Question

The derivative of $\sin^{-1}x$ exists in the interval

• A. $[-1,1]$
• B. $(-1,1)$
• C. $\mathbb{R}$
• D. $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

Hint:

Derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$, so it exists only when $|x|<1$.

Answer 1

The Correct Option is B

To determine the interval in which the derivative of the inverse sine function \( \sin^{-1} x \) exists, we first need to understand its domain and its derivative.

Understanding the Domain of \( \sin^{-1} x \) 

The function \( \sin^{-1} x \), also known as \( \text{arcsin}(x) \), is defined for \( x \) in the interval \([-1, 1]\). This is because sine values range only from -1 to 1, and inverse sine can only take inputs within this range.

Calculating the Derivative of \( \sin^{-1} x \)

The derivative of \( \sin^{-1} x \) is given by the formula:

\(\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}\)

For this derivative to exist, the denominator should not be zero and the expression under the square root should be non-negative:

• \(1 - x^2 \geq 0\) implies \(x^2 \leq 1\) which means \( -1 \leq x \leq 1 \).
• \(\sqrt{1-x^2} \neq 0\) implies \( x \neq -1 \) and \( x \neq 1 \).

Conclusion: Interval of Existence

Therefore, the derivative exists for \( x \) in the open interval (-1, 1), excluding the endpoints where the denominator becomes zero. Hence, the correct answer is:

Correct Answer: \( (-1, 1) \)

The derivative of \( \sin^{-1}x \) is undefined at \(x = -1\) and \(x = 1\) due to division by zero at these points.