To find the interval in which the derivative of the inverse sine function, \(\sin^{-1}x\), exists, we must first understand the definition and properties of the function and its derivative.
The derivative of \(\sin^{-1}x\) is given by:
\(\frac{d}{dx}[\sin^{-1}x] = \frac{1}{\sqrt{1-x^2}}\)
The expression \(\sqrt{1-x^2}\) is defined only when the value inside the square root is non-negative because we cannot take the square root of a negative number in the real number system. Therefore, for the derivative \(\frac{1}{\sqrt{1-x^2}}\) to exist, we require:
\(1-x^2 \geq 0\)
Solving this inequality:
\(1 \geq x^2\)
This further simplifies to:
\(-1 \leq x \leq 1\)
Since the function becomes undefined when \(x = \pm1\) because it would result in division by zero, the derivative is only defined in the interval:
\(-1 < x < 1\)
Thus, the interval in which the derivative of \(\sin^{-1}x\) exists is \((-1,1)\).
Therefore, the correct answer is: \((-1,1)\).