Answer
$\frac{dy}{dt}$ = $\frac{cos(\sqrt {1+\sqrt t})}{{\sqrt {t+t\sqrt t})}}$
Work Step by Step
$y$ = $4sin(\sqrt {1+\sqrt t})$
$\frac{dy}{dt}$ = $\frac{d}{dt}$$4sin(\sqrt {1+\sqrt t})$
$\frac{dy}{dt}$ = $4cos(\sqrt {1+\sqrt t})$ $\frac{d}{dt}$$(\sqrt {1+\sqrt t})$
$\frac{dy}{dt}$ = $4cos(\sqrt {1+\sqrt t})$[$\frac{1}{2\sqrt {1+\sqrt t})}$] $\frac{d}{dt}$ ($1+\sqrt t$)
$\frac{dy}{dt}$ = $4cos(\sqrt {1+\sqrt t})$[$\frac{1}{2\sqrt {1+\sqrt t})}$][$\frac{1}{2\sqrt t}$]
$\frac{dy}{dt}$ = $cos(\sqrt {1+\sqrt t})$[$\frac{1}{\sqrt {t+t\sqrt t})}$]
$\frac{dy}{dt}$ = $\frac{cos(\sqrt {1+\sqrt t})}{{\sqrt {t+t\sqrt t})}}$
