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: Evaluate \( f(0) \)

First, we need to find \( f(0) \):

\[ f(0) = \sqrt{\frac{1 + \sin^{-1}(0)}{1 - \tan^{-1}(0)}} \]

Since \( \sin^{-1}(0) = 0 \) and \( \tan^{-1}(0) = 0 \):

\[ f(0) = \sqrt{\frac{1 + 0}{1 - 0}} = \sqrt{\frac{1}{1}} = \sqrt{1} = 1 \]

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

Next, we will differentiate \( f(x) \). We can rewrite \( f(x) \) as:

\[ f(x) = \left( \frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)} \right)^{1/2} \]

Using the chain rule, we have:

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

### Step 3: Differentiate the inside function

Now we need to differentiate the inside function \( \frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)} \) using the quotient rule:

Let \( u = 1 + \sin^{-1}(x) \) and \( v = 1 - \tan^{-1}(x) \).

Using the quotient rule:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]

Where:

- \( u' = \frac{d}{dx}(1 + \sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} \)
- \( v' = \frac{d}{dx}(1 - \tan^{-1}(x)) = -\frac{1}{1 + x^2} \)

So we have:

\[ \frac{d}{dx}\left(\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}\right) = \frac{\left(\frac{1}{\sqrt{1 - x^2}}\right)(1 - \tan^{-1}(x)) - (1 + \sin^{-1}(x))\left(-\frac{1}{1 + x^2}\right)}{(1 - \tan^{-1}(x))^2} \]

### Step 4: Evaluate \( f'(0) \)

Now we need to evaluate \( f'(0) \):

1. Substitute \( x = 0 \) into \( u' \) and \( v' \):

- \( u'(0) = \frac{1}{\sqrt{1 - 0^2}} = 1 \)
- \( v'(0) = -\frac{1}{1 + 0^2} = -1 \)

2. Substitute \( x = 0 \) into \( u \) and \( v \):

- \( u(0) = 1 + \sin^{-1}(0) = 1 \)
- \( v(0) = 1 - \tan^{-1}(0) = 1 \)

Now we can find:

\[ \frac{d}{dx}\left(\frac{1 + \sin^{-1}(x)}{1 - \tan^{-1}(x)}\right) \bigg|_{x=0} = \frac{(1)(1) - (1)(-1)}{(1)^2} = \frac{1 + 1}{1} = 2 \]

Substituting back into \( f'(0) \):

\[ f'(0) = \frac{1}{2} \left( \frac{1 + 0}{1 - 0} \right)^{-1/2} \cdot 2 = \frac{1}{2} \cdot 1 \cdot 2 = 1 \]

### Final Answer

Thus, \( f'(0) = 1 \).

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