If `y=sin^(-1)(sin x),` then `dy/dx` at `x =pi/2` is

Options

A. 1
B. -1
C. non-exisent
D. none of these

Explanation

To find \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \) for the function \( y = \sin^{-1}(\sin x) \), we will analyze the function based on the value of \( x \).

### Step 1: Determine the range of \( y \)
The function \( y = \sin^{-1}(\sin x) \) behaves differently depending on the value of \( x \):
- For \( x \) in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( y = x \).
- For \( x \) in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), \( y = \pi - x \).
- For \( x \) in the interval \( \left[\frac{3\pi}{2}, 2\pi\right] \), \( y = x - 2\pi \).

### Step 2: Evaluate \( y \) at \( x = \frac{\pi}{2} \)
At \( x = \frac{\pi}{2} \):
- Since \( \frac{\pi}{2} \) is in the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), we use the second case:
\[ y = \pi - x = \pi - \frac{\pi}{2} = \frac{\pi}{2} \]

### Step 3: Find \( \frac{dy}{dx} \) for each case
1. For \( x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \):
\[ \frac{dy}{dx} = 1 \]

2. For \( x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \):
\[ \frac{dy}{dx} = -1 \]

### Step 4: Analyze the derivative at \( x = \frac{\pi}{2} \)
At \( x = \frac{\pi}{2} \), we have two different derivatives:
- From the left (approaching \( \frac{\pi}{2} \) from the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)): \( \frac{dy}{dx} = 1 \)
- From the right (approaching \( \frac{\pi}{2} \) from the interval \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \)): \( \frac{dy}{dx} = -1 \)

Since the left-hand derivative and the right-hand derivative at \( x = \frac{\pi}{2} \) are not equal, the derivative \( \frac{dy}{dx} \) does not exist at this point.

### Final Answer
Thus, \( \frac{dy}{dx} \) at \( x = \frac{\pi}{2} \) does not exist.

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