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}.
