Sure, let's determine [tex]\(\frac{dy}{dx}\)[/tex] for the given parametric equations [tex]\(x = \frac{t}{1 + t}\)[/tex] and [tex]\(y = \frac{t^2}{1 + t}\)[/tex].
### Step 1: Differentiate [tex]\(x\)[/tex] with respect to [tex]\(t\)[/tex]
First, we find [tex]\(\frac{dx}{dt}\)[/tex]:
[tex]\[ x = \frac{t}{1 + t} \][/tex]
Using the quotient rule:
[tex]\[ \frac{dx}{dt} = \frac{(1 + t) \cdot \frac{d}{dt}[t] - t \cdot \frac{d}{dt}[1 + t]}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{(1 + t) \cdot 1 - t \cdot 1}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1 + t - t}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dx}{dt} = \frac{1}{(1 + t)^2} \][/tex]
### Step 2: Differentiate [tex]\(y\)[/tex] with respect to [tex]\(t\)[/tex]
Next, we find [tex]\(\frac{dy}{dt}\)[/tex]:
[tex]\[ y = \frac{t^2}{1 + t} \][/tex]
Again, using the quotient rule:
[tex]\[ \frac{dy}{dt} = \frac{(1 + t) \cdot \frac{d}{dt}[t^2] - t^2 \cdot \frac{d}{dt}[1 + t]}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dy}{dt} = \frac{(1 + t) \cdot 2t - t^2 \cdot 1}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dy}{dt} = \frac{2t(1 + t) - t^2}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dy}{dt} = \frac{2t + 2t^2 - t^2}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dy}{dt} = \frac{2t + t^2}{(1 + t)^2} \][/tex]
### Step 3: Find [tex]\(\frac{dy}{dx}\)[/tex]
Now, we need to find [tex]\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{\frac{2t + t^2}{(1 + t)^2}}{\frac{1}{(1 + t)^2}} \][/tex]
[tex]\[ \frac{dy}{dx} = (2t + t^2) \cdot \frac{(1 + t)^2}{1} \cdot \frac{1}{(1 + t)^2} \][/tex]
[tex]\[ \frac{dy}{dx} = 2t + t^2 \][/tex]
Thus, [tex]\(\frac{dy}{dx}\)[/tex] for the given parametric equations is [tex]\(t^2 + 2t\)[/tex].
