If `f(x)=log_x(lnx)` then `f'(x)` at x=e is

Options

A. e
B. `-e`
C. `e^(2)`
D. `e^(-1)`

Explanation

To find \( f'(x) \) at \( x = e \) for the function \( f(x) = \log_x(\ln x) \), we will follow these steps:

### Step 1: Rewrite the function using the change of base formula
The logarithm can be rewritten using the change of base formula:
\[ f(x) = \log_x(\ln x) = \frac{\ln(\ln x)}{\ln(x)} \]

### Step 2: Differentiate using the quotient rule
To differentiate \( f(x) \), we will use the quotient rule. If \( u = \ln(\ln x) \) and \( v = \ln(x) \), then:
\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \).

### Step 3: Compute \( u' \) and \( v' \)
1. Differentiate \( u = \ln(\ln x) \):
\[ u' = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x} \]
2. Differentiate \( v = \ln(x) \):
\[ v' = \frac{1}{x} \]

### Step 4: Substitute \( u, u', v, v' \) into the quotient rule formula
Now substituting these into the quotient rule gives:
\[ f'(x) = \frac{\ln(x) \cdot \frac{1}{x \ln x} - \ln(\ln x) \cdot \frac{1}{x}}{(\ln x)^2} \]
This simplifies to:
\[ f'(x) = \frac{1/x - \ln(\ln x)/x}{(\ln x)^2} = \frac{1 - \ln(\ln x)}{x (\ln x)^2} \]

### Step 5: Evaluate \( f'(x) \) at \( x = e \)
Now we need to evaluate \( f'(e) \):
1. Calculate \( \ln(e) = 1 \)
2. Calculate \( \ln(\ln(e)) = \ln(1) = 0 \)

Substituting these values into \( f'(x) \):
\[ f'(e) = \frac{1 - 0}{e \cdot (1)^2} = \frac{1}{e} \]

### Final Answer
Thus, the value of \( f'(x) \) at \( x = e \) is:
\[ \boxed{\frac{1}{e}} \]