Answer
$y' = \dfrac{1}{4\sqrt{x}+2x}$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make an «u» substitution to make it easier
$f(u) = \ln(u)$
$u = 2+\sqrt{x}$
Derivate the function:
$f'(u) = \dfrac{1}{u} \times u'$
Now let's find u'
$u' = \dfrac{1}{2\sqrt{x}}$
Then undo the substitution:
$f'(x) = \dfrac{1}{2+\sqrt{x}} \times \dfrac{1}{2\sqrt{x}}$
Simplify and get the answer:
$y' = \dfrac{1}{4\sqrt{x}+2x}$
