If \[y=\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\], then \[\frac{dy}{dx}=\] [Roorkee 1971]

\[\frac{2{{(x-\sin x)}^{1/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{x-\sin x}-\frac{1}{2x} \right]\]

\[\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{1-\sin x}-\frac{1}{2x} \right]\]

\[\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{x-\sin x}-\frac{1}{2x} \right]\]

If \[y=\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\], then \[\frac{dy}{dx}=

1.	If \[y=\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\], then \[\frac{dy}{dx}=\] [Roorkee 1971]
A.	\[\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{1-\sin x}-\frac{1}{2x} \right]\]
B.	\[\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{x-\sin x}-\frac{1}{2x} \right]\]
C.	\[\frac{2{{(x-\sin x)}^{1/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{x-\sin x}-\frac{1}{2x} \right]\]
D.	None of these
Answer» C. \[\frac{2{{(x-\sin x)}^{1/2}}}{\sqrt{x}}\left[ \frac{3}{2}.\frac{1-\cos x}{x-\sin x}-\frac{1}{2x} \right]\]

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If \[y=\frac{2{{(x-\sin x)}^{3/2}}}{\sqrt{x}}\], then \[\frac{dy}{dx}=\] [Roorkee 1971]

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