`x=(1+logt)/(t^(2)), y=(3+2logt)/(t)`

Explanation

To solve the problem, we need to find \(\frac{dy}{dx}\) given the parametric equations:

\[ x = \frac{1 + \log t}{t^2}, \quad y = \frac{3 + 2 \log t}{t} \]

### Step 1: Differentiate \(x\) with respect to \(t\)

We will use the quotient rule for differentiation, which states that if \(u\) and \(v\) are functions of \(t\), then:

\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Here, let \(u = 1 + \log t\) and \(v = t^2\).

1. Differentiate \(u\):
\[ \frac{du}{dt} = \frac{d}{dt}(1 + \log t) = \frac{1}{t} \]

2. Differentiate \(v\):
\[ \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t \]

Now, applying the quotient rule:

\[ \frac{dx}{dt} = \frac{t^2 \cdot \frac{1}{t} - (1 + \log t) \cdot 2t}{(t^2)^2} \]

Simplifying this:

\[ \frac{dx}{dt} = \frac{t - 2t(1 + \log t)}{t^4} = \frac{t - 2t - 2t \log t}{t^4} = \frac{-t - 2t \log t}{t^4} = \frac{-1 - 2 \log t}{t^3} \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Using the same quotient rule, let \(u = 3 + 2 \log t\) and \(v = t\).

1. Differentiate \(u\):
\[ \frac{du}{dt} = \frac{d}{dt}(3 + 2 \log t) = \frac{2}{t} \]

2. Differentiate \(v\):
\[ \frac{dv}{dt} = \frac{d}{dt}(t) = 1 \]

Now, applying the quotient rule:

\[ \frac{dy}{dt} = \frac{t \cdot \frac{2}{t} - (3 + 2 \log t) \cdot 1}{t^2} \]

Simplifying this:

\[ \frac{dy}{dt} = \frac{2 - (3 + 2 \log t)}{t^2} = \frac{2 - 3 - 2 \log t}{t^2} = \frac{-1 - 2 \log t}{t^2} \]

### Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{\frac{-1 - 2 \log t}{t^2}}{\frac{-1 - 2 \log t}{t^3}} = \frac{-1 - 2 \log t}{t^2} \cdot \frac{t^3}{-1 - 2 \log t} \]

The terms \(-1 - 2 \log t\) cancel out:

\[ \frac{dy}{dx} = t \]

### Final Answer

Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = t \]