When dealing with derivatives, nn-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.
