If \[y=\sin (2{{\sin }^{-1}}x),\]then \[\frac{dy}{dx}=\] [AI CBSE 1983]

\[\frac{2+4{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}\]

\[\frac{2-4{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}\]

\[\frac{2-4{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}\]

\[\frac{2+4{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}\]

If \[y=\sin (2{{\sin }^{-1}}x),\]then \[\frac{dy}{dx}=\]

1.	If \[y=\sin (2{{\sin }^{-1}}x),\]then \[\frac{dy}{dx}=\] [AI CBSE 1983]
A.	\[\frac{2-4{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}\]
B.	\[\frac{2+4{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}\]
C.	\[\frac{2-4{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}\]
D.	\[\frac{2+4{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}\]
Answer» B. \[\frac{2+4{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}\]

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If \[y=\sin (2{{\sin }^{-1}}x),\]then \[\frac{dy}{dx}=\] [AI CBSE 1983]

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