Answer
$\dfrac{dy}{dx}=\dfrac{1}{3}(1+4x)^{-2/3}$
Work Step by Step
Identify the inner function and the outer function
$y = \sqrt[3] {1+4x}=(1+4x)^{1/3}$
Outer function: $y=u^{1/3}$
Inner function: $u=1+4x$
$y'=\dfrac{1}{3}u^{-2/3}$
$u'=4$
$\dfrac{dy}{dx}=y'u'=\dfrac{1}{3}(1+4x)^{-2/3}(4)$
$\dfrac{dy}{dx}=\dfrac{1}{3}(1+4x)^{-2/3}$
