The answer is #dy/dx=1/(2sqrt(x))#. This is valid for #x>0#.
If you know the power rule #d/dx(x^{n})=nx^{n-1}#, then you can derive this by recalling that #sqrt(x)=x^{1/2}# so that #dy/dx=1/2 x^{-1/2)=1/(2x^{1/2})=1/(2sqrt(x))#.
If you don't know the power rule, it can be derived from the limit definition of the derivative as follows:
#d/dx(x^{1/2))=lim_{h->0}(\sqrt(x+h}-\sqrt{x})/h#
#=lim_{h->0}(x+h-x)/(h(\sqrt{x+h}+\sqrt{x}))=lim_{h->0}1/(sqrt{x+h}+\sqrt{x})=\frac{1}{2\sqrt{x}}# when #x>0#.
When #x=0#, #f(x)=\sqrt{x}# is defined, but #f'(0)# is undefined. Technically, it's the non-existence of the following limit that confirms this most rigorously:
#lim_{h->0^{+}}(\sqrt{h}-\sqrt{0})/h=lim_{h->0^{+}}1/\sqrt{h}\mbox{ DNE}#.
