Answer
$f'(x) = \dfrac{3-8\sin(4x)\cos(4x)}{2\sqrt{3x-\sin^2(4x)}}$
Work Step by Step
First, lets make an «u» substitution in order to make it easier.
$f(u) = \sqrt{u}$
$u = 3x - \sin^2(4x)$
Then lets derivate using the chain rule
$f'(u) = \dfrac{u'}{2\sqrt{u}}$
Now let's find u'
$u' = 3 -2(\sin(4x))\times 4\cos(4x)$
$u' = 3 - 8\sin(4x)\cos(4x)$
Now let's undo the substitution and simplify and you got the answer:
$f'(x) = \dfrac{3-8\sin(4x)\cos(4x)}{2\sqrt{3x-\sin^2(4x)}}$
