`f(x) = sin^(-1)(sin x)` then `(d)/(dx)f(x)` at `x = (7pi)/(2)` is

Options

A. 1
B. `-1`
C. Not differentiable
D. None of these

Explanation

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(\sin x) \) and find its derivative at the point \( x = \frac{7\pi}{2} \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = \sin^{-1}(\sin x) \) is defined as the inverse sine of the sine function. The output of \( \sin^{-1}(y) \) is restricted to the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

2. **Determining the Domain**:
The function \( \sin x \) is periodic with a period of \( 2\pi \). Therefore, we need to find the equivalent angle of \( x = \frac{7\pi}{2} \) within the principal range of the inverse sine function.

\[ \frac{7\pi}{2} = 3\pi + \frac{\pi}{2} \quad \text{(which is equivalent to } \frac{\pi}{2} \text{ in the range of } [-\pi, \pi]) \]
To find the equivalent angle, we can subtract \( 2\pi \) until we fall within the range:
\[ \frac{7\pi}{2} - 3\pi = \frac{7\pi}{2} - \frac{6\pi}{2} = \frac{\pi}{2} \]

3. **Evaluating \( f(x) \)**:
Now we can evaluate \( f\left(\frac{7\pi}{2}\right) \):
\[ f\left(\frac{7\pi}{2}\right) = \sin^{-1}\left(\sin\left(\frac{7\pi}{2}\right)\right) = \sin^{-1}\left(\sin\left(\frac{\pi}{2}\right)\right) = \sin^{-1}(1) = \frac{\pi}{2} \]

4. **Finding the Derivative**:
The derivative of \( f(x) = \sin^{-1}(\sin x) \) is not straightforward at points outside the principal range. The function is not differentiable at points where \( x \) is outside the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

5. **Conclusion**:
Since \( \frac{7\pi}{2} \) is outside the domain of \( f(x) \) (which is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)), we conclude that \( f(x) \) is not differentiable at \( x = \frac{7\pi}{2} \).

### Final Answer:
The derivative \( \frac{d}{dx} f(x) \) at \( x = \frac{7\pi}{2} \) is **not differentiable**.

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