Question

Find the derivative of the function \( f(x) = \sin^{-1} \big( 2x \sqrt{1 - x^2} \big) \) with respect to \( x \), for \( x \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).

• A. \( \frac{2}{\sqrt{1-4x^2}} \)
• B. \( \frac{1}{\sqrt{1-x^2}} \)
• C. \( \frac{2}{\sqrt{1-x^2}} \)
• D. \( \frac{1}{\sqrt{1-4x^2}}

Hint:

\textbf{Key Fact:} Use trigonometric substitution to simplify inverse functions, and always verify the derivative with the given domain.

Answer 1

The Correct Option is A

• Simplify the Expression: Let \( x = \sin \theta \), where \( \theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \), so \[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta. \] Then, \[ f(x) = \sin^{-1}(2x \sqrt{1 - x^2}) = \sin^{-1}(2 \sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta). \]
• Evaluate the Inverse Sine: Since \( \theta \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \), we have \[ 2\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \] so \[ \sin^{-1}(\sin 2\theta) = 2\theta. \] Thus, \[ f(x) = 2\theta = 2 \sin^{-1} x. \]
• Differentiate: Now, \[ f(x) = 2 \sin^{-1} x. \] Differentiating, \[ f'(x) = 2 \cdot \frac{d}{dx} \left(\sin^{-1} x\right) = 2 \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}}. \]
• Alternative Approach: Alternatively, one can differentiate directly using the chain rule on \[ f(x) = \sin^{-1}(2x \sqrt{1 - x^2}), \] but the simplified form above is easier and valid for \( x \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \). Note that the expression under the square root in the denominator could appear as \( \sqrt{1 - 4x^2(1-x^2)} = \sqrt{1 - 4x^2 + 4x^4} \), which corresponds to the domain restriction.
• Conclusion: The correct derivative is \[ \boxed{ \frac{2}{\sqrt{1 - x^2}} } \] and matches the simplified approach.