Answer
$k'(x) = \dfrac{-1}{(2+x)^{2}}$ and $k'(2)=-\dfrac{1}{16}$
Work Step by Step
Consider $k(x) = \dfrac{1}{2+x} \implies k(2) = \dfrac{1}{4}$
Therefore,
$k'(x) = \lim\limits_{h \to 0}\dfrac{\dfrac{1}{2+(x+h)}-\dfrac{1}{2+x}}{h}= \lim\limits_{h \to 0}\dfrac{-1}{(2+x+h)(2+x)}$
$k'(x) = \dfrac{-1}{(2+x)^{2}} \implies k'(2) = \dfrac{-1}{(2+2)^{2}}= -\dfrac{1}{16}$
Hence, we have $k'(x) = \dfrac{-1}{(2+x)^{2}}$ and $k'(2)=-\dfrac{1}{16}$
