If `x=t*logt,y=(logt)/(t)," then the value of "(dy)/(dx)" at "t=1` is

Options

A. 1
B. e
C. 0
D. `1+logt`

Answer

To find the value of \(\frac{dy}{dx}\) at \(t=1\) given the equations \(x = t \log t\) and \(y = \frac{\log t}{t}\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)
Given:
\[ x = t \log t \]
Using the product rule, where \(u = t\) and \(v = \log t\):
\[ \frac{dx}{dt} = u \frac{dv}{dt} + v \frac{du}{dt} \]
Calculating the derivatives:
- \(\frac{du}{dt} = 1\)
- \(\frac{dv}{dt} = \frac{1}{t}\)

Now substituting:
\[ \frac{dx}{dt} = t \cdot \frac{1}{t} + \log t \cdot 1 = 1 + \log t \]

### Step 2: Differentiate \(y\) with respect to \(t\)
Given:
\[ y = \frac{\log t}{t} \]
Using the quotient rule, where \(u = \log t\) and \(v = t\):
\[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]
Calculating the derivatives:
- \(\frac{du}{dt} = \frac{1}{t}\)
- \(\frac{dv}{dt} = 1\)

Now substituting:
\[ \frac{dy}{dt} = \frac{t \cdot \frac{1}{t} - \log t \cdot 1}{t^2} = \frac{1 - \log t}{t^2} \]

### Step 3: Find \(\frac{dy}{dx}\)
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = \frac{\frac{1 - \log t}{t^2}}{1 + \log t} \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at \(t = 1\)
Now we substitute \(t = 1\):
- \(\log 1 = 0\)

Thus:
\[ \frac{dy}{dx} = \frac{1 - 0}{1^2} \cdot \frac{1}{1 + 0} = \frac{1}{1} = 1 \]

### Final Answer
The value of \(\frac{dy}{dx}\) at \(t = 1\) is:
\[ \boxed{1} \]