#### Answer

$f'(u) = 24x^2(3x^3+5)^3$

#### Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$.
Original expression: $y = (2x^3 + 5)^4$
$u = g(x) = 2x^3 + 5$
$y = f(u) = x^4$
Apply the chain rule to find the derivative:
$f'(g(x)) \times g'(x) $ = > $f'(u) \times u'$
$\frac{dy}{du}(u)^4 \times\frac{du}{dx}(2x^3+5) $
Apply the power rule
$4(2x^3 + 5)^3\times6x^2$
$24x^2(2x^3+5)^3$