Answer
$f'(x) = \dfrac{1}{6\sqrt{x}\sqrt[3] {(12+\sqrt{x})^2}}$
Work Step by Step
First, lets make an «u» substitution in order to make it easier.
$f(u) = \sqrt[3] u$
$u =12+\sqrt{x}$
Then lets derivate using the chain rule
$f'(u)=\dfrac{1}{3\sqrt[3] {u^2}} \times u'$
Now let's find u'
$u' = \dfrac{1}{2\sqrt{x}}$
Now let's undo the substitution and simplify
$f'(x) = \dfrac{1}{3\sqrt[3] {(12+\sqrt{x})^2}} \times \dfrac{1}{2\sqrt{x}} $
And you got the answer:
$f'(x) = \dfrac{1}{6\sqrt{x}\sqrt[3] {(12+\sqrt{x})^2}}$
