The maximum value of `((log x)/(x))` is

Options

A. `((1)/(e))`
B. `(2)/(e)`
C. `e`
D. 1

Explanation

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

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) using the quotient rule. The quotient rule states that if \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \).

Here, let:
- \( u = \log x \) and \( u' = \frac{1}{x} \)
- \( v = x \) and \( v' = 1 \)

Thus, we can differentiate \( f(x) \):
\[ f'(x) = \frac{\left(\frac{1}{x}\right) \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2} \]

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

### Step 3: Solve for \( x \)
From \( \log x = 1 \), we find:
\[ x = e \]

### Step 4: Determine the nature of the critical point
To confirm that this point is a maximum, we will check the second derivative \( f''(x) \).

First, we differentiate \( f'(x) \):
\[ f'(x) = \frac{1 - \log x}{x^2} \]
Using the quotient rule again:
Let \( u = 1 - \log x \) and \( v = x^2 \):
\[ f''(x) = \frac{(0 - \frac{1}{x}) \cdot x^2 - (1 - \log x) \cdot 2x}{x^4} \]
Simplifying this, we have:
\[ f''(x) = \frac{-x + 2(1 - \log x)}{x^3} \]

### Step 5: Evaluate the second derivative at \( x = e \)
Now we substitute \( x = e \):
\[ f''(e) = \frac{-e + 2(1 - 1)}{e^3} = \frac{-e}{e^3} = -\frac{1}{e^2} \]
Since \( f''(e) < 0 \), this indicates that \( x = e \) is a maximum point.

### Step 6: Find the maximum value
Now we substitute \( x = e \) back into the original function to find the maximum value:
\[ f(e) = \frac{\log e}{e} = \frac{1}{e} \]

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

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