Question:medium
If \( y = \sin^{-1} \left( 2x \sqrt{1 - x^2} \right) \), \( -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \), then \( \frac{dy}{dx} \) is equal to:
If \( y = \sin^{-1} \left( 2x \sqrt{1 - x^2} \right) \), \( -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \), then \( \frac{dy}{dx} \) is equal to:
Show Hint
Memorize the identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$. It appears frequently in calculus exams and saves significant time over the Chain Rule.
Updated On: May 6, 2026
- \( \frac{x}{\sqrt{1 - x^2}} \)
- \( \frac{1}{\sqrt{1 - x^2}} \)
- \( \frac{2}{\sqrt{1 - x^2}} \)
- \( \frac{2x}{\sqrt{1 - x^2}} \)
- \( \frac{-2x}{\sqrt{1 - x^2}} \)
Show Solution
The Correct Option is C
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