If `f(x)=log_(e)(log_(e)x)/log_(e)x` then `f'(x)` at x = e is

Options

A. 0
B. 1
C. e
D. `1//2`

Explanation

To find \( f'(x) \) for the function \( f(x) = \frac{\log_e(\log_e x)}{\log_e x} \) at \( x = e \), we will use the quotient rule for differentiation. The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then:

\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]

### Step 1: Identify \( u(x) \) and \( v(x) \)

Let:
- \( u(x) = \log_e(\log_e x) \)
- \( v(x) = \log_e x \)

### Step 2: Differentiate \( u(x) \) and \( v(x) \)

1. **Differentiate \( u(x) \)**:
\[ u'(x) = \frac{d}{dx} \log_e(\log_e x) = \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{\log_e x} \cdot \frac{1}{x} = \frac{1}{x \log_e x} \]

2. **Differentiate \( v(x) \)**:
\[ v'(x) = \frac{d}{dx} \log_e x = \frac{1}{x} \]

### Step 3: Apply the Quotient Rule

Now, we can apply the quotient rule:

\[ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]

Substituting \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \):

\[ f'(x) = \frac{\log_e x \cdot \frac{1}{x \log_e x} - \log_e(\log_e x) \cdot \frac{1}{x}}{(\log_e x)^2} \]

This simplifies to:

\[ f'(x) = \frac{\frac{1}{x} - \frac{\log_e(\log_e x)}{x}}{(\log_e x)^2} \]

### Step 4: Simplify the Expression

Factor out \( \frac{1}{x} \):

\[ f'(x) = \frac{1}{x} \cdot \frac{1 - \log_e(\log_e x)}{(\log_e x)^2} \]

### Step 5: Evaluate at \( x = e \)

Now, we substitute \( x = e \):

1. Calculate \( \log_e(e) = 1 \).
2. Calculate \( \log_e(\log_e(e)) = \log_e(1) = 0 \).

Substituting these values into the derivative:

\[ f'(e) = \frac{1}{e} \cdot \frac{1 - 0}{(1)^2} = \frac{1}{e} \]

### Final Answer

Thus, the value of \( f'(x) \) at \( x = e \) is:

\[ \boxed{\frac{1}{e}} \]