If `f(x)=cos^(-1){(1-(log_(e)x)^(2))/(1+(log_(e)x)^(2))},` then `f'((1)/( e ))` is equal to

Options

A. e
B. `-e`
C. 1
D. `2e`

Explanation

To find \( f'\left(\frac{1}{e}\right) \) for the function

\[ f(x) = \cos^{-1}\left(\frac{1 - (\log_e x)^2}{1 + (\log_e x)^2}\right), \]

we will follow these steps:

### Step 1: Rewrite the function in terms of \( \theta \)

We can relate this function to the double angle formula for cosine. We know that:

\[ \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}. \]

Let \( \log_e x = \tan \theta \). Then, we can rewrite \( f(x) \) as:

\[ f(x) = \cos^{-1}(\cos(2\theta)) = 2\theta. \]

### Step 2: Express \( \theta \) in terms of \( x \)

Since \( \theta = \tan^{-1}(\log_e x) \), we have:

\[ f(x) = 2 \tan^{-1}(\log_e x). \]

### Step 3: Differentiate \( f(x) \)

Now we differentiate \( f(x) \):

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

Using the chain rule, we have:

\[ f'(x) = 2 \cdot \frac{1}{1 + (\log_e x)^2} \cdot \frac{d}{dx}[\log_e x]. \]

The derivative of \( \log_e x \) is \( \frac{1}{x} \). Thus,

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

### Step 4: Substitute \( x = \frac{1}{e} \)

Now we need to evaluate \( f'\left(\frac{1}{e}\right) \):

\[ f'\left(\frac{1}{e}\right) = 2 \cdot \frac{1}{1 + (\log_e \frac{1}{e})^2} \cdot \frac{1}{\frac{1}{e}}. \]

Calculating \( \log_e \frac{1}{e} \):

\[ \log_e \frac{1}{e} = \log_e e^{-1} = -1. \]

Thus,

\[ f'\left(\frac{1}{e}\right) = 2 \cdot \frac{1}{1 + (-1)^2} \cdot e = 2 \cdot \frac{1}{1 + 1} \cdot e = 2 \cdot \frac{1}{2} \cdot e = e. \]

### Final Answer

Therefore,

\[ f'\left(\frac{1}{e}\right) = e. \]

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