\[\frac{d}{dx}\,\,\left[ {{\tan }^{-1}}\left( \frac{\sqrt{x}(3-x)}{1-3x} \right) \right]\]= [Kerala (Engg.) 2005]

\[\frac{2\sqrt{2}y-3}{2\sqrt{2}}=\frac{-3\sqrt{2}\times +3}{2\sqrt{2}}\]

\[\frac{d}{dx}\,\,\left[ {{\tan }^{-1}}\left( \frac{\sqrt{x}(3-x)}{1-3

1.	\[\frac{d}{dx}\,\,\left[ {{\tan }^{-1}}\left( \frac{\sqrt{x}(3-x)}{1-3x} \right) \right]\]= [Kerala (Engg.) 2005]
A.	\[\frac{1}{2(1+x)\,\sqrt{x}}\]
B.	\[\frac{3}{(1+x)\,\sqrt{x}}\]
C.	\[\frac{2}{(1+x)\,\sqrt{x}}\]
D.	\[\frac{2\sqrt{2}y-3}{2\sqrt{2}}=\frac{-3\sqrt{2}\times +3}{2\sqrt{2}}\]
E.	\[\frac{3}{2(1+x)\sqrt{x}}\]
Answer» F.

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\[\frac{d}{dx}\,\,\left[ {{\tan }^{-1}}\left( \frac{\sqrt{x}(3-x)}{1-3x} \right) \right]\]= [Kerala (Engg.) 2005]

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