Answer
$ \dfrac{1}{2\sqrt x}$
Work Step by Step
$g(x) = 1+\sqrt x$
We will apply definition of derivative, such as: $g'(x) = \lim\limits_{z \to x}\dfrac{1 + \sqrt z - 1-\sqrt x}{(z-x)}$
Then, we have $g'(x) = \lim\limits_{z \to x}\dfrac{ \sqrt z -\sqrt x}{(z-x)}= \lim\limits_{z \to x}\dfrac{1}{\sqrt z+\sqrt x}=\dfrac{1}{\sqrt x+\sqrt x}$
Hence, $g'(x) = \dfrac{1}{2\sqrt x}$