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

Options

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

Answer

To find the derivative of the function \( f(x) = \log_x(\log x) \) at \( x = e \), we will follow these steps:

### Step 1: Rewrite the function using the change of base formula
The logarithm \( \log_x(\log x) \) can be rewritten using the change of base formula:
\[ f(x) = \frac{\log(\log x)}{\log x} \]
Here, we can choose the base of the logarithm as \( e \) (natural logarithm).

### Step 2: Differentiate the function using the quotient rule
To differentiate \( f(x) \), we will use the quotient rule, which states that if \( f(x) = \frac{u}{v} \), then:
\[ f'(x) = \frac{u'v - uv'}{v^2} \]
In our case, let:
- \( u = \log(\log x) \)
- \( v = \log x \)

Now, we need to find \( u' \) and \( v' \).

### Step 3: Differentiate \( u \) and \( v \)
1. Differentiate \( u = \log(\log x) \):
\[ u' = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \]
(using the chain rule)

2. Differentiate \( v = \log x \):
\[ v' = \frac{1}{x} \]

### Step 4: Apply the quotient rule
Now we can substitute \( u \), \( u' \), \( v \), and \( v' \) into the quotient rule:
\[ f'(x) = \frac{\left(\frac{1}{x \log x}\right)(\log x) - (\log(\log x))\left(\frac{1}{x}\right)}{(\log x)^2} \]
This simplifies to:
\[ f'(x) = \frac{\frac{1}{x} - \frac{\log(\log x)}{x}}{(\log x)^2} \]
\[ f'(x) = \frac{1 - \log(\log x)}{x (\log x)^2} \]

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

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

### Final Answer
Thus, the derivative \( f'(e) \) is:
\[ \boxed{\frac{1}{e}} \]