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

Options

A. does not exist
B. is equal to `(2)/( e )`
C. is equal to `(1)/( e )`
D. is equal to 1

Answer

To find \( f'(e) \) 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: Substitute \( \log_e x \) with \( \tan \theta \)

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

\[ x = e^{\tan \theta}. \]

### Step 2: Rewrite \( f(x) \)

Using the identity for cosine of double angle, we can rewrite \( f(x) \):

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

### Step 3: Find \( \theta \) in terms of \( x \)

Since \( \theta = \tan^{-1}(\log_e x) \), we can express \( f(x) \) as:

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

### Step 4: Differentiate \( f(x) \)

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

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

Using the chain rule:

\[ \frac{d}{dx} \left( \tan^{-1}(u) \right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}, \]

where \( u = \log_e x \) and \( \frac{du}{dx} = \frac{1}{x} \).

Thus, we have:

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

### Step 5: Evaluate \( f'(e) \)

Now we will evaluate \( f'(e) \):

1. Calculate \( \log_e e \):
\[ \log_e e = 1. \]

2. Substitute \( x = e \) into \( f'(x) \):
\[ f'(e) = 2 \cdot \frac{1}{1 + (1)^2} \cdot \frac{1}{e} = 2 \cdot \frac{1}{2} \cdot \frac{1}{e} = \frac{1}{e}. \]

### Final Answer

Thus,

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

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