If \[y={{\sin }^{-1}}\sqrt{x}\], then \[\frac{dy}{dx}=\] [MP PET 1995]

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

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

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

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

1.	If \[y={{\sin }^{-1}}\sqrt{x}\], then \[\frac{dy}{dx}=\] [MP PET 1995]
A.	\[\frac{2}{\sqrt{x}\sqrt{1-x}}\]
B.	\[\frac{-2}{\sqrt{x}\sqrt{1-x}}\]
C.	\[\frac{1}{2\sqrt{x}\sqrt{1-x}}\]
D.	\[\frac{1}{\sqrt{1-x}}\]
Answer» D. \[\frac{1}{\sqrt{1-x}}\]

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If \[y={{\sin }^{-1}}\sqrt{x}\], then \[\frac{dy}{dx}=\] [MP PET 1995]

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