If `f(x)=cos^(-1)(2x^(2)-1), x in [-1,1].` Then

Options

A. f(x) is differentiable on (-1,1) such that `f'(x)=(-2)/(sqrt(1-x)^(2))`
B. f(x) is differentiable on `(-1,0) uu (0,1)` such that `f'(x)=(-2)/(sqrt(1-x^(2))`
C. f(x) is differentiable on `(-1,0) uu (0,1)` such that `f'(x)={{:(,(-2)/(sqrt(1-x^(2))),0 lt x lt 1),(,(2)/(sqrt(1-x^(2))),-1 lt x lt 0):}`
D. f(x) is differentiable on (-1,1) such that `f'(x)={{:(,(-2)/(sqrt(1-x^(2))),0 le xlt1),(,(2)/(sqrt(1-x^(2))),-1 lt x le 0):}`

Explanation

To solve the problem, we need to find the derivative \( f'(x) \) of the function \( f(x) = \cos^{-1}(2x^2 - 1) \) for \( x \) in the interval \([-1, 1]\).

### Step-by-Step Solution:

1. **Identify the function**:
\[ f(x) = \cos^{-1}(2x^2 - 1) \]

2. **Determine the range of \( 2x^2 - 1 \)**:
- The expression \( 2x^2 - 1 \) varies as \( x \) goes from -1 to 1.
- At \( x = -1 \): \( 2(-1)^2 - 1 = 1 \)
- At \( x = 0 \): \( 2(0)^2 - 1 = -1 \)
- At \( x = 1 \): \( 2(1)^2 - 1 = 1 \)
- Therefore, \( 2x^2 - 1 \) ranges from -1 to 1.

3. **Split the function based on the intervals**:
- For \( x \in [0, 1] \): \( f(x) = \cos^{-1}(2x^2 - 1) \)
- For \( x \in [-1, 0] \): \( f(x) = 2\pi - \cos^{-1}(2x^2 - 1) \)

4. **Find the derivative \( f'(x) \)**:
- Using the derivative of \( \cos^{-1}(u) \), which is \( -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \):
- Let \( u = 2x^2 - 1 \), then \( \frac{du}{dx} = 4x \).
- Thus, for \( x \in [0, 1] \):
\[ f'(x) = -\frac{1}{\sqrt{1 - (2x^2 - 1)^2}} \cdot 4x \]
- For \( x \in [-1, 0] \):
\[ f'(x) = \frac{4x}{\sqrt{1 - (2x^2 - 1)^2}} \]

5. **Evaluate the expression \( 1 - (2x^2 - 1)^2 \)**:
- Simplifying \( 1 - (2x^2 - 1)^2 \):
\[ = 1 - (4x^4 - 4x^2 + 1) = 4x^2 - 4x^4 \]
- Thus, \( \sqrt{1 - (2x^2 - 1)^2} = \sqrt{4x^2(1 - x^2)} = 2|x|\sqrt{1 - x^2} \).

6. **Final form of \( f'(x) \)**:
- For \( x \in [0, 1] \):
\[ f'(x) = -\frac{4x}{2x\sqrt{1 - x^2}} = -\frac{2}{\sqrt{1 - x^2}} \]
- For \( x \in [-1, 0] \):
\[ f'(x) = \frac{4x}{2(-x)\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}} \]

7. **Check differentiability at \( x = 0 \)**:
- The left-hand derivative at \( x = 0 \) is \( \frac{2}{\sqrt{1 - 0^2}} = 2 \).
- The right-hand derivative at \( x = 0 \) is \( -\frac{2}{\sqrt{1 - 0^2}} = -2 \).
- Since the left-hand and right-hand derivatives do not match, \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion:
The derivative \( f'(x) \) is given by:
\[ f'(x) = \begin{cases} -\frac{2}{\sqrt{1 - x^2}} & \text{if } x \in (0, 1] \\ \frac{2}{\sqrt{1 - x^2}} & \text{if } x \in [-1, 0) \end{cases} \]
And \( f(x) \) is not differentiable at \( x = 0 \).