Answer
$- \dfrac{1}{(x+2)^{2}}$
Work Step by Step
Consider $f(x) = \dfrac{1}{x+2}$
we will apply definition of derivative, such as:
Now, $f'(x) = \lim\limits_{z \to x}\dfrac{\dfrac{1}{z+2} - \dfrac{1}{x+2}}{(z-x)}=\lim\limits_{z \to x}\dfrac{1}{z-x}\dfrac{(x-z)}{(z+2)(x+2)}$
$\implies \lim\limits_{z \to x}\dfrac{-1}{(z+2)(x+2)}=\dfrac{-1}{(x+2)(x+2)}$
Hence, $f'(x) =- \dfrac{1}{(x+2)^{2}}$