Answer
$\displaystyle \frac{dy}{dt}=( 0.5x^{-0.5}+1.5x^{0.5})\cdot\frac{dx}{dt}$
Work Step by Step
$x=x(t)$ so, by the chain rule,
$\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$
$\displaystyle \frac{dy}{dx}=$... product rule...
$=\displaystyle \frac{d}{dx}[x^{0.5}](1+x)+x^{0.5}(\frac{d}{dx}[1+x])$
$=0.5x^{-0.5}(1+x)+x^{0.5}\cdot 1$
$=0.5x^{-0.5}(1+x)+x^{0.5}$
$=0.5x^{-0.5}+0.5x^{0.5}+x^{0.5}$
$=0.5x^{-0.5}+1.5x^{0.5}$
$\displaystyle \frac{dy}{dt}=( 0.5x^{-0.5}+1.5x^{0.5})\cdot\frac{dx}{dt}$
