The function `f(x) = sin ^(-1)(cos x)`is

Options

A. not different at `x=(pi)/(2)`
B. differentiable at `(3pi)/(2)`
C. differentiable at x=0
D. differentiable at `x=2pi`

Answer

To determine the properties of the function \( f(x) = \sin^{-1}(\cos x) \), we will analyze its continuity and differentiability step by step.

### Step 1: Check Continuity
The function \( f(x) = \sin^{-1}(\cos x) \) is a composition of two functions: \( \sin^{-1}(x) \) and \( \cos(x) \).

1. **Continuity of \( \cos(x) \)**: The cosine function is continuous for all \( x \in \mathbb{R} \).
2. **Continuity of \( \sin^{-1}(x) \)**: The inverse sine function \( \sin^{-1}(x) \) is continuous for \( x \) in the interval \([-1, 1]\).

Since \( \cos(x) \) takes values in the range \([-1, 1]\), the composition \( \sin^{-1}(\cos x) \) is continuous for all \( x \in \mathbb{R} \).

### Step 2: Check Differentiability
To check the differentiability of \( f(x) \), we need to find its derivative \( f'(x) \).

1. **Differentiate \( f(x) \)**:
\[ f'(x) = \frac{d}{dx} \left( \sin^{-1}(\cos x) \right) \]
Using the chain rule:
\[ f'(x) = \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot (-\sin x) \]
Simplifying this gives:
\[ f'(x) = \frac{-\sin x}{\sqrt{1 - \cos^2 x}} = \frac{-\sin x}{\sqrt{\sin^2 x}} = \frac{-\sin x}{|\sin x|} \]

2. **Analyze \( f'(x) \)**:
The expression \( \frac{-\sin x}{|\sin x|} \) is defined except where \( \sin x = 0 \). This occurs at:
\[ x = n\pi \quad \text{for } n \in \mathbb{Z} \]
At these points, \( f'(x) \) is not defined, which means \( f(x) \) is not differentiable at \( x = n\pi \).

### Conclusion
1. **Continuity**: The function \( f(x) = \sin^{-1}(\cos x) \) is continuous for all \( x \in \mathbb{R} \).
2. **Differentiability**: The function \( f(x) \) is not differentiable at points \( x = n\pi \).

### Final Answer
- The function \( f(x) = \sin^{-1}(\cos x) \) is continuous for all \( x \in \mathbb{R} \) but not differentiable at points \( x = n\pi \) where \( n \) is an integer.