Answer
$f'(x)=\frac{1}{2(\sqrt {x+4})}$
Work Step by Step
$f(x)=\sqrt {x+4}$
$f'(x)=\lim\limits_{h \to 0}\frac{\sqrt {(x+h)+4}-\sqrt {x+4}}{h}\times\frac{\sqrt {(x+h)+4}+\sqrt {x+4}}{\sqrt {(x+h)+4}+\sqrt {x+4}}$
$f'(x)=\lim\limits_{h \to 0}\frac{x+h+4-x-4}{(h)(\sqrt {x+h+4}+\sqrt {x+4})}$
$f'(x)=\lim\limits_{h \to 0}\frac{h}{(h)(\sqrt {x+h+4}+\sqrt {x+4})}$
$f'(x)=\lim\limits_{h \to 0}\frac{1}{(\sqrt {x+h+4}+\sqrt {x+4})}$
$f'(x)=\frac{1}{(\sqrt {x+0+4}+\sqrt {x+4})}$
$f'(x)=\frac{1}{(\sqrt {x+4}+\sqrt {x+4})}$
$f'(x)=\frac{1}{2(\sqrt {x+4})}$
