Find $\frac{dy}{dx}$ if $y = \sin^{-1}(2x\sqrt{1-x^2})$ where $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$.
- A$\frac{2}{\sqrt{1-x^2}}$
- B$-\frac{2}{\sqrt{1-x^2}}$
- C$\frac{1}{\sqrt{1-x^2}}$
- D$-\frac{1}{\sqrt{1-x^2}}$
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