If `f(x)=(log_(cotx)tanx)(log_(tanx)cotx)^(-1)`
`+tan^(-1)((x)/(sqrt(4-x^(2))))`, then f'(0) is equal to

Options

A. 2
B. 0
C. `1//2`
D. -2

Answer

To find \( f'(0) \) for the function

\[ f(x) = \left( \log_{\cot x} \tan x \right) \left( \log_{\tan x} \cot x \right)^{-1} + \tan^{-1} \left( \frac{x}{\sqrt{4 - x^2}} \right), \]

we will break it down step by step.

### Step 1: Simplifying the logarithmic part

We start with the logarithmic expressions:

\[ \log_{\cot x} \tan x = \frac{\log \tan x}{\log \cot x} = \frac{\log \tan x}{\log \left( \frac{1}{\tan x} \right)} = \frac{\log \tan x}{-\log \tan x} = -1. \]

Similarly,

\[ \log_{\tan x} \cot x = \frac{\log \cot x}{\log \tan x} = \frac{-\log \tan x}{\log \tan x} = -1. \]

Thus,

\[ \left( \log_{\tan x} \cot x \right)^{-1} = -1. \]

So, we have:

\[ \left( \log_{\cot x} \tan x \right) \left( \log_{\tan x} \cot x \right)^{-1} = -1 \cdot -1 = 1. \]

### Step 2: Simplifying the second term

Now we simplify the second term:

\[ \tan^{-1} \left( \frac{x}{\sqrt{4 - x^2}} \right). \]

### Step 3: Combining the results

Thus, we can rewrite \( f(x) \):

\[ f(x) = 1 + \tan^{-1} \left( \frac{x}{\sqrt{4 - x^2}} \right). \]

### Step 4: Finding the derivative \( f'(x) \)

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

\[ f'(x) = 0 + \frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{\sqrt{4 - x^2}} \right) \right). \]

Using the chain rule for differentiation:

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

where \( u = \frac{x}{\sqrt{4 - x^2}} \).

### Step 5: Finding \( \frac{du}{dx} \)

We need to differentiate \( u \):

\[ u = \frac{x}{\sqrt{4 - x^2}}. \]

Using the quotient rule:

\[ \frac{du}{dx} = \frac{\sqrt{4 - x^2} \cdot 1 - x \cdot \frac{-x}{\sqrt{4 - x^2}}}{(4 - x^2)} = \frac{\sqrt{4 - x^2} + \frac{x^2}{\sqrt{4 - x^2}}}{4 - x^2}. \]

This simplifies to:

\[ \frac{du}{dx} = \frac{(4 - x^2) + x^2}{\sqrt{4 - x^2}(4 - x^2)} = \frac{4}{\sqrt{4 - x^2}(4 - x^2)}. \]

### Step 6: Evaluating \( f'(0) \)

Now we substitute \( x = 0 \):

\[ u(0) = \frac{0}{\sqrt{4}} = 0, \]

and

\[ \frac{du}{dx} \bigg|_{x=0} = \frac{4}{\sqrt{4}(4)} = \frac{4}{2 \cdot 4} = \frac{1}{2}. \]

Thus,

\[ f'(0) = \frac{1}{1 + 0^2} \cdot \frac{1}{2} = \frac{1}{2}. \]

### Final Answer

Therefore,

\[ f'(0) = \frac{1}{2}. \]