Answer
$\frac{dy}{dx}$= $\frac{(1+t)^{1.5}}{2}$
Work Step by Step
x=$\frac{t}{1+t}$
Use Quotient Rule
$\frac{dx}{dt}$ = $\frac{(1+t) * (t)' - (t) * (1+t)' }{(1+t)^{2}}$
$\frac{dx}{dt}$ = $\frac{1}{(1+t)^2}$
y=$\sqrt {1+t}$
Use Power Rule
y=$(1+t)^{1/2}$
$\frac{dy}{dt}$ = $(1/2)(1+t)^{-1/2}$
$\frac{dy}{dt}$ = $\frac{1}{2 \sqrt{1+t} }$
$\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx}$ = $\frac{\frac{1}{2 \sqrt{1+t} }}{\frac{1}{(1+t)^2}}$
$\frac{dy}{dx}$ = $\frac{(1+t)^2}{2 \sqrt{1+t}}$
$\frac{dy}{dx}$ = $\frac{(1+t)^{1.5}}{2}$
