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: Simplify the function using trigonometric identities
We recognize that

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

if we let \( \tan \theta = \log_e x \). Thus, we can rewrite the function as:

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

### Step 2: Differentiate the function
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, we get:

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

### Step 3: Evaluate \( f'(e) \)
Now we need to evaluate \( f'(e) \):

1. First, calculate \( \log_e e \):
\[ \log_e e = 1. \]

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

### Final Result
Thus,

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

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