Answer
$\frac{dy}{dx}=\frac{e^x}{3\sqrt[3]{(e^x+1)^2}}$
Work Step by Step
Considering $u=e^x+1$, the function can be written as $y=\sqrt[3]{u}$ or $y=u^{1/3}$.
We have the composite function $y=f(g(x))$ where $g(x)=e^x+1$ and $f(u)=u^{1/3}$.
Then,
$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ (By the chain rule)
$=\frac{d}{du}(u^{1/3})\cdot \frac{d}{dx}(e^x+1)$
$=\frac{1}{3}u^{-2/3}\cdot (e^x+0)$
$=\frac{1}{3\sqrt[3]{u^2}}\cdot e^x$
$=\frac{1}{3\sqrt[3]{(e^x+1)^2}}\cdot e^x$
$=\frac{e^x}{3\sqrt[3]{(e^x+1)^2}}$
Thus,
$\frac{dy}{dx}=\frac{e^x}{3\sqrt[3]{(e^x+1)^2}}$
