## Calculus: Early Transcendentals 8th Edition

$\frac{1}{2}(1+t)^{\frac{3}{2}}$
Given: $$x=\frac{t}{1+t}$$ and $$y=\sqrt 1+t$$ We can solve for dy/dx: $$\frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ So, we find the derivatives of each of the parametric equations with respect to t. $$\frac{dx}{dt}=\frac{dt}{dt}[\frac{1\times(1+t)-t\times(1)}{(1+t)^2}]=1[\frac{1}{(1+t)^2}]=\frac{1}{(1+t)^2}$$ and $$\frac{dy}{dt}=\frac{dt}{dt}\frac{1}{2}(1+t)^{-\frac{1}{2}}=1\frac{1}{2}(1+t)^{-\frac{1}{2}}=\frac{1}{2}(1+t)^{-\frac{1}{2}}$$ Then, $$\frac{dy}{dt}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{1}{2}(1+t)^{-\frac{1}{2}}}{\frac{1}{(1+t)^2}}=\frac{1}{2}(1+t)^{-\frac{1}{2}}\times(1+t)^{2}=\frac{1}{2}(1+t)^\frac{3}{2}$$