If ` f(x) = log_(x^(2)) (log_(e) x) "then f' (x) at x= e"` is

Options

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

Explanation

To find \( f'(x) \) for the function \( f(x) = \log_{x^2}(\log_e x) \) at \( x = e \), we will follow these steps:

### Step 1: Rewrite the logarithm using the change of base formula
Using the change of base formula, we can express \( f(x) \) as:
\[ f(x) = \frac{\log_e(\log_e x)}{\log_e(x^2)} \]
Since \( \log_e(x^2) = 2\log_e x \), we can simplify this to:
\[ f(x) = \frac{\log_e(\log_e x)}{2\log_e x} \]

### Step 2: Differentiate \( f(x) \)
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, let:
- \( g(x) = \log_e(\log_e x) \)
- \( h(x) = 2\log_e x \)

Now we need to find \( g'(x) \) and \( h'(x) \).

### Step 3: Find \( g'(x) \)
Using the chain rule:
\[ g'(x) = \frac{1}{\log_e x} \cdot \frac{1}{x} = \frac{1}{x \log_e x} \]

### Step 4: Find \( h'(x) \)
Differentiating \( h(x) \):
\[ h'(x) = 2 \cdot \frac{1}{x} = \frac{2}{x} \]

### Step 5: Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \) into the quotient rule
Now substituting into the quotient rule:
\[ f'(x) = \frac{\left(\frac{1}{x \log_e x}\right)(2 \log_e x) - \log_e(\log_e x)\left(\frac{2}{x}\right)}{(2 \log_e x)^2} \]
This simplifies to:
\[ f'(x) = \frac{\frac{2}{x} - \frac{2 \log_e(\log_e x)}{x}}{4 (\log_e x)^2} \]
\[ = \frac{2(1 - \log_e(\log_e x))}{4x(\log_e x)^2} = \frac{1 - \log_e(\log_e x)}{2x(\log_e x)^2} \]

### Step 6: Evaluate \( f'(x) \) at \( x = e \)
Now we evaluate \( f'(x) \) at \( x = e \):
\[ f'(e) = \frac{1 - \log_e(\log_e e)}{2e(\log_e e)^2} \]
Since \( \log_e e = 1 \), we have:
\[ f'(e) = \frac{1 - \log_e(1)}{2e(1)^2} = \frac{1 - 0}{2e} = \frac{1}{2e} \]

### Final Answer
Thus, the value of \( f'(x) \) at \( x = e \) is:
\[ \boxed{\frac{1}{2e}} \]