The function `f (x)` given by `f(x) =sin^(-1)((2x)/(1+x^2))` is

Options

A. everywhere differentiable such that `f'(x)=-(2)/(1+x^(2))`
B. `"such that f'(x)"={{:(,(2)/(1+x^(2)),-1 lt x lt 1),(,(-2)/(1+x^(2)),|x| gt 1):}`
C. `"such that f'(x)"={{:(,(-2)/(1+x^(2)),-1 lt x lt 1),(,(+2)/(1+x^(2)),|x| gt 1):}`
D. not differentiable at infinitely many points.

Explanation

To determine the properties of the function \( f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \), we will analyze its differentiability and find its derivative.

### Step 1: Identify the function and its domain
The function 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 range \([-1, 1]\) for \( f(x) \) to be defined.

### Step 2: Determine the range of \( \frac{2x}{1+x^2} \)
To find the range of \( \frac{2x}{1+x^2} \), we can analyze its behavior:
- The function \( \frac{2x}{1+x^2} \) can be rewritten as \( 2 \tan(\theta) \) where \( x = \tan(\theta) \).
- As \( x \) varies from \(-\infty\) to \(+\infty\), \( \frac{2x}{1+x^2} \) varies between \(-1\) and \(1\).

### Step 3: Differentiate \( f(x) \)
Using the chain rule, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\right) = \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) \]

### Step 4: Differentiate \( \frac{2x}{1+x^2} \)
To differentiate \( \frac{2x}{1+x^2} \), we use the quotient rule:
\[ \frac{d}{dx}\left(\frac{2x}{1+x^2}\right) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

### Step 5: Substitute back into the derivative of \( f(x) \)
Now substituting back:
\[ f'(x) = \frac{1}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \]

### Step 6: Simplify the derivative
To simplify \( f'(x) \):
1. Calculate \( 1 - \left(\frac{2x}{1+x^2}\right)^2 \):
\[ = 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} = \frac{(1-x^2)^2}{(1+x^2)^2} \]
2. Thus,
\[ f'(x) = \frac{2(1-x^2)}{(1+x^2)^2} \cdot \frac{(1+x^2)}{1-x^2} = \frac{2(1-x^2)}{(1+x^2)(1-x^2)} = \frac{2}{1+x^2} \]

### Step 7: Analyze the differentiability
- For \( |x| < 1 \), \( f'(x) = \frac{2}{1+x^2} \).
- For \( |x| > 1 \), we have \( f'(x) = -\frac{2}{1+x^2} \).

### Conclusion
The function \( f(x) \) is differentiable everywhere except at points where \( \frac{2x}{1+x^2} \) is not defined or where the derivative changes sign, which happens at infinitely many points.