Let's find [tex]\(\frac{dy}{dx}\)[/tex] for the functions defined by the parametric equations [tex]\(x = \frac{t}{1+t}\)[/tex] and [tex]\(y = \frac{t^2}{1+t}\)[/tex].
1. First, we need to find the derivatives of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] with respect to [tex]\(t\)[/tex].
For [tex]\(x = \frac{t}{1+t}\)[/tex]:
[tex]\[
\frac{dx}{dt} = \frac{(1+t) \cdot 1 - t \cdot 1}{(1+t)^2} = \frac{1 + t - t}{(1+t)^2} = \frac{1}{(1+t)^2}
\][/tex]
For [tex]\(y = \frac{t^2}{1+t}\)[/tex]:
[tex]\[
\frac{dy}{dt} = \frac{(1+t) \cdot 2t - t^2 \cdot 1}{(1+t)^2} = \frac{2t(1+t) - t^2}{(1+t)^2} = \frac{2t + 2t^2 - t^2}{(1+t)^2} = \frac{2t + t^2}{(1+t)^2}
\][/tex]
2. Next, we apply the chain rule to find [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\][/tex]
Given [tex]\(\frac{dx}{dt} = \frac{1}{(1+t)^2}\)[/tex] and [tex]\(\frac{dy}{dt} = \frac{2t + t^2}{(1+t)^2}\)[/tex], we divide [tex]\(\frac{dy}{dt}\)[/tex] by [tex]\(\frac{dx}{dt}\)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{\frac{2t + t^2}{(1+t)^2}}{\frac{1}{(1+t)^2}} = (2t + t^2) \cdot (1+t)^2 \cdot \frac{1}{(1+t)^2} = 2t + t^2
\][/tex]
Therefore, the derivative [tex]\(\frac{dy}{dx}\)[/tex] for the parametric equations [tex]\(x = \frac{t}{1+t}\)[/tex] and [tex]\(y = \frac{t^2}{1+t}\)[/tex] simplifies nicely to:
[tex]\[
\boxed{t(t + 2)}
\][/tex]
