When dealing with derivatives, #n#-roots (#n in mathbb{N}#) are better treated as *rational powers*: #x^{1/n}#.

In case of square roots:

#sqrt(x)=x^{1/2}#

To compute the derivative of an #n#-root, we can apply the derivation formula for real powers (see the proof at the end of this post):

#d/dx (x^{alpha})=alpha x^{alpha -1}#, where #alpha in mathbb{R}#, #forall x > 0#

Note that in some cases, the validity of this formula can be extendend to #x ne 0# or to #x in mathbb{R}#.

In case of square roots #alpha=1/2# the domain is #x>0#. We get:

#d/dx (sqrt(x))= d/dx (x^{1/2}) =1/2 x^{1/2-1}=1/2 x^{-1/2}=1/2 1/x^{1/2}=1/(2sqrt(x))#

In our specific case, #f(x)=2 sqrt(x)#. We get:

#d/dx f(x)=d/dx(2sqrt(x))=2 d/dx(sqrt(x))=2 1/(2 sqrt(x))=1/sqrt(x)#

To prove that #d/dx x^{alpha}=alpha x^{alpha -1}# for #alpha in mathbb{R}# and #x >0#, we can use *implicit differentiation* and logarithms.

Let #g(x)=x^alpha#. Note that #x>0 => x^alpha >0 => g(x) >0#, so we can write the equality using logarithms and rewrite it using the "logarithm of a power" property:

#ln g(x) =ln x^alpha <=> ln g(x) = alpha ln x#

Now, we differentiate on both sides (implicit differentiation):

#d/dx (ln g(x))=d/dx (alpha ln x) <=> {d/dx g(x)}/{g(x)}=alpha 1/x#

So we get the following result (recall that #g(x)=x^alpha#):

#d/dx g(x)=alpha/x g(x)=alpha/x x^alpha = alpha x^{-1} x^{alpha}=alpha x^{alpha-1}#

That's the result we were looking for.