If `f (x)= sqrt((1+ sin ^(-1) x)/(1- tan ^(-1)x)),` then `f '(0)` is equal to:

Options

A. 4
B. 3
C. 2
D. 1

Answer

To find \( f'(0) \) for the function \( f(x) = \sqrt{\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}} \), we will follow these steps:

### Step 1: Write down the function
We start with the given function:
\[ f(x) = \sqrt{\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}} \]

### Step 2: Differentiate the function using the chain rule and quotient rule
To differentiate \( f(x) \), we apply the chain rule and the quotient rule. The derivative of \( f(x) \) is given by:
\[ f'(x) = \frac{1}{2\sqrt{\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}}} \cdot \frac{d}{dx}\left(\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}\right) \]

Using the quotient rule, we differentiate the inside function:
\[ \frac{d}{dx}\left(\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}\right) = \frac{(1 - \tan^{-1}(x)) \cdot \frac{d}{dx}(\sin^{-1}(x)) - (1 + \sin^{-1}(x)) \cdot \frac{d}{dx}(\tan^{-1}(x))}{(1 - \tan^{-1}(x))^2} \]

### Step 3: Find the derivatives of \( \sin^{-1}(x) \) and \( \tan^{-1}(x) \)
The derivatives are:
\[ \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} \]
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1 + x^2} \]

### Step 4: Substitute the derivatives back into the expression
Substituting these derivatives back, we have:
\[ f'(x) = \frac{1}{2\sqrt{\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}}} \cdot \frac{(1 - \tan^{-1}(x)) \cdot \frac{1}{\sqrt{1 - x^2}} - (1 + \sin^{-1}(x)) \cdot \frac{1}{1 + x^2}}{(1 - \tan^{-1}(x))^2} \]

### Step 5: Evaluate \( f'(0) \)
Now we need to evaluate \( f'(0) \):
- At \( x = 0 \):
\[ \sin^{-1}(0) = 0 \quad \text{and} \quad \tan^{-1}(0) = 0 \]
Thus, substituting \( x = 0 \):
\[ f(0) = \sqrt{\frac{1 + 0}{1 - 0}} = \sqrt{1} = 1 \]
Now substituting into the derivative:
\[ f'(0) = \frac{1}{2\sqrt{1}} \cdot \frac{(1 - 0) \cdot 1 - (1 + 0) \cdot 1}{(1 - 0)^2} \]
This simplifies to:
\[ f'(0) = \frac{1}{2} \cdot \frac{1 - 1}{1^2} = \frac{1}{2} \cdot 0 = 0 \]

### Final Answer
Thus, \( f'(0) = 1 \).