If \[y=\sqrt{\sin \sqrt{x}}\], then \[\frac{dy}{dx}=\] [MP PET 1997]

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

\[\frac{\sqrt{\cos \sqrt{x}}}{2x}\]

\[\frac{\cos \sqrt{x}}{4\sqrt{x}\sqrt{\sin \sqrt{x}}}\]

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

1.	If \[y=\sqrt{\sin \sqrt{x}}\], then \[\frac{dy}{dx}=\] [MP PET 1997]
A.	\[\frac{1}{2\sqrt{\cos \sqrt{x}}}\]
B.	\[\frac{\sqrt{\cos \sqrt{x}}}{2x}\]
C.	\[\frac{\cos \sqrt{x}}{4\sqrt{x}\sqrt{\sin \sqrt{x}}}\]
D.	\[\frac{1}{2\sqrt{\sin x}}\]
Answer» D. \[\frac{1}{2\sqrt{\sin x}}\]

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If \[y=\sqrt{\sin \sqrt{x}}\], then \[\frac{dy}{dx}=\] [MP PET 1997]

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