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

Step 1: Recall the derivative formula.
The derivative of the inverse sine function is given by \[ \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} \]
Step 2: Determine where the derivative exists.
For the derivative to exist, the denominator must be defined and non-zero. Thus \[ 1-x^2>0 \]
Step 3: Solve the inequality.
\[ 1-x^2>0 \] \[ x^2& Lt;1 \] \[ -1& Lt;x& Lt;1 \]
Step 4: Interpret the result.
Therefore the derivative exists only inside the open interval \[ (-1,1) \] At \(x = \pm 1\), the denominator becomes zero and the derivative is undefined.

Step 5: Conclusion.
Thus the derivative of \(\sin^{-1}x\) exists in \[ (-1,1) \] Final Answer: $\boxed{(-1,1)}$