## Calculus 8th Edition

$\dfrac{dy}{dx}=\dfrac{1}{3}(1+4x)^{-2/3}$
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: $u=1+4x$ $y'=\dfrac{1}{3}u^{-2/3}$ $u'=4$ $\dfrac{dy}{dx}=y'u'=\dfrac{1}{3}(1+4x)^{-2/3}(4)$ $\dfrac{dy}{dx}=\dfrac{1}{3}(1+4x)^{-2/3}$