Find the maximum value of the function `(log x)/x " when "x gt 0`.

Answer

To find the maximum value of the function \( f(x) = \frac{\log x}{x} \) for \( x > 0 \), we will follow these steps:

### Step 1: Find the derivative of the function
We will use the quotient rule to differentiate \( f(x) \). The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then:
\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]
Here, \( g(x) = \log x \) and \( h(x) = x \).

Calculating the derivatives:
- \( g'(x) = \frac{1}{x} \)
- \( h'(x) = 1 \)

Now applying the quotient rule:
\[ f'(x) = \frac{\left(\frac{1}{x}\right)x - \log x \cdot 1}{x^2} \]
Simplifying this:
\[ f'(x) = \frac{1 - \log x}{x^2} \]

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

### Step 3: Determine if this critical point is a maximum
To confirm whether this critical point is a maximum, we will find the second derivative \( f''(x) \).

### Step 4: Find the second derivative
We differentiate \( f'(x) = \frac{1 - \log x}{x^2} \) again using the quotient rule:
Let \( g(x) = 1 - \log x \) and \( h(x) = x^2 \).

Calculating the derivatives:
- \( g'(x) = -\frac{1}{x} \)
- \( h'(x) = 2x \)

Now applying the quotient rule:
\[ f''(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]
Substituting the values:
\[ f''(x) = \frac{-\frac{1}{x} \cdot x^2 - (1 - \log x) \cdot 2x}{(x^2)^2} \]
Simplifying this:
\[ f''(x) = \frac{-x - 2x + 2x \log x}{x^4} = \frac{-3x + 2x \log x}{x^4} \]
\[ f''(x) = \frac{x(2 \log x - 3)}{x^4} = \frac{2 \log x - 3}{x^3} \]

### Step 5: Evaluate the second derivative at \( x = e \)
Substituting \( x = e \):
\[ f''(e) = \frac{2 \log e - 3}{e^3} = \frac{2 \cdot 1 - 3}{e^3} = \frac{-1}{e^3} \]
Since \( f''(e) < 0 \), this indicates that \( f(x) \) has a local maximum at \( x = e \).

### 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 the function \( \frac{\log x}{x} \) when \( x > 0 \) is:
\[ \boxed{\frac{1}{e}} \]