$
Equation of tangent line: $y = \frac{1}{4}- \frac{1}{16}(x-2)$

#### Work Step by Step

$k(x) = \frac{1}{2+x}$
$k(2) = \frac{1}{4}$
Therefore,
$k'(x) = \lim\limits_{h \to 0}\frac{\frac{1}{2+(x+h)}-\frac{1}{2+x}}{h}$
$k'(x) = \lim\limits_{h \to 0}\frac{-1}{(2+x+h)(2+x)}$
$k'(x) = \frac{-1}{(2+x)^{2}}$
Slope at x = 2:
$slope = -\frac{1}{16}$
Equation of tangent line:
$\frac{y-\frac{1}{4}}{x - 2} = -\frac{1}{16}$
$y = \frac{1}{4}- \frac{1}{16}(x-2)$