If `x=(1+logt)/(t^2),\ \ y=(3+2logt)/t ,\ \ ` find `(dy)/(dx)` .

Explanation

To find \(\frac{dy}{dx}\) when \(x = \frac{1 + \log t}{t^2}\) and \(y = \frac{3 + 2 \log t}{t}\), we will use the chain rule and implicit differentiation. Let's go through the steps one by one.

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

Given:
\[ x = \frac{1 + \log t}{t^2} \]

We will use the quotient rule to differentiate \(x\):
\[ \frac{dx}{dt} = \frac{(t^2)(\frac{d}{dt}(1 + \log t)) - (1 + \log t)(\frac{d}{dt}(t^2))}{(t^2)^2} \]

Calculating the derivatives:
- \(\frac{d}{dt}(1 + \log t) = \frac{1}{t}\)
- \(\frac{d}{dt}(t^2) = 2t\)

Substituting these into the quotient rule:
\[ \frac{dx}{dt} = \frac{t^2 \cdot \frac{1}{t} - (1 + \log t) \cdot 2t}{t^4} \]
\[ = \frac{t - 2t(1 + \log t)}{t^4} \]
\[ = \frac{t - 2t - 2t \log t}{t^4} \]
\[ = \frac{-t - 2t \log t}{t^4} \]
\[ = \frac{-(1 + 2 \log t)}{t^3} \]

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

Given:
\[ y = \frac{3 + 2 \log t}{t} \]

Using the quotient rule again:
\[ \frac{dy}{dt} = \frac{(t)(\frac{d}{dt}(3 + 2 \log t)) - (3 + 2 \log t)(\frac{d}{dt}(t))}{t^2} \]

Calculating the derivatives:
- \(\frac{d}{dt}(3 + 2 \log t) = \frac{2}{t}\)
- \(\frac{d}{dt}(t) = 1\)

Substituting these into the quotient rule:
\[ \frac{dy}{dt} = \frac{t \cdot \frac{2}{t} - (3 + 2 \log t) \cdot 1}{t^2} \]
\[ = \frac{2 - (3 + 2 \log t)}{t^2} \]
\[ = \frac{2 - 3 - 2 \log t}{t^2} \]
\[ = \frac{-1 - 2 \log t}{t^2} \]

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

Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\frac{-1 - 2 \log t}{t^2}}{\frac{-(1 + 2 \log t)}{t^3}} \]
This simplifies to:
\[ = \frac{-1 - 2 \log t}{t^2} \cdot \frac{t^3}{-(1 + 2 \log t)} \]
\[ = \frac{t^3}{t^2} = t \]

### Final Answer
Thus, we have:
\[ \frac{dy}{dx} = t \]