If `f(x)=sin^(-1) ((2x)/(1+x^2))` then `f(x)` is differentiable on

Options

A. `[-1 , 11]`
B. R - { -1, 1}
C. R - (- 1, 1 )
D. None of these

Answer

To determine where the function \( f(x) = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \) is differentiable, we will follow these steps:

### Step 1: Identify the function and its domain
The function \( f(x) \) is defined as:
\[ f(x) = \sin^{-1} \left( \frac{2x}{1+x^2} \right) \]
The expression inside the inverse sine function, \( \frac{2x}{1+x^2} \), must lie within the interval \([-1, 1]\) for \( f(x) \) to be defined.

### Step 2: Find the range of \( \frac{2x}{1+x^2} \)
To find where \( \frac{2x}{1+x^2} \) lies between -1 and 1, we analyze the expression:
1. Set \( \frac{2x}{1+x^2} = 1 \):
\[ 2x = 1 + x^2 \implies x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1 \]
2. Set \( \frac{2x}{1+x^2} = -1 \):
\[ 2x = -1 - x^2 \implies x^2 + 2x + 1 = 0 \implies (x+1)^2 = 0 \implies x = -1 \]

### Step 3: Determine the intervals
The function \( \frac{2x}{1+x^2} \) reaches its maximum at \( x = 1 \) and minimum at \( x = -1 \). We need to check the behavior of \( \frac{2x}{1+x^2} \) in the intervals:
- For \( x < -1 \)
- For \( -1 < x < 1 \)
- For \( x > 1 \)

### Step 4: Analyze the derivative
We can differentiate \( f(x) \) using the chain rule:
\[ f'(x) = \frac{d}{dx} \left( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \right) = \frac{1}{\sqrt{1 - \left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) \]
Calculating \( \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) \):
Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} \]

### Step 5: Determine where the derivative is defined
The derivative \( f'(x) \) is defined as long as:
1. \( 1 - \left( \frac{2x}{1+x^2} \right)^2 > 0 \)
2. \( 1+x^2 \neq 0 \) (which is always true for real \( x \))

### Step 6: Analyze the critical points
The critical points are \( x = -1 \) and \( x = 1 \). We check the left-hand and right-hand derivatives at these points:
- \( f'(-1) \) and \( f'(1) \) must be checked for continuity.

### Conclusion
From the analysis, we find that \( f(x) \) is differentiable everywhere except at the points \( x = -1 \) and \( x = 1 \).

Thus, the function \( f(x) \) is differentiable on:
\[ \mathbb{R} \setminus (-1, 1) \]