The maximum value of `(ln x)/x` is:

Options

A. e
B. `1/e`
C. `1/e`
D. 1

Answer

To find the maximum value of the function \( y = \frac{\ln x}{x} \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = \frac{\ln x}{x} \) with respect to \( x \).

Using the quotient rule, we have:
\[ \frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(\ln x) - \ln x \cdot \frac{d}{dx}(x)}{x^2} \]
Since \( \frac{d}{dx}(\ln x) = \frac{1}{x} \) and \( \frac{d}{dx}(x) = 1 \), we can substitute these derivatives into our equation:
\[ \frac{dy}{dx} = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2} \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ \frac{1 - \ln x}{x^2} = 0 \]
This implies:
\[ 1 - \ln x = 0 \quad \Rightarrow \quad \ln x = 1 \]
Taking the exponential of both sides, we find:
\[ x = e \]

### Step 3: Determine the nature of the critical point
Next, we need to determine whether this critical point is a maximum or minimum. We can do this by analyzing the sign of the derivative \( \frac{dy}{dx} \) around \( x = e \).

- For \( x < e \): \( \ln x < 1 \) implies \( 1 - \ln x > 0 \), so \( \frac{dy}{dx} > 0 \) (function is increasing).
- For \( x > e \): \( \ln x > 1 \) implies \( 1 - \ln x < 0 \), so \( \frac{dy}{dx} < 0 \) (function is decreasing).

Since the function increases before \( x = e \) and decreases after \( x = e \), we conclude that \( x = e \) is a maximum.

### Step 4: Find the maximum value
Now we substitute \( x = e \) back into the original function to find the maximum value:
\[ y = \frac{\ln e}{e} = \frac{1}{e} \]

Thus, the maximum value of \( \frac{\ln x}{x} \) is \( \frac{1}{e} \).

### Conclusion
The maximum value of \( \frac{\ln x}{x} \) is \( \frac{1}{e} \).

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